A Course of Mathematics for Engineers and Scientists. Volume 6: Advanced Theoretical Mechanics

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Vol. 6 PLIMPTON & CHIRGWIN M Mai

—1 =

rn

M ZIP ■ I

CO =mit

0 i mil MI 2 C' c3 n a m

A Course of MATHEMATICS for ENGINEERSand SCIENTISTS VOLUME 6

rna

m

=0 Cf) -h a)

= 12. Cl) 0 m 2

C. Plumpton A. H. Chirgwin Department of Mathematics, Queen Mary College, University of London

2

C0

.1

CO)

PERGAMON

PERGAMON PRESS

A Course of MATHEMATICS for ENGINEERS and SCIENTISTS VOLUME 6

C. Plumpton anti B. H. Chirgwin Department of Mathematics, Queen Mary College, University of London

A Course of Mathematics for Engineers and Scientists is a seven volume work which covers the mathematics required by science and engineering students during their undergraduate studies. The earlier volumes lead on from senior school mathematics while the later volumes are concerned with the more advanced topics and methods used i n chem ist ry, physics an d engineering. The authors continually keep in mind the fact that the scientist requires answers and needs to apply mathematics to actual problems. Thus these volumes besides giving basic knowledge, introduce techniques and provide manipulative practice; the basic underlying methods for special or new mathematical techniques in science and engineering are fully examined. Numerous, well arranged and well graded examples have been worked in the text and many more are available as exercises. A Course of Mathematics for Engineers and Scientists

will appeal to all physicists, chemists, engineers and mathematicians, from the university or technical student to the highly advanced research or practical worker. After using the appropriate volume the reader will have developed sufficient basic knowledge and experience of techniques to follow and develop the use of mathematics in all specialised branches of study and research. This volume is devoted to mechanics of three dimensions. The contents are dealt with under the following main headings: Sets of forces; Equilibrium; The dynamics of a particle; The motion of a system of particles; Gyroscopic motion; Free rotation and steady motion; Lagrange's equations; Stability of motion; Impulsive motion; The oscillations of a dynamical system with a finite number of degrees of freedom;Normal modes; The vibrations of strings; Analytical dynamics; Answers to the exercises.

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A Course of Mathematics for Engineers and Scientists C. PLUMPTON and B. H. CHIRGWIN Department of Mathematics, Queen Mary College, University of London A Course of Mathematics for Engineers and Scientists is a seven volume work to cover the mathematics required by science and engineering students during their undergraduate studies. The earlier volumes lead on from senior school mathematics while the later volumes are concerned with the more advanced topics and methods used in chemistry, physics and engineering. Volumes 1 to 5 cover the following: Volume 1. The technique and application of differentiations and integration of one variable, geometry of two dimensions and complex numbers. 25s. net $4.50 Volume 2. Ordinary differential equations (excluding series solutions), linear equations and determinants, vector algebra and solid geometry, and differentiation and integration of two or more variables. 30s. net $5.00 Volume 3. The foundation of mechanics, plane statics and virtual work, continuously distributed forces, kinematics, particle dynamics, systems of particles, the uniplanar motion of a rigid body, impulsive motion and variable mass. 42s. net $6.50 Volume 4. Vector analysis, the solution of some differential equations, some special functions, the differential equations of field lines and level surfaces, matrices. 35s. net $5.00 Volume 5. Fourier series, the Laplace transformation, functions of a complex variable, Fourier integrals and transformations: the inversion integral for the Laplace transformation.

25s. net $3.75 ". . . advanced students or qualified engineers who wish to gain knowledge and manipulative experience of the mathematical techniques developed will find these books of value. Further volumes in the series will be awaited with interest." Journal of the Institution of Electrical Engineers

PERGAMON PRESS Headington Hill Hall, Oxford 44-01 21st Street, Long Island City, New York 11101 2& 3 Teviot Place, Edinburgh 1 4& 5 Fitzroy Square, London WI 24 Rue des Ecoles, Paris 5e 6 Adelaide Street East, Toronto

A COURSE OF

MATHEMATICS FOR ENGINEERS AND SCIENTISTS Volume 6 ADVANCED THEORETICAL MECHANICS

A COURSE OF

MATHEMATICS FOR ENGINEERS AND SCIENTISTS Volume 6 ADVANCED THEORETICAL MECHANICS

BRIAN H. CHIRGWIN AND

CHARLES PLUMPTON DEPARTMENT OF MATHEMATICS QUEEN MARY COLLEGE MILE END ROAD, LONDON E.1

PERGAMON PRESS OXFORD • LONDON • EDINBURGH • NEW YORK PARIS • FRANKFURT

Pergamon Press Ltd., Headington Hill Hall, Oxford 4 & 5 Fitzroy Square, London W.1 Pergamon Press (Scotland) Ltd., 2 & 3 Teviot Place, Edinburgh 1 Pergamon Press Inc., 44-01 21st Street, Long Island City, New York 11101 Pergamon Press S.A.R.L., 24 rue des Ecoles, Paris 5e Pergamon Press GmbH, Kaiserstrasse 75, Frankfurt-am-Main

Copyright CO 1966 Pergamon Press Ltd.

First edition 1966

Library of Congress Catalog Card No. 60-13894

1835/66

CONTENTS PREFACE

CHAPTER I. KINEMATICS IN THREE DIMENSIONS

vii

1

Introduction. The transformation law for vectors. Finite rotations. Successive rotations : Euler's angles. Angular velocity. Relative motion. Moving frames of reference. The acceleration of a particle. The general motion of a rigid body. Angular velocities about non-intersecting axes. CHAPTER II. SETS OF FORCES: EQUILIBRIUM

48

Introduction. Equilibrium. Equivalent sets of forces. The principle of virtual work. Other sets of line vectors. CHAPTER III. THE DYNAMICS OF A PARTICLE

82

General principles. A particle with one degree of freedom. The use of rotating and accelerated axes. The spherical pendulum. Motion on a surface of revolution. Motion relative to the rotating earth. The motion of a charged particle. CHAPTER IV. THE MOTION OF A SYSTEM OF PARTICLES

129

Description of the system. The dynamical variables. Conservation laws. The inertia matrix. Principal axes of inertia. Dynamical variables for rigid systems. The motion of a sphere. CHAPTER V. GYROSCOPIC MOTION, FREE ROTATION AND STEADY MOTION

177

Introduction. Rotation under no forces of bodies with kinetic symmetry. The steady motion of a gyroscope or top. The general motion of a top. Euler's dynamical equations. Free rotation. More general motions. CHAPTER VI. LAGRANGE'S EQUATIONS Generalised methods. The dynamical variables. Generalised forces. Classification of constraints. Application of the principle of virtual work. Conservation laws. Ignoration of coordinates. The motion of a charged particle.

232

vi CHAPTER VII.

CONTENTS

STABILITY OF MOTION

281

Introduction. Steady motion with two degrees of freedom. The stability of free rotation of a rigid body. The stability of a top. The gyro-compass. The stability of a rolling wheel. CHAPTER VIII. IMPULSIVE MOTION Elementary discussion. Generalised methods. General theorems.

306

CHAPTER IX. THE OSCILLATIONS OF A DYNAMICAL SYSTEM WITH A FINITE NUMBER OF DEGREES OF FREEDOM— NORMAL MODES 347 Introduction. Systems with two degrees of freedom. Stability of equilibrium : free oscillations of a system with n degrees of freedom. The oscillations of a linearly constrained system—Rayleigh's principle. A reciprocal theorem, CHAPTER X. THE VIBRATIONS OF STRINGS

386

The fundamental concepts of wave motion. Transverse vibrations. Normal modes. Forced vibrations and damping. Reflection and transmission at a discontinuity. Longitudinal vibrations. Application of Rayleigh's principle. Miscellaneous problems. CHAPTER XI. ANALYTICAL DYNAMICS

456

Introduction. Hamilton's principle. The principle of least action. Hamilton's equations of motion. Transformation theory : contact transformations. Infinitesimal contact transformations. The Hamilton—Jacobi equation. ANSWERS TO THE EXERCISES

497

INDEX

507

PREFACE THIS volume in our series is devoted to mechanics of three dimensions. In the discussion of problems and in the development of the theory we make use of the concepts and methods, such as matrices, vectors, tensors, and transformation methods, introduced in Volumes IV and V. In addition to generalising the results obtained in Volume III we include discussions of some topics which we find inadequately covered in existing texts and have kept in mind the need to provide the basis for further developments. We hope that the reader will find here a treatment which will enable him to go on to study more advanced topics such as analytical dynamics, the applications of Rayleigh's principle, wave motion, quantum theory, etc. From the start the definition of a vector via the transformation law obeyed by its components is emphasised and matrix methods are used to handle sets of components. Special attention is given to the definition of angular velocity and the proof that it can be represented by a vector. Before generalised methods, Lagrange's equations and analytical dynamics are introduced, the simpler problems of three-dimensional dynamics are discussed, often with the help of rotating axes. The use of rotating axes is stressed because it is difficult, if not impossible, to obtain the kinetic energy expressions needed in generalised methods without the use of rotating axes. After the introduction of Lagrange's equations a full discussion is given of stability and small oscillations, considered as a question of dynamics rather than statics, calling on the properties of matrices, eigenvalues, etc. The inclusion of a chapter on wave motion following that on normal modes is somewhat unusual in a volume on discrete mechanics, but the opportunity is taken to extend the ideas of normal modes to systems with an infinite number of degrees of freedom. The opportunity is also taken here to introduce Rayleigh's principle for determining eigenvalues or approximating to them. This is one of the points where a method which is extensively used beyond the boundaries of discrete classical mechanics is introduced to the reader. We hope this volume will be suitable for students taking honours courses in mathematics and also for students of physics and engineering vi i

Viii

PREFACE

who meet problems of theoretical mechanics in three dimensions and of analytical mechanics in their studies. Once again we take the opportunity to acknowledge our indebtedness to many colleagues for discussion and advice, particularly to Dr. I. E. Farquhar and Professor V. C. A. Ferraro. We also wish to thank Oxford University, the Syndics of the Cambridge University Press, and the Senate of the University of London for permission to use questions set in their examinations. Queen Mary College, LONDON, E.1

B. H. CHIRGWIN C. PLUMPTON

CHAPTER I

KINEMATICS IN THREE DIMENSIONS 1:1 Introduction Before we can develop the dynamics of particles and rigid bodies moving in three dimensions we must have a means of specifying the position and motions of these particles and bodies. To describe the position and motion of a particle (or point) we can use any of the usual systems of coordinates. A description of the position and motion of a rigid body requires more elaborate means for we must include both position, orientation and changes in both. When discussing the (two-dimensional) motion of a lamina (Vol. III) we considered the motion of an infinite sheet of glass, to which the lamina was supposed to be attached, on a table top. To show the motion of the body we drew a pair of axes on the glass and compared its position with that of a similar pair of fixed axes on the table top. To discuss three-dimensional motions we use sets of rectangular, three-dimensional axes, one set moving with the body and one set "fixed in space " ; and we specify position and orientation by the relations between the moving and stationary axes. In short, we use a set of rectangular axes as the representative rigid body. In our working we shall introduce vectors to represent not only positions, velocities and accelerations but also angular displacements and velocities. The definition of a vector (Vol. IV Chapter I) requires that its three resolutes in any frame of axes are obtained from those in any other frame by a certain law of transformation. Consequently, when we wish to prove that a set of three quantities constitutes a vector we shall usually introduce an additional, arbitrary set of axes in order to investigate the law of transformation. Because we shall be using sets of quantities it will be very convenient to employ matrix methods for the manipulations. This may introduce a slight confusion of terminology since the word "vector" may be used not only in a geometrical or mechanical sense, e.g., position vector, velocity vector, force vector, but also in a matrix sense, e.g., column la

CATES

1

2

A COURSE OF MATHEMATICS

vector or row vector. We shall always keep the word vector for use in the former sense and use the symbols a, r, ... (upright, or Roman, boldface) to denote a vector, which has components or resolutes {a, a2 a,}, {x1 x2 x3} , . . . in one frame and resolutes {a, a, a,}, e2 e3} ••• another frame. We shall keep the term matrix to denote a row or column matrix; thus the resolutes of a will constitute the column matrix a = {a, a2a„} in one frame and the column matrix a = {a, as a,} in another frame. We shall use italic bold-face symbols to denote a matrix, which may either be square or rectangular. (In general, a bold-face symbol, following the convention adopted by Jeffreys and Jeffreys, stands for a collection of elements, distinguished among themselves by suffixes, in which the suffixes have been omitted.) We use braces { } to denote a column matrix which (in order to save space) has been printed horizontally.

1:2 The transformation law for vectors In this section we express in matrix form some of the results concerning vectors which were established in Vol. IV, and familiarise the reader with the use of the transformation matrix in obtaining the resolutes of a vector in different frames of reference. The notation is summarised in the table:

Vector

Resolution in 0 x1x2 x3

Resolution in 0 6. S 23

a

{al a2 a3} = a

{alococa} = a

b

(bi b2 b3)

c

{ci. C2 C3) = C

r (position vector)

{xi x2x,} = x

=

b

(i1132133)

=

is

{711 V2 Y3} = V {616 $3} = g

We regard the resolutes of a vector a as forming a column matrix a (of order 3 x 1) in the frame Ox, x2 x3 . This relation between the two frames is given by X1 =111 e1 + 1122+ille3; el =111 x1 + /21X2 + /3 X3 , X2 = 121 e1 + 122 e2 + 123 e3, e2 = 112 X1+ 122 x2 + 132 X3 , x3

= 13 ei +132 e2

133 e3

3 =

113X1 + 12 3 X2 + 133 X3

(1.1)

§ 1: 2]

3

KINEMATICS IN THREE DIMENSIONS

We can write these relations either in suffix notation, using the summation convention, as (1.1a) li; xi ; &), xi = or we can use matrix notation,

x=

Lg, g = L'x, L = (10)•

(1.2)

The matrix L is an orthogonal matrix satisfying LL' = I = L'L.

(1.3)

The elements are direction cosines, loitself being the cosine of the angle between 0; and O. The definition of a vector implies that the resolutes given in the table above satisfy c:tf =

a, , /3.„ = /, b,

= /,i

or the equivalent matrix relations

= L'b, y = L'c.

= L'a,

(1.4)

The scalar product is a . b = ai bi =al bs a 2b 2 +a3b3 =

b = b'a.

The corresponding product in the frame 0e, e2e3 is = oe r i3 = P'cc• The invariance of the product is shown, in matrix notation, by ce = a' LL'b = a' b,

b'LL'a = b'a.

The reader should become familiar with the use of the transformation matrix L, or its transpose L', to obtain resolutes in one frame from those in another frame. An important relation, for our purposes, is the matrix formulation of a vector product. Suppose that c=axb; then c, = a2b3— a3b2 , c2 = a, — a, b3 , c, = a1b2 — a2b1 ,

and

Y1 = (X2/3 — a3 32 ,Y2 = °COI. — ( 1)33 ,Y3 = °‘1,2 — °C2/1 •

These equations can be written

(c1

c,

c, 1 a*

C

0— 'x3 el C.

2) (1) 0 — a3 a a3 0 — a, b2 , Y2 = ocs 0— NI b, — a2 al 0 Y3)1 (— a2 a1 0

I2 , 13

4


or, using matrix symbols, c = A b, y = A1i3.

(1.5)

Here the matrices A, Alare skew-symmetric. We now investigate the relation between A and A1. From eqn. (1.4), c = Ly, b = L(3, so that Ly = AL13.

y = L' ALi3 = Ai t . Hence we identify Al = L' AL, A =

(1.6)

Therefore a matrix equation (1.5) stands for a vector product provided that the matrices A and Alcorresponding to different frames transform according to eqn. (1.6). Then the vector relation e = a xb may be expressed in terms of resolutes in any frame. Example. When the frame 0$1 is obtained from Ox1 x2 x3by a rotation about one of the coordinate axes, L takes the following forms : (i) Rotation through a positive angle a about Ox1 ; (ii) Rotation through about Ox, ; (iii) Rotation through y about Ox3 ;

1

0

0

cos/

(i)L = (0

cos a

— sin a)

0

sing

cosoc

0

1 (ii) L =( 0 \ —sini3 0

sin 0 cosji

/cos y — siny 0 (iii) L --- (siny 0

cosy 0 . 0

1

1:3 Finite rotations We now use the ideas of the last section to specify the orientation of a rigid body which has one point fixed. To indicate the orientation of the body we take the frame 0$1 e2 e, to be fixed in the body, and the frame Ox1 x2 x3to be any other frame fixed in space, the common origin 0 being the fixed point of the body. There are two methods available for specifying the displacement of the body from the orientation Ox1 x2 x3 to the orientation 6% illustrated in Fig. 1 (which shows a special, two-dimensional, case of rotation about Ox3).

§ 1: 31


5

(i) An arbitrary point P has coordinates x and 4 in the respective frames (P is not, in general, a material particle of the body), and these coordinates are related by an orthogonal matrix R, where

x = R.

(i

(1.7)

)

Fm. 1. We use the matrix R to represent the orientation of the body, i.e. of the frame Wi ,e3, in the frame Oxi x2 x,. (ii) We imagine that the displacement of the body, carrying a point with it, takes a typical point originally at P to its final position P. Then the coordinates x of P are related to those of P by Rx, x

(1.8)

where R' is the transpose of R in eqn. (1.7). Although the representation (ii) is in some ways more natural, it is less useful than representation (i) for several reasons. Later we must consider problems in which a point P moves not only in space but also relative to the moving body; method (ii) is not suitable for such problems. When we consider vectors we are at liberty to resolve them in any frame we choose ; the use of method (i) enables us to make resolutions in the frame 0$, 2 3 easily. This latter resolution is frequently useful as it expresses the state of affairs as seen by an observer moving with the body. By using an othogonal matrix R to represent any displacement of a rigid body about 0 we can prove Euler's theorem that any such displacement is a rotation about some axis through 0. From eqn. (1.8), if any point is unaltered in position by the displacement, :Ye = :71

, X =

fix.

6


This shows that x is an eigenvector of R corresponding to the eigenvalue 1. Now any (real) 3 x 3 orthogonal matrix, whose characteristic equation is a real cubic equation, must have at least one real eigenvalue, since every real cubic equation has one real root. Being an orthogonal matrix, R has eigenvalues of modulus 1; and since det R = 1 (the frames being right-handed), the product of the three eigenvalues is 1. Hence, the eigenvalues are 1, eig', e-iT, where 0 S < r. The eigenvector corresponding to the eigenvalue 1 is a real vector x and consequently all those points whose coordinates are k x are unaltered in position by the displacement and constitute a line which is the axis of rotation. There can be only one such axis. If the remaining eigenvalues a ' 9)are real they must either be both + 1 or both —1; in the first case R = I and there is no displacement; in the second R is one of the matrices of the example on p. 4 with cc, ,8 or y equal to n. When the eigenvalues g' are complex the corresponding eigenvectors are not real; hence the axis of rotation is unique. We show, in an example below, that the eigenvalues e-± "Pgive the angle pthrough which the body has been turned about the axis of rotation. Since the sum of the eigenvalues of a matrix is equal to the sum of the diagonal elements, the trace of the matrix, then tr R = 1 + eiT

e--1P = 1 + 2 cosq)

(1.9)

This result affords the quickest way of finding the angle of rotation. Example. We give a geometrical proof that 9, is the angle of rotation. We suppose that R has the following eigenvalues and eigenvectors : = 1, Ra = a; ei

, R(b ic) = et cP(b

ic);

=

R(b — ic) =

(b — ic).

where a = {a1 a2 a3), b = (b1 b2 b3), c = {c1 c2 c3) and all the elements are real. First we prove that the vectors a, b, c having the respective matrices a, b, c for their resolutes are orthogonal. b

= R'R(b ic) = a' (b

R' (b ic), a' = a' R'

ic) = ei 9a' R' (b

ic) = of g' a' (b

ic).

Since eiT 1 we deduce that a' (b i c) = 0, a'b =0, a'c= 0, i.e., a . b = 0, Multiplying b

i c = ei R' (b i c) on the left by b' b'

I

- -

(b

ie)

a.c= O.

i c', we obtain

ei T (b' -I- ie') R' (b I- ic).

§ 1: 3]

7


But (b' + ic') R' = ei99(b' + ic'), (b' + ic') (b + ic) = e2l9 (b' + ic') (b + ic). Since e21 9'

1, 0 = (b' + ic') (b + ic) = (b'b — c'c) + i(b'c + c' b). .•.b'b = ec, b'c=

—

c'b.

But b' c = c' b = b, c, + b2 c2+ b, c, = b . c. b c = O. Hence the vectors a, b, c are mutually perpendicular. By considering the real and imaginary parts of R(b ic) = eic° (b + ic)

we obtain Rb = b cosp

—

c sing), Rc b sin()) + c cow.

(1)

Consider now the interpretation (ii) of p. 5 applied to the points whose coordinates are b, c carried with the body in the rotation R. According to (1) these points take up positions whose coordinates are b, e where 6 = b cosy!, — c sing), = b simp + c cosp. The geometrical interpretation of this result is that the two perpendicular vectors b, e are rotated in their own plane through an angle go. Since a is perpendicular to this plane we see that the effect of the rotation R is to turn the body through an angle 99 about a as the axis.

We now find the displacement produced at an arbitrary point of the body by a rotation and show that we can represent this by a suitable vector. We write eqn. (1.8) as —

x = (R

—

I) x ,

x (R 1) ,

where I is the 3 x 3 unit matrix. We combine these two equations in the relation — x = (R — I) (R I)- (X x) . X — x = 0 (x

x) ,

(1.10)

1) -1 .

(1.11)

where = (R — I) (R

Since R-1= R' and the order of the factors in the reciprocal of a product is reversed we find another form for 0, as follows : 0 = (R — R' R)(R R' R)-1= (I — R')R[(I R')11]-1

= (I — R')R11' (1 + R')-1= (I — R')(I

R')-1.

(1.12)

8


Taking the transpose of eqn. (1.11) we obtain = [(R

—

I)(R

I)-1y = [(R

I) -1]' (R

—

1)' .

But taking the transpose and the reciprocal of a matrix are commutative operations, hence 0' = (R' 1)-1(R' — I) = (R' R1 R)-1(R' — R' R) = (I ± R)-1R R' (I — R) = (I ± R)-1(I — R) = —O.

Hence 0 is skew-symmetric. This implies that we can write the r.h. side of eqn. (1.10) in the form of a vector product. If we can prove that 0 satisfies the transformation law (1.6) then eqn. (1.10) can be replaced by a vector equation, valid in all frames of reference. We use a frame 0 X, X, X, related to Ox1 x2 x3 by

X = /Ix, X = L'x where x, x; are the coordinates of P, P in Ox1 x2 x3. Hence X = L'Rx = L'RLX = R1 X,

where R, = L'RL

or

11,' = L' R' L.

We construct the matrix 01from R1according to eqn. (1.11). 01 = (R1— I)(R, + 1)-1= (L' RL — L' L)(L' RL L'L)-1 = L'(R — I)L[L' (R I)L]-1= L'(R — I) L[L' (R 1)-1L] = L' (R — I)(R 1)-1 L,

since LL' = I = L'L. .*. el = L'OL.

This agrees with eqn. (1.6) so that the three non-zero elements of 0 — 03 02 03 0 - 01 - 02 01 0

=

(1.13)

transform as the components of a vector 0, and —

is the vector form of (1.10).

r

0 x (f

r)

(1.14)

§1: 3]

9


To obtain a geometrical interpretation of 0 we note that if 1. = r, 0 x r = 0, i.e., 0 is parallel to the axis of rotation, on which r = r. We consider a special case; the fact that eqn. (1.14) is a vector equation means that a geometrical result established in one frame of reference holds for all frames related by the transformation (1.2). When 1 0 0 0 cosy — sina ( R 0 sina cosy —

the rotation is through a positive angle a about the axis Ox1. Now

R

—

0 I = (0 ,0

0 — 2 sin2 ja 2sin1-a cosa 0

0 sinia —cost a

=— 2 sin )s a 0 0 (R

+ 1) -1 =

1

— 2

(0 .'. 0 = — sin !-,oc 0 0 0 — (0 0

0 0 tan 1 a

— 2sin ?-,a costa — 2sinqa.

1 0 0

0 cos , sin -oc

0 1 —tan ;a

0 0 sin 2a costa —cost a sin 1a

0 tanl a 1 1 0 0

0 1 — tanl-a

0 tani oc 1

)

0 —tan2cx . 0 .*. 0 = (tan ia 0 0} .

Hence 0 is a vector directed along the axis of rotation of magnitude tan -ia when the angle of rotation is a, the sense of the rotation being given by the r.h. rule. The reader should note that, whilst the direction of 0 is along the axis of rotation, the magnitude of 0 is not equal to the angle of rotation. Since = (R

f)(R

I)-1,

10


we obtain R and take the transpose, remembering that 0' = — 19, showing that R= (I + 0)(1

—

0)-1 , R' = (I + 0)-1(I

—

0).

(1.15)

Hence we can construct R to correspond to any angle of rotation about any axis by first constructing 0 from the vector 0, using eqn. (1.13). Example. Show that, under a rotation 0 about the axis u (a unit vector), the vector r is transformed into f

cos8 r + (1

—

cog) (r u) u

—

sine r x u.

If this transformation is written Xi = hi mx„„ (summation convention) determine the elements himof the rotation matrix. Prove that h„,„,= 2 cog -I- 1, ekim 41,n

—

2 sin() Uk •

Given that the matrix (

3

-

3

1)

3

1 3

1

is a rotation matrix, find the angle and the axis of rotation.

FIG. 2. In Fig. 2 the rotation takes P into P and PN is perpendicular to LP; both LP and LP are perpendicular to the axis of rotation. The vector u x (u x r) is in the direction of PL and has magnitude LP; also — r = PP = PN NP.

§ 1: 3]


11

PN = u x (u x r) (PN/PL), PN = P L(1 — cos0), —)NP = (u x r) (NP/PL); NP = PL sine.

.*. r — r = (u x r) sine + u x (u x r) (1 — cose) = (u x r) sine + [u(u . r) — r] (1 — cose). = r cos0 + (1 — cos0) (r . u) u — sine (r x u). In terms of coordinates in a frame 0 x1 x2 x3this equation is xt = xt cos0 + (1 — cos0) (x,u1)u, — sine slkmxkum = 61„,x„, cose + (1 — cos()) uiu„,x„, — sine Ein, k uk xn, = [oi„, cos0 + (1 — cos0) aci u. — sine ei „, kuk] x„,• .• , hi. = 8l m cos° + (1 — cos0) ui u„, — si10 si„, kuk •

From this, on summing by putting / = m, we obtain h nn = 3 cos° + (1 — cos0),

since u is a unit vector and Eimk is skew-symmetric. .• h„,„, = 1 + 2 cos 0. [See eqn. (1.9)]. Also ski„,h1„, = ek.. cos() + (1 — cos0) skim ul un, — sin° ek i„,ei„, iui = —sine 26, k u, = —2 sine uk . The first two terms vanish because Eki m is skew-symmetric while 6,,„ and u, um are symmetric in l and m; the last term is obtained from the relation skimeiii = — eikmezit = kams + aki6mi

in which we put j = m, (see Vol. IV Chapter I, p. 103) giving 2612. For the given matrix sine =

h.. = 1 + 2 cos0 = 2, cos°

.". 0 = n/3. Also

611m him = h23 h32

—3

=

=

1,

him — h31 — hi3=

— 3 = —1,

83imhim = hi2 — h2i=

= —1 •

821in

U3 2

Nigh u3) = (

-

1

—

1

1 1 1 u = — — — 1/ 3 1/3 V3 •

—

1).

12


1:4 Successive rotations: Euler's angles In general, the order in which finite rotations are applied to a rigid body affects the final position of the body; this property is shown by the noncommutation of the products of the corresponding matrices. Suppose the rotations corresponding to R1 and R2are applied successively in that order to a rigid body. Then the representation (i) of §1: 3 requires an intermediate frame Or n, ri3(see Fig. 3) and the coordinates of a point P in the three frames are the respective matrices x, which are related by = R1 ,, X

= RiR2 g.

Fia, 3. Hence the resultant rotation is represented by R = 11,112 .

(1.16)

(The product of two orthogonal matrices is itself an orthogonal matrix.) It is important to note that the second rotation occurs as the second member of the product in eqn. (1.16). If the rotations are applied in the reverse order the resultant rotation is represented by S = R2 Ri , and, in general, R S

Euler's angles There is a standard method, due to Euler, in which the orientation of a rigid body is expressed by means of three angles 0, q,, e. The reader is advised to study the figures carefully and be certain to understand the sequence of operations by which the frame OE, E, E is obtained from

§ 1:4]

13


Ox1 x2 x3using the intermediate positions shown in Fig. 4. The two intermediate frames required are 0121172173and OC, C, C2 ; each of these is reached by rotations specified below.

2 (i)

(iv) FIG. 4.

(i) oxix2 x3 on/ 172Th . Rotation through an angle y about Ox, (or 0173): Fig. 4 (i) to (ii): matrix R1 . (ii) Ox11 n, n3 oc i b2 b3 • Rotation through an angle 0 about On, (or matrix R2 . 0 : Fig. 4 (ii) to (iii) OC, C2 > Oei e2 : Rotation through an angle g) about O (or WO : Fig. 4 (iii) to (iv): matrix /13 . In matrix representation [of type (i) § 1:3] these stages are given by: — -

cosy — sin y 0

(i) x = R, ;

Rl = sinyp 0

(ii)

= R2 g ;

cosh 0 ;

0

1

cos 0 0 sin0 0 0 1 R2 = — sin 0 0 cos 0

14


(cosy (iii) ;

=

R3g ;

R, =-- sin 92

— sin g) 0 cos 99 0).

0

0

1

Hence x = Rg, where R=

Ri R2 R3

(COS 0

=

cos 99 cosy — sin 99 sin v

cos 0 cos y sin v + sin y cosy — sin 0 cos y

— cos 0 since cosy) — cos 99 sin v

sin0 cos v

cos0 sin y sin v + cos y cos v

sin e sin v

—

sin0 sin ce

COS 0

(1.17) By choosing values 0 _< 0 _< r, 0 2n, 0 < v < 2n, the octant Oei e, of the sphere in Fig. 4 (iv) can be made to take any arbitrary position. The rigid body being "attached" to the octant can therefore take any orientation, which may be represented in terms of three parameters. Although an orthogonal matrix has nine elements, these elements are dependent upon three independent parameters. [This is also shown by eqn. (1.15) in which 0 has three arbitrary, non-zero elements.] Therefore, a rigid body with one point fixed has three degrees of freedom. The interpretation in § 1:2 of the elements of a rotation matrix as direction cosines gives the columns of (1.17) as the direction cosines of 0e1e, e, referred to Ox1 x2 x3 , and the rows of (1.17) as the direction cosines of Ox1 x2 x3referred to W, • (Unfortunately, some authors use co where we have used o and vice-versa. There is no definite convention on this point, and in this book we keep to the nomenclature adopted above.) Examples. (i) Hamilton's theorem. The use of a sphere to represent the orientation gives an alternative, geometrical treatment of some of the above results. The fixed point of the body is taken as 0 and the motion of any point P of the body is mapped on a sphere with centre at 0 by the intersection of 0 P and the sphere in a point "representing" P. The body is fixed if two representative points on the sphere are fixed ; the body rotates about OA if a point A on the sphere is fixed, A being the intersection of the sphere and the axis of rotation. Suppose a body is given a rotation through an angle a about OA followed by a rotation about 0 B through an angle ie. (In Fig. 5 all the lines are great circles on the surface of the sphere ; for convenience the centre 0 is not shown.) We join AB by a great circle and construct points C, C1as the intersections of great circles whose planes make angles a, ,0 with the plane of A B, as shown in Fig. 5. The rotation about OA takes C to Cland the subsequent rotation about

§ 1 :4]


15

OB takes C1back to C. Hence the line OC is the axis of the resultant rotation. If the rotations are given in the opposite order, 0 C1is the axis of rotation : this demonstrates the non-commutativity of rotations. If the angle between the planes of the great circles AC, BC is f y (0 _s y s 70, we can construct the spherical triangle BC A„ from great circles, so that ACB and Al CB are symmetrical w.r. to B C. The angle between the planes AC and A1C is 2 (yr — y). Now the first rotation leaves A unaltered in position and the second rotation, about OB, takes A to A1, since ABA,= An additional rotation about OC through an angle — (27r — y) leaves C unaltered and restores A to its original position. Hence after three rotations, through a about OA, through 13

FIG. 5. about OB, and through y about OC, both A and C have returned to their original positions and therefore the body is restored to its original position. (The rotation about OC through — 27r does not affect the final position of the body.) This result is Hamilton's theorem. The angles a, jl, y are twice the angles between the respective pairs of planes which define the spherical triangle ABC. (ii) A rigid body is successively rotated in the same sense through an angle Tc/3 about each of two axes which intersect at an angle yr/3. Determine the axis and angle of the equivalent single rotation. Choose the axis of the first rotation to be Ox1, then

1 R,

0

0

0

=(

—

0 11/3) •

1113

The axis of the second rotation is taken to be the vector

16


Then since the body is turned through an angle x/3, and tan c/6 = 1/1/3, 02 = (1/1/3)

o 0

i –*1/3

+3 /3

0

02 = (o 0 — i.

/

-

02

=

1 (0

i1/3 1

I + 02 =

0

1

0

1-1/3

—a1/3 1

0 det (/ — 02) = t.

1

1

-

a 13

is

7W3

3

6-1/3

-11/3

= Al/3

–a1

— 02)-1 =4

—*

1

s

i13

... R2 = (I -I-02) (I — 02)-1

0 =( 0 1

I )

sAV3

1 - ili3 A-1/3

-I- -413

1

--R-

1

I

a 1-1/3

/ a'y 3

--11/3 = *1/3 i i -t *1/3

which is an orthogonal matrix.

.43 1

.-. ft= R, R2 = (R -;÷d/3

_3 - 16 A1/3

-03 . -1

But tr R = 1+;= 1 + 2 cos 7 5 V39 .•. cos Tr = — 32— , sin co — 32

(0 < < ;2).

The characteristic equation of the matrix R reduces to 16/13— 9A2+ 9A — 16 = 0.

.•

—1

,

cos go = — as before.

—7 ±i51139 _ e 32 7 32 ,

sin cp

51/39 32

-3 ,f

-IP

§ 1 : 4]


17

The eigenvector corresponding to A = 1 is {3 j/3 1} which gives the direction of the axis of the resultant rotation. In order to relate the sense of the rotation with the direction of the axis we must either determine the vector 0 of eqn. (1.14) or consider the displacement of a suitable point. In this case the vector {3 j/3 1} does give the positive direction of the vector 0.

Exercises 1:4 1. Obtain the formula — r = x (P + r) by assuming the existence of the axis of rotation and using Fig. 2. 2. Use the method of representing the motion of a body on a sphere to prove the existence of the axis of rotation. 3. Obtain the direction cosines of the matrix (1.17) by elementary geometry from a consideration of the angles in Fig. 4. 4. Prove the formula of the example of p. 10 assuming the formula — r = (tan

u x + r)

0)

Construct the matrix O from the rotation matrix given, and verify the angle of rotation and the axis of rotation. 5. Find the angle of rotation a and axis of rotation corresponding to the matrix ( 1 2 1/ 2 R

10 1 2 2V2

1/3

1

1/3

21/2

2

21/2 1

1 1/22

0_

6. Find the matrix R corresponding to a rotation through an angle a, where cos a = 7/25, sin a = 24/25, about the axis a 1 — I). 7. P is a point of a rigid body, and (x, y, z) are its coordinates referred to fixed right handed rectangular axes Ox, Oy, O z. If the body is given a positive (i.e. right handed) quarter turn about 0.x, show that the new coordinates of P are the components of M, p, where p is the column vector {x y z} and Mx is the matrix with rows (1 0 0), (0 0 — 1), (0 1 0). Find the matrices My, Mx corresponding to positive quarter turns about Oy and 0 z, and show that Mz My Mx = M. Express Mxin terms of M; 1, M;1and My, and interpret the result.

18


1:5 Angular velocity When considering the motion of a lamina in its own plane we took the angular velocity to be a scalar quantity defined as the rate of increase of an angle. To describe the three-dimensional motion of a rigid body we need a vector to specify change of orientation: we have now to devise a suitable definition of angular velocity. We can start from eqn. (1.14) and assume that the displacement i — r takes place in a short interval of time bt so that lim

f—r

61

v=

1 [20 — hm — (f x t 6t->0

Now the limiting value of I (i r) is r, the position vector of that particle of the body whose velocity v we are finding. It is tempting to define an angular velocity vector by

w = lim

20

v=w x r.

,

Si->O at

However, we must prove that this limit exists and that the resolutes transform according to the correct law; to prove this necessitates a different approach. Suppose that the frame 0 1 2 S 3 is moving and that we choose the point P to move with this frame. Then the coordinates g will not vary with time whilst the coordinates x will vary. At any instant

g

x, 0 = it' x

= — RA' x.

R'

(RR' = I)

(1.18)

[When A = (au), we denote the derivative by A = (daii/dt) = (au), i.e., we differentiate every element.] The matrix X is made up of the resolutes of the velocity of P (moving with the body) in the frame Oxi x2 x3. Eqn. (1.18) shows that = Sx, where

S=

—

RR' .

The transpose of S is S' = — (R )' = — 11R' .

(1.19)

§ 1 : 5]


19

Since R is orthogonal we obtain, by differentiating I = RR', 0 =

RA'.

S' = — RR' =

= — S.

Hence S is skew-symmetric and (1.18) takes the form, in the frame

Ox, x2 x3, of a vector product. To show that we can write (1.18) as a vector equation we prove that S transforms according to eqn. (1.6). We introduce a new fixed frame 0 X1X2 X3 such that x = LX, where L does not vary with time; then = R' LX = X, RI = R' L.

The matrix Sicorresponding to S and constructed from R1according to eqn. (1.19) is S1 = Ri k = ( L' R)(it'L) = L' S L. —

—

This agrees with eqn. (1.6) and shows that the non-vanishing elements of S transform like the components of a vector. Hence we introduce an angular velocity vector co whose resolutes {.(21 Q2 ..(23) in the frame Oxi x2 x,are defined by ( S=

0 —S2,

D2

S23

S21

0

— Q2 Q1

=—

= RR' .

(1.20)

0

Then eqn. (1.18) takes the vector form v=wxr

(1.21)

,

where v is the velocity vector of a point P, whose instantaneous position vector is r, P moving with the body. It is often convenient to use the resolutes of w in the frame 0 1e2 3 . Denoting these resolutes by {co, co, w3} we have according to (1.6), with L = R,

0

— ()3

w3

0

\—co2

col

a)2 —col

0

=

R'SR=

= (1.22)

We obtain the final entries of (1.22) by differentiating I = R' R. Eqns. (1.20) and (1.22) show how we obtain the angular velocity vector

20


of the rigid body from the matrix R, giving its instantaneous orientation, and R. To obtain a geometrical interpretation of w we consider a special case 1 =

(0 0

0 cosy cos sincc

0) — sinoc , cosoc

0 R = d, 0 0

0 — sina cos cx

0 — cos oc , — sin oc

where R stands for a rotation about the axis Oxi . Then

S =

0 R1 = (0 0

0 0 1

0 —1 0

{Q1 Q2 Q3} = {o< 0

0}.

(1.23)

This result shows that the angular velocity vector is directed along the axis of rotation at that instant and has the angular speed of the body for its magnitude. It must be emphasised that, apart from certain exceptional arrangements, such as that for eqn. (1.23), none of the resolutes {co, co, co,} or 421 Q2 Q3} is the time derivative of a single angle. In general, there is no angle which can be constructed whose rate of increase is col , or 0,, etc. (This fact is of importance later in connection with "holonomic " mechanical systems.) When the phrase "the angular velocity of a body about an axis" is used, it is to be interpreted as the resolute of co along that axis. When speaking of linear motion we usually keep the word velocity for the vector and use the word speed or velocity component for the magnitude or resolute of the vector. It would, possibly, be desirable to use the words angular speed to denote the magnitude of co, but there is no general convention to do this. The phrase "angular velocity of a line" is also used sometimes; this phrase always implies the angular velocity of a line about an axis. The phrase commonly occurs in connection with the rolling of one cone on another when it may be necessary to consider the motion of a generator, or of an axis of symmetry of one of the cones. This angular velocity is interpreted as the angular velocity (as defined in two-dimensional motion) of a line drawn perpendicular to the fixed axis and intersecting the moving line.

§ 1 : 5]


21

When the motion of a rigid body about a fixed point is represented by the motion of points on a sphere [see Ex. (i) p. 14], the instantaneous axis of rotation intersects the sphere in a point of the body which is instantaneously stationary. This is analogous to the instantaneous centre of the plane motion of a lamina. Corresponding to the centrode curves of plane motion the movement of the instantaneous axis of rotation traces out one cone in space, the polhode cone, and another cone in the body, the herpolhode cone. It can be proved that the motion of a body with one point fixed is given by the rolling of the herpolhode cone, taking the body with it, on the polhode cone ; the vertices of the two cones, of course, coincide at the fixed point of the body. A very important application of the formulae (1.20) and (1.22) gives the angular velocity of a rigid body in terms of Euler's angles and their derivatives. In the general case when R = (kJ) formula (1.20) shows that = /kJ (using the summation convention) since

[Fe]; k = lki . [S]23

•• Ql =—

6 siny

122 132 t 123 133)

(121 131

(1.24a)

q*;, sine cosy ,

the values for lobeing given by (1.17). Similarly, S22 = 0 cosy, + sin() sin y ,

p

S23 = c COS 0 +

(1.24 h)

From eqn. (1.22) (summation convention).

[It'

= 1, j 1k,

••• col —

[ nr R]32 = 112 113 -

= 6 sing9 — y, sine

122 123 ± 132 133

(1.25a)

COS99.

Similarly

w, = 0 cosq9

y sine since, co, =

cos°.

(1.25b)

(Note that in obtaining (1.24) the free suffixes in io /kJ are both first suffixes, i.e., refer to the frame Oxl x2 x3 ; in obtaining (1.25) the free suffixes in /kJ ik , refer to 0 1 e2 e3.)

22


Examples. (i) If i, j, k denote unit vectors along the axes O11, W2, respectively, then, resolved in the frame 0 x,x,x,, 1 iS 1111 121

d dt

131),

„ is

(11 1 12 1 63 1)

j1S {42 122 132), dj dt

k

is

{113 123

133) .

dk is ( 6 1 2 62 2 63 2)

d t

is ( 113 123 63 3) •

refers to the frame (To obtain these we remember that the second suffix in and the first suffix to the frame 0 x3 x2 x3.) Hence from the expression 0 $, for co, in eqn. (1.25 a) we see that wl

— 112 113 + 122 123 + 132 133

=d.4 • k•

Similarly, W2 —

dk (.03

dt / dj

(') =

C°1

C°2 k CD3 = 1

di dt

l• /dk .\

.kl+jl

1)

+ k

/ di \ it

This latter expression is a useful formula for co. (ii) In a Hooke's joint connecting two shafts A and B, the line XX' of the pivots fixed in A makes an angle fl with the axis of B and the line YY' of the pivots fixed in B makes an angle a with the axis of A. If the shaft A is given a spin w about its own axis, show that the spin co' imparted to the shaft B is (co sin a)/sin /3 and that the relative spins of the moving parts about X X' and YY' are co sin a cot j9and co cos a respectively. The joint has three parts (see Fig. 6) : the two rods X' 0 X, Y' 0 Y rigidly fixed at right angles, bisecting each other at 0 ; the shaft A, terminating in a semicircle with XX' as diameter ; the shaft B terminating in a semicircle with Y Y'

as diameter. The shaft A can pivot freely about XX' and B can pivot freely about Y Y' ; each shaft, if produced, would pass through 0. We let a, b be unit vectors along the respective shafts, as shown, and i, j, k a set of mutually orthogonal unit vectors with i along 0 X and j along 0 Y. The angular velocities of the three parts,

§ 1: 5]


23

viz., the shaft along a, X Y X' Y', the shaft along b, are, respectively , 52=col i+co2 j+co3 k, Because of the construction of the joint

. = 0,

j

0;

and we are given that a j = cos oc, b

= cos/3.

We obtain the conditions that must be satisfied from the fact that X (and X') is a point both of the shaft A and of the intermediate piece ; similarly Y (and Y') is a point both of the intermediate piece and of B. The velocity of X is given from A and the intermediate piece, coaxl= faxi=jco,—kco„

(1)

Similarly, from the velocity of Y, co'bxj=52xj--ico3 +kco1.

(2)

To obtain these we have taken 0 X and 0 Y as of unit length. From (1) coax i.j= co( a k) = co„ wa x 1. k — co( a . j) = — co2. . = 0, a j = co,/co = cosec,

k = cos/co.

Similarly, from (2) dbxj.1--co'(b. k) = — w3, co' b x j.k= co' (b . i) = co,. .•. b. = co,/co' = cos/3,

j = 0, b k = co,/co'.

.•. co, = co' cos/3, w2= w cosoc. Since a b are unit vectors ,

1 = cos2 o + (w3/w)29 w2 w3 3 = (012 sin2/3. 1 = cos21 + (coda ?, (.02 coz 3 Hence, with an appropriate choice of directions co slim = co' sinfl. The remaining components of angular velocity of X Y X' Y' are co cosoc, co' cosi3 = co co-0 sinoc. (iii) Two conical shells of semi-vertical angles fl and /3 + 2oc respectively have a common axis and vertex. Between them is pressed a solid cone of semi-vertical

24


angle a, which touches each of them along a generator. The shells are made to rotate in opposite senses with angular velocity w about their common axis. Show that, if there is no slipping, the speed of a point on the axis of the solid cone at a distance r from the vertex is w r cos (a + fl) tan a and that the angular velocity of the solid cone about its axis is co sin (a + /5') cot oc. Determine the instantaneous axis of rotation of the solid cone. Since all the cones are right circular cones we use a frame of axes 0 x, x2 x, (see Fig. 7) such that the generators of contact OA, OB both lie in the plane x2 = 0. If 0 C = r, OA = OB = r sec a. A is {r sec a sin/3 0 r seta cosfl), B is {r seccx sin(j3 + 2oc) 0 r seta cos (/3 + 2a)).

FIG. 7. The velocity of A, a point on the inner cone is, {0 0 co} x {r secoc sinfl 0 r seca cos/3} = {0 co r seccx shit', 0). Similarly the velocity of the point B on the outer cone is {0 0 —co} x {r secoc sin (i3 + 2a) 0 r sec a cos(fl + 2a); {0

co r sec oc sin(13 + 2 a) 0}.

But the velocity of C is vc = (vA+ vB) = {vci vc2 vc3}. vc2 = co r

— sin (3 + 2oc)] seta = --co r cos (a.1-- /3) Lana.

§ 1 : 5]


25

If Qcis the resolute of the angular velocity of the solid cone in the direction OC i.e., the "angular velocity of the solid cone about its axis", then

vA2 = vc 2 — A C . t2c , co r sec a sinfl = — co r cos (a -F f3) tan a — Qc r • • • QC tan a = — co [sec oc sin /3 + cos (a F /3) tan a] . pc = — co sin (a + fl) cotes. If the point J is stationary then OJ is the instantaneous axis. The condition for J to be stationary is

since

VA

JA

JB

VA 2

VB 2

and vBhave opposite directions. Hence tana — tamp tang + tang) sin/3

sin (/3 + 2a) •

tang) = tan2 a cot (a + /I). Exercises 1:5 1. If two points A and B of a rigid body, having respective position vectors rA and rBreferred to 0, have velocities YA, TB respectively prove that, if the point 0 of the body is fixed, the angular velocity is VAX VB YA •

VB X YA

rB

VB •

ra

2. A wheel of radius a rolls on a horizontal plane and makes a constant angle f3 with the plane, the centre of the wheel describing a circle of radius Aa with velocity V. Show that the angular velocity of the wheel about the instantaneous axis is

in2 s fl )1' 1 I-

a(

2.2

.

and find the component angular velocities about the axis of the wheel and the diameter through the point of contact. 3. A right circular cone of vertical angle 2a rolls without slipping on the outside of another fixed right circular cone of vertical angle 2/9, the vertices of the cones being always in contact. If co is the angular velocity of the former cone about its own axis, find (i) its angular velocity about its instantaneous axis, and (ii) the angular velocity of its own axis about the axis of the fixed cone. Also find (iii) the angular velocity of the axis of the fixed cone relative to the moving cone. Consider also the cases in which (1) the moving cone rolls on the inside of the fixed cone (a < M, (2) the moving cone surrounds the fixed cone (a > ff), the vertices being always in contact. 2

CMES

26


1:6 Relative motion We have seen that the compounding of two finite rotations is non-. commutative ; we now investigate the result of compounding two angular velocities. We suppose that there are three frames of reference, Ox,x,x, which is fixed, 0 1$2 $3 and On, n2 n3which are both moving. At any instant these frames are related by

= R2g • = R1 R2 g = R4, R = R1 R2 . (1.26) The elements of all three matrices vary with time, R1being made up from cosines of angles between 0; , Oni, R2being made up from cosines between On;and Opp . We can therefore construct three angular velocity matrices, such as (1.20) or (1.22). We imagine an observer A associated with the frame 0 x,x2 x3, B with On, n2 713 , and C with 0$, $3 . When we construct an angular velocity matrix we must be quite sure in which frame the corresponding angular velocity vectors are resolved. From R1we construct the matrices RI R; made from resolutes in 0 xi x2 x, , (1.27) R;.11, made from resolutes in 0111" 2n3 These resolutes refer to the angular velocity of B's frame as evaluated by A, Awn , since A's frame is stationary. From R2 we construct matrices : X

112.1r2 made from resolutes in On,n,n,, ./i1/2made from resolutes in 0e,$2$3.

(1.28)

The resolutes in these matrices are those referring to the angular velocity that B ascribes to C's frame, ignoring the motion of B's own frame. This is the angular velocity of Oe, 2 $3relative to On, n2 n3 , Bc.oc • Finally, from R we construct matrices

R' made from resolutes in 0 xi x2x,, R'R made from resolutes in 0 el$2$3.

(1.29)

The resolutes refer to the angular velocity of C's frame as evaluated by A, Acoc . To investigate the relation between these vectors we resolve them all in Ox1 x2 x3. We then have the following three angular velocity matrices, consisting of resolutes in the frame Ox1 x2 x3:

R1Ri giving the resolutes of R1.112 R2R1giving the resolutes of

B°C

AR'

AO C•

giving the resolutes of

AW B ,

(1.30)

§ 1 : 6]

27


[To obtain the second line of (1.30) we have quoted eqn. (1.6).] But R = R, R2 . R = 11,R, + R1/12 . RR' = (111112+ R1R2)1:4111 = 141R; + R1 R211A.

This is equivalent to the equation or

Atoc = Atos + 134-0c

BWC

= AWC AWB •

(1.31)

Eqn. (1.31) shows that when a rigid body has two angular velocities simultaneously we add the angular velocity vectors to obtain the resultant angular velocity. Alternatively, the angular velocity of one moving frame relative to another moving frame is obtained by a formula similar to that for linear velocities. Example. Find the angular velocity of A's frame relative to B's frame, i.e., find BcoA . The relation between the frames is =

x.

The observer B will calculate the angular velocity of A's frame using 11, whereas A would use R1. Thus the angular velocity ascribed by B to A's frame is

(ii;.)(R;)1 = RiR1 =

—

10?„

the resolutes being those in B's frame. This matrix has the opposite sign to that in (1.30). Hence BWA = AWB•

Infinitesimal rotations If the frame 0e, e, $3receives a small displacement from Ox, x, x,, then R = I + h, where h is a matrix whose elements are small. Since R is orthogonal = R' R = (1 + ) (I + h) = 1 + h + h' + hr h= — h' ,

correct to the first order. The skew-symmetric matrix 0 corresponding to this rotation is 0 = (R 2*

—

1)(R + 1)-1= h(2 1 + h)-1= 1-h

28


correct to the first order. The displacement caused by this infinitesimal rotation has resolutes E•x = — x, where

Bac = 0('&•

x) = 1-h(2x ± 8x) = hx,

correct to the first order. This corresponds to the vector equation

Sr = 80 x r,

(1.32)

where

0 —603 h — ( 603 0 — 603 601

60, — 601 , 0

60i Se = 602 , 603

the vector Se being directed along the axis of the infinitesimal rotation and having the (small) angle of rotation for its magnitude. Corresponding to two successive small rotations R1 = I +

, R2 =

h2

the resultant rotation is given, correct to the first order, by

R1 R2 = I

+ h2 = R2 R1 .

This shows that, correct to the first order, the infinitesimal rotations commute. In fact, they are just the displacements occurring in a rigid body, which is rotating about a fixed point for an infinitesimal interval of time. The simple addition rule for the small rotations is the counterpart of the result proved above for the addition of angular velocities.

1:7 Moving frames of reference Suppose that a vector a has resolutes a = {a1 a2 a3} and a = {csi oc2 cx,} in the respective frames of reference Ox1 x2 x3 and 0 1e2 e3, where

a = Ra

(1.33)

at any instant. Now if a is a vector which varies with time, then there are matrices a = a2 a3} and de = {dc, Cx2(5C3) obtained by differentiating the resolutes. If the frame Ox1 x2 x3is fixed, we regard a as the matrix whose elements are the resolutes of d a/dt in Oxi x2 x3. To an observer moving with 0$1e2 e3 the matrix acan be regarded as the resolutes, in the frame O 1 e2 e3, of a vector we call a a/at. The partial differentiation sign is used because, in effect, we are ignoring the motion

X1.:• 8]

29


of the frame in obtaining the vector a ale t, which has the resolutes de = {a1 a2 6c3} in the frame OBI 2 3. The matrix Rec consists of the resolutes ofa s/at in the frame Ox1 x2 x3, because the transformation (1.33) between the two frames applies to all vectors. If we differentiate (1.33) we obtain a = Rex + . it =

ilRi a .

But 11 R' is the angular velocity matrix, so that this equation expresses, in terms of resolutes in the frame Ox1 x2 x3 , the vector equation da dt

=

aa at

CO X

a.

(1.34)

This formula is fundamental to all processes in which a rotating frame of reference is used. Example. If a stands for the position vector of P and the variation ;1 is due to a rotation of P with angular velocity to' relative to the moving frame, then (3 a da — = co' x a and — = (co' + co) x a. dt t This is an alternative proof that angular velocities can be added vectorially.

1:8 The acceleration of a particle

We now apply the result of the last section to the motion of a point (or particle) when its position is specified in coordinate systems other than rectangular cartesian. In Vol. IV we explained that the "curvilinear" components of a vector are the resolutes of that vector taken along certain, mutually perpendicular, directions. These directions are those of the coordinate lines at that point. In particular, the velocity and acceleration of a particle are specified, in such coordinate systems, by giving the corresponding resolutes. We illustrate the method by calculating the spherical polar resolutes of the velocity and acceleration of a particle. We use two frames of axes, as shown in Fig. 8. The directions of 0 eie2 s are defined as follows:

30


0$3lies along the radius to the particle P; Wilies in the "vertical" plane containing 0$3 and Ox, and is perpendicular to 0$3;

OE, makes a r.h. set with 0$1and 0$3, and therefore lies in the "horizontal" plane Ox, x,. The polar coordinates (r, 0,v) of P are shown on the figure where OP = r, angle x, W, is 0, and v is the angle between the xraxis and the

x2

FIG. 8.

"meridian" plane W, containing P. The position vector r has components in the respective frames

{x1 x2 x,}, {0 0 r}. As the particle moves so the frame 0$1 $2 $3 rotates about 0. At any instant the orthogonal matrix relating coordinates of a point in the two frames is (cos e cosy, — sin? sin0 cosy cosy) sin0 silly) . x = ii 4 R = cos0 sinzp (1.35) — sine 0 cos 0 ,

(This matrix can be obtained from the product of the matrices corresponding to a rotation through v about 0 x, followed by a rotation through 0 about 0$2; or it can be obtained directly by elementary trigonometry.) When 0, 7,v vary, the angular velocity of the frame resolved along the axes 0E1E2E3is given by the elements of the matrix R' R eqn. (1.22); ,

§ 1 : 8]

31


the elements are

col = — sine, co, = 0, w3= yJ cos°. (For practical purposes the easiest way to obtain this particular result is to resolve the sum of the angular velocities, 0 about 0e2 and ti) about Ox,, in the required directions.) To find the resolutes of the velocity of P in the directions 0e, e, e, we use the equation, from (1.34),

8r dr v == — dt 8t

x

T.

But

8r co x r = — sine el

= (0 0 ,

cos01 x {0 0 r} = {r0 rti, sin0 0}. v = {7.0 rye sine r}.

(1.36)

A second application of the formula (1.22) gives the acceleration dv v dt = at Now

8v — = + 7.0 hp sine + sine + r/p cose , at x v = — sine 6 1p coso} x {r0 r/j, sine r} = fr — r/p2 sine cose r cos° sin° — sin2 O — 7.021. Hence the resolutes of f are f, = r6 — ryi2 sin0 cos0, /2 = ref) sine + 2i sin0 2r4cos0, fa

_ — 2.02 — r,02 sih20 .

(1.37)

The resolute f, is the radial component; f, is the longitudinal component; 12is the transverse (azimuthal) component.

The motion of a particle on a curve Before we discuss the motion of a particle on a curve in three dimensions we must consider the curvature properties of such a (twisted) curve. We recall first the definition of the curvature x of a plane curve. A pair of axes drawn at a point P, Fig. 9 (i), changes direction as P moves along

32


the curve; the curvature is defined to be x = dO/ds. If P is assumed to be moving with speed v(= ds/dt), the angular velocity of the axes is given by dO ds dO xv. dt = as dt (Strictly, the angular velocity vector is perpendicular to the plane of the curve.) For a curve in space we use a frame whose axes are in the directions of the tangent, principal normal, and binormal. (See Vol. II § 4:12.) As

FIG. 9.

P progresses along the curve this frame has an angular velocity which is used to define the curvature of the curve. For this frame the unit vectors i, n, b take the place of i, j, k of Ex. (i) p. 22. From that example it follows that Wit

d

_

di

— • n. co, = v — d • o co, = v —b d • t co3 = vd s s ' 8 If the position vector of P is given as a function r (s) of arc length, s, along the curve then dr — -t= ds by definition. The vector, b, being normal to the osculating plane, is in the direction of r dr (see Vol. II § 4 :12). x ds ds2 Hence = (r' x r'1 ) f(s),

§ 1 : 81


where primes denote differentiation w.r. to s and / (s) = 1/1 (r' is a function of s, making b of unit length.

33 r")1

db = / (s)(r' x r"') f' (s)(r' x r"), ds r" x r" 0.

since d •i ds

/ (s) (r' x r'") . r'

f (8)(r' x r") . r' = 0,

(1.38)

since both triple products contain two identical members r'. The angular velocity vector of the frame is hence a vector in the plane of i and b, and we may write w = v(ri (1.39) This equation defines t and x. The form of (1.39) holds because the plane containing fi and i, the osculating plane, is the plane for which three intersections with the curve coincide at P. Hence, when P moves a short distance along the curve, this plane does not tend to rotate about the principal normal, i.e., co, = 0. The component v x of co gives the rate at which the tangent changes direction and so, by comparison with a plane curve, x is called the curvature. The component v r of Co gives the rate at which the osculating plane rotates about the tangent and so r is called the torsion of the curve at P. The reciprocals of these are called the radius of curvature e = 1/x, and the radius of torsion a = 1/r. The relations between i, n , b and x, r are given by the Serret—Frenet formulae, which we now derive. The unit tangent, in a frame which moves with P, has resolutes (1 0 0} ••

di dt

at

di v

•

But x i = v(ri

xb) x i =

di •—=x. • • ds

(1.40a)

Similarly, since ft has resolutes (0 1 0} which do not alter in the moving frame, do d ft v(tb — xi). v co x = v(ti + xb) x dt = ds =

do • • ds 2 a OMES

= —xi +rb.

(1.40 b)

34


Finally, db db dt =v d b

co x b = v(ri + xb) x b = v(--rii). db " ds

(1.40 c)

Eqns. (1.40a), (1.40b) and (1.40c) are the Serret–Frenet formulae. We also use the formula for moving axes to find the resolutes of the acceleration of P along the moving frame of Fig. 9 (ii). In this frame the velocity of P is always directed along the tangent and the resolutes of v are therefore {v 0 0}, i.e., v = v t. The acceleration of Pis the vector f, where dv v f = + co x v. dt at But av dv 0 01 = dv dt t. at j dt dv .*. f = — i + v(ti + xb) dt

xvt,

dv

f = — t + v2 xfi. dt

(1.41)

Eqn. (1.41) shows that P has a component of acceleration dv/dt ( = d2s/dt2 = v dv/ds) along the tangent, and a component x v2 = v2/e directed along the principal normal; there is no component along the binormal. This result is more general than that for a particle on a plane curve although it is formally the same. (A plane curve is a three-dimensional curve whose torsion is zero at all points on the curve.) Example. If the time rate of change of acceleration of a point moving along a space curve is in the direction of the binormal show that the velocity is proportional to el , and that esatisfies d2

d82

e

2.

2/3

Since

dv f = -1 dt

df dt

d2 v dt2

v2 +

e

dv di v3 dfi 2v2 dv v3 de t + — v — + — — + — — fi – — — Il• dt ds e ds e ds e2 ds

§ 1: 8]

KINEMATICS TN THREE DIMENSIONS

35

Here we have used the relation between the operators d/dt = v d/ds.

••

df =t dt

d2v _ v3 ) dt2

v3 ea

( 3v2 dv e ds

D 11

v3 dp I. p2 ds

-

Since df/dt is directed along the binormal v3

d2 v dt2

3v3 dv 7

e2'

v3 dp • — ds e2 ds

The second of these is a separable equation giving v in terms of e, and integrates to v = A e'13. The first equation now gives

dt2

d r1 d

d iv dv \

d2 v

= v ds _ 1

as)

A2v d

ds

d

d2

d s2

e%

1

(Az e

2• -c

2/31 = v3

•

ds ds e

2

3

2v2 A2e2

— 2 —h.

This is the required result. Exercises 1: 1. Show that there are three points on the curve r = au3 + bu2 + + d the osculating planes at which pass through the origin, and that they lie in the plane (r x b). c = 3(r x a). d. 2. A point P moves along a curve in space, the arc from some fixed point P, of the curve up to the point P being s. Prove that, if r is the position vector relative to an origin 0, then r = ,§T; and, if dT/ds = xN, where x z 0, find an expression for the acceleration of P, explaining the significance of the unit vectors T, N and the scalar x. 3. The cylindrical coordinates of a point P relative to an origin 0 are r, 0, z. If It, S, Z are the corresponding mutually orthogonal unit vectors, prove that 1 d r dt—(r2B)S+zZ. OP = (Y^ — r 62) R + — If OP = a R 2)3Z x OP, where cc and j9are constants, show that if OP is initially in the plane z = 0, then the path remains in that plane, and r2 =h + 13 r2 , 2.2 = k 2ocr — /32 2.2— h2r-2, where h and k are constants. 2 a*

[see over]

36


If, further, when 0 = 0, r = 2cx//32, r = 0 and 6 = 13, show that the path is the cardioid r = (42) (1 + COS 0) . 4. If a point moves with uniform speed v on the surface of a right circular cone, with vertical angle 2a, so that its path cuts all the generators at the same angle fl, find expressions for the accelerations of the point when it is at a distance r from the vertex and show that the component acceleration in the tangent plane to the )3 to the generator in the cone through the point is inclined at an angle tangent plane. 5. A particle moves on a sphere so that its path cuts all meridians at the same angle and its longitude increases uniformly. Prove that the acceleration has magnitude proportional to the cosine of the latitude and direction making an angle equal to the latitude with the radial direction. 6. The earth rotates about its polar axis with constant angular speed co. A frame of reference Oxyz has its origin 0 at a point whose latitude is A and is directed so that Ox is due S, Oy is due E, and Oz along the outward radius. Find the resolutes along Ox, Oy, Oz of the angular velocity of the frame. Assuming that the centre of the earth is at rest and that the earth is a sphere of radius a, prove that the resolutes {u v w} of the velocity and {f g h} of the acceleration of a point at {x y z} are given by u = — yco sinA, v = 9 + xco

(z + a) co cosA, w = — yco cos2 ;

f = X — 2gco sinA — xco2 sin2 A — (z + a) co2sinA cos A, g

g + 2tco sin2 + 2ico cosA — yco2,

h = z — 2#co cosA — xco2sing cosA — (z + a) co2 cos2A.

1:9 The general motion of a rigid body So far we have proved the existence of an angular velocity vector for the motion of a rigid body having one point fixed. We now consider the motion of a rigid body which has no point fixed. The rigid body is represented by a frame A. $1e2 e3 whose origin A may be moving (see Fig. 10). The coordinates of a point P fixed in the body are x in the frame 0; x2 x3 and g in the frame A S 1 e2 e3 . The orientation of the moving frame is given at any instant by the matrix R, whose elements are the cosines of the angles between the directions of the axes of the two frames. The coordinates are related by

x = xA R .

(

1.42)

On differentiating this equation, remembering that g does not vary with time, we obtain the resolutes of the velocity of P in the fixed frame 0x1 x2 x3from the relation =

itg =

+ RR' (x — xA).

(1.43)

§ 1: 9]

37


The matrix R R' is the angular velocity matrix of the body, giving the resolutes in the frame Oxi x2 x3 . Hence eqn. (1.43) is equivalent to the vector equation (1.44) v = vA + co x (r — rA). This is the fundamental equation for the motion of a rigid body and gives the velocity of an arbitrary point P as the sum of the velocity of x3

'x2

FIG. 10.

a "base point" A and that of a rotation co about A. If we use another base point B, about which the angular velocity is (Di , vB = vA + to x (rB rA) v vB + co, x (r — rB).

and

.•. r = VA

co x (r5— rA) + co x

— rB)

= VA ± co x (r — rA). .'.w x (r — rB) = co, x (r — rB). Since r is arbitrary we conclude that w = cot . We therefore refer to w as "the angular velocity of the body" without specifying the base point. The vector 4) is invariant for changes of base point. Those points of the body where v is parallel to w must satisfy the condition 0=v xto=vA xto+ [cox(r—rA)]xco =VA x r — rA —

(r — rA) co2 — [(r — rA) . co] co W

•

r — rA

4.0 -

— rA) .

vA

4) X 2

1

X VA (02

0)2

[x co =0.

co .

38


Since r may vary we have

r — rA = w

63 X VA

(1.45)

where A is a (variable) scalar. This equation is the vector equation of a straight line. At each point on this line the velocity of the particle of the body is w

.

VA

co2

= pc.),

where p = (w . VA )/0. The motion of the body is a screw motion having this line as axis; the angular velocity of the body is w and the body is advancing along the axis with velocity p co; the scalar p is the pitch of the screw motion. The line is called the central axis (sometimes the instantaneous axis) of the motion. The value of p is not dependent upon the choice of "base point", for, if v is the velocity of any other point of the body, v.

= vA .co + [co x (r — r,)] .

= vA

.

.

(1.46)

The scalar product v . w is an invariant for the change of base point. This result concerning the equivalent screw motion is the threedimensional counterpart of the theorem of the existence of the instantaneous centre in plane motion. As the body moves the change of position of the central axis generates one surface in space and another surface in the body; the motion of the body is given by the combination of rolling of the "body-surface" on the space-surface with sliding along the direction of the central axis. The two extreme cases of the screw motion are: (a) p = 0; v . w = 0, in which case the motion is rotation about an axis without advance along the axis, i.e., rotation with one point fixed; (b) p infinite; co' = 0, i.e., w = 0, in which case the motion is one of translation, without rotation; every point of the body has the same velocity. If a rigid body is given an arbitrary small displacement, the change in position of any point is given, from eqn. (1.42), by ox =

oxA

(R g).

The term (5 (114) is a small rotation which can be written as 60 x (r — rA), using eqn. (1.32). Hence, for an arbitrary displacement

§1: 9]


39

of a rigid body, the change in position of any point of the body whose instantaneous position has position vector r is (1.44a) Sr = SrA + 60 x (r rA). —

[This result may also be obtained from eqn. (1.44) by taking Sr = v at, SrA =V Aat, 6 0 = w at, and regarding the displacement as taking place in a short interval of time at.] Examples. (i) Two wheels, of radii a and b and centres A and B respectively, are freely mounted at the ends of an axle AB with their planes at right angles to the axle. They are placed with their rims resting on a rough horizontal plane on which they roll without slipping, the distance apart of their points of contact being c. If the spins of the wheels about the axle are w, w' respectively, find the angular velocity of the vertical plane containing the axle.

We use the frame of reference A l23shown in Fig. 11 with A S3 along the axle and A Sl in a vertical plane passing through C, where the wheel with centre A touches the horizontal plane. The phrase "spin about an axis" means the resolute of the angular velocity vector along this axis. The three "rigid bodies" here have angular velocities : the frame of reference, angular velocity 0 = {01 02 03} the wheel at A, angular velocity w = {01 02co} ; angular velocity co' = {61 02 col. the wheel at B, [Since the wheels are mounted on the axle their angular velocities can differ from 0 only in the resolute along A It is intuitively apparent that the angular velocity of the frame of reference is directed along the vertical, which is given by the unit vector A = {cos oc 0 — sin oc}. Hence we may write 0 = {Q cos cc 0 — D sin oc}, where is the angular speed with which the frame rotates about a vertical axis. Since the point of the wheel at C is stationary, ;

0 vA +

x {a 0 0).

••• TA = (a 0 0) x {Q cos oc

0

co) = {0 — a 0.) 0).

40


Similarly,

1,8 = { 0

— bw' 0).

From the motion of the frame (regarded as a rigid body) VB = VA

+ 0 x {0 0

C

cosa}.

.'. {0 — bco' 0} = {0 — a co 0}+ {0 — cS2 cos2 a bw' — aw c costa This is the required angular speed. (ii) Two concentric spherical shells of radii a and b rotate with constant angular velocities co and co' respectively about fixed diameters. A sphere placed between them rolls in contact with both. Prove that the centre of the sphere describes a circle with constant angular velocity (aw + bw')/(a + b) about the centre of this circle, and find the angular velocity of the sphere.

We represent the direction OACB (Fig. 12) by the unit vector P. The velocities of A and B are VA = a (o.) x , vB = b (co' x t). Hence the velocity of C is vc = EVA + vB) = ia(co x t) + ( a co + b co' \ x ( 1 a+b )

2

(a + b)i) =

x t) x rc,

where rcis the position vector of C. Hence C moves as if it were a point of a rigid body rotating about 0 with angular velocity

St

-

aw + bw' a+b

§ 1: 10]


41

Since S2 is constant, w and co' being constants, C moves in a circle. We suppose that the angular velocity of the sphere is 0, then = VA

e

(b — a) + (b — a) (0 x

b(co' x 1.) = a(w x

[bco' — a co — (b — a) 0] x = 0. Since i is variable in direction, 0—

bco' — ao, b— a

Exercises 1:9 1. A sphere of radius a rolls without slipping between two parallel plane boards, distant 2a apart, which rotate with uniform angular velocities co, and co2 about non-coincident normals 01N1and 02 N2, respectively. Prove that, in general, the centre of the sphere describes a circle, and determine its angular velocity. Discuss the particular case when the planes rotate with equal and opposite angular velocities. 2. A rigid body S has spin co and a particle A of S has velocity v. Show that every particle P of S with velocity vector parallel to co lies on the line AP

xv (0 2

+ uw,

,a being an arbitrary scalar. The instantaneous velocities of particles at the points (a, 0, 0), (0, a/y3, 0), (0, 0, 2a) of a rigid body are (u, 0, 0), (u, 0, v), (u + v, —v 1/3, v/2) respectively, referred to a rectangular cartesian frame. Find the magnitude and direction of the spin of the body and the point at which the central axis cuts the xz-plane. 3. A body is pivoted about an axis AB and makes n/27r revolutions per second relative to its bearing. The latter is itself made to revolve with spin co about a fixed axis OZ inclined at an angle 0 to AB, the shortest distance OX between OZ and AB being a. Show that the motion of the body at any instant can be represented by a right-handed screw motion, of pitch (anco sing)/(w2 + n2 + 2con cos0), the axis of this screw intersecting OX at a distance from 0 equal to an (n + co cos 0)/(co2 + n2 + 2wn cos 0).

1:10 Angular velocities about non-intersecting axes If a body is rotating, the angular velocity can be represented by a directed segment of the axis of rotation, the length of the segment being proportional to the angular speed; the sense of rotation and the direction of the segment are related by the r.h. rule. It is immaterial which segment of the axis is used, i.e., angular velocity is a line vector. So far we have considered the motion of a body when it has angular velocities about

42


each of two intersecting axes simultaneously. We now consider the motion of a body which has two angular velocities, simultaneously, about each of two non-intersecting axes. We must define precisely how a body can have two such angular velocities and we do this by employing three frames of reference : a frame A$1 e2 e, which is fixed in the body; a frame Ox1 x2 x3in which the point A is fixed; and a frame 0 X, X2 X3which we shall take as "fixed in space". The orientations of these frames are given by the two matrices R1, R2: the matrix R1gives the orientation of Ox1 x2 x, in the fixed frame 0 X, X2 X, ; the matrix R2gives the orientation of A el e, e, in Ox1 x2 x3and is the same as the matrix R of (1.42). If both R, and R2 vary with time, the "body" A l e2 e, has two angular velocities, one about an axis through 0 and another about an axis through A. Corresponding to eqn. (1.42) we have, for the coordinates of an arbitrary point P in the body, x = xA I Rz (1.47) where xAdoes not vary with time. Coordinates in the frame OX, X2 X3 are related to those in Ox1 x2 x3by the matrix R, through the equations

X = R,x ,

XA = Ri xA .

(1.48)

In eqn. (1.47), XA varies with the time because of the variations in R1. The coordinates of P in 0X1 X2 X3 are

X = Ri xA R i Rg • Differentiating this relation gives the resolutes of the velocity of P in the frame 0X1 X2 X3 . = iti XA (h i R2 =

R1 xA (R1 R2

=Rl Ri`Y A = it,R;XA

RA)g Rik)R2

(x — x A)

+ R1 112) — XA) (kRj R1ll2R2R1) (X — XA).

(1.49)

All the terms on the r.h. side of (1.49) have a vector interpretation since Rl RI and R, R, R21R1 are both angular velocity matrices [see (1.30)] giving resolutes in the frame 0 X1 X2 X3. The vector form of (1.49) is v = wl x rA v = vA

w2) x (r — rA), x (r — rA), (w1

(1.50) (1.51)

where w =w1 co,. The form (1.51) shows that the angular velocity of the body is the vector sum of the angular velocity co, of the frame

§ 1: 10]


43

0x1 x2 x3and the angular velocity co, of the body relative to the frame, even when the axes do not intersect. We can also write (1.50) in the form v = co, x r

032 x (r — rA).

(1.52)

Using Fig. 13 we see from (1.52) that the velocity of P is the sum of the separate velocities it would have if the body were given each of the angular velocities alone. There are two special cases we consider in more detail.

1. Angular velocities with parallel axes In this case we write wl = wie .*. v = e x [ay.

= we e.

a)2 (r — rA)] = e x [(a), + co2) (r — re)] = (co, + w2) x (r — re ) ,

(1.53)

where (col. + co2) rG = co, rA . This definition shows that 0 is the centroid of masses co, at 0 and co, at A. (In Fig. 13 the direction of the angular velocities, e, is not necessarily perpendicular to the plane of the triangle OP A.) Hence the two parallel angular velocities are equivalent to a velocity equal to their sum about an axis through the centroid. Clearly, this result can be generalised to give the sum of any number of parallel angular velocities. The special case for which a), = — w1 , gives v

— x co2 rA = — co, x rA

44


This velocity is independent of r, showing that every point of the body has the same velocity, i.e., the motion produced is a translation without rotation. 2. Angular velocities about skew axes We have seen that angular velocity is a line vector so we can choose the points 0 and A to be anywhere along the respective axes. It is convenient to choose them to be at the feet of the common perpendicular. We use a frame of reference which simplifies the handling of skew lines. (See Vol. II § 4:7.) The arrangement is shown in Fig. 14, where we use alphabetical notation. In this frame j sincc), w2= co, (i cosoc — j sinoc).

rA = a k, co, = co, (i cos oc

The resultant angular velocity of the body is = Wl + 632 = (col + w2) i Cosa + (co, — w2) j sin ce. co2 = co +

w2 + 2co1co2cos2oc.

The velocity of the point of the body instantaneously coincident with 0 is vo = w2 x ( — ak) = — aco2 (i cosoc — j sing) x k = aco2 (i sing

j cosoc).

Hence the invariant v . w of this motion is

vo . (co, + 032) raw, (i since

j cosoc) • [(co, + w2) i cosoc

= aco, co, sin2oc.

(co, — co2)j sin oc] (1.54)

Ex.


45

If oc 7r/2, this does not vanish and the pitch of the equivalent screw motion is not zero. The position of a point on the central axis is given by (co x vo)/co2, [see eqn. (1.46)]. But w x vo a)

[(co, + cot) i cosa + (w1 co2)j sing] x aco2 (i sing + jcosoc) wi + 4 + 2co1co2 costa —

a co2 [(co, + w2) costa — (co, — co2) sin2a] k wi + 4 + 2co1co2 cos2a i.e.,

co x v, (02

aw2 (w2 + a), costa) k o4. + 4 2ohco2 cos2a

(1.55)

This gives a point on the z-axis and is shown at C in Fig. 14. Therefore the central axis intersects the common perpendicular 0A. From eqn. (1.55) we obtain co2 (co2 +co, cos2a) 0C (1.56) CA w1 (w1 w2cos2a) When a = 0, n/2 this result is equivalent to eqn. (1.53) above. Miscellaneous Exercises 1 1. Prove that the angle of rotation a of the resultant rotation defined by Euler's angles, eqn. (1.17), is given by cos ia = coslO cos i(q) p). 2. A right circular cone of vertical angle 2a rolls without slipping on a rough plane, its vertex being at a fixed point in the plane. If .12 is the angular velocity of the cone about its axis, find the angular velocities of (a) the generator of contact, (b) the cone about this generator. 3. If u, v, w are the simultaneous velocities of any three points of a rigid body, and a, /3, y the angles between the directions of these velocities taken in pairs respectively, prove that u2 — V2 , uv cosy — V2, uw cos/3 — V2 = 0,

uv cosy — V2,

v2 — V2, vw cosa — V2 w

uw cosfl — V2 , vw cosa — V2,

2 ___ V2

V being the lmear velocity of the equivalent screw motion. 4. A solid rotates about a fixed axis OA of unit vector u through an angle 99, displacing a marked point P by d to the point P'. Q is the midpoint of PP', and OQ = r. Prove that d = 2u x r tan f-

The axis OA makes an angle a with the vertical 0 Z,LAOP=i3(>17c— a), and 0 P = a. If the solid is constrained to rotate at a rate co about OA, show that when P crosses the horizontal plane through 0, the horizontal and vertical resolutes of velocity of P are aco cos a and a co (sin2 a — cos213)'I.

46


5. Two concentric cylindrical surfaces of radii r and r' rotate with constant angular velocities co and co' respectively about their common axis and measured in the same sense. A sphere of centre C just fits between the cylinders and rolls without slipping upon them. Prove that C describes a circle with angular velocity (r w + r' cor)1(r + r') about its centre, and find the angular velocity and the position of the axis of rotation of the sphere at a given instant, there being no spin about the diameter of contact. 6. A rigid body is rotating about a fixed point with respect to a given frame S. Show that there exists a unique vector 52 such that for each vector r fixed in the body its rate of change t with respect to S is given by t = 52 x r. A sphere is in contact with two horizontal plane surfaces which are rotating about non-coincident vertical axes with angular velocities co, co'. Show that if the sphere does not slip its centre describes a circle with angular velocity (co + co')/2. Find the centre of the circle. 7. A sphere of radius b and centre C rolls in the space between two rough concentric spheres of radii a + b and centre 0, which are constrained to rotate with constant angular velocities 521and 522 respectively. Show that the centre of the rolling sphere describes a circle uniformly with angular velocity co about an axis through 0, of unit vector such that a + ba — b 52 ± s22, 2a 1 2a and that the angular velocity 52 of the rolling sphere is given by a+b 522b

SLl

a—b 2b

--->-

522 + AOC ,

where 2 is a scalar, whose physical interpretation should be given. 8. A point P describes a fixed circle of centre C and radius a so that C P rotates at a constant rate co. If a fixed point 0 on the circumference of the circle is the origin of moving axes which rotate in the plane of the circle at a constant rate co' in the same sense as co, show that the polar equation of the path of P (r, 0) relative to these moving axes is coo r = 2a cos ( co — 2co' given that when t = 0, 0 = 0 and 0, C, P are collinear. Show also that the ratio of the radial resolute of acceleration of P relative to fixed axes at 0 to the apparent radial resolute of acceleration of P in the moving axes is co2I(co2

2,00)/

2w'2).

9. A rigid body consists of two straight wires AC, BC soldered at right angles at C, and free to rotate about the fixed point C, whose coordinates are (0, 0, a).

Ex. I]


47

The wires A C, BC are compelled to slide in slots along the lines y = — a, z = 0 and z = 0, x = — a respectively. Show that the locus in space of the instantaneous axis is 2x2 + 2y2 + (z — a)2 + 3(x + y) (z — a) + x y = 0. 10. The coordinates of a point on the surface of a torus (anchor-ring) are given by x = (a + b sine) cosy', y = (a + b sine) Any), z = b cos y . Obtain expressions for the resolutes of velocity and acceleration of the point in the normal, meridian, and transverse directions. 11. S' is a frame rotating with angular velocity SI about a point 0 fixed in a frame S. If a', a are the accelerations of a point r and v', v are its velocities in the frames S', S show that v = -1/1 + S2 x r and a = a' + a, d- ac, where a, = S2 x r + S2 (52 r) — r (S2)2 and a, = 2S2 x v'. An insect crawls around the rim of a disc of radius a with uniform speed v relative to the disc. The disc is constrained to rotate with constant angular velocity w about a line through its centre fixed in a frame S and making an angle a with a normal to the disc. Taking 5' to be a frame fixed in the disc, calculate the radial components of a' and a, and show that the radial component of a, varies between — co2a and — o.)2 a cost a.

CHAPTER II

SETS OF FORCES: EQUILIBRIUM 2:1 Introduction We assume that the reader is familiar with the definitions and concepts of elementary mechanics, in particular with force. (See Vol. III Chapter I.) In Vol. III we confined our attention to forces whose lines of action were parallel to one plane and to the motions of bodies parallel to this plane, the typical rigid body being a lamina. Following on our considerations of the motion of a rigid body in three dimensions we now discuss forces in three dimensions whose lines of action, in general, do not intersect. The results given in Vol. III are all special cases of those we shall obtain here.

2:2 Equilibrium When a particle is subject to the action of a number of forces F., the necessary and sufficient condition for equilibrium is that F=

F, = O.

(2.1)

This condition can be expressed, in three dimensions, by the equations X, = 0,

17, = 0,

Z, = 0,

(2.2)

where {X, Y1 Z,} are the resolutes of the force FL in an arbitrary frame of reference. The conditions for equilibrium of a rigid body must include the turning effects of all the forces acting on the body. In plane systems the turning effect of a force P about a point A was measured by its moment P . AN, where AN is the perpendicular distance of the line of action of P from A, and P = I P 1. If r is the position vector of a point Q (see Fig. 15) on the line of action of P, then the vector product AQ x P = (r rA) x P has the magnitude P . AN, and when 0, A, Q and N lie in the same plane it —

48

§2:2

SETS OF FORCES: EQUILIBRIUM

49

is directed perpendicular to that plane. For a three-dimensional set of forces F. acting on a rigid body the vector product (r, — rA) x F, is a vector perpendicular to the plane containing r, — rA and F, . (The vector r, is the position vector of a point on the line of action of F1.) The resolute of this vector product, the vector moment of F, about A, along any line passing through A in the direction is the turning effect of F, about this line. The resolute is (r, — rA) x F, . t.

0 FIG. 15.

A rigid body is in equilibrium under the action of a set of forces if (roughly speaking) there is no tendency for the forces to move the body in any way, i.e., no tendency to cause a translation or to cause a rotation about any axis through any point. This condition is fulfilled if =

Fi =o, T(A) = z (r, — rA) x Fi = 0,

(2.3)

where A is an arbitrary point. [It can be proved that if F = 0, and 11 (A) = 0, then T(B) = 0 where B is any other point. See eqn. (2.9).] When the forces satisfy these conditions the set is, by definition, in equilibrium. If the resolutes {X Y Z} of F and IL (A) M (A) N (A)} of r (A) are taken in the directions of an arbitrary frame of reference, eqns. (2.3) lead to six equations

X = 2' X, = 0, Y = Y, = 0, Z = Z, = 0, L (A) = 0, M (A) = 0, N (A) = 0,

(2.4)

as the necessary and sufficient conditions for equilibrium. [Sometimes we use suffix notation with {X1 X2 X3} for the resolutes of F and 101 (A) G2(A) 03(A)} for the resolutes of I' (A); the drawback of this

50


notation is a possible confusion of the suffix with the label i enumerating the forces of the set.] The reader will recall, from the last chapter, that a rigid body has 6 degrees of freedom and so needs 6 conditions to fix its position. Examples. (i) A uniform rod of length 2a and weight M rests with one end B on a smooth wall and the other end A on the ground where it is smoothly pivoted ; it is kept in equilibrium by a force F acting at B in the plane of the wall. The perpendicular from A to the wall meets it at M and the angle BA M is a. The inclination of MB to the vertical is O. If the normal reaction at the wall is R find the direction of F which makes IF I/IR I a minimum. The forces acting on the rod, resolved in the frame M xyz, are ; at A, {0 2a cosa 0), the reaction {X Y Z}; at B, {2a since sin0 0 2a since cos0}, the forces F, R, {F cosA R F sing} ; at G, {a sinoc sin° a cosa a sing cos0}, the

FIG. 16. weight {0 0 — W). This set of forces is in equilibrium. (The force F is shown in Fig. 16 as two resolutes F cosA = F1and F sin A = F3). Since we do not need to know X, Y, Z we use only the equilibrium condition r (A) = 0.

r(A)= AB x (F R)

AB x W = 0,

{2a since sin° —2a cosa 2a sina cos 0} x {F cos A R F sin A} + + {a sin a sin0 —a cosa a sin a cosO} x (0 0 — W} = 0. .'.{a sing sin0 — a cosa a sin g cos 0} x {2F cosA 2R 2F sin A — W} = 0. In terms of resolutes this leads to the three equations, (2F sinA — W) a cosoc + 2Ra sinoc cos° = 0, 2F a cosA since cos° — (2F sinA — W) a since sin0 = 0, 2Ra sing sin° + 2Fa cosA cosa = 0.

§ 2 : 2]

SETS OF FORCES: EQUILIBRIUM

51

To solve the given problem we must find the value of A which makes I 1? VIR I a minimum. From the third of the resolutes above F R

sing sing cos A cos a

Since FIR > 0, F, R being scalar magnitudes and therefore positive, the minimum value is given by cos A = — 1, A= n and in this case FIR =IFIIIRI= tancx sin°. Therefore F acts in the negative direction of the x-axis. (The reader should note a point of technique : when specifying the position and resolutes of a force it is advisable to write the coordinates first, as in the above working, so that the vectors are in the correct order for taking moments. Following this procedure helps to eliminate mistakes in sign). (ii) A tripod consists of three equal, uniform rods OA, OB, and 0 C, of weight W and length 2a, smoothly jointed at their common extremity 0. When the feet are held at the vertices of an equilateral triangle ABC, a light spring attachment exerts on each rod a couple of moment Wig+ in the plane through the rod in question and the centroid G, of ABC, where 0= LAOG and the tendency of the couple is to reduce 0. If 2/7c < < 1, show that the tripod can stand in equilibrium on a smooth horizontal table with A, B, and C at the vertices of an equilateral triangle. If weights W1, 1V2 , and W3 are now hung at the mid-points of OA, OB, and 0 C, and if A, B, and C are kept in the same positions by inextensible strings BC, CA, and A B, show that the tensions in the strings are equal and that their common value is + W2 + W3) tan0}/(61/3). As in the solution of two-dimensional statics problems it is usually necessary to consider the conditions of equilibrium of parts of the system as well as of the whole. It is also necessary to choose suitable directions for resolution and for taking moments. For the first part of this problem the forces T1, T2 , T3and the weights W1, W2 , W3 are absent (see Fig. 17). Each rod is subject to the normal reaction from the floor, R, the weight W, the couple Wk0, whose axis is perpendicular to the appropriate plane, and the action from the smooth joint at 0. Since the arrangement is symmetrical we consider only one rod. We do not need to know the action at 0 and so we take moments for the rod OA, about an axis through 0 perpendicular to the plane AOG.

R . AG — W AG — Wk0 = O.

( 1)

If we consider the equilibrium of the whole structure we see that the only external forces are the three weights W, and the three reactions R. Hence, vertical resolution

52


for the whole system gives (2)

R= W. .•. W AG = Wk0, a sine = kO.

(3)

This equation has a root 0 = 0, which we reject as a solution of this problem. If equilibrium is possible (3) must have another solution for 0 in the range 0 < 0 < n.

Since 1 > sin 0/0 > 2/n (see Vol. V p. 139), a solution is possible if, and only if,

k 2 — < 1, and — > — , a a n 2 k — < — 2ga the particle leaves the surface at once. If u2 < 2ga, prove that it remains on the surface and that, when the particle is in the plane z = h, it exerts on the surface a thrust of magnitude lm(2ga — u2) (c + a) a_'/'(h ay". 4. A smooth surface of revolution has the equation zr = c2 in cylindrical coordinates r, q z. If the axis of z is vertically downward and a particle is projected horizontally along the inner surface at depth z = a, with a speed due to a free fall ,

110


from rest through a distance n a, prove that the particle rises initially if n > and that its coordinate r is bounded if n < 1. If I < n < 1, and the reaction B makes an acute angle a with the z-axis when the particle is at the highest point of its path, show that B is then given by

R = mg cos a (1 + 2(1 — n) tans). 5. The z-axis of the smooth spheroid whose equation is x2/b2

y2/b2

z2/a2

1

is taken vertically downwards. Show that if a particle of mass m is projected horizontally along the inside of the spheroid at a depth c below the centre with speed a (2g/c)V2, it will move always between the horizontal circles z = c, z = 0, and show that the reaction of the spheroid upon the particle when it touches the circle z = 0 is 2mg (a2— c2)/(b c). 6. A particle is projected with velocity V horizontally along the interior surface of a conical cup, of semi-angle a, at a height 3 V2/g above the vertex; the axis of the cup is vertical. Show that the path of the particle lies between two horizontal circles, whose heights above the vertex are 3 V2Ig and 3 V2/2g. Show that the ratio of the reactions exerted by the cup on the particle when it is at these heights is 1 + 3 tan2a 8 + 3 tan2a

3:6 Motion relative to the rotating earth The simplest model of the earth is a sphere which is rotating about a diameter, the axis of the rotation being the diameter joining the N and S poles. In fact, the earth is not exactly a sphere, nor does the axis of rotation stay exactly in the same position relative to the material of the earth. But for our present purposes we shall ignore these deviations and use the simple model. To describe the motion of a particle an observer on the earth, moving with the earth, would normally use a set of rectangular axes which are fixed relative to the earth. To an observer outside the earth and not moving with it, say on the sun, such a set of axes has an angular velocity and also a translation of the origin ; it is not a Newtonian frame. Consequently, a particle observed by an observer on the earth will show deviations from "Newtonian" behaviour. Roughly, this deviation occurs because the earth "turns underneath" the particle while it is in flight near the earth, and it is this turning which causes the apparent deviations. We consider some special examples.

§ 3: 6]

THE DYNAMICS OF A PARTICLE

1. Apparent gravity There are two possible definitions of "vertical": one geometrical, in which the line joining the centre of the earth to a point on the surface gives the vertical at that point; the other dynamical, in which the vertical is the direction of the acceleration of a free particle released from rest near the surface. (The direction of a plumb line is the downward vertical in the dynamical sense.) We use an origin of vectors 0 which is fixed on the surface of the earth, and the angular velocity of the earth is taken to be the constant

FIG. 29.

vector S2 (see Fig. 29). We also assume that the velocity of C in its motion in the orbit around the sun can be taken as uniform, i.e., the time intervals considered must be small compared with one year. From eqn. (3.5) the acceleration vector of P is az x ax

f = + at 2 + 2S2 x — + S2 x (S2 x x), Ot

where f, is the acceleration of 0 and the angular velocity S2 is constant. Since a aiat = 0, both 0 and C being fixed in the earth,

vo = S2 x a, fo = .*. f =

+ 2S2 x

at

(S2 x a) +

x (S2 x a) = S2 x (S2 x a).

S2 x (S2 x x) + S2 x (S2 x a). (3.23)

The "actual" acceleration due to gravity is caused by the earth's gravitational attraction which is the attraction of a sphere; we

112


call this g, for points close to the earth's surface. The "apparent" acceleration due to gravity is the vector 02 x/ate observed by 0 for a particle accelerating freely near 0. In (3.23) we put f = go and a2xiat2 = g, with x = 0, the particle being at the earth's surface, and a x/at = 0, the particle being released from rest. Therefore go = g + SZ x (5/ x a),

(3.24)

(i.e., in the equation of motion (3.5a) P = m go , fA = fo , x = 0 = ax/a t.) The difference between the two accelerations arises because of the centrifugal force. If a particle is sufficiently close to the earth's surface, and is moving sufficiently slowly, so that we may take x = 0, ax/at = 0, then its equation of motion under the action of a force P is [see eqn. (3.5a)] az x m ate = P + mg. (3.25) Hence, we use "apparent" gravity instead of "actual" gravity.

2. The Coriolis force A free particle in motion relative to the earth is subject to the Coriolis force and so deviates from a straight path, in the eyes of a terrestrial observer. The Coriolis force — 2mS2 x a x/e t has a resolute in the earth's surface which tends to deflect the velocity ax/at to the left in the N hemisphere (right in the S hemisphere). It has been suggested that this is one factor which causes ocean currents in the N hemisphere to circulate predominantly in the anti-clockwise sense.

3. The weight Suppose a particle is suspended from P so that it is at rest at 0 relative to the earth. Then a2 x/Ot2 = 0 and T, the tension in the string, is given by T + mg = 0 from eqn. (3.25). This shows that the plumb line gives "apparent" gravity. In Fig. 30 the angle 0 (the geocentric colatitude) and 2 (the geographical colatitude) give the directions of the two verticals mentioned earlier. Because the particle moves in a circle it has an acceleration S22AO towards A. The "real" gravitational attraction, F, acts

§ 3 : 6]

113


along OC and the tension along KO, their resultant being directed along OA. Hence using CB 0 as a "triangle of forces",

F mS-22 A 0 mg BO CO CB Using the sine formula gives

F cos

But a Q2/g

mg maS22sin0 sin (0 — A) cos 0

1/290, a small quantity.

ctS22 0 — A — sine cos0. g

mg F

cos0 cos [A -I— 2.)] * 1 cos A cos A aS22 aS2 2 . o -* 1 — — SIR`A 1 g 9

— A) sink cost

In the following examples we assume that S2 is small, take a Q2 to be of the first order (since a is large compared with the coordinates xi), and neglect products such as xiS22. "Vertical" refers to the apparent vertical and g denotes apparent gravity. Examples. (i) A shell is fired with velocity v relative to the earth from a point which is taken as the origin of vectors. Neglecting air resistance, local variation of gravity, and terms of order Q2, show that its position at time t is given by x = vt

gH v x 52] t2

(gx 52)t3

.

114


Show also that the rotation of the earth diminishes the time of flight on the horizontal plane in the ratio 2[17 x 52 . g] g2 • In solving this type of problem we must recognise which are the small terms. We do this by comparison with the elementary solution with 52 = 0. Then we make successive approximations by substituting approximate values into small correction terms to give a closer approximation. The equation of motion is, writing ax/a t = X, etc., + 252 x = g. The use of apparent gravity g includes the term 52 x (52 x a). This can be integrated to give I+ 252 xx=v+gt. (1) If we denote the zero order approximation by capital letters, the motion is X = g,

= v + gt, X = vt + Igt2.

We substitute the approximate value X in the small term 2 (52 x x) to give — (v + gt) = -252 x (vt + Integrating this leads to x = vt +g t2 — 52 x (vt2 +

(2)

giving the required expression for x. The end of the flight occurs when x is horizontal, i.e., when x . g = 0, and in the zero-order approximation X . g = 0 at time T = —2(v . g)/g2. The corrected time of flight is given by 0 = (v g) t + g2t2— (Q x v g) t2. t [1

2 (52 xv.g)1 — g2

t=T T

t

•

T

[1

2 (v g) g2

—

T.

2 (52 x v . g) g2 2(52 x v g) g2

2(v x g2

g)

as required. (ii) A particle, in latitude A, is projected vertically upwards from the surface of the earth with velocity v. Prove that, to the first approximation, it returns to the earth's surface at a point distant (1 Ova cos .1)1g2 to the west of the point of projection, where S2 is the angular velocity of the earth. We use a rotating frame, see Fig. 31, with Oxi.due E, Ox, due N, Ox, vertical at 0. The origin of this frame is moving due E. In the standard arrangment:

§ 3: 6]


115 5

Motion of the frame Velocity of the origin vo = {aS2 cosA 0 0); Angular velocity of the frame 0 = {0 Q cosA S2 sinA) . xa 13

S FIG. 31.

Motion of the particle Position vector x = {x, x, x3}; 0x Velocity vector v = vo + — + 0 x x,

at

= {aS2 cos). 0 0} + {th„ + {0 S2 cosA S2 sinA} x {x1 x2 x3} = (; — x,S2 sinA + x3S2 cosA + a S2 cosA t, + x, S2 sinA ±3— xiS2 cosAl. a IT Acceleration vector f = — + 0 x v, a .•. f = {Xi— 2x2 52 sinA + 2x352 cosA 22 + 2±1Q sinA

x3 — 2±152 cos.}

+ {0 a 02sin A cos A — a Q2COS2 where we have omitted all terms containing 02, except a 02.

The forces The weight is a force of magnitude mg0acting towards the geometric centre of the earth (see Fig. 30) and, resolved correct to the first order in the given frame, is {0 m a D2sinA cosA — m go}.

The equations of motion ;— 2 ±,S2 sinA + 2 3,12 cos A = 0, X2 + 2tin sinA + a S22 sin A cos A = a S22 sin A cos). , — 2;S2 cos A — a S22cos2 A = — go. After cancellations, each of these can be integrated once to give — 2x212 sinA + 2x352 cosA = 0, x2 + 2x152 sinA = 0, — 2;52 cosA = v — go t [1 — (a Q2Igo) cos2A] = v — g t . [We have incorporated the initial conditions x1= x2 = x, = 0, t3 = x2 = 0, x3 = v. It should also be noted that A here is the complement of the angle A in Fig. 30 and that CO . Q2 (= aQ2) and KO . S22 differ only by a quantity of the second order].

116

A COTJE,SE OF MATHEMATICS

The zero-order approximation is, expressed in capital letters, X1 = 0, .2C2 = 0, .2(3= v — gt ; Xi = 0, X2 = 0, X,— vt — igt2.

The next approximation is, therefore, on substituting these values into the terms containing Q, = — 2(vt — igt2).Q cosA, t2 = 0, = v — gt. These can be integrated, subject to the initial conditions, giving

x1 = — (v t2— i-gt3) S2 cos A,

x, = 0, x3 = vt — igt2.

When the particle returns to the earth x3 = 0, t = 2v/g. At this time 4v3

8 v3 \

4v3

x, = — (— g2 — 3— —) g .S2 cos A = —3—12 g2 COS A ,

= , x3 = .

Since x1< 0 the particle falls to the west of 0. (iii) Foucault's pendulum We suppose that a particle P, of mass m, is hung from a point A {0 0 l} referred to the axes of Fig. 31 by a light inextensible string of length 1 and AP oscillates near the vertical Ox3. The coordinates of P at time t are {x1 x2 x3} and T is the tension in the string. The acceleration components are as in the previous example. The forces are: the weight {0 0 — m g), the tension T {— ;11 — x2/l 1 — x311}. F=

T x, 1

—

T x,

— mg + T

lx3 I

Because the string is inextensible /2 =

+

+ (l — x3)2.

.• . 21x3= xi + x2 + Assuming that small oscillations take place, i.e., that x1, x2, x3are small, we see that x3is of the second order and may be neglected. Hence, the equations of motion are

x2 + 2412 sinA = — T x21(m1). 0=—g+

x, — 2:t2 2 sin A = — T xil(m1),

We now use the complex combination z = x1+ i x2, so that, with T = mg,

+ Qi sin A + (g11) z = 0. In this equation we write co = 2 sin A, n2 = g/l, giving

2iwz: v 2 z — 0.

§3: 6]


117

Now the position of the bob is represented by the complex number z in an Argand diagram with the real axis due E and the imaginary axis due N. The solution of the differential equation has the form z = Z e-ic", where 2,-I- (c02 + n2)

= 0.

The factor e -1a)t shows that the path traced out by z is obtained by rotating the path traced out by Z about the vertical with angular speed — w = — Q sin A. We have seen [Vol. [II § 6:5, and example (iii) p. 103] that Z traces out an ellipse with centre at 0 in the periodic time, (neglecting co2), 27c/n. Hence, we conclude that the bob traces out the small ellipse of a spherical pendulum "fixed in space", while the earth "rotates underneath it" with the angular speed S2 sin A. This result was first verified experimentally by Foucault in 1851 and the steady rotation, in the sense opposite to the rotation of the earth, of the plane of oscillation of a long pendulum gives experimental evidence of the rotation of the earth. Exercises 3:6 1. A particle is released from relative rest at a height h above the earth. Show that when it reaches the ground it has deviated from the apparent vertical by a distance Q cos cc y(8h3/g) approximately, where S2 is the earth's angular velocity, a. is the latitude, and air resistance is neglected. In which direction is this deviation? 2. A projectile is fired eastward in a northern latitude A with speed v at an angle of elevation oc. Neglecting terms in Q2, where Q is the angular speed of the earth's diurnal rotation, show that the projectile will strike the horizontal plane through the point of projection at a point whose distance from the plane of projection is 4S2 v3 sin2 oc cos a sin A g

towards the south. 3. A particle is projected in latitude A with speed V at an elevation a in the vertical plane through the direction 0 to the east of north. Show that owing to the Earth's rotation (angular velocity S2) the time of flight is increased by (2.(2172sin 2 a cos A sing)/g2 , where powers of S2 beyond the first are neglected, g is assumed uniform, and air resistance is ignored. Find the distance from the vertical plane of projection at which the particle falls. Show also that there is a certain value of a for which the range is unaffected to the first order in Q. 4. A shot is fired from a point on the earth's surface so that the horizontal component of the initial velocity points due north. Show that, when the shot again reaches the earth's surface, its deviation from the northerly direction due to the earth's rotation is to the east or to the west according as 3 tan A — tan cc is positive or negative. Here A is the latitude, and oc is the angle of projection above the horizontal. The earth's curvature and variations in gravity may be neglected; it is also assumed that the time of flight is short compared with the length of a day.

118


5. A tidal current runs in a channel of constant breadth b in latitude 2, with uniform speed V in a direction between E and N. Prove that if Q V is very small compared with g, the tide is higher on the eastern side of the channel by approximately (2.12b V sin 2)/g. 6. A simple pendulum of length 1 is suspended from a point on the earth's surface in north latitude 2. The bob of the pendulum is released from rest relative to the earth in a position slightly displaced from its equilibrium position. If x and y are the displacements of the bob in the apparent horizontal plane to the south and east respectively at any time t, show that

x iy = a ei("""){p cosp t + in sinpt}, where n = Q sing, p2 = g 11 and a and a are real constants depending on the initial position of the pendulum. 7. A simple pendulum of length 1 is allowed to describe a circular cone of small angle; show that the period depends on the direction of rotation about the vertical, and that the difference of the periods is approximately (4n ,521 sin 2.)/g, where .S2 is the angular velocity of the earth.

3:7 The motion of a charged particle We considered the motion of a particle under the action of a central force in Vol. III Chapter VI. Such a motion is in fact a plane motion, although the particle is not constrained in any way; since the force acts through the fixed centre the angular momentum vector mh is constant, so that the motion takes place in a fixed plane which contains the centre of force. We do not consider this motion here. As pointed out in Vol. III the inverse square law gives the force between two particles carrying electric charges. Here we consider the motion of a charged particle in magnetic, or combined electric and magnetic, fields. This motion is important in connection with the behaviour of ionised gases in a magnetic field, or with the motion of charged particles, emitted by the sun, in the earth's magnetic field or in interstellar space. An understanding of these motions is also necessary for the design of particle-accelerators for use in experiments of nuclear physics in which the nuclei of atoms are bombarded with fast-moving atomic particles. We shall not discuss such detailed applications but point out the salient features of such motions. Note, however, that we discuss the motion of a single particle only. In astrophysical and other applications the interactions between charged particles, when there are many moving particles, may cause significant differences from the behaviour discussed below.

§ 3: 7]


119

The theory of electromagnetism (see Vol. VII) indicates that a charged particle moving in an electromagnetic field having E as its electric and H as its magnetic component experiences a force, due to the electromagnetic field, e [E (v x II)/c], where e is the charge and v the velocity vector of the particle. The factor 1/c (c is equal to the velocity of light), arises from the scheme of definition of units (Gaussian, here) in electromagnetic theory: with different schemes of units the coefficients are different. The contribution e E is very similar to the force of a gravitational field which is proportional to the mass of the particle. It is the contribution e x II)/c, called the Lorentz force, which is basically responsible for the characteristics of the motion. The Lorentz force acts at right angles to the velocity. Hence the general effect of a magnetic field is to alter the direction of motion without altering the speed of the particle, and so adds a circular component to whatever motion is due to other agencies. This circular motion takes place in a plane perpendicular to H and its sense is given from that of H by a left hand (corkscrew) rule. The result of the action of a magnetic field on a moving charged particle is to produce a spiral path, or sometimes a cycloidal motion. On a large scale, as in the motion of charged particles in the neighbourhood of the earth, the particles tend to follow spiral paths which wind closely around the lines of magnetic force; in fact, the particles may even appear to be following the lines of magnetic force, so closely does the spiral fit. Usually, the electric and magnetic vectors of an electromagnetic field are perpendicular to each other but are neither uniform in space nor constant in time. However, such uniform, constant fields give an approximation to the motion of a particle for short distances and times and, for slowly varying fields, provide a qualitative idea of the shape of the path. The magnetic field, as will be seen in the following examples, is always associated with a rotation or circular effect in the path. The electric field is usually the negative gradient of a scalar potential function, E = — grad 99. Examples. (i) A particle carrying an electric charge moves in a uniform magnetic field. Show that in general its path is a circular helix, and consider the special case when the particle is projected in a direction perpendicular to the field. What is the path of the particle in a uniform electric field?

120


The vector equation of motion is mf = e(v x Wk. We choose fixed axes of coordinates with Ox, parallel to the vector H, leaving other details of the axes to be decided later. In such a frame the equation of motion eH

m

-{±1'x2 ±3} X {0 0 1). •

me

With co = eH/(mc), this becomes — co 2 = 0, i2 + co±3. = 0, i3 = 0. We solve these equations by writing = x1+ ix2obtaining +iw = 0, x3=0. These integrate to give = A + Be' , x3 = Ct D, where A, B are arbitrary complex constants, and C, D are arbitrary real constants. We choose the origin so that x3 = 0 when t = 0. Therefore D = 0 and x, = Ct.

— A=

Since I — AI = IBI = constant, the point represented by traces out a circle with angular speed — w. The constant a) is called the cyclotron frequency. Since x3 increases uniformly with time the path is a circular helix. When the particle is projected at right angles to the magnetic field ±3 = 0 initially and C =0. Hence x3= 0 throughout the motion, which is the circle given by IC — AI = In a uniform electric field the equation of motion is mf = eE. Since f is constant the acceleration is also constant and the path is a plane parabola, as in the motion of a projectile. (ii) The potential of an electrostatic field is 99 = bxy, where b is a constant. An electron of charge e and mass m passes through the origin of coordinates with velocity (u, v, w). If eb > 0, prove that the path of the electron is given by 2a w x = (u + v) sin (az) + (u — v) sinh (az), 2a wy = (u + v) sin (az) — (u — v) sinh (az), where eb )V2 mw2 In this problem the coordinate axes are prescribed and the electric field is E = — grad 99 = — (by bx 0).

§ 3 : 7]

121


The equation of motion m f = eE has resolutes mx = —eby, mg = —ebx, 920. = 0. = _ oc2 w2y, = _ c 2 w2 x, 2= 0. The initial conditions are, t = 0: y=v, i=w.

x=y=z=0; • z = wt.

The differential equations for x and y can be integrated by any standard method, e.g., writing = x iy, or using Laplace transformations, to give x = {(u

v) sin (ocwt) + (u — v) sinh (awt))1(2ocw),

y = {(u

v) sin (a wt) — (u — v) sinh (awt))I(2aw).

The required result follows. The equations we have obtained represent two surfaces; the path of the particle is their intersection. (iii) An axially symmetric magnetic field has components (H (r) 0 0) referred to cylindrical polar coordinates {r 0 z). At time t = 0 a particle of mass m and charge e is at the point (a 0 0) and has velocity (0 0 V}. An electric field (E (r, t) 0 0) acts on the particle and is such that in the subsequent motion r stays equal to a. Find expressions for z and 0 as functions of t, and deduce that m V2 E (a, t) = -- — sin2 t, e a where (.0 —

eH(a) MC

In cylindrical polar coordinates the following vectors have resolutes in the directions r-increasing, 0-increasing, z-increasing given respectively by v = {i

f = jr

—

relz

d

(r2 0)

—

d

t

The force on the particle is eE e(v x II)/c = e {E 0 0) + (e/c)

r0

x (H 0 0)

= e{E iHIc —rOH/c}. The resolutes of the equation of motion are m(Y. — r 02)= e E (r, t),

d eZ — — (r2 0) = — H(r), m r dt

=—

er0

H (r).

In the special motion given, r = a, so that ma02= — eE (a, t), ma& = 5

(-IMES

eH(a) . z c

2.7

—

eH(a) c

ael. (1) (2) (3)

122


We write eqns (2), (3) as

a6 = wz, z = —aco6, where o) = [eH(a)]l(m,e). Solving these equations by writing = z + ia0, we find z = B sincot + A coscot + zo, a0 = A sincot — B coscot + a0,, where A, B, 0,, z, are arbitrary constants. The initial conditions are, when t = 0,

0 = 0, z = 0; cce+ = 0,

= V.

.•. A = 0, B = Vico, a00 = Vico, zo = 0. .• a0 = (V la)) (1 — coswt), z = (V/w) sinwt. But from eqn. (1)

E(a,t) = —

ma

62= —

e

ma V2 — sin2co t a2 e

(iv) A particle of charge e and mass m moves in a uniform magnetic field H and in an axially symmetrical electric field of potential V (r), where r is measured from an axis that is in the direction of H. Given that the particle has zero velocity at a point where r = a, find expressions for its subsequent radial and transverse velocities in terms of r. Show that if V (r) = V (a) — A log (r la), A being constant, the particle will never cross the cylindrical surface r = A a, where A is given by 8A logyl.

eH2 a2 /

1 \2

A

m c2

Here we use cylindrical polars {r 0 z} with the z-axis parallel to the vector H. The field vectors are

E = {— 0 0), H = {0 0 H}, where V' = d171c1r, and v = {r rB i}. The force acting on the particle is

F = eE + e(v x 11)/c = e {— V' 0

+ (e/c)

TO i} x {0 0 H}

= e{— V' + rOH/c

0}.

The equation of motion has resolutes — rO 2= —eV'/m, + corO,

1 d

7 dt

= — cot, z = 0,

(1) (2) (3)

where co = (eH)/(mc). Eqn. (2) can be written d fr2O, dt

co

r2 dt 2 )

which integrates, subject to the initial conditions El = 0 where r = a, to give

r20 =

(a2 — r2) .

(4)

53:7]


123

Eqn. (1) now gives (a2

r2)2

e V'

co2 (a2 — r2)

4r3

9

2r

i.e., e V' —

+ cot (a4 7.4)

m

4r3

This integrates to give e v(r)

1 2

(02

a4

r2

4

2r2

2

+

C.

The condition v = 0 where r = a implies that r = 0 there. •

.

1 • 2

t2 =

rrc

e V (a) 0=

co2a2 — C. 4 +

r2 co2 a4 — V(?)] + — {a2— — 4 2r2 2

[ V (a)

0) 82

1 [17(a) —

2

V (r)]

a2

[ 7. — r

2 -

(5)

Eqns. (4), (5) give the required velocity resolutes. When V (a) — V (r) = A log (r/a) we see from (5) that r = 0 when r = Aa. (A > 1), if co2 a2 1 — A logA (— — A.)2 . m 8 A r A sketch of the graphs of log (7

a2 r 2 shows that they intersect at r = a r and at one other point only (for r > a). Only between these points does the graph of a2 2 r log 7t-) lie above that of (—r— — r) ; hence r2 0 only if a r Aa. We conclude

(

that the particle never crosses the cylinder r = Aa.

Exercises 3:7 1. A charged particle, of mass m and charge e, when moving in an electric field E acting in the direction of the x-axis and a magnetic field H acting in the direction of the y-axis, has the following equations of motion: mx = e(E — fH), mg = 0, r/0 = exH , where the dots denote differentiation with respect to time. By solving these equations, show that if the particle is at the point (0, 0, 0) at time t = 0 and if 9 = 0 at t = 0, then the path of the particle cuts the z-axis at a distance (27tmE)1(eH2) from the origin, no matter what are the values of t and z at time t = 0. 5*

124


2. A particle of mass m and charge e moves under the action of a central force — mk2r, where r is the vector from the origin 0 to P, the position of the particle. It is also acted on by a uniform magnetic field of magnitude H whose lines of force run parallel to the z-axis. Given any circle in the plane z = 0 with centre at 0, show that the particle can be set in motion so that its path is this circle described with constant speed; also that the periodic time has one of two values for all such motions, one value for each sense of circulation, these values not depending on the radius. y2eH Find the ratio of the two periodic times when k — 971c

3. A particle of unit mass moves in the plane z = 0, referred to a right-handed Cartesian set of axes Oxyz, under the action of two forces; one force is proportional to the velocity of the particle and acts in a direction opposite to the velocity, while the other force is also proportional to the velocity but acts in a direction perpendicular to the velocity and perpendicular to 0 z (the velocity, the direction of this force and O z forming a right-handed set). Show that the equations of motion are

= -Hg - k±,

g= fit— kg, where H and k are constants. Show that, if the particle is projected from the origin with velocity V, then, as cc, the particle will approach a point distant V/y(k2 + H2) from the origin. Prove further that the total distance travelled tends to V/k. 4. A particle of mass m carries electrostatic charge e and is initially situated at the origin 0 of rectangular axes Ox, Oy, 0 z, where 0 z is vertically upwards. A uniform magnetic field of intensity H acts parallel to the axis 0 y, and the particle is allowed to fall from rest under the action of gravity, g. Prove that in the subsequent motion its greatest depth d below the plane 0 x y is 2g m2 c2/e2 H2 and that it meets Ox again at a distance ird from the origin after describing a path of length 4d. 5. An electrified particle of mass m and charge q is projected with velocity v, in a medium which opposes the motion by a force equal to k times the velocity. In addition there is a uniform magnetic field H which produces on the particle a transverse force qv x H, where v is the instantaneous velocity. From energy considerations, or otherwise, show that I v I = Ivoj eIf the particle is projected from the origin in a direction normal to H, if the x-axis is taken parallel to v0 , and the z-axis parallel to H, show that the particle will ultimately come to rest at the point k2

m vo q2H2 (k, —qH, 0),

where vo = vo',H = Hi. 6. The equation of motion of an electron of mass m and charge e moving in an electric field E and magnetic field H is m a, = eE

v x H, where v is the veloc-

ity of the electron and c is numerically equal to the velocity of light. Show that,

Ex. III


125

relative to a set of axes rotating about the direction of the magnetic field with a

II, the form of the equation of motion be2me comes m a2 = eE, provided that terms proportional to the square of the magnetic field can be neglected. constant angular velocity SZ = —

Miscellaneous Exercises III 1. A rigid smooth fixed wire is in the form of a helix drawn on a circular cylinder of radius a whose axis is vertical. The tangent to the helix makes the angle a with the horizontal. A bead slides down the helix from rest. Prove that the time taken to descend the vertical distance Is is (2h/g)'I' coseca. Find also the reaction of the wire when the speed of the particle is v. 2. A heavy particle moves on a smooth bowl in the form of a paraboloid of revolution which has its axis vertical and vertex downwards. The latus rectum of an axial section of the bowl is 4a. The particle is projected horizontally along the surface with velocity 1/(2gy2) at a height ylabove the vertex. Show that the path of the particle touches alternately the two horizontal circles at heights yi and y2 above the vertex. Determine the angle a through which the axial plane through the particle turns between one contact with the circle y = yiand the next contact with the circle y = y2 . Show in particular that, if yi = a and y2 = 3a,

v ( 3 _ cos0) TS

a

—

V3 r

J

2 j

2 — cos°

dO.

0

3. A particle of mass m moves on the smooth inside surface of the paraboloid of revolution x2 + y2 = 4a z, whose axis is vertical and vertex downwards. The path of the particle lies between the horizontal circles z = p and z = q. Show that the angular momentum of the particle about the axis is mh, where h2 = 8ag pq, and that the velocity of the particle is {2 g (p q— z)yb. Show also that the reaction between the surface and the particle when the particle is at z = p is mg (a + q) {a(a p)}112 • 4. A particle moves under gravity and without friction on the surface formed by the rotation of the curve a3 Y = az + a2 about the y-axis, which is vertically downwards. It is projected horizontally with velocity v from a point on the surface at distance a from the axis. Show that if v2= ga/2 the particle continues to move in a circle of radius a. Show further that if v2 < g a the motion takes place between two circles of radii a and ay/1/(ga — v2), and that if v2 > ga the particle rises steadily.

126


5. A fixed smooth circular cone of angle 2a has its axis vertical and its vertex downwards. A particle of mass M hangs below the cone by a long string that passes through a small hole at the vertex, the upper end of the string being attached to a particle of mass m that is held on the inner surface of the cone at distance d from the vertex. The system is set in motion by projecting m horizontally along the cone with velocity V. Show that, provided mV2> gd(M + m cosoc), the particle m never approaches nearer than d to the vertex. Show also, in this case, that the greatest distance of the particle m from the vertex is a solution of the equation 2g (M + m cos a) x2— Tn,V 2(d + x) = 0. Determine the tension in the string at the moment when m is at the greatest distance from the vertex. 6. (i) The bob P of a simple pendulum moves in a vertical circle of radius 1, the speed at the lowest point A being (2g d)'/2. Taking the height of P above A as 21 sin2v, show that 2i)2 = g (1 — k2 Sin2V)

l k2

(k2 = 21Id).

(ii) The bob P of a spherical pendulum moves on a sphere of radius a between planes cutting the sphere at depths b f c below the point of suspension 0. If the depth of P below 0 is z, show that a2 Z2= 2g(z — b — c) (z — b + c) (f — z). where 2b f = c2— a2— b2. Putting z = b + c cos 2v, show that v varies synchronously with the angular motion of a simple pendulum of length a2/c, the speed of the bob at the lowest point being d

b

(a2 + (b + c) (3b —

7. A particle of unit mass is released from rest at a point of latitude and falls under gravity in a medium which produces a retardation it times the velocity. Show that the horizontal deviation at time t due to the rotation of the earth is 2gSl cosA k3

[kt(1 + e-m) — 2(1 — e-kt)]

eastwards, to the first order in the earth's angular velocity Q. 8. A shell is fired with speed V from a gun in north latitude A, and is observed to strike the horizontal plane through the point of projection in a direction 0 to the west of south. If a is the elevation angle of projection, and Sl the angular velocity

Ex. III


127

of the Earth, and atmospheric resistance is neglected, show that to the first order in Q the direction of projection must be

0 — 212 V tan a. (3 sin A cos cc + cos A cos0 sin g)/3g to the west of south. 9. Give the theory of the simple pendulum of length L for a place of apparent latitude A, taking account of the rotation of the earth. In particular show that the motion can be described as simple harmonic motion in an ellipse, with period 2n y(L/g) relative to axes which rotate about the apparent vertical with angular velocity Q sin A, where g is the apparent acceleration under gravity and Q is the angular velocity of the earth. Show that if the particle is released from rest relative to the earth at a distance d to the East of its position of relative equilibrium, then the semi-axes of the ellipse are of lengths d and d.1211(LIg) sin), 10. An electron of charge e and mass m moves under the influence of constant uniform electric and magnetic fields E = (E, 0, 0) and H = (0, 0, II). Show that the electron's motion is made up of (i) a uniform motion with an arbitrary speed parallel to H, (ii) a circular motion with angular velocity eHlmc in a plane perpendicular to H, (iii) a uniform motion with constant velocity c(E x H)/H. 11. A charged particle of mass m and charge e (e.m.u.) moves in the magnetic field of a current j (e.m.u.) flowing in an infinite straight wire. Show that the angular momentum of the particle about the axis of the wire is constant and that if, when the particle is at P, its direction of motion makes angles 0, tp with the direction of the current and the magnetic field at P, then

(i) cos0 —

2ej m

log r = constant, (ii) r cow = constant,

where v is the speed of the particle and r its distance from the axis of the wire. Show further that, if the particle is projected from a point P distant a from the wire in the direction of the magnetic field at P, the trajectory will be confined to where b is the positive root of the equation the region of space defined by a < r 1

an

2ej 2 \ mv

log

LI

a other than a. 12. An electron of mass m, charge — e and velocity v, moving in a constant uniform magnetic field Hk, experiences a force 2

—(ellIc) v x k, where k is a fixed unit vector and c is the velocity of light. Prove that if the electron is subject also to a force F = f(r) r, where r is the position vector of the electron

128


relative to a fixed origin 0, the equation of its motion relative to a frame rotating with constant angular velocity w k about 0 is 9??,

eH 2(ok x + co2 k x (k x r)} = F — — c

xk

eHco

(k x r) x k .

Hence show that if eHlm,c is small the effect of the magnetic field is, to the first order in eH/mc, to cause electron orbits to precess with angular velocity (eH/2mc) k. When there is no magnetic field an electron moves in simple harmonic motion with frequency n along a straight line 1. Show that in a constant uniform magnetic field, inclined at an acute angle a to 1 and of magnitude H such that eHlmc is small, the motion of the electron is composed of vibrations with frequencies n, n + n' and it - n', where n' = e 111(47cmc).

CHAPTER IV

THE MOTION OF A SYSTEM OF PARTICLES 4:1 Description of the system We consider now mechanical systems which consist of a number of particles. Such systems include particles connected together by light, elastic or inelastic, strings, rigid bodies and systems of connected rigid bodies. We suppose that a particle of the system, of mass mi , is at position ri , has velocity vi and acceleration fi, the origin of vectors being a fixed point. Although dri d2 r, dv. f. = vi dt "

dt

dt2

we shall usually keep the separate symbols v, f to denote the velocity and acceleration of a material particle of the system. When we have occasion to refer to some other point A, B, ..., which may or may not coincide with a material particle, we shall denote its position by rB , ... and generally use derivatives TA ,rA,etc., for the velocity and acceleration of such points in preference to the symbols v, f. We shall make considerable use of the centre of mass 0 of a system and of position and motion relative to G. We shall use f, V, f, for the centre of mass and Ri , 1k, R, for quantities relative to O. Therefore ,

ri =

Ri

fi =

Ri , vi =

(4.1)

and, from the definition of 0,

mi = mi Ri — 0, /' mi ft, — 0,

,

rn

—0•

(4.2)

Each of the particles of the system is subject to the action of forces which arise from three causes: (i) applied forces or external forces such as gravity, electromagnetic action and so on which arise from causes outside the system; 5a

CMES

129

130


(ii) internal forces which are forces acting on one particle of the system due to interactions with the remaining particles of the system; (iii) forces of constraint which arise from constraints such as smooth hinges, inextensible strings or rods, surfaces in contact. We denote the force acting on the particle labelled i by F, F, being the applied or external force, F: being the internal force. Forces of constraint are included in F, or F: according to the type of constraint. Thus, if two particles of the system are joined by an inextensible string the tension in the string is an internal force of constraint and is included in . On the other hand, if the system is constrained to rotate about a fixed point, or remain in contact with some surface, the force from this constraint is an external force and is included in F, . We make one important assumption concerning the internal forces, viz., that they conform to the Law of Action and Reaction. This means that if particle i exerts a force on particle j, then the latter exerts on particle i, and a force =— The internal force is given by F; =

F. Because of this detailed

balancing of the internal forces in pairs we assume that, considered as a set of forces taken together, F: are in equilibrium. .*.

= 0,

r (A) = 2' (ri— rA)

x

= 0.

These are the conditions of equilibrium (2.3) and A is an arbitrary point. A rigid body is regarded as made up of continuous matter which can in principle be subdivided into elements which are as small as we please. (Of course, this assumption is not in strict accordance with the molecular theory of matter but serves as a good approximation to the solid bodies of everyday experience. This theory is a macroscopic theory, not a microscopic one.) To apply the results of our theory of systems of particles to rigid bodies we assume that the "elements" (of volume or mass) of a rigid body can be treated as identifiable particles. Thus e dr, where e is the density at the volume element d r, plays the part of m, and summations over "all the particles of the system" can be made by integration. We assume that the internal stresses (which must be such as to maintain rigidity in all circumstances) exerted by one element on another, are in equilibrium for a rigid body just as for a system of discrete particles. Subject to these qualifications the results obtained in

§ 4 : 2]

THE MOTION OF A SYSTEM OF PARTICLES

131

the following sections are independent of the particular law which gives the magnitude of the action between two particles or elements of the system. In practice, of course, solid bodies are not "rigid" in the strict sense (see Vol. III § 1 :2) but our theory nevertheless gives considerable insight into the motions of solid bodies. The theory developed in this chapter has much that is formally identical with Chapter VII of Vol. III; here, however, we are not limited to plane systems or motions. The consequences of the removal of this limitation are chiefly to be found in results concerning rigid bodies. 4:2 The dynamical variables Following the lines of Chapter VII of Vol. III we give first the definitions of the variables which are of significance in dynamics, and consider a "cloud of particles", each particle having its own motion.

1. Kinetic energy This is defined to be the sum of the kinetic energies of the individual particles,

T=

Mi

=

27 m,v7.

Making the transformation given in (4.1) we obtain

T = z 27 mi(v2 + 2V . + = i MV2 + V • (Z

T = M V 2 T,., where T,. =

Mi

(4.3)

fq is the kinetic energy of motion relative to G and

M (= Z mi) is the total mass of the system. 2. Linear and angular momentum Each particle of the system has its momentum m,v, and all these momenta constitute a set of line-vectors since each momentum is localised in a line passing through the point r,. We employ here a treatment similar to that used in Chapter II for a set of forces and obtain similar results in this case. Associated with an arbitrary "base point" A are two vectors; one of these vectors, the resultant sum taking no account of position, is (4.4) p -= 7 mi vi 5a*

132


This is defined to be the linear momentum of the system. The other vector, a moment vector, takes account of position and is h(A) =

(4.5)

(ri — rA) x mi vi .

This vector is defined to be the angular momentum, or moment of momentum, of a system about A. As for other sets of line vectors, the vector p is independent of the choice of base point A, but the angular momentum vector is given by h (A) = h (0) — rA x p.

(4.6)

Here 0 is the origin of vectors, and (4.6) enables us to find the angular momentum about any point A from h (0) and p. From eqn. (4.5) h (0) =

ri x mi vi .

(4.7)

There is a central axis for this system but it does not play an important part in dynamics. The use of transformation (4.1) shows that (4.8)

p = MV, h (A) =(P— rA) x MV

Ri x mi Ri

(4.9)

We can define a vector angular momentum relative to A by (ri — rA) x mi (vi — tA).

h,.(A)

(4.10)

This is important only in the case for which A is coincident with G. Then h,. (G)

(ri — =

x mi (vi— V)

(ri —P) x mi vi —

R, x mi V

(ri— 1.) x mi vi since the term (/` miRi) x V vanishes by (4.2). .*. h(G)

(ri —

x mi vi = h,.(G).

The formulae (4.3) and (4.9) illustrate the Independence of Translation and Rotation because one term in each formula refers to the motion of G, viz., 2MV2, — rA) x MV, and the other term refers to the motion relative to G, viz.,

Tr = 2 E mi

h,.(G) =

Ri x mi fi i =h (G).

§ 4: 2]


133

3. Equations of motion Each particle of the system moves according to the equation

m, f, = F, (4.11) there being one such equation for every particle. These equations can first be added vectorially for all the particles of the system, giving /miff = F, = F, = F, (4.12) where F is the resultant of the applied forces. [Since we assume that the F: form a set in equilibrium we put = 0.] Since d

miff =— dt

_,

dp

m, vi) = —

we can write eqn. (4.12) in the form dp dt

F

(4.13)

This is the equation of linear motion of the system. The internal forces F: can also be eliminated by taking moments about an arbitrary base point A. To do this we multiply each of eqns. (4.11) vectorially on the left by (r, — rA) and sum over all particles of the system.

(r, — rA) x m,f, =

(r, — rA) x (F,

F;).

Since the Friare taken to be in equilibrium

(r, — rA) x F; = r (A) = 0, and

(r, — rA) x F, = T (A)

—

(r, — rA) x mi ff .

(4.14)

The transformations (4.1) can be used to modify the r.h. side of eqn. (4.14) thus: (r, — rA) x mi f, = f rA) x mi (i ft,) —

= (f — rA) x ./144 R, x mi ni , (4.15) illustrating again the independence of translation and rotation. The equation of motion (4.14) now becomes (f. — rA) x /141+ 7R, x (F

—

rA) x Mi

= T(A),

h (0) — F(A). d dt

(4.16)

134


This is the general equation of angular motion and applies for moments about any point A, moving or stationary. By differentiating eqn. (4.5) we obtain dh (A) dt =

FA) x inivi

(vi x

=

MiVi

„L"

rA) x mi ff

It(A).

dh (A) + rA x p = T (A). • dt

(4.17)

This equation for angular motion about A holds whether A is moving or stationary. There are two cases of special importance, in which eqn. (4.17) assumes a specially simple form. (i) A is stationary, r 4 = 0.

(A) • • dhdt

— T(A).

(4.18)

(ii) A coincides with 0, but is not, in general, stationary. In this case rA = .'.rA xp=VxMV= 0. ••

dh(G) dt

_

T(G).

(4.19)

In these two cases, and when rA is parallel to p, the equation of angular motion has a rate of change of angular momentum for the 1.h. side. One of eqns. (4.17), (4.18) or (4.19) together with eqn. (4.13) gives two vector equations, or six component equations, of motion governing the evolu• tions of a system of particles. Eqns. (4.13) and (4.8) show that the motion of 0 is identical with that of a particle of mass M under the action of the resultant of the applied forces. Eqn. (4.19) shows that motion relative to 0 is independent of the motion of G. Eqns. (4.19) and (4.13) are commonly used because of their simple forms. 4:3 Conservation laws

1. If '6 is a fixed direction (i.e., e has unit magnitude and constant direction) such that F .e = 0, then dp 0 F . = . —= (p . dt .•. p . e — constant . (4.20)

§ 4: 3]


135

Hence the resolute of the linear momentum in the direction e is constant. This is the conservation of linear momentum. 2. If 41 is a fixed direction such that e 11 (0) = 0, then, as above, .

h ( G ) = constant.

(4.21)

Hence the angular momentum about an axis through G in the direction "e' is constant, whether G is stationary or not. 3. If e is a fixed direction such that e' . F (A) = 0 and A is stationary, then the moment of the applied forces about a stationary axis is zero. We deduce that

6 . h (A) = constant,

(4.22)

i.e., the angular momentum about this stationary axis is constant. Eqns. (4.21) and (4.22) express the conservation of angular momentum.

4. Conservation of energy When the particles of the system move, all the forces Fi and F: do work, and when the system moves so that it returns to its initial configuration the net work done may or may not be zero. If the forces are such that the net work is zero when the system undergoes an arbitrary "closed" displacement (i.e., one which returns the system to its original configuration) which is consistent with the constraints, then the forces are a conservative set and a function V exists such that the work done by the forces in a small displacement of the system is the perfect differential — d V. (See § 2:4.) Hence the rate at which the forces, both internal and external, do work is — dV/dt. But mi fi = Fi F: , and the rate of working of the forces is (Fi F:) • vi

'.•

d. V dt

d / 1 2) d T m. dt , - `717 2 v '

milt

(T V) = T+V=E, where E is a constant, the total energy.

(4.23)

136


4:4 The inertia matrix We now consider the modifications necessary in the preceding results when the particles of the system under consideration constitute a rigid body. The modifications result from the use of formulae such as (1.21) and (1.44) giving the velocity of a typical point of a body in terms of the angular velocity vector. In Chapter I we identified the resolutes of a vector product as the matrix product of an anti-symmetric matrix and a column matrix. There we used the anti-symmetric matrix to give the angular velocity resolutes; here we use an anti-symmetric matrix made up from the resolutes x of the position vector. Thus, instead of (0 Sx =

co3 —CO2

a),

0

x3

we use

;Vito = ( —x3 x2

W2 X3 — a)3 X2

W2) (1 0

X2

—Col

0 — X2

0 — x1 X

x1

%Xi — C91X3

x,

co,x, — co,x,

NI \

a), x, —

02

0

,w,

(

0 x3

= al 'X = (a)1 W2 a)3)

=

— x,

Oh X2

= co3x1 — coix3 col.X2 — 0)2 X/ — x, 0 x1

x2 — xi . 0

(4.24 a)

(4.24b)

Since the resolutes xi of the position vector transform according to equation (1.1) the anti-symmetric matrix X transforms according to equation (1.6) i.e. ( 0 — X

=

L4 ,

X1 =

S3

0 — 1

=L'XL.

(4.25)

0 We now use the expressions (4.24) for the velocity of a point of a rigid body rotating about 0 to obtain the expressions for kinetic energy and angular momentum. T = 127 m (63 x r)2 = 2 Z mw'XX'w = z w'I(0) w,

(4.26)

which is a quadratic form in col , co,, co,. Here the matrix I(0) is defined by /(0) = mXX'. (4.27) Similarly the resolutes of the angular momentum h (0) = 27 r x m(w x r) form the column matrix 27 = i(o) w . (4.28)

§ 4: 4]


137

To obtain this matrix representation of the triple vector product we use equation (4.24) twice. Now 0 XX' =

— x3

x3 - X2

0 x1

x2 0 - x1)(— x3 0 x2

(4 + 4 - x1x2 =_

x3

- X2

0 0

-xi X3

- X1X2

4 + Xj.

- X2X3

- XiX3

- X2 X3

Xj.4- X2

.

(4.29)

Hence the elements of the inertia matrix I(0) are

=

m(x2 + 4) = A, 133 =

/22 =

M(X2

X?) =

B,

m (4 + 4) = C ,

(4.30)

123 = — mx2 x3 = — D , 131 = — mx3 x1= — E , 112 =

i.e.

MX1 X2 = -F,

/ ik (0) = 21 m(r2 6,, — xi xk) (r2 = xl + 4

+

showing that the elements are moments or products of inertia. In Vol. III § 7:7 we introduced moments A, B, and a product H, of inertia of a rigid lamina. The six quantities defined in (4.30) are the complete set of coefficients of inertia for a rigid body. The matrix refers to the point 0 which is the origin of the coordinates being used. The point 0 being arbitrary, there is an inertia matrix for every point of a body. (Some authors use F, G, H respectively for D, E, F; we used H instead of F in Vol. III. We avoid the use of G because of possible confusion with the centre of mass.) The definitions imply that

B C > A > 0, C + A > B > 0, A + B > C > O. (4.31) (There is no theorem of perpendicular axes in three dimensions except when the rigid body happens to be a lamina.)

The theorem of parallel axes We noted above that there is a different inertia matrix for every point of the body. The theorem of parallel axes gives the relation between 1 (0) and the corresponding matrix I (0) calculated from a set of parallel

138


axes through the centre of mass G. The coordinates of a point are given by xi = y, or x = y, where LT my, = 0 or Limy = O. Hence 1(0) = mXX' = m(X Y) (X' Y'), the matrices X, Y being constructed from xi , yiaccording to the formula (4.24). On performing the summation we obtain

/(0) =

mXX' X(Z mY') (1' mY) X' + = MXX' .1(0), I(G) = mYY'.

mYY' (4.32)

Expressed in terms of the coefficients of inertia this relation is equivalent to three relations similar to each of

A = M +

A0 , D = M x2 x, D0 .

(4.32 a)

Hence the moment of inertia about an axis through G is less than that about any parallel axis not passing through G.

The transformation of the inertia matrix If we rotate the frame of reference so that the coordinates, i of a point in the new frame are given by eqn. (1.2) ,

Xi -

$11,

or

x = 14,

then, by eqn. (1.6), X1 = L'XL. The definition of the inertia matrix gives

i(0) = =

mX,X; =

mL'XLL'X'L

XX'L = L'I(0)L.

Expressed in terms of the elements Iik , LH this relation is Iii = /k J 4./

m•

(4.33)

This is the transformation law for a second order tensor; hence the set /kmis sometimes called the inertia tensor. Example. We find the moment of inertia of the rigid body about a line drawn in an arbitrary direction. The direction is specified in the frame Ox1 x2 x3 by the unit vector A = {11 1213).

§ 4: 5]

139


The definition of moment of inertia (see Vol. III) is = E mp2 where p is the perpendicular distance of the particle of mass m from the axis of rotation. The moment of inertia is then

= E mp2 = m[r2

—

(r . e)2]

= E m[(4. + 4 + 4) - (11 x1+ 12 x3 + 13x3)2] = E nt[ac;. (C + = A IT + BC +

+

4(l3 + q) + 4 (/;. +

- 213 13; x, —

213 12x i.x2

— 2D121, — 2 E1,11 — 2 F1,12,

i.e. /I k (0) il ls.

(4.33 a)

If we choose the direction of the axis 6% used above to coincide with 6, then (/, 12 13) = {1„12, 133) and = 711. The relation just proved therefore is one case of the transformation law (4.33).

4:5 Principal axes of inertia If we take an origin and frame of axes at any point 0 of a rigid body, eqns. (4.30) define the elements of the inertia matrix. We define the momental ellipsoid associated with 0 to be the quadric surface Iik XiXlc = M 84 •

(4.34a)

Expressed in more familiar terms this equation is Ax2

+ Cx3 — 2Dx2 x3 — 2Ex3 x1— 2Fx1 x2 = Ms4 .

(4.34)

The 1.h. side of (4.34) is the "quadratic form" of the matrix /(0). The r.h. side is chosen in the form Me4to make the equation dimensionally homogeneous with e a length which can be given a convenient value. This quadric surface provides a geometrical representation of the inertial properties of the given rigid body for rotations about 0. The reader should distinguish between the current coordinates xi in eqn. (4.34) and the coordinates xi which occur in eqns. (4.30) defining the coefficients of inertia. In the latter they are dummy variables of integration; in the former they are the current coordinates in the equation of a surface. This surface is not a material surface; it is an abstract surface used to give a geometrical representation of the inertia properties of the system of particles. It is no more real than, say, a velocity time curve which gives a geometrical representation of the motion of a particle. ,

140


If a radius of the momental ellipsoid in the direction 13 = {/i} is of length r, then the point ; = rl, lies on the ellipsoid and

1,„(r1,)(r1k) = 11184 r2— Ilk /ilk

s4 J

(4.35)

where J.is the moment of inertia about an axis through 0 in the direction e. Since .1 is neither infinite nor zero and is positive (apart from certain special, degenerate cases such as a thin heavy rod, which has zero moment of inertia about its own length) r2is always finite and positive. Hence we deduce that the surface must be an ellipsoid. Any ellipsoid has three mutually perpendicular principal axes. If these axes are used as coordinate axes, the equation of the surface takes the simple form which has no product terms. In the case of the momental ellipsoid these axes are called the principal axes of inertia at 0. Calling these axes 0 1 2 3, related to 0 x1 x2 x3by a transformation (1.1), the principal moments of inertia at 0 are Al = m(a + a), B1 = m + eT),

01 =

m(ri + n).

(4.36)

The corresponding products of inertia vanish,

Di =

= 0, .E, =

= 0, F1 =

=0.

Clearly, if there is a plane of symmetry of the body passing through

0 then one of the principal axes is perpendicular to this plane. The determination of principal moments and principal axes is just the determination of the eigenvalues and eigenvectors of the inertia matrix (see Vol. IV § 5:8). It is possible to choose arbitrary values of yi (i = 1, 2, 3) and write them in the form yi =cli , where II + 4 + l3 = 1. Using this form for arbitrary y, we can write eqn. (4.33a) in the form kYi ✓

= C2 Iiklilk = c25 > 0.

Hence the quadratic form of the matrix k) is positive definite and the coefficients of inertia must satisfy the appropriate conditions (see Vol. IV § 5:9). Since (iik) is a symmetric matrix it also follows that the eigenvalues, which are the principal moments of inertia, must be positive.

§ 4: 5]


141

The momenta! ellipsoid at 0 is the most important of the momental ellipsoids, sometimes being called the central ellipsoid. It has the equation a2 x2 + b2 y2 + c2 z2 = 1, (4.37) where Mae, Mb2, Mc2are the principal moments of inertia at G, and a, b, c are the radii of gyration about the principal axes there. (Here we use an alphabetical notation and have chosen e4 =1/M.) From a knowledge of a, b, c using the theorem of parallel axes and the tensor transformation law [or eqn. (4.34)] we can find the inertia coefficients and principal axes at any other point. This is usually sufficient for most practical problems. However, there is a geometrical method of describing the relation between principal directions at an arbitrary point P and those at G. The ellipse whose equation referred to principal axes at G is x2 _4_ y2 z2 (4.38) a2 b2 c2 =1 is called the ellipsoid of gyration. It has the same axes as the central ellipsoid (4.37) and its semi-axes are equal to the radii of gyration about the coordinate axes. A quadric whose equation is x2

a2 +0

y2 b2 + 0

Z2 1 2 H._ C 0 —

is confocal with the ellipsoid of gyration for any value of 0. There are three values 01, 02 , 03of the parameter which give members of the confocal family passing through P. These three surfaces are mutually orthogonal at their intersection, the lines of intersection giving the directions of the principal axes of inertia at P. If r2 =0P2, then the principal moments of inertia at P have the values M (r2 — 01), M (r2— 02), M (r2 — 00. Although this is a simple geometrical picture, the determinations of 01, 02, 03and the directions of the lines of intersection are no simpler than the use of the theorems discussed above. We do not prove these results here but indicate in questions 8-12 of Exercises 4:5 how the results just stated can be proved. The momental ellipsoid of a body which has two equal principal moments of inertia at 0 is a surface of revolution. Hence any pair of perpendicular axes which are at right angles at 0 to the axis of symmetry will serve as principal axes; the axis of symmetry must pass

142


through the centre of mass of a uniform body and is therefore a principal axis at every point of its length. [See Exercises 4:5 no. 4.] A body which has two equal principal moments of inertia is said to have kinetic

symmetry. Equimomental systems As for plane distributions of mass there are arrangements of particles which have the same inertial properties as a rigid body (see Vol. III § 7:7). A set of necessary and sufficient conditions for two systems to be equimomental are: ( i) they have the same total mass and the same centre of mass; (ii) they have the same principal axes and principal moments at their common centre of mass. The proof, that these conditions are necessary and sufficient for the two systems to have the same moment of inertia about an arbitrary line in space, follows the same argument as that given in Vol. III. We do not repeat it here. The chief use of equimomental systems is for calculating moments of inertia and one in frequent use is the equimomental system for a triangular lamina, viz., three particles of mass a M one at the midpoint of each side. ,

Examples. (i) A uniform thin sheet of mass M is in the form of a lune of a sphere (one of the regions of the surface bounded by two great circles) of radius a and angle a. Prove that its principal moments of inertia at the centre of the sphere are 2 — M a2 , 3.

(2 3

sina 3a

Ma .

The points on the surface are given by x = a sin0 cosy!), y = a sine sing), z = a cos0,

a

a, 0

An element of area on the surface is dS = a2 sin 0 d0 dcp, (Fig. 32). Hence, writing a for the surface density of the material, M = affdS = 2cra2a. Also

A = f f a(y2 + z2) dS = aff a2sine (a2 sin20 sin2 = cr a4

I o 4

a/2

n

a2 cos2 0) d0 dg) a/.2

sin3 0 d0 f sing) clg9 + aa4 J sine cos2 0 d0 f clyo -a/2

(2a — sina).

0

-c/2

§ 4 : 5]


Similarly

B=off (z2 + x2) dS = a

143

J I a2sin0 (a2cos2 0 a2 sin2 0 cost q2)d0 dgo

= s a a4(2a. + sina),

C=of (x2 +y2) dS = offal sin3 0 d0 dq) = o- al cx.

Fia. 32. Since the plane Ozx is a plane of symmetry,

D = off yzdS = 0,

F = cif xy as

= o.

Since the plane 0 x y is a plane of symmetry

E = f f zx as

= o.

Hence the coordinate axes are principal axes and the principal moments are

A=M

2

sina

3

3a

a 25

B _ m 2

sin a

3

3a

c_2 maz . 3

(ii) Find the principal moments of inertia of a uniform solid hemisphere of mass M and radius a at its centre of mass. With a point 0 on the circular rim as origin and with right-handed rectangular axes Ox a diameter, Oy the tangent to the circular rim, and Oz perpendicular to the plane face and on the same side as the centre of mass, calculate the moments and products of inertia at 0. Hence find the positions of the principal axes at 0, and show that the moments of inertia about them are Mae

'

2M a2 , 5

Ma2.

144


The moment of inertia of a complete sphere of mass M, about any axis through K, its centre, is * M, a2. If we consider the sphere made up of two identical hemispheres, each of mass M, in contact, the moments of inertia of each of the hemisperes about the axis through K are equal, so that 25 = (2M) a2.

=

me.

Hence we conclude that the moment of inertia of a hemisphere about any axis through K is * Ma2. Since K G, Fig. 33, is an axis of symmetry, any pair of orthogonal axes through G which are perpendicular to KG are principal axes, together with KG itself. We choose a pair parallel to Ox, Oy of Fig. 33. Using the theorem of parallel axes we find that the principal moments at G are

Ma2 (2— —9 , 5

64

Ma2 (2— — 9 , — 2 Ma2, 5

64

5

since KG = I a. A second application of the theorem relates these moments to the moments about the coordinate axes Oxyz. ••• A = ppia2 9 = maz(s _ + 1) mazy C = Ma2 (1 + 1) = IMa2 . The products of inertia are

D = 0, E = 0 M (a) (

3a 8

—

3 Ma2 8

since the coordinates of G are {a 0 a). The principal axes are obtained from Oxyz by a rotation about Oy through an angle a, where C— A 3 tan 2a — — . 2E 4 cos 2a = , sin2a = *. (C.f. the rotation of axes to give the principal axes of a lamina, Vol. III § 7:7.) The principal moments of inertia at 0 are A, = A cos2 C sin2oc — 2E sin a cos a = MOW Ma2 2(1 — t) —

= B =5 Ma2,

8

Ma2 • 35 = 1.401Ma2

Co = A sin2oc C costa + 2E sin a. cos oc = 161 Ma2

§ 4: 5]


145

(iii) The meridian lines of a uniform solid paraboloid of revolution, of height are parabolas of latus rectum 4a. Prove that the equation of a momental ellipsoid for the paraboloid at any point Q on the rim of its base, referred to rectangular axes Q x y z, where Qy is the diameter of the base through Q and Qz is parallel to the axis of the paraboloid, is (28a + h) x2 + (4a + h) y2 + 32az2 ± 8 1/(ah) yz = constant. We draw the frame CiTiC at the vertex parallel to Qxyz respectively (see Fig. 34). The coordinate axes from symmetry, are principal axes at 0 with moments A1, A1, C1. By dividing the solid (of density e into circular discs and integrating, the mass )

is given by M = ne f 4a dC = 2neah2. Also

Ai= nef4aC(f

+

c1C = 47rea(kah3 + 1h4) = 111(lah ih2)

Similarly

C1= ne f 4ct ..-14ac1C = 3,a2h3 = a 111 ah The height of G is given by -

MC = nef 4a0dC. ,*, =3h. Hence the principal moments of inertia at G are Mb2, Mb2, Mc2, where _ th2,) M(3ah 2-8 h2), mb2 = m (l ah

Mc2= 3Mah.

146


Referred to the axes Qxyz the coordinates of G are = 0, g = 2 V (ah), z = fh. The theorem of parallel axes now gives 2 2 h2 14 A = M ( —ah + + + 4ah)= M (- ah + h2) 9 3 18 9 3 6

B=M

2

a+

h2

h2

+ ) = M(3 ah +

1

2 ),

C = M (lah + 4ah) = . The products of inertia are

D = 0 + M(-1h) [21/(a h)] = —1M V(ah3), E = 0 = F. The momental ellipsoid has the equation (28a + h) x2 + (4a + h) y2 + 32a z2 + 8 1/(a h) yz = constant which is obtained from the standard form on cancelling the factor Mh/6. Exercises 4:5 1. Find the directions of the principal axes of inertia of a square lamina of side 2a through a corner, and prove that the moment of inertia of the square about an axis through this corner perpendicular to one pair of edges and at 45° to the other pair is 2Ma2. 2. Prove that the moment of inertia about an arbitrary line through a point 0 of a mass-system is inversely proportional to the square of the radius, in the direction of the line, of the momental ellipsoid at 0. The edges at the corner 0 of a uniform cube are taken as rectangular axes Ox, Oy, Oz. Show that the equation of the momental ellipsoid at this corner is of the form 4 (x2 + y2 + z2‘) 3 (yz + z x + xy) = constant . If the cube is of mass M and side 2a, find the moment of inertia about a line through O in the plane Oxy making an angle 0 with Ox. 3. Prove that the moment of inertia of a thin uniform spherical shell, of mass M and radius a, about a diameter is -1 Mae. Prove that the principal axes of inertia of a thin uniform hemispherical shell, of mass M and radius a, at a point of its rim make angles 3n/8, nI8, n/2 with the axis of the hemisphere; and find the principal moments of inertia at the point. 4. Prove that a principal axis of inertia at the centre of mass of a material system is a principal axis at all points of its length. A uniform solid right circular cylinder is of radius a and length a I/3. Show that one of its principal axes at any point A on the circumference of one end passes through the centre of mass of the cylinder, and find the principal moments of inertia of the cylinder at A.

§ 4: 5]


147

5. A uniform solid has the form of the portion of the cylinder x2 + y2 = a2 between the planes z = 0, z = h + y tang, (h > a tang, 0 < a < ei a), the axes being rectangular. Prove that the momental ellipsoid of the solid at the origin is given by the equation (4h2 + 3a2 sec2a) (x2 + y2) + 6a2 z2 — 6a2y z tan a = constant. 6. The base of a uniform solid cone is the circle x2 + (y — a)2 + z2 = a2 , z = 0, and the vertex is the point (0, a, h). Show that the equation of the momental ellipsoid at the origin is (23a2 + 2h2) x2 + (3a2 + 2h2) y2 + 26 a2 z2— 10 a hyz = constant. 7. Prove that for a rigid body, in general, A > 0, AB — F2 > 0, ABC — 2DEF — AD2— BE2— CF2 >0; (see Vol. IV § 5:9). 8. Show that the principal moments of inertia of a body whose inertia coefficients at 0 are A, B, C, D, E, F are the stationary values of 5 subject to the condition 12 ± m2 ± a2 = 1. 9. A rigid body has Mae, Mb2, Mc2as principal moments of inertia at G, the coordinate axes Gx, Gy, Oz being the principal axes of inertia at O. Find the inertia tensor at any point P 77, 10. Use the result of question 8 to show that the principal moments of inertia at P are Mg, Mkt, mkR, where k1, k2 , k3are the roots of 712 a2

r2

k2

b2

r2

k2

c2

r2

1, 1c2

where r2 = 2 + 172 11. If Or , 02 , 03are the parameters which determine the three quadrics of the confocal family X2

y2

z2

a2 +0

b2 +0

c2+ 0

—

1

which pass through P n, 0, show that the principal moments of inertia of the rigid body at P are M (r2— Or), M (r2— 02), M (r2— 03), where r2 = op2 _ + n2 ± 2. 12. Show that the respective directions of the principal axes of inertia at P are given by the vectors

Ja

2 + 72 k2

b2 + 72 k2

e2 + 72 k2

when k takes the three values given in question 10. Deduce that these directions are given by the lines of intersection of the three confocal quadrics in question 11.

148


4:6 Dynamical variables for rigid systems We now obtain expressions for the variables of § 4:2 when the system of particles is a rigid body. To specify the motion of the body we give the velocity of a particle A, or of the centre of mass 0, of the body and the angular velocity, i.e., v = vA + co x (r

v---Trd co x(r -

—

—

r„),

f)=V±o3

x R,

v being the velocity of the particle whose position vector is r.

1. Kinetic energy We use the expression (4.3)

T = 1-MV2 +Tr , where

Tr = z mR2=

z

m(co x R)2

We deduce from (4.26) and § 4 : 4 that

Tr = T = 1 "2 +

ik(G) Nicol, • ocoi B

C °a-4

2Dow2co3— 2E0 (03(01— 2 Fow1w2)•

(4.39)

Example. The ends of a rod AB of length 2a and mass m have velocities vA , Prove that the kinetic energy is m(vi + vA . vB + The velocity of the centre of mass is V = (vA + TB), and vB = vA + co x AB. If the vector co makes an angle B with AB, then T,. =

mat co2 sin2 0. But

s 20 = (vB_ VA)2. (w x AB)2= co24a2in •T—

In (VA + VB)2 2 4

m(VB VA)2 4 6

1

6

nz(v2A-I 4A .

vg).

2. Linear momentum The fact that the system of particles is a rigid body does not affect the formulae (4.4), (4.8) for linear momentum. p= Zmy = MV.

§ 4 : 6]


149

3. Angular momentum When the body is rotating about the origin 0, formula (4.28) shows that

hi(0) = I ik (0) wk, i.e.,

hi = Awl— F co, — E co, , h2= —Fah B co, — D co, , h3 = — E co, — D o),

(4.40)

Cc0 3 .

This result indicates that the angular momentum vector of a rotating body has a different direction from the angular velocity vector (unlike the relation for plane motion). The only exceptions to this occur when the body is rotating about a principal axis of inertia. Eqns. (4.40) show

that if a radius OJ of the momenta' ellipsoid has the direction of co, the angular momentum vector has the direction of the normal to the momenta' ellipsoid at the end J of this radius (see Fig. 35). If the body is not rotating about a fixed point, we use eqn. (4.9) which gives (4.41) h(G), h(A) = — rA) x p and the resolutes of h (G) are given by relations similar to (4.40),

h1 (0) = Aow, — F0co2— E „co, , h2 (0) = — F „col B „co, — D0w3, 11,3 (0) = — E0co — Dow, + Cocos . The calculation of the angular momentum of a system of rigid bodies connected to each other is often a repeated application of (4.41) to subsystems in the given system.

150

A COT.TRSE OF MATHEMATICS

Example. Find the angular momentum h (G) of a system which consists of the two identical, uniform rods AB, AC which are freely jointed at A, when the system is free to move in the horizontal plane and its position is specified by the coordinates {xA yA} of A together with the angles 0, q' (see Fig. 36). The line LAG bisects the angle CAB. Since h (G) = h,.(0) [see eqn. (4.10)] we can calculate the angular momentum by reducing G to rest. (We assume a z- or (-coordinate axis perpendicular to the figure.) The frame A Sri has angular velocity (0 0 0) and the coordinates of E and F are respectively (0 a sin q) 0), {0 — a sin()) 0),

FIG. 36. and their velocities relative to G, are given by r

= -+

at

co x r,

i.e., {0 ± a cp cow 0) + (0 0 0) x {0 ±a sing) 0) =

a() sing) fag) cos 9) 0).

To find the contribution from each rod to h (G) we use the independence of translation and rotation in the form (4.9) for AC and AB separately in their motions relative to G. The contribution from the motion of the centre of mass E of AB is

m (0 a sing) 0) x { — aO sing) a 49 cosg9 0), and the contribution from the rotation of AB about its own centre of mass is

}mat {0 0 0 + 4)). Hence h (G) = m {0 a sing) 0) x { — aO sing)

a go cosq) 0) + lma2 (0 0 6 + 49)

+ m {0 —a sin()) 0} x (a0 sing) — acp cow 0) + 3ma2 (0 0 0 — 9,)} = {0 0 2ma2 O sin2 q) + 3ma2 0}. In this simple case, where the motion is in one plane, the results could be written down almost at sight by the methods of Vol. III; nevertheless, the reader is advised

§ 4: 61


151

to become thoroughly familiar with the use of eqn. (4.9) and the many possible applications of the independence of translation and rotation. The general principles underlying the above calculation apply to all motions, not just to plane motion.

4. Equations of motion The equation of linear motion (4.13) has the same form when the system is a rigid body, or a number of connected rigid bodies. When a rigid body is in motion, in general, it moves relative to the coordinate frame and so the coefficients of inertia vary with the motion and the derivative dh/dt in either eqn. (4.16) or (4.17) requires a knowledge of dA/d t, etc. These latter derivatives would be cumbersome to find, and, in any case, cannot be found until the motion of the body is known. Therefore the solution of the (differential) equations of motion is virtually impossible. This difficulty is overcome by using special frames of reference and is considered in the next chapter. Here and in § 4:7 we give a number of examples which are simple, direct applications of the preceding results, or can be solved because the symmetry of the bodies considered, usually spheres, is such that dA/dt, etc. are zero. Examples. (i) A cube of edge a and mass M is free to turn about one edge which is vertical, and a smooth hole is bored along a diagonal not intersecting the axis of rotation. A particle of mass m is placed at the upper end of the diagonal and the system is released from rest. Show that the cube rotates finally at a rate ma2) (3M k2 2 ma {2g al(M ma2))'/2, where k is the radius of gyration of the cube about the axis. This is a system in which the applied forces are conservative and have no moment about the vertical axis of rotation and the constraints are smooth. We use the laws of conservation of energy and of angular momentum. In order to calculate these

152


quantities it is helpful to use a rotating frame O l Z3which coincides with the edges of the cube as shown in Fig. 37. At an arbitrary time during the motion the cube has turned through an angle 0 and the particle has descended a distance x along the hole relative to the cube.

Motion of the frame The angular velocity 0 = {0 0 0).

Motion of the system (1) the cube. The position of the centre of mass k = {a a a}. The velocity of the centre of mass V=

Or

at

+ 0 x r = {0 0 0} + {0 0 0} x

a a} = 4a0 ( 1 1 0). -

.•. pi = iMaO {-1 1 0}. Since any axis through G is a principal axis of inertia with moment M kg, k(G) = IYI kg {0 0 0}. (The suffix 1 denotes the contribution from the cube only.) 111(0) = t x Pi +

(G)

= Ma2 0 {1 1 1} x {-1 1 0) + /1/146 {0 0 1} + MO {-1a2 --4a2 k2}, = MO {-1a2—1-a2 where k2 = ko + z a2. The kinetic energy of the cube is

T, = fMt 2+ 1M 7402 =131a262 i.311,(2)62 = I.M1C262. (2) The particle. The position vector is r2 = {0 a a} + x (1/y3 —1/j/3 —1/y3), since the direction of the diagonal is that of the unit vector {143 — 1/y3 — 1/y3}. ,*. r2 =

3a—

a—

V

V3

1.

T3

The velocity of the particle is vz = =

IV 3 IV 3

_

+ 0 x r2

x 01 _X +6 a+ x V3 V 3I V3 0 6 ict x \ th + Ox t1 V3 V3 1 . V3 • th

th

th

r2 Ot

—

§ 4 : 6]

153


Hence the angular momentum of the particle about 0 is

h2(0) = r2 x m = m {— V3 -m{

x X a - .7ff a- — x V3 V

x

6 (a

X

]13f

V 3) ax - — 3

'0

2

(a V3 \ 1/3

- X + Ox y3

- (a - -L) V3 \ V3/

3

0 (a - )21. ,

V3 /

The kinetic energy of /the whole system is 2

T = iMk202 +

- 0 (a V3

)ll

t 612 ±2 v3 +

V3

The potential energy of the system arises from the position of the particle, .•, V = mg (a -

113

Hence

li Mk2 02 +

+ mg (a -

x V3 I =

E.

Initially 0 = 0 = v2 , x= 0, giving E = mga.

Mk26 2+1m4= mgx1113. When the particle leaves the hole, x = a V3,

1- Mk202 +

.•.

2 [-t (-± - a0) +

3

V3

.±2

---j= mga. 3

(1)

The initial value of the angular momentum about the axis of rotation 0 2 is zero,

.• .mk20 + ,,r;_ .1a73. When x

0 1a -

= O.

+

a V3, Al k2 + m [a2O —

0,

= V 3 (Mk2 + m a2) 01(m a) . Substitution in (1) gives

1(Mk2 + ma2) (3Mk2 2m,a2)O2 = m2 ga3, which gives the stated value for O. (ii) Two uniform rods AB, BC each of mass m and length 2a are smoothly jointed at B. The rod AB is free to rotate about A which is fixed, C can slide freely along a horizontal line through A. and the plane of the rods is vertical and is free 6

CMES

154


to rotate about a vertical axis through A. When the rods are perpendicular to each other and B is below A C, the plane of the rods is set rotating with angular velocity Q about the vertical through A. Show that when each rod makes an angle 0 with the horizontal 4a 02 (1 + 3 sin2 0) = 4aD2(2 — sec2 0) — 3g (1/2 — 2 sin0), and that B rises or falls initially according as S22 > or < 3 1/ 2g/(16a). Since the system is free to rotate about the vertical AV (Fig. 38) the laws of conservation of energy and of angular momentum about AV apply. We solve the problem by obtaining these quantities for each rod separately, using a special set of axes for each rod in the calculations.

Fro. 38. 1. The rod AB The axes E 1 e2e3 shown in Fig. 38 have E 1along the rod and E E e3 forming a r.h. set. The point E is the centre of mass of this rod.

horizontal,

The motion of the frame The velocity of the origin is the velocity of the centre TE ; the angular velocity of the frame is OA g = ( -11, sin 0 0 j, cos 0). The motion of the rod AB The velocity vEis obtained from the condition that A is fixed and that the angular velocity of the rod is the same as that of the frame. TE

= 0,413 x AE = {-0 sin° 0 y cose} x {a 0 0}

= {0 alp cos° —a0}. (This can also be seen directly from the figure.)

§ 4: 6]


155

The axes E e, e2 e3are principal axes of inertia for the rod AB with moments 0, 1-ma2, ima2 respectively

hAB(E) = 3ma2 {0 6 V) cos 0} . Since

PAB= MTE = M {0 alp cos0 — a0}, --)hAB (A) = AE x PAR + hAB (E) = m {a 0 0) x {0 aye cos0 — a0} + ima2 (0 6 tfr 4 nia2{0

3

COO)

0 fp COSO).

(Since A is fixed this could also be written down directly.) The unit vector k in the vertical direction is k = {— sin 0 0 cos 0}. .•. k

hAB(A) = 3ma2 cos° .

Also

g

TAD = m a2 [02 + 62 +

m a2(6 2+ 1p2cos20) = 3 ma2 (62

cos0)2] +

c0,20),

and the potential energy is

VAR= — mg a sin0 2. The rod BC

The motion of the frame (F 711 n2 no The velocity of the origin is the velocity of the centre of mass; the angular velocity is ()BC = ( 11) sin 0 —0 1p cos 0}.

The motion of the rod BC Since B is a point of the rod A B, TB= {—IP sine

6 1j) cos()} x {2a 0 0) = (0 2a/p cos° —2a6),

where the resolutes are given in the frame E eie2 TF

The velocity of F is given by

YB °BC x BF.

We now resolve vectors in the frame F r7 1772 Th. Since F122 and E 2 are parallel

TB = {-2a sin20 2a tj, cos° —2a 61 cos26}. TF

= {-2a6 sine

3alp cos0

a6(1 — 2 cos20)),

The axes F r7117 2 n, are principal axes, so that hsc(F) = 6*

rca2 (0 —6 y, cos0).

Pac = mvf.

156


Referred to the frame F

Th973the point A has the coordinates

{—a(1 + 2 cos 20) 0 2a sin 20}.

hBc (A) = AF x PBC + hBc(F) — m {a(1 + 2 cos20) 0 —2a sin20} x {-2a6sin20 3a7pcos° a6 (1 — 2 cos20)}

+ ma2 {0 —6 y cos0} = ma2{6V, cos0 sin20 go ,pcose(y+ 6 cos20)}. The vertical direction k has resolutes {sin 0 0 cos 0}, so that k . hBc(A) = ma2 y[6 sine cos0 sin20 + cos2 0 (V + 6 cos20)] = Vma2V, cost 0 . The kinetic energy is IBC = 6 ma2(62 +?P2 cos2 0)

ima2[402+ 9V)2 cos2 0 + 62 (1 — 4 cos20)]

*ma2[62 (1 + 6 sin2 0) + 71p2 cos2 0], and

VBc = — mga sine. The energy equation for the whole system is Ima2[62 (2 + 6 sin2 0) + 8/P2 cos2 0] — 2mga sine = E, and the angular momentum about the vertical A V is $3 ma2 cos2 0 = h. These are equivalent to 262 (1 + 3 sin2 0) + 81,2 cos2 0 — 3 (g la) sine = E', From the initial conditions, viz. 0 =

4cos20 = h' .

6 = 0, zp =

E' = 4522— (3g)I (a 1/2), h' = D/2. Q

2 cos2 0 3g 1 202 (1 + 3 sin2 0) = —(sine — a V2

+ D2 (4

2 , cos20

i.e., 4a02 (1 + 3 sin2 0) = 4aQ2 (2 — sec2 0) — 3g(1/2 — 2 sine). If we differentiate this equation, we obtain 8a0(1 + 3 sin2 0) + 24a02 sine cos° = — 8aD2 sec2 0 tang + 6g cos0. Hence initially, when 6 = 0, 0 = n/4, 8a0(2) = —16aD2 + 3 I2g . If. B rises initially 0 < 0 and

Q2 > 3 1/2g/(16a).

§ 4: 6]


157

(A number of steps could be shortened in the above working by using the figure to write down components of velocity, angular momentum, etc., so obtaining eqns. (1) and (2) more quickly.) (iii) A plane lamina in the form of a triangle ABC moves on a fixed table with only the side BC in contact with it. If BC makes an angle q) with a fixed line in the table and the plane of the lamina is inclined at an angle 0 to the table, show that the angular velocity co of the lamina is given by w = (j)k — Oil, it being a unit vector along BC and k a unit vector perpendicular to the table. Show that the resolutes of co along rectangular axes of which one is parallel to BC and a second is normal to the triangle are {- 6 cp cos 0 cp sin 0). If the lamina is uniform and equilateral of side 2a, show that its kinetic energy is iMV2 + 1 2 Ma2[(1 + 2 cos2 0) 62 + (1 + cos2 0) 492], where M is its mass and V is the horizontal component of the velocity of the centre of mass.

Fier. 39. We suppose that the origin (Fig. 39) coincides with the instantaneous position of the mid-point of BC and that the directions of the axes Ox, x2 x3are fixed with Ox3vertical. The frame 012 3has the directions shown in Fig. 39 with 0n3 perpendicular to the plane of ABC, and 0 in the plane of the triangle. (The vertex A is nearer to the reader than B and C and is above the plane 0 x1 x2.) Since the angle q) is measured in the horizontal plane 0 x1 x2the lamina has an angular velocity q7 k. Since the angle 0 is measured in a plane perpendicular to 0 i.e., the direction ti, the lamina also has an angular velocity 6 n. (The minus sign occurs because the sense of 0 increasing is opposite to the r.h. rotation about the direction n.) to = (j(k — —

To obtain resolutes along the directions stated we note that, referred to k has resolutes {0 cos 0 sin 0) and fi has resolutes 0 0), so that = (j) {0 cos() sin(9) — 6 {1 0 0) = (— 0 17 cost) q7 sin0).

158


This result can also be obtained (with greater labour) by the methods of Chapter I. The rotation required to bring a frame from the position 0 x„ x2 x3 to that of 0 $, es esis made by a rotation about Ox, through an angle co, followed by a rotation about 0$1through an angle n - 0.

cosy° R = (siny./ 0

-sing)

0)(1

0

COS (I)

0

0

sin g

0

1

0

cos 0

0 -cosy • sin g

The resolutes of the angular velocity co are obtained from the matrix R' R. We leave the reader to verify that the resolutes are as stated. In the special case when ABC is equilateral the axis 0$3passes through G and A. We find the principal moments of inertia at G about axes parallel to 0 el es es are -11-3/a2, aMae, *Ma2. (To find these, we use the equimomental system of three particles each of mass sM at the mid-points of the sides.) Since the height of G above the horizontal plane is = (a/V3) sins and V is the horizontal component of G's velocity,

22 = V2 + -3a2 02 cos2 0 We use formula (4.39) to give the kinetic energy

T = 1-M (V 2 + 02 0204382 0) + Ma2 (02 +

cos2 0 + *02 sin2 0)

= iM V2 +12ma202(1+ 2 cos2 0) + 02 (1 + cos2 0)]. Exercises 4:6 1. A solid in the form of a cube of edge 2a with a fine smooth hole drilled through it along a diagonal can turn freely about a fixed vertical axis coinciding with an edge which does not intersect this diagonal, and is capable of no other motion. When the solid is at rest a particle of mass m is placed gently in the hole at its upper end. Show that at the instant at which it emerges the solid has turned through an angle a tan-1(— a)

and has acquired an angular velocity w, given by the equation (a2 b2) (a2 + 3b2) w2 = 4a3g, where b is a length such that the moment of inertia of the solid about the axis is 2 m (b2 - a2) . 2. A uniform heavy rod of length a is smoothly pivoted at one end to a fixed support. The rod rests in equilibrium with the other end in contact with a smooth vertical wall, the distance of the fixed support from the wall being d (< a). Show

§ 4: 6]


159

that, if the rod is gently displaced from this position of equilibrium, contact with the wall will cease when the rod makes an acute angle cos-1(2 y(a2— d2)/3a) with the upward vertical. 3. The sum of moments, about a point Q, of the momenta of the particles of a dynamical system relative to a Newtonian frame is h. The centre of mass of the system is G and the moment about Q of the external forces on the system is N. If h = N throughout the motion, show that either (i) Q is a fixed point, or (ii) QG is a constant vector (which may be zero), or else (iii) Q and G move so that their velocities are always parallel. A circular ring of radius 2a and centre A rotates in its own plane on a smooth horizontal table with constant spin co about a fixed point 0 of its circumference. A second uniform thin ring of radius a and centre B, coplanar with the first, rolls without slipping on the inside of the larger ring. If 0 is the angle made by AB with OA, show that 6 + co2sin° = 0. If initially 0, A, B are collinear and in that order and the initial angular velocity of the inner ring is equal to that of the outer but in the opposite sense, show that cosh co t = sec 0 so long as the rings remain in contact, i.e., for a time co-1cosh-12(1 + y3). 4. A system of particles of total mass M and centre of mass G moves in a rigid body motion with angular velocity w, the velocity of G being V. Prove that the kinetic energy T of the motion is given by 2T = M V2 + A + B + C where A, B, C are the principal moments of inertia at G and co,, co„ co3 are the resolutes of co along the principal axes at G. A thin uniform disc of mass M, radius a and centre G rolls without slipping on a horizontal plane surface with the plane of the disc making a constant angle with the vertical and G moving with constant speed V in a horizontal circle of radius 2a. If T is the kinetic energy of the disc, prove that 3 4

25 T a). Obtain the equations of motion and derive the following results: (1) the spin Q of the vertical plane through OA is constant, where A is the point of contact; (2) the component of spin n of the

§ 4: 7]


167

sphere about OA is in general proportional to the height of 0 above some fixed level; (3) the vertical motion of 0 is in general simple harmonic, with period (n V14)/Q. We use a moving frame 0$1 z e3in which we resolve all vectors (see Fig. 42). The origin 0 is at the centre of the sphere and 0 r lies along the radius to the point of contact A. 0 3is vertically upwards, and 0 $2 completes the r.h. set. The direction of OA is given by the angle y, and the position of M, where A0 intersects the axis of the cylinder, is given by z (= LM), L being a fixed point on the axis of cylinder. Motion of the frame Velocity of the origin, v, = {0 (b — a) y Z}. Angular velocity of the frame, 0 = {0 0 y).

Motion of the sphere The velocity of the centre 0 is V = v0 . V = {0

(b — a)ti,

The angular velocity of the sphere is w = {coi co, w3}. Therefore the velocity of the particle at A is vA = vo + w x OA = {0 (b — a) p Z} {o co, co3} x {a 0 0}. The condition for rolling is that the particle at A is stationary. 0 = {0

(b — a) y aco, z— awe}.

co, = i/a, co3= — (b — a) y/a. Hence p = m {0

(b — a) y i}, h (G) = f mat {co, i/a — (b — a) tpla}.

168


The farces At the centre 10 0 0), the weight (0 0 -- mg) ; at A {a 0 0}, the reaction {X Y Z}.

{X

- mg) , r(G) = {0 -aZ a Y}.

I

The equations of motion The equation of motion dp/dt = F, when referred to a rotating frame becomes

0p 4 Ot m {0

(b - a)

0X

p = F.

+ {0 0 tp} x m (0 (b - a)lp Z) = {X Y

.* - m (b - a) 1P2= X , m (b - a) 1.1; = Y , = Z - mg .

- m g} . (1) (2) (3)

The equation of angular motion is obtained by taking moments about the origin, which is also the centre of mass.

•• .•.

0,60 {co,

dh(G)

oh(G)

dt

at

+A x h(G) = r(G).

'qa - (b - a) i; /a} + {0 0 /p) x ma2 {a

/a - (b a) a)

= {0 - aZ a 11 . ma2w1 *maizp = 0,

*ma + inta2 coi /p

=

-aZ,

(4) (5)

- Ima(b - a)lp-= a Y.

(6)

The first step in the manipulations, the elimination of the unknown forces X, Y, Z gives acid=

7 ,3-F 2aco1qp1 - 5g, -

- a) 1,7)

0.

Hence V) =12 = constant, and

awl =Qi, 7z + 2anco1= - 5g.

(7) (8)

Integration of (7) gives aco, = S2(z - z0), where z, is a constant of integration. + 2(22 (z - zo) = -5g, 2 35 g + — S22 (z - zo + —) = 0. 7

2 /22

(9)

Eqn (9) shows that the vertical motion of 0 is simple harmonic motion with period n y14 V( 7 \ Eqns. (1), (2) show that, since ?,b = Q, the normal reaction 12 2) S2 from the cylinder is X = -- m (b - a) 02; also Y = 0, Z = mg - 7 m,(22 (z - z0). When (9) is integrated, subject to initial conditions, the total reaction of the cylinder on the sphere is determined.

§4:7]

169


(ii) A solid uniform sphere of mass M and radius a rolls on the rough inner surface of a fixed cylinder of radius c ± a whose axis is horizontal. The half-plane through the axis of the cylinder and the centre of the sphere makes an angle 0 with the downward vertical. Establish the equations of motion, and deduce that

6 + n2 sin() = 0, where n2 = 5g/7c. Prove also that d2 u de2

p2 u = 0,

d2co3 d 02

+ P2 (03

where u is the component of the velocity of the centre of the sphere along the cylinder, co3is the spin of the sphere about the radius from the point of contact, and p2 = 2/7. Consider in particular the motion in which the conditions at the instant t = 0 are 0 = 0, 0 = n, u = U and w3= 0. Prove that during the subsequent motion oh varies between the values ± {U sin (p n13)} 1(a p).

We use a moving frame of reference A 1z a in which A lies on the axis of the cylinder and moves along the axis as the sphere moves. The axis A S3goes through the point of contact B, and A llies along the axis of the cylinder; A completes the r.h. set (see Fig. 43).

e2

Motion of the frame The velocity of the origin TA = u 0 0}. The angular velocity of the frame 0 = {-0 0 0}. {

Motion of the sphere The position of C is V = {0 0 c}. t The velocity of C is V = TA t

— a

0

x r.

17 = {u 0 0) + (0 0 0) + {- 0 0 0) x {0 0 c} = (u c0 0). -

170


V The acceleration of C is I = — + 0 x v.

at

I=

c0 0} + {— B 0 0} x {u cO 0} = {d cO —002}.

(These results could, of course, almost be written down at sight.) The angular velocity of the sphere is co = {co, cue oh}, and the condition for rolling at B is vp = 0.

0 = {u c0 0} + {col oh co3} x {0 0 a} = {u + aw2 cO — awl 0}. co, = cO/a, (2),= — u/a. The linear momentum is p = M (u c 0 0). The angular momentum about C, the centre of mass, is

h(G) =

a2 {a w3co3} = M a2(cella — a (2)3).

The forces At {0 0 c} the weight {0 — Mg sing Mg cos0}; at {0 0 c + a} the reaction {X Y Z).

,•, F (B)

{X

Y — Mg sin0 Z + Mg cos0},

{0 0 — a} x (0 — Mg sine

M g cosO) = {— M g a sine 0 0}.

The equations of motion Since we are not required to find X, Y, Z we formulate the equations of motion, without involving these quantities, by taking moments about B, a moving point not coincident with G, using eqn. (4.16). This is dh(G)

dt

+ (t — rB) x Mf = r(B).

[We resolve all vectors in the frame Al

dh(G)

0 h(G)

dt

at

using the formula

+ 0 x h(G).]

The resolutes of this equation of motion are

* Ma2 (cOla — Ufa 63) + {— 0 0 0} x sMae {c 6/a co,) + {0 0 — a} x M c6 —c)2) = {— M g a sin0 0 0}. :.*Mac0 + MacO = —Mga sing, — *Mai + %Ma20w3— Mau = 0, tMa2 a)3 +*MauO = 0. 7c0 + 5g sin° = 0, 711, = 2a0w3, ath, + u0 = 0. + n2 sin° = 0, where n2 = 5g/(7 c).

(1)

§ 4: 7]

171


The remaining equations lead to (since du/d0 = ii/0, dco3/d0 = cb3/0) du dO

2a

d co, dO

7 c°3'

=_ a

(2) (3)

Elimination, in turn, of a), and u gives d2 u d 02

+ p2u = 0,

d2co, d02

+ p2co, = O.

(4) (5)

In the particular case given, from (1) and the initial conditions, do )2 at =

1

2n2 (cos0 — — 2.

Therefore the variations in 0 are exactly synchronous with a simple pendulum of length gln2 (= 7c/5) oscillating through an angle n/3 on either side of the vertical. The solution of (4) is u=Cl cosp 0 +02 sinpO. Therefore, from (2), (03 = s (Pia) (— Cl sinp 0 + 02 cosp 0) = (—Cl sinp 0 + C, cosp 0)/(aP) • From the initial conditions Cl = U, C, = 0. U u= U cospO, w3 = --sin p0. ap The maximum and minimum values of 0 occur when dO/dt = 0, i.e., when cos 0 = or 0 = ±n/3. The values of co, therefore vary between the limits — U sin (" ap 3 (iii) A uniform solid sphere of centre C rolls without slipping on the inside of a fixed sphere of centre 0 so that C describes a horizontal circle at a depth d below 0 with angular velocity co. If the angular velocity of the sphere has no component along OC, show that co2= 5g/7d. We use a rotating frame of reference with 0 at the centre of the fixed sphere in which the axis 0 E, passes through the point of contact A, 0 zis horizontal, and 0 3 completes the r.h. set and makes an angle a with the upward vertical (see Fig. 44). The plane 01 S3 rotates about the vertical with angular velocity w. Motion of the frame The angular velocity of the frame is 6 = {— co sin a 0 co cosy}. Motion of the sphere The position of the centre C is P = {a — b 0 0}. t + 0 x P. The velocity of the centre is v = t

172


V {— co sina 0 w cosa.} x (a — b

0 0) = (0 (a — b) w cosa 0).

f = {—w sina 0 w cosa} x (0 (a — b) w cosa. 0) = {— (a — b) w2cos2a 0 — (a

b) co2 cosa sina).

Since C is at a depth d below 0, (a — b) sina = d. Also

p = m {0 (a — b) co cosa 0}. The angular velocity of the sphere is Ci) = (0 w2 w3).

h(G) = *mb2{0 w2 w3). (It is given that w, = 0.) The condition for rolling is that V = co x — rA), since the sphere is rotating about A. (0 (a — b) w cosoc 0) = (0 w2 co3) x {— b 0 0}. 0 = 0,

—bco, = (a — b) co coso

b co2= 0 .

The forces At C {a — b 0 0}, the weight {mg sing 0 — mg cos a}; at A {a 0 0}, the reaction {X Y Z} . r(A)—mbg cosoc 0). The equations of motion In order to exclude the unknowns X, Y, Z we take moments about the moving point A using eqn. (4.16), dh(G)

at

+ (t rA) x MI —

ah(G)

at

+0 x h(G)+ (t — rA) x Mf =

r (A).

.*. m b2 0 + — co sina 0 w cos a} x * mb2{0 0 — (a — b) cosoc/b} {—

b 0 0} x m {— (a — b) co2 cos2a 0 — (a — b) co2cosa sina) = (0 — mbg cos a 0).

§ 4: 7]


173

Equating the components on the two sides of this equation gives only one relation 2mb2 (b — a) w2cosa sina — mb (b — a) w2cosa sina = — mbg cosa. 5 b .• — (b — a) w2sincx = 5g 7 co2= 591(7c1). Exercises 4:7 1. A uniform sphere rolls without slipping on a horizontal plane which is kept rotating with constant angular velocity Q about a fixed vertical axis. Show that the centre of the sphere describes a circle fixed in space, and that it makes a complete revolution in time 7702. 2. A uniform sphere rolls under gravity on a rough plane of inclination a, which rotates with constant angular velocity co in its own plane about a fixed axis. Prove that the mean velocity of the centre of the sphere is horizontal and of magnitude (5g sin a)/(2co). 3. A perfectly rough plane rotates uniformly about a vertical axis in its own plane with angular velocity Q. A sphere of mass m is placed in contact with the plane and rolls on it under the action of gravity. Prove that the sphere never descends a distance more than 5g/(22below its original position. 4. A homogenous sphere of radius a is projected on a rough horizontal plane with velocity v and spin co about the horizontal diameter making an angle a with the direction of projection. Show that, while slipping occurs, the direction of friction is constant and that, when pure rolling begins, the direction of the velocity of the centre has turned through an angle 2acocosa tan-1 ( 5v — 2aco sin a ) • 5. A uniform solid sphere rolls without slipping inside a stationary inverted cone of semi-angle 7r/4 with its axis vertical. If the centre of the sphere describes a horizontal circle of radius c with angular velocity (5g/9c)11E, prove that the sphere will have no component angular velocity about the vertical. Show that the coefficient of friction between sphere and cone is not less than 2/7. 6. A uniform solid sphere of radius a rolls without slipping on the inner surface of a rough hollow circular cylinder fixed with its axis horizontal. If u is the component, parallel to the axis of the cylinder, of the velocity of the centre of the sphere, and 12 the component, about the radius to the point of contact, of the angular velocity of the sphere, prove that 2 a2122 + 7 u2 = constant. If initially u = U and SZ = 0, prove that at any subsequent instant u = U cos (17; (0 — a)),

174


where 0 is the angle between the downward vertical and the plane through the axis of the cylinder and the centre of the sphere, and a is the initial value of 0. 7. A uniform solid sphere of radius a rolls on the outside of a fixed rough cylinder of the same radius whose axis is horizontal. Initially the sphere stands on top of the cylinder and spins about its vertical axis with angular velocity S2 when it is slightly disturbed. When the normal at the point of contact makes an angle 0 with the vertical, show that 7a 62 = 59(1 — cos 0). Show also that the angular velocity of the sphere about the normal is then S2 cos nO, where 7n2 = 2.

Miscellaneous Exercises IV 1. A thin ring is of mass m and radius b. An axis parallel to its plane intersects the line through the centre of the ring perpendicular to its plane at a distance z from the centre. Prove, from first principles, that the moment of inertia of the ring about this axis is m(' b2 + z2) . A solid anchor-ring is obtained by rotating a circle of radius a about an axis in its own plane such that its centre describes a circle of radius 4a and centre 0. Prove that the principal moments of inertia at 0 are 67 Ma2/4, 69 Mae/8, 69 Ma2/8. Indicate clearly the directions of these principal axes. 2. The principal moments of inertia of a body at its centre of mass are A, B, C where A > B C. Prove that there exist points P at which the momental ellipsoid is a sphere if, and only if, B = C, and determine the positions of these points. A uniform solid right circular cone is of mass M; the vertical angle is 2a and the length of a slant side is 1. Show that the moment of inertia about the axis of symmetry is (3M12 sine a)/10 and about a line through the vertex perpendicular to the axis is Imp (1 — Determine a in order that the momental ellipsoid at the vertex should be a sphere. 3. Find the principal axes at a point 0 on its rim for a uniform thin hemispherical shell, and prove that a momental ellipsoid at 0, referred to its principal axes, is given by the equation 10x2+ (7 — 31/2) y2 + (7 + 3 1/2) z2 = constant. 4. Two equal uniform rods AB, BC, each of mass M and length 2a, are hinged freely at B and move freely on a smooth horizontal table with the centre of gravity of the system at rest. If 20 is the angle between the rods and 0 is the angle made by AC with a fixed direction in the plane, show that the kinetic energy T and the angular momentum h about the centre of gravity are given by

T = +Ma2{(1 + 3 cos20) 62 + (1 + 3 sin20) 02), h = f M a2(1 + 3 sin2 0) The ends A and C are joined by a light inextensible string whose length is slightly less than 4a. Initially the system rotates with angular velocity w in the plane of

Ex. IV


175

the table. Show that if the string breaks the subsequent relative motion of the rods is approximately simply harmonic with period 2n/(co y3). 5. A uniform circular hoop of radius a and mass m rolls along a line of greatest slope of a plane inclined at oc to the horizontal. An insect of mass m, which may be treated as a particle, moves on the inside of the hoop so as always to be at the point where the hoop touches the plane. If the system starts from rest, prove that the angular velocity of the hoop after time t is (gt since)/(2a), and show that in this interval the work done on the insect by the reaction of the hoop is — mgt i2 sin2 a. 6. A uniform rigid arm of mass M and length a is free to rotate in a horizontal plane about a smooth vertical axis through one end. A uniform rod of mass m and length b is free to swing about the other end of the arm in a plane perpendicular to the arm. If the rod is held at an angle a with the downward vertical and the system released from rest, show that the arm moves to and fro through an angle 3mb sing/{(M + 3m) a}. 7. A plane lamina of mass M can swing freely under gravity about a horizontal axis perpendicular to its plane. The axis intersects the lamina at the point 0, G is the centre of mass of the lamina, Mkt is its moment of inertia about a horizontal axis through G and GO = h. If R is the force exerted by the axis through 0 and co is the angular velocity of the lamina, show that the equations of motion of the lamina are Ma = R + Mg, Mk2 i.o = h x R, where a is the acceleration of G. The axis through 0 is made to move backwards and forwards in a horizontal plane and in a direction perpendicular to itself, and the displacement of the axis from its mean position at time t is ah cos nt, where a is a constant whose square can be neglected. If p2 = gh/(h2 k 2), prove that the (small) inclination, 0, of GO to the vertical at time t is pz h 0 g (n2 p2) (cospt — cospt), provided that p n and that 0 0, 0 = 0 when t = 0. Discuss the case n = p. 8. A uniform solid sphere is projected on a rough horizontal plane with which the coefficient of friction is ke. The sphere is initially in contact with the plane at the origin of fixed rectangular coordinates Ox, Oy on the plane, and its centre is set in motion with speed 'a, parallel to Ox, while the point of contact initially slips relative to the table with speed Vc, in a direction making an angle a with Ox. Show that the velocity of slip remains constant in direction, and that for a time 2 Vol(7 mg) the centre of the sphere moves in a parabola, of semi-latus rectum uo sin2al(,ug). 9. A uniform sphere of radius a rolls without slipping on a horizontal turntable which is rotating with variable angular velocity S2 k about the vertical axis OZ through a fixed point 0 of the turntable. If r is the position vector of the point of

176


contact from the origin 0, prove that 7 i• = k x (2S2r — c), where c is a constant vector perpendicular to k, depending upon initial conditions. If the centre of the sphere moves in a straight line passing through the axis OZ making an angle a with c, show that it does so with constant speed (c sin a)/7 and that S2 = (c cos a)/2r. 10. A uniform sphere of mass m rolls without slipping on one face of a wedge of mass M and angle cc, the other face of the wedge being in contact with a smooth horizontal plane. Rotation of the wedge, and translational motion of the wedge parallel to the line of intersection of the faces, are prevented by frictionless guides. Prove that, in general, the point of contact of the sphere traces out a parabola on the inclined face of the wedge, and that the acceleration of the wedge is 5mg sing cosa 7 M + 2m + 5m sin2a

11. A uniform sphere of radius a rolls inside a fixed hollow sphere of radius b in such a way that its centre describes a circle in a horizontal plane with constant angular speed Q. Show that the angular velocity w of the rolling sphere can be expressed in the form w= ni — S2 (- b — 1)i x (k x i) \a where n is a scalar, k is the unit vector vertically upwards and i is the unit vector normal to the two spheres at the point of contact and directed towards their centres. Show that n is a constant and that / b S22 -— a

a2 ga 1)(1 —) cos 0 + Dv, — =0 K2 K2

where 0 is the angle between i and k, and K is the radius of gyration of the moving sphere about a diameter. 12. A uniform solid sphere of radius a rolls without slipping on the outside of a fixed sphere of radius b — a. Initially the moving sphere is spinning with angular velocity n at the highest point of the fixed sphere. It is then slightly disturbed, and at the lowest point of the subsequent motion the magnitude of the velocity of its centre is v and the acute angle between the diameter of contact and the upward vertical is O. If the spheres are just on the point of losing contact there, show that v sine = .1an(1 — cos0), i76 v2 = bg (1 — 000), v2 = bg cos0, and deduce that 34a2 n2 = 945b g.

CHAPTER V

GYROSCOPIC MOTION, FREE ROTATION AND STEADY MOTION 5:1 Introduction In this chapter we consider some examples of the motion of a rigid body in which the difficulty of the variation of the inertia coefficients in the equations of motion can be overcome. First we consider bodies with kinetic symmetry and use a moving frame of reference which has one axis along the axis of symmetry of the body (usually chosen to be With this arrangement, the axis of symmetry, which is fixed in the body, is a principal axis with moment Co and any two perpendicular axes at right angles to the axis of symmetry are principal axes with moments A,. Thus, although the body may move both in space and relative to the axes, nevertheless in this case the moments of inertia referred to the coordinate (principal) axes at any instant are A0 , A0 , Co . (For the rest of this chapter we omit the suffix and use A, A, C.) Second, for bodies without kinetic symmetry the only way of obtaining constant coefficients of inertia in the equations of motion is to use axes which are fixed in the body. In addition we usually choose these axes to be principal axes of inertia at the chosen origin. Because we use rotating axes with, in general, a moving origin, we use the equations of motion in the general forms of (4.13) and (4.16) or (4.17) and resolve all vectors in the moving frame. The equation of linear motion (which we seldom need) becomes

a p +0 x p = F, at

(5.1)

where 0 is the angular velocity of the frame. The equation of angular motion in the form (4.16) becomes

ali)

at

+x

h(G) 177

x

=

r(o) ;

(5.2)

178


and the form (4.17) becomes Oh (0)

at

+e x h(0) vo x p = T (0).

(5.3)

We now use 0 instead of A to denote the origin moving with velocity v0 . (This avoids confusion with the moment of inertia; here we shall seldom need to refer to a fixed origin even if 0 is moving.) The subject matter of this chapter is the use of the above equations in various contexts.

5:2 Rotation under no forces of bodies with kinetic symmetry The phrase "under no forces" used in this and in a subsequent section (§ 5:6) seems to indicate a highly artificial situation. However, we have seen that the motion of a system relative to its centre of mass G is independent of the translatory motion so that if we choose 0 to coincide with G the results of this section (and § 5:6) apply to the motion of the body relative to G when the applied forces are equivalent to a single resultant force through 0, i.e., r (0) = 0. The results also apply to the motion of a satellite in space where the gravitational couple produced by other bodies about its centre of mass can be neglected. We consider then, a body with kinetic symmetry rotating about a fixed origin 0 when the moment I' (0) of the applied forces is zero. We note first that, since the applied forces do no work, the kinetic energy of the body is constant, and second, that the angular momentum h (0) is also constant. Figure 45 is a diagram of the case in which the body is an oblate spheroid with centre 0. The axis 0 3is the axis of symmetry perpendicular to the "equatorial" plane 0 The line OH is drawn in the direction of h(0), which is fixed, and is called the invariable line. We choose the direction of 0 to lie in the plane HO and 0 $2 completes the r.h. set. We denote the angle HO e3 by 0, one of Euler's angles, and give the position of the plane HO e, by the angle y, another of Euler's angles, which it makes with a fixed plane through OH. We also show in Fig. 45 the direction OJ of the angular velocity, the instantaneous axis referred to later, and the third of Euler's angles y. This is the angle between OQ, where Q is some marked particle of the body fixed in the equatorial plane, and the axis 0 We now consider the motion under the usual headings.

§ 5: 2]

179

FREE ROTATION AND STEADY MOTION

The motion of the frame sin0 6 V) cos0}.

The angular velocity of the frame is 0 = {

The motion of the body Since the only possible motion of the body relative to the frame is a rotation about 0 3, the angular velocity to differs from 0 only in the third component. . W = {co, w2 co,} =

sin0 O w3}

Ez

FIG. 45

Since the coordinate axes are principal axes,

h (0) = {h1 h2 h3 } = { Awl Aco, Co),} = {

—

Alp sin0 AO

C(03},

and the kinetic energy is

T = 2 A (col +

+ 2C co2 = z A (02 + b2 sin20)

= z W . h (0) = constant .

The forces It is given that T (0) = 0.

C (5.4)

180


The equation of motion We have noted above that, because dh(0)Idt = 0, the angular momentum vector h (0) is a constant vector. Because of our choice of axes we can write h(0) = { — hsin0 0 hcos0}

{ —Ay, sine

AO Cco,},

where h = I h (0)1 = constant.

O = 0, i.e., 0 = a

=

constant.

(5.5)

From this result we deduce the remaining characteristics of the motion. Since C w3 =h cosoc,

w3= n = (h/C) cosoc = constant,

(5.6)

where n is the spin of the body about its axis of symmetry. From the expressions for hiwe find

hi = — h sin a = —A /j) sin a . 1p = h/A = constant.

(5.7)

From (5.6) and (5.7) we see that (.1.) = { — (hIA)sinoc

0

(hIC)coscxl =

w sing 0 co cos/3), (5.8)

where Ito' = w = constant, and gis a constant angle (see Fig. 45). From these results we see that the vectors h (0), co and k, where k is a unit vector {0 0 1} along 0 $3, all have zero for their second component. i.e., all lie in the plane HO $3 ; moreover since the angles a and g are fixed these vectors are fixed relative to this plane, OJ making the angle a~g with OH. We can sum up the results obtained so far as follows: the three lines in Fig. 45, viz., 0 3the axis of symmetry, OJ the instantaneous axis, OH the invariable line, are fixed in the plane 0 $, $3which rotates about OH with constant angular velocity //), where

=

h Cn A A cosoc

(5.9)

[Since h (0) is constant and w is a vector with constant magnitude, the relation T = 2w . h (0) implies that the angle HOJ is also constant because T is constant.] Since HOJ is a fixed angle, the instantaneous axis traces out a right circular cone in space having OH as its axis. Also, since 30J is a fixed

§ 5 : 2]

FREE ROTATION AND STEADY

MOTION

1.81

angle, the instantaneous axis traces out a right circular cone in the body having 0 as its axis. These two cones are the herpolhode cone and the polhode cone respectively, discussed in § 1 :5, and the motion of the body is given by the rolling of the polhode cone, taking the body with it, upon the herpolhode cone which is fixed in space. The two cases, (i) A > C, and (ii) A C are illustrated in Fig. 46. By equating the resolutes of eqn. (5.8) we deduce tang — (A/C) tanf .

(5.10)

Fro. 46. The kinematically possible motion in which the polhode cone rolls inside the herpolhode cone is a motion which cannot take place as a free rotation. A couple must act on the body to produce such a motion. Referring to Fig. 45 we see that the angle 99 is measured from OQ, a line fixed in the body, to the plane 0 j. e, in which OJ is fixed. Hence —9'9 is the angular speed with which OJ traces out the polhode cone, i.e., it completes the cone in time 2n/99 and moves around 0 in the negative sense. (Note that when OJ has completed one circuit around the polhode cone neither OJ nor the body are in their original positions but OJ again intersects the particles along its original position in the body.) The resolute co, ( = n) of the angular velocity is related to Euler's angles by [eqn. (1.25b)] (.03 = — 9.) =

q

+

IP COS 0 .

coscx — n.

182


If we denote the angular velocity of OJ about 0 by p, then, from eqns. (5.9), (5.10), ( tan/3

n

1) )

tanc,c Since n = w cosfl,

relative to the body

co sin 0(3 — cx) sin a

P

(5.11) (5.12)

In case (i) of Fig. 46, a > 13 and p is negative; in case (ii) a < and p is positive. Although OH is usually drawn "vertically" in diagrams, its direction is not related in any way to gravity. The direction of 0 and the initial angular velocity determine the initial position of 0 13i the kinetic symmetry implies that OH lies in this plane. The subsequent motion is one of uniform precession about OH, as described above. The term precession is the name given to the motion of a plane or an axis about a fixed line as above. We shall meet other examples of precessional motion subsequently. Example. A uniform solid cube is free to turn about a corner 0 under no forces except the reaction at 0. The cube is set rotating about an edge through 0 with angular velocity D. Show that the diagonal of the cube describes in space a cone of semi-vertical angle tan-1(11/y2). Find the angular velocity with which the diagonal describes the cone. Although a cube is not a solid of revolution, for rotations about a corner it has kinetic symmetry about the diagonal through that corner. Since the moment of inertia of a cube, of mass M and edge 2a, about any axis through its centre of mass is f Mae, the moments about the coordinate axes of Fig. 47 are

A=I3= Ma2(3 +3) = 11Ma2/3,

C = a Ma2.

We suppose that the cube is set rotating about the edge OE; then the plane BOG is the plane containing the invariable line and 0t1. The resolutes of the angular velocity along and perpendicular to OG are n = Dif 3 ,

col= S211 (2/3) .

The corresponding resolutes of the angular momentum are 2Ma2 3 1/ 3

S2,

llf 2Ma2 3 1/ 3

Q.

The latter resolutes are Is cos oc, Is sin a respectively, where a is the angle between

§ 5 : 2]


183

OG and the invariable line. tana = 11/j/2.

Hence the diagonal describes the cone in space as stated. The angular velocity of the precession is given by Cn 11 Q A cosoc. 2 1/3 cosoc • But tan g = 1142, 123 \ 20 1 41

11

V

6 )

11

V

(

2)•

Exercises 5:2 1. A thin uniform circular disc can turn freely about its centre and is set spinning about a line through its centre inclined to the axis of the disc at an angle n/4. Show that the axis of the disc describes a fixed circular cone of semi-vertical angle tan-1Z, and find the angle of the cone that the instantaneous axis of rotation describes relative to the disc. 2. A uniform right circular cone, of height h and radius a, has its centre of gravity G fixed, and is set rotating about the line GA, where A is a point on the circumference of the base. Prove that the angle between the axis of the cone and the invariable line is constant and equal to tan-1{(h2 4a2)/(2ah)}, and describe the motion of the instantaneous axis. 3. A uniform double right circular cone of semi-vertical angle oc, symmetrical about the common vertex, about which it is free to rotate, is set rotating about a generator. Prove that the axis of the body describes in space a cone of semi-vertical angle tan-1( tanoc + 2 cotoc).

184


5:3 The steady motion of a gyroscope or top When we refer to a gyroscope we usually mean a mechanical system consisting of a heavy body spinning about its axis of symmetry and so mounted that the direction of its angular velocity, and its axis, may change freely. A simple top, with a point of its axis fixed, see Fig. 48, is "-3

FIG. 49

an elementary example of such a system; we now consider in detail the motion of a top. The chief feature of gyroscopic motion is the precession, which can be illustrated quite simply as follows. Suppose a heavy body such as the flywheel of Fig. 49 is spinning about an (horizontal) axis; then the angular momentum vector is directed along OH. If a couple r acts on the system about the axis OK, the angular momentum receives an

§ 5: 3]


185

increment HH' in a short interval at, where HH' = Pat and its direction is parallel to OK. The angular momentum vector, after time 6t, therefore lies along OH', which is the new direction of the axis of the flywheel. Thus, by exerting a couple about a "horizontal" axis a rotation of the wheel about a "vertical" axis is produced. If the angle HOH' = Sy, then

HH' =hay =Pat,

and, in the limit as at

-- 0, =

Tlh.

(5.13)

Here we have used "vertical" and "horizontal" to describe directions, but these directions are not related to gravity. For any r.h. set OHKZ, a couple along OK acting on a system with angular momentum along OH produces a precession qp, given by (5.13) about OZ. It is this fact that the axes of the couple and of the resulting precession are different which provides the characteristic features of gyroscopic motion. Example. A rectangular plate of sides 2a and 2b and mass m spins with constant angular velocity w about a diagonal. Show that the couple which must act on the plate in order to maintain this motion has magnitude nico2ab

a2

b2

a2

b2 ,

(a > b)

and is perpendicular to the instantaneous plane of the plate.

FIG. 50. With reference to Fig. 50 the moments of inertia about Osl 0;2 are lina2, *mb2respectively and the angular momentum is ,

h (0) = im {—a2 w sing 0 b2 w cos = {h1 h2 he), where tan a = b/a. 7

OMES

186


We resolve the angular momentum into a horizontal component ho and a vertical component ho, where

h,„ =imco (a2 sin2 oc + b2 cos2 ho= 3mco(b2

—

a2) sin a cos oc

mco (b2 — a2) ab 3 (a2 + b2)

The component ho is stationary, but the component h0 is precessing about the vertical diagonal with angular velocity w. The couple necessary to cause this precession has, from eqn. (5.13), magnitude m (02 (az — b2) ab

r = Ihood

3 (a2 + b2)

Its axis is perpendicular to the diagonal and to h0 , i.e., perpendicular to the plane of the rectangle, at any instant.

We consider now the motion of the top of Fig. 48 when the only forces acting are the weight, acting through 0, (OG = 1), and the reaction from the smooth pivot at 0. We use a rotating frame of reference with 0 3as the axis of symmetry and the plane W1$3containing the vertical 0 V. (C.f. the axes of Fig. 45 with OH vertical.) We use our customary arrangement of working. The motion of the frame The angular velocity of the frame is 0 = { — tp sin° 0 bcos0} (Note that 101 is NOT the same as 0 here and in this chapter generally.) The motion of the body The angular velocity of the body is w = { — sin() 0 (DJ ; the angular momentum is h (0) = {—A v; sin0 A 0 C w3}. (Here A, A, C are the moments of inertia about axes through 0, not through G.) The kinetic energy is T = A (02 + 2 sin20) Cco:23 . The forces At

0, {0 0 0}, the reaction {X Y Z};

at

0, (0 0 1), the weight

11 (0) = (0

(Mg sin() 0 Mgl sin° 0}.

—Mg cos0).

§ 5: 3]


187

This is the couple which causes the top to precess. Since the pivot is smooth and gravity is a conservative force, we take the potential energy to be V = M gl cos0

The conservation laws The energy equation is

A (02 +

sin2 0) -iCco3

Mgl cos0 = E = constant. (5.14)

A unit vector in the (fixed) vertical direction is n = — sin° 0 cos0} and n . r (0) = 0. Hence angular momentum about this axis is conserved. h (0) = Ay, sin2 0 Cco, cos° = K = constant. (5.15) Although the first and third resolutes of r (0) are zero, we cannot deduce that the corresponding resolutes of h (0) are constant, since the directions concerned are not fixed.

The equation of motion The eqn. (5.3), with v, = 0, has resolutes

d — A — (lp sinO) Cco30 — A tp° cos° dt

= —Ayi sin° Cco30 — 2A p0 cos° = 0 (5.16) AO — A p2 sinO cos° + Ccoop sin° = Mgl sin°, (5.17)

Cth, — Ay,6 cos° + Ay,6 coo = c th3 =0.

(5.18)

From eqn. (5.18) we see that 63 = 0, i.e., co3 = n = constant.

(5.19)

This apparent conservation of angular momentum arises from the kinetic symmetry. We now have three first integrals of the equations of motion, viz., the energy equation (5.14), the angular momentum equation (5.15), and eqn. (5.19). We develop these equations further in §5:4; here we consider only steady motion. Because of the symmetry of the top, the significant information concerning its behaviour is not its actual position but the position and motion of the axis and the spin about this axis. The latter is constant, and the motion is steady if 0 = a = constant, so that the axis precesses about the vertical with angular velocity V, = S2 = constant. In order to deter7*

188


mine what steady motions are possible, we substitute these values into eqns. (5.16), (5.17). Only eqn. (5.17) gives any information, viz., — AS22sinoc cosy

CnS2 sing = Mglsinoc

Therefore, for steady motion, either sing = 0

or

AS22cosy — CnS2 Mgl = 0.

(5.20)

We conclude that the steady motions can only persist when either (1) the top spins about the vertical with G either above 0, (cx = 0), or below 0, = n), or (2) the top precesses steadily with the axis inclined to the vertical, the velocity of precession for given n and a being determined by eqn. (5.20). Such a precessional motion can occur if, and only if, S2 is real, i.e., if and only if C2 n2 >4AMg1 cosoc. (5.21) Therefore, if the top is not spinning fast enough, the only possible steady motions are these given by (1). Examples. (i) We give a vector treatment of the steady precession. We consider first the motion of the point with position vector k = {0 0 1} on the axis of symmetry. Since it is a point of the top its velocity is dk = to x k.

at

dk ,•. k x dt But

k x (co x k) = co — k(o) k). w.k=n. dk = nk + k x dt •

This equation gives co as the sum of one component nk along the axis and another component k x (dk/dt) perpendicular to the axis. Hence h(0) = A (k x

dk

+ Cnk.

We now restrict the motion to be one of steady precession about the vertical fi with angular velocity 0 fi. Since the point k is a point of the frame dk dt = 0(11 x k), d2 k

dk = S22 x (n x k) = .122[11 (fi . k) — k]. dt2 =( S2 11 x at


But

dh(0)

—

dt

189

d2 k dk A (k x —) + Cn dt2 dt

= .A122 (k x n) (ft . k) + CS2n(il x k)

= (AS22 cosoc — CS2n) (k x n). The weight of the top —Mgt"' acting at G (position /k) is the only force which has a moment about 0 viz., Mg1(11 x k). The equation of the motion is dh (0) (AS22cosa — CS2n) (k x = Mg1(11 x k). dt AS22cosoc — CS2n+ Mgl= 0, unless ri x k = 0, i.e., oc = 0 or n. (ii) A top consists of a uniform circular disc of mass 4m and radius 3a, at the centre of which is rigidly attached the end 0 of a uniform rod OX, of length 4a and mass m, the rod being perpendicular to the disc. Show that, if it can precess in steady motion with the end X fixed and the lowest point of the disc at the same level as X, then the spin about the axis must be not less than y{94g/(9a)}.

The moments of inertia are 9

A = 4m (- 2 -I- 16a2) + me 16 — 4a

3

235ma2 3

9 a2 C = 4m = 18ma2 . 2 The inclination of the axis is given by cos fl = 3/5 (see Fig. 51). The distance / = XG is given by 5m1 = 4m . 4a + m . 2a = 18ma. The condition for steady precession is 3 235 18g S22. — 5 — 1812n + — = 0, a 3 47122 — 18S2n + 18(g/a) = 0.

190


The roots for Q are real, i.e., steady precession is possible, if and only if 81n2 z 47 . 18 (g la), i.e., n2 j (94g)/(9a). (iii) A rod, of length / and of negligible weight, is freely pivoted at one end to a fixed point, and at the other end carries a wheel of mass M, rotating with angular momentum h about the rod as axis. The moment of inertia of the wheel about its axis is C. Show that the rod can revolve as a conical pendulum at an inclination a to the vertical in a period 2n/w, where a) is given by

(C + 2M12) w2cosa + 2hw — 21141= 0, the two roots corresponding to opposite directions of motion. The arrangement described is, in effect, a top which is moving with its centre of mass below the point of suspension 0. The moment of inertia of the wheel about a diameter is IC (we regard the wheel as a thin disc) and about a parallel axis through 0 is A = iC + M12. Also the spin n is given by h = Cn. This top is precessing steadily with 0 = n — a; the condition for this is

(1C + 11112) co2cos (a — oc) — h co + M y 1 = 0 , (C + 211112) w2cosa + 2hw — 21141 = 0. This equation has real roots and therefore two steady precessions are possible. Also, since the product of the roots of this equation is negative, the two possible precessions are in opposite senses. Further, since the sum of the roots is negative, the precession in the negative sense is faster, i.e., viewed from above 0 the wheel and this precession have the same sense. (iv) We verify that the steady precession of a top is given by the rolling of the polhode cone inside the herpolhode cone. (This is the motion which is impossible as a free rotation.) Fig. 52 represents the state of affairs with the usual position of the axes. The point G is the centre of mass of the top.

The motion of the frame Angular velocity 0 = {-1,2 sin a 0 Q cosa).

The motion of the body The angular velocity of the body co = {—Q sing 0 n}. .•. 11(0) = {— AS2 sina 0 Cn}.

The forces As on p. 186 the moment of the applied forces is

F(0) = (0 Mgt sin a 0) .

§ 5: 4]


191

The equations of motion Only the second resolute of the equation O h (0)

t

+ 0 x h(0) = T(0)

does not vanish identically and leads to

— A S22sing Qom + CnS2 sing = Mgl sing.

Hence, sin a = 0, or

A 02cosa — Cnil + Mgl = 0.

The possible motions of this sort are, therefore, the steady precessions. It should be noted that the positive direction of the angular velocity vector is JO so that {— Q sing 0 n} = {— co sin/3 0 — co cos/3}. It is clear from the figure that Q, n must have opposite senses.

5:4 The general motion of a top In order that a top shall perform one of the precessing steady motions considered in § 5:3, the conditions of projection (or initial conditions) must be such that E, K, n have exactly the correct values. We now investigate the motion of a top with arbitrary initial conditions so that, in general, the motion is not steady. Since the spin n is constant, we investigate the variations of 0 and ? (or 6 and v) to exhibit the motion of the axis; we represent this motion

192


diagrammatically by the path traced by the axis as it intersects a fixed sphere with its centre at 0 (see Fig. 54). We shall refer to the equations of motion (5.16-17) and use the energy and angular momentum equations (5.14) and (5.15) in the forms

A(02 +

sin2 e) + 2M gl cos0 = 2E — Cn2 =2E', Ay, sin2 0

Cn cos0 = K.

(5.22) (5.23)

We transform these equations by writing z = cos 0, z = —0 sine and eliminating /f) between equations (5.22) and (5.23) We thus obtain ti)—

Az 2=

K — Cnz A (1 — z2) '

1(z) = 2 (E' Mglz) (1 —

(5.24)

z2) — A -1(K — Cnz)2 . (5.25)

Eqn. (5.25) is an equation of the type discussed in Vol. III § 5:2 and in connection with the spherical pendulum (§ 3 :4 of this volume). We obtain our results concerning the variations of 0 (or z) by investigating the zeros of f (z), thereby obtaining the "vertical" (meridianal or longitudinal) motion of the axis on the sphere ; then eqn. (5.24) gives the "horizontal" (azimuthal or latitudinal) motion. The combination of the two motions indicates the general shape of the path of the axis, displayed in Fig. 54. The quantities E', K, n are determined entirely by the initial conditions, and we can assume, without loss of generality, that n > 0. Apart from this there are no relations between or restrictions on these quantities except that E' — Mgl. [See eqn. (5.22); no real motion is possible for which E' < — Mgt] Since z = cos 0, only I z1 5 1 corresponds to real motion ; nevertheless we consider values of z outside these limits when investigating the sign of f (z). For reasons which will become clear later we consider first motion in which K Cn. If K = On a motion is possible in which the top passes through, or tends towards, the vertical position 0 = 0 provided that E' M gl also; we exclude such motions from consideration at this stage. Similarly we also exclude motions in which K = — On. To locate the zeros of f (z), a cubic polynomial, we tabulate its sign for certain values of z, assuming that initially z = z, and 0 + 0.

z /(z)

— 00

—1 z, 1

+—

00

+•

§ 5 : 4]


193

The table shows that f (z) has three real zeros z1, z2 , z3where —1 < zi < zo , zo < z2 < 1 , 1 < z3. The graph of / (z) has the general shape shown in Fig. 53. Since

1(z) = A z2 >0, the possible values of z lie in the interval z1 z < z2 (the shaded portion of the graph), the variation in z is a libration, and the path of the axis is confined between two circles (of latitude) on the sphere, which the path touches alternately.

We now consider the implications of eqn. (5.24). Since lz < 1, changes sign when z = K (Cn). Three cases arise (a) K/(Cn) < z1, (b) z1 < K/(Cn) < z2 , (c) z2 < K/(Cn). In case (a) yi has a negative sign for all permissible values of z; consequently the plane containing the axis precesses continuously, but not uniformly, in the negative sense, the axis rising and falling at the same time. In case (b) changes sign for a value of z within the permissible range; consequently, while the axis is in the lower part of the permissible zone > 0, and in the upper part tp < 0. The path touches the meridian where z = K (C n). In case (c), has a positive sign for all permissible values of z and the motion is similar to (a) but in the opposite (positive) sense. Cases (a) and (c) are illustrated in Fig. 54 (i) and case (b) in Fig. 54 (ii). We omitted above the particular cases (d) zl =K/(Cn),

(e) z2 = K/(Cn).

These cases correspond to very special circumstances of projection. 7 a CMES

194


In fact, case (d) is impossible to achieve in practice since it implies that when z = z1, j = 0 and 0 = 0, i.e., the axis is stationary; since the graph of Fig. 53 has a positive gradient at z = z1, in subsequent motion the axis rises, i.e., 0 < 0 when z = z1. However, on substituting = = 0 into the equation of motion (5.17) we see that 0 > 0. We are thus led to a contradiction in supposing that z1can equal KI(Cn).

FIG. 55. The case (e), although it implies a special set of initial conditions, is easy to achieve in practice. As in case (d), when z2 = K/(Cn) the axis is stationary and, at this point tp = 0, 0 > 0. Hence the axis falls from the position on the circle z = z2 at right angles, a motion achieved physically by releasing the axis of the top from rest when the spin is n. The path is shown in Fig. 54 (iii) with cusps lying on the upper circle. (This case also covers the case excluded above when 0 = 0 initially.) While, in general, arbitrary initial conditions produce a graph as in Fig. 53, there may be certain special conditions which cause the zeros

§ 5 : 41


195

z1, z2to coincide. The graph of / (z) is then as in Fig.55. Since there is no part of this graph above the axis in the range —1 z < 1, the only possible motion is one for which z retains the fixed value z = z1 . The path of the axis is a horizontal circle and the motion is a steady precession. [This is verified in example (ii) below.] Examples. (i) A top consists of two equal uniform circular discs of radius a, with their centres P and Q rigidly connected by a thin light rod PQ of length 4a. The planes of the discs are perpendicular to PQ, and the top turns freely about a fixed point 0 of the rod distant a from P and 3a from Q. If at one stage of the motion the axis PQ is at rest instantaneously, inclined at an angle of 60° to the upward vertical, and the spin about this axis is 3 y(7g/2a), prove that the inclination of PQ to the vertical always lies between 60° and 120°.

The moments of inertia are (see Fig. 56) A = gna2(i +1 + 9) = 21ma2/2,

C = ma2 .

In the usual notation the energy and angular momentum equations [(5.22), (5.23)] can be written 2g 21 (02 + 1P2 sin2 0) + 2 . — cos° = 2E', a 2 21 sin20 -I 3 /(7g ) cos0= K. 2 2a When°=60°, 0= 0=y. .•. 21(02 + yi2 sin2°) + 8 (g/a)cos 0 = 4 (g/a), 21 tp sin2 O + 3 V (14g/a) cos0 = V (14g/a). Elimination of Vi gives (1 — 2 cos0)2 ). 2102 + 9 7g \ + 8 () 0cos = 4 (1 a k a) 21 sin2O 2a .•. 2102 sin2O = (g/a) [4(1 — 2 cos9) sin2 O — 2 (1 — 2 cos0)2], i.e.,

4202 sin2 0 = (gla) (1 — 2 cos0) (1 + 2 cos0) (5 — 4 cosO)

Hence 0 must lie between the limits

01= cos-1(— i) = 120°, 02 = cos-1(i) = 60°. 7 a*

196


(ii) Steady precession with angular velocity Q. The conditions for a double zero z = z1of the function f(z) in eqn. (5.25) ara f(z1) = 0, f(z1) = 0. Now

If

f (z) = 2(E' — Mglz) (1 — z2) — (K — C nz)2IA,

(1)

f (z) = —2Mg1(1 — z2) — 4z(E' — Mglz) 2(CnIA)(K — Cnz).

(2)

= Q when z = z1and f (z,) = 0 = f (z,), then

S2—

K — Cnz1 A(1 — 4) •

Substitution for K — Cnz1into (1) and (2) gives 2(E' — M g 1 zi) (1 — 4) = A S22(1 — 4)2 2(Cn/A) AS2(1 — 4) = 2Mg1(1 — 4) + 4z, (E' — Mglz,). Elimination of E' — Mglz, from these equations leads to 2 (CnS2 — M gl) (1 — 4) = 2AS22 z1(1 — i.e., AS22 z1— Cn12 + Mgt = 0. This is the condition (5.20) for a steady precession with angular speed Q.

The case K = Cn Although it is unlikely to be realised exactly in practice we consider now the case in which K = Cn. This is the case in which the axis may pass exactly through the vertical position 0 = 0; it does not necessarily do so, however. Eqns. (5.24) and (5.25) become V—

Cn(1 — z) A(1 — z2) C2 n2

Cn A(1 + z)

AZ 2= f(z) = —2Mg1(1 — z)[ 2Amgi (1 z)

(5.26)

E' ( mgi z) (1 + z)]

= —2Mgl(1 — z) q(z), where

q(z) = 2b2 (1 — z) — (x + 1 — z) (1 + z) = z2 — z(2b2 + x) + (2b2 — x — 1).

(5.27)

§ 5 : 4]


197

To simplify the expressions we have written b2 =

C2 n2 E' x — 1. 4AMgl ' Mgl

The zeros of the quadratic factor q (z) give the zeros of 1(z) other than z = 1. We locate the zeros of q(z) by tabulating the sign. — 00 q(z)

-1

1

00

4b2 (> 0) —2x +.

This table shows that the sign of x is the crucial factor in locating the zeros of q (z). (This is the reason for choosing the above form for x.) We know that E' > — Mgl, so that x > — 2; the table shows that we can distinguish the following cases. (f) If x > 0, i.e., E' > Mgl, then q(z) has two real zeros, one on each side of z = 1. Since q(-1) = 4b2 >0, for all values of x, the smaller of these zeros must lie in the range —1 < z < 1. (g) If x = 0, i.e., .E' = Mgl, then q(z) = (1 — z)(2b2— 1 — z). Then the zero z = —1 + 2 b2 lies within the range —1 < z < 1 provided that b2 < 1, i.e., if C2n2 < 4Amgl. (h) If x = 0, i.e., E' = Mgl, then the zero z = —1 + 2b2 lies to the right of z = 1 if b2 > 1, i.e., if C2 n2 > 4A Mgl. When x < 0 the table does not help in the initial location of the zeros of q (z) since all the entries in the second line are positive. The condition that q(z) has real zeros is, from eqn. (5.27), x

—2(b +1)2 or x

— 2 (b — 1)2 .

Since x > — 2, the first of these cannot be satisfied so that the remaining possibility is —2(b — 1)2 5x 5 0. (5.28) When this condition is satisfied the zeros, which are real, must lie one on each side of z = z,„,„ = b2 ix, for which value q(z) has a minimum. The condition (5.28) shows that ;nit, > — 1 2b > — 1. Therefore no zero lies to the left of z = — 1, and we are left with two possibilities, when x < 0.

198


(j) Both zeros lie in the range —1 < z < 1, in which case b must satisfy the inequality — 1 + 2 b < 1 or b 1. b2 >1-2 x . Since x < 0, this implies that b2 > 1, i.e., C2 n2 > 4AMgl. All these cases are illustrated in the graphs of Fig. 57. Condition

f (z)

q(z)

(f) r>Mg1 , (x>o).

(g)r Mg' (x=o), cznz 4A Mgl, a steady precession is impossible and the nutation becomes so large that the top falls over. In general, the larger the value of n the smaller are the effects of nutation and the slower is the precession; for extremely large values of n the precession about the vertical is imperceptible and the loop motion of case (b) becomes effectively the circular precession about the invariable line of free rotation. When n is extremely large, the gravitational effects are negligible. Having considered in detail the motion of a top we shall not discuss other gyroscopic arrangements here. The motion of a top on a smooth table, or the spinning of a solid of revolution on a smooth table, all exhibit similar characteristics of precession and nutation. The investigation of the motions follows similar lines, but usually involves more complicated geometrical conditions. The essential ideas are those described above. Examples. (i) The axis OP of a symmetrical top is of length 1. The top is precessing steadily with the axis horizontal and the end 0 fixed when a particle of mass m is attached suddenly at P. The velocity of the particle just before attachment is such that the top receives no impulse. Show that in the ensuing motion the maximum depth of P below 0 is the positive value of d that satisfies the quadratic

2A (C nS2 + ingl) d 2 + (A2s22 csn2) ld — 2 Amg13 = 0 where A, A, C are the principal moments of inertia at 0 subsequent to the attachment of the particle, S2 is the initial angular velocity of precession, and n is the spin about the axis. The motion after attaching the particle is a general motion of a top in which the initial conditions are those of the preceding steady precession. The energy and angular momentum equations are

A(62+ .02sin20) + 2(Mb + ml) g cos0 = Ai) sin2 0 Cn cos0 = K,

,

(1)

(2)

202


where b = 0G and G is the centre of mass of the top alone. The condition for precession of the top alone is

(A — m12) S22cosn/2 — CnS2 + M gb = 0. .•. Mgb = CnS2, M being the mass of the top alone. Hence A(62

.02 sins) + 2(C n,S2 + mlg) toss = 2E'.

(3)

From the initial conditions, viz., 0 = n/2, 0 = 0, = S2,

2E' = AD2, K = AS2. ••• 02 = D2 ,p2sin20

— (2/A) (Cn,S2 + mig) toss,

sins = S2 — (C nl A) toss. Elimination of hgives, after some reduction,

A 02sins = (— dI13) {— 2 A m13g + (A2S22 + 02 n2) dl + 2 A (C n S2 + mlg) d2], where we have written d = 1 sin (0 — a), i.e., cos 0 = — d/1. The maximum value of d occurs when the second factor vanishes. If

g)(d) = 2 A (C n S2 + mlg) d2+ (A2S22 + 02 n2) dl — 2 A m13g , we locate the zeros by tabluation:

d

— 00 —1 0

1 00

g)(d) + — — + + except that q) (— 1) = — (AS2 — C n)2 = 0, if AS2 = C n. Hence the only zero of g)(d) which is numerically less than 1 is positive. The other zero between — oo and — 1 is of no physical significance. (ii) A solid uniform prolate spheroid whose axes arc 2a, 2b, 2b spins steadily on a smooth horizontal table. It has angular velocity n about its axis of figure, that axis has angular velocity w about the vertical and c is the constant height of the centre above the table. Show that n2

b2) (c2— b2) g • b4c

5 (a2

and that, if n has its least value, then w2

5 g (a2— b2) c(a2 +b2)

•

We employ the usual arrangement of rotating axes (Fig. 59), but instead of 0 being fixed we have to use the condition, in the evaluation of r(0), that P is on the surface and that the tangent plane is horizontal.

§ 5: 4]


203

The moments of inertia are A = m (a2 b2), C = m b2 and, since 0 does not accelerate vertically, the force at P is mg vertically upwards. We shall not write out the full details but summarise the solution as follows. 0={—w sina 0 co cos oc}, h(0) = m {— (a2 + b2) co sina. 0 2 b2 n} and I'm= mg {0 PN 0}. The second component equation of angular motion about 0 leads to — k(a2 + b2) sina cos a + f b2nco sinoc = gPN.

(1)

This corresponds to the usual condition for steady precession.

Flo. 59. The equation of the section of the surface is

a2 1 -4 - +4 a2

and we choose P to be {b sin A 0 —a cos 2), where A is the eccentric angle. Then 1 c2

cos2A

sin2A

a2

b2

c being the perpendicular distance ON. The condition that the direction of the normal to the tangent plane at P is vertical is sin A b sin a

cos A a cosy

(2)

204


From (2) we deduce that c2 = a2 cos2a + b2 sin2 oc,

PN = [(a2— b2) sina cosa]/c.

Hence eqn. (1) becomes (a2 + b2) co2sina cosoc — 2 b2 nco shim + 5 (g 1c) (a2— b2) sina cosoc = 0.

(3)

The condition that (3) has real roots leads to the condition on n; the given value of w is then the (repeated) root of (3) when n has the smallest allowed value. (iii) A top with mass m and centre of mass G consists of a rod 0 G N with 00 = h, ON = 4a rigidly attached to the centre N of a circular disc of radius 3a. The plane of the disc is perpendicular to ON, which is the axis of dynamical symmetry, and the principal moments of inertia at 0 are mkt, mkt, Amk2. The point 0 is freely attached to a fixed point of a smooth inelastic horizontal table. N is vertically above 0 and the body has spin D. Show that if D2< 16g h/(5A2 k2) a slight disturbance will cause the disc to strike the table and the subsequent motion of ON has period 167c/(5 A Q), assuming that contact of the disc with the table is maintained.

no. 60. Since this top is slightly disturbed from the vertical position, the subsequent motion must correspond to case (g) above with the axis falling away from the vertical, Fig. 60. If the top is to hit the table, then the lowest position of the axis would, in the absence of the table, bring the circumference of the disc below the horizontal through 0, i.e., z1< 3/5. As usual k2 (62 + sine) + 2gh cos0 = 2E' , sin20 AD cos0 = K. Since the top starts from the vertical 6 = 0 when 0 = 0. ;. E' = gh, K = 62= 2g(h1k2) (1 — cos0) — 1p2 sin20, tj) sin2 0 = AD(' — cog).

(1) (2)

§ 5 : 4]

205


Elimination of gives 62 sin2 0 = (1 — cos 0)2 [2g (h/k2) (1 + cos()) — 22.122]. The zero of the last factor is z, where k222Q2 z1 =—l+

2g h

The disc strikes the table top if z, < 3/5, i.e., if 5k222,122 < 16gh. Because the table is inelastic the subsequent motion of the axis is initially horizontal, and because the table is smooth both K (= S2) and n (= 12) are unaltered and eqn. (2) continues to hold. Hence the initial conditions for the subsequent motion are that 6 = 0 when cos0 = 3/5, and that has the final value of the previous motion, [obtained from eqn. (2)], viz. 52 12/8. Hence k2(02

.'. 62

y:,2 sin2 0)

k2A2 s22 + 2g h cos 0 = 2E1

gh(6 + A2s22 1 — k2 — 5 — 2cos0 ) 4

4

6 gh 4-

5

•

1— cos° 1 + cos°)•

If the motion corresponds to a rising axis, 6 < 0 when cos 0 = 3/5. On differentiating we obtain 02 gh — sin°. (1 + cos0)2 [k2 Hence, if contact is to be maintained, 6 > 0 when cos 0 = 3/5, s22

i.e.,

64gh 25.12 k2

If contact is maintained, the subsequent motion is not a nutating gyroscopic motion, but a steady precession in which the disc slides or rolls on the table. The period of the precession is 27c 16n —= 5212 If 12 lies between the limits given by 64hg 2522k2

122 < 16gh 522 k2

then the axis of the gyroscope rises after impact. In fact, the motion is given by 2g h =

5k2

(3 — 5z) (1 — z) [z + 1

5k2 ,1292 8gh

206


Therefore the axis rises until 5 k2 22Q2 z

1+

8gh

which is less than 1. The subsequent motion, then, is an example of case (j) above.

Exercises 5:4 1. The extremities of a diameter of a thin uniform spherical shell, of radius a, are rigidly fixed to an axle of negligible mass, and one end of the axle, which is distant 4a from the centre of the shell, is pivoted to a fixed point. The system is spun as a top and set free with its axle instantaneously at rest at an angle of 45° to the upward vertical. If the axle descends until it is at an angle of 60° to the upward vertical, show that the angular velocity of the spin is 151/ {2 (1 + 1/2) Oa}. 2. A heavy symmetrical top of mass M spins on a fixed point 0 of its axis. Its centre of mass G is distant h from 0 and its principal moments at 0 are A, A, C. The shape of the top is such that it can spin freely only if 0 G makes an angle less than a with the upward vertical. The top is started with spin p and its axis at rest making an angle j9with the upward vertical. Prove that the rate of precession increases with the angle between 0 G and the upward vertical, and hence show that for free movement to persist p must exceed 2MghA cos,8 — cos a

sina C •

3. A symmetrical body of mass M, centre of mass G, with a point 0 on its axis fixed, has moments of inertia C about its axis and A about any perpendicular axis at 0. A light elastic string of natural length 1, with one end fastened at 0, passes through a smooth ring vertically above 0 at a height 1 and the other end is fastened to a point C of the axis on the same side of 0 as G, where OC = 1. If the modulus of elasticity of the string is Mghll, show that the axis OC can rotate at any constant rate 12 around the vertical, at any constant angle a with it, if the body is given a spin about its axis equal to AQ cos a/C and the axis is projected in a suitable manner. When the axis, at a given angle a, is steadily precessing at a prescribed rate = 2 y{(Mghcosa)/A), the elastic string is cut; prove that G descends a distance h sin a before rising again. 4. A symmetrical top of mass M, whose principal moments at the mass centre G are A, A, C, spins on a smooth horizontal plane. The end of the peg in contact with the plane is part of a sphere whose centre is at a distance h from G. The top is released with spin n about its axis of symmetry from a position in which that axis is stationary and inclined at an angle a to the vertical. Prove that if 2p cos cc > 1

§ 5: 5]


207

the inclination of the axis oscillates between a and where cos 13 = p — y(p2 — 2p cos a + 1), p standing for n2 02/(4MghA). Prove that, when the spin is large enough for 13 — a to be taken as a first order small quantity, the approximate rate of change of azimuth of the axis of symmetry after time t is given by Mgh Cn

n2 C2 (1 COSDt),

where D2—

A (A + Mh2sin2a)

5. A top, free to move about a fixed point 0 and having spin n, consists of a uniform solid sphere of radius a rigidly attached to the end A of a light rod OA. The centre of the sphere lies in the line of the rod and is distant 1 (> a) from 0. If at some instant OA is horizontal and momentarily at rest with the sphere in contact with a smooth vertical wall at a perpendicular distance a + 1 cos a from 0, < n/2) show that in the subsequent motion the sphere will remain in contact with the wall until the rod next becomes horizontal, provided that n is in the right sense, and 8 a4 n2sin a >= 45 (5 /2 + 2a2) g 1 cos2a . 6. A symmetrical top has mass M and principal moments A, A, C at a point 0 of its axis which is at distance Is from the centre of mass G. 0 is fixed and the top moves with its axis nearly vertical and with spin n. Prove that the projection of 0 on a horizontal plane has a motion which is approximately the superposition of two motions of periods 21-c/(2, ,u), where

2 = Cn/(2A) and it = y(C2n2— 4MghA)/(2A). 7. A top consists of a thin uniform shell of radius a and centre C and is free to move about the fixed point 0 of its surface. The radius 0 C makes an angle 0 with the upward vertical 0 Z, and 9) is the angle between the plane ZOC and a fixed vertical plane. Initially, 0 = in, 0 = 0, 117 = 2n/5, where n, the spin of the top about its 3 g )'12 15g y/2 axis, is ( — . Show that subsequently cos 0 = tanh2 — t, and that C traces 2a 10a out a spiral path on a fixed sphere of centre 0 and radius a. Sketch the path of C.

5:5 Euler's Dynamical Equations We now consider the motions of bodies without kinetic symmetry. We choose the origin to coincide either with a fixed point of the body or with the centre of mass, and take the principal axes of inertia at the origin as coordinate axes. We denote these moments of inertia by A,

208


B, C (in this section) and assume, unless otherwise stated, that A > B > C. Since the axes are fixed in the body, A, B and C do not vary as the body moves. The usual arrangement of working now follows.

The motion of the frame The angular velocity of the frame is that of the body. °=w=

cot ( 03) •

The motion of the body The angular momentum and kinetic energy are h(0) = {Awl Bw2C co,} , T = z (A + Bo.4, ± C

The forces We assume that the applied forces have been reduced to F and at the origin, where F = {X1 X2

r (0),

T(0) = {L1 L2 L3}.

The equation of motion We do not usually need to consider the equation of linear motion; the equation of angular motion is dh(0) dt

h(0)

at

+ 0 x h (0) =

(0),

having resolutes

{Ao:), Bth, Cw3} + {co, w2co,} x {Aw, Bw2 Cw3} =

L, L3}.

Written out in full these resolutes give Euler's Dynamical Equations

Ad), — (B — C) w,co, = Li , Bth, — (C — A) w3w1 =L2,

(5.30)

Cth, — (A — B) coi w, = L3 . Few general rules can be laid down for the integration of eqns. (5.30). It will often be necessary to use Euler's angles to specify the position of the body and to express co,, co3 , co3in terms of these angles [see § 1:5, eqn. (1.25)]. As usual, in suitable cases the conservation of energy and of angular momentum provide two first integrals of the equations.

§ 5: 6]


209

5:6 Free rotation We continue here the investigation started in § 5:2. Choosing the origin 0 to coincide with a fixed point of the body, or with G, as appropriate, we use Euler's Dynamical Equations to investigate the motion. As before, the kinetic energy and the angular momentum vector are constant, a line parallel to the latter drawn through 0 being the invariable

line. For convenience, in this section we shall replace i , z , a respectively by x, y, z. In Fig. 61 the frame Oxyz is formed by the principal axes

at 0 and the (central, if 0 = G) momental ellipsoid has the equation A xe + By2 Cz2 = Ms4

(5.31)

and is the surface represented in that figure. The angular momentum vector is h(0) = {Aco, Bw2Cco,} and the coordinates of J, where the instantaneous axis cuts the momental ellipsoid (5.31), are

{x y

= (rico) {co, w2co,},

where r = OJ and w = Ito 1. Since J lies on (5.31), (r/w)2 (A w? /34 CcoD = MO. r2 = (M84co2)/(2 T),

(5.32)

i.e., r is proportional to w. The tangent plane to the ellipsoid at J has the equation A wi z Bw2 y Gov = ME4 wIr (5.33)

210


and its normal ON is parallel to the invariable line. But the expression on the 1.h. side of (5.33) is the scalar product OJ . h (0) = ON . h.

.• . ON

Meow hr

2 T _V (2T .114 e4 = constant . Mh84 V( 111 84)

(5.34) h2 This result shows that the tangent plane at J is fixed since ON is in a fixed direction and is of fixed length. (The plane is sometimes called an

Fm. 62.

invariable plane.) Since 0 J is the instantaneous axis, we have proved Poinsot's construction of the motion, viz., the free rotation of a rigid body is given by the rolling of the momental ellipsoid on the invariable plane at the distance ON from 0 given by (5.34), with angular velocity proportional to the radius OJ [eqn. (5.32)]. Fig. 62 is a diagrammatic representation of the situation showing the curves traced out by J, the polhode curves, on the surface of the momental ellipsoid in various possible motions. It is clear that, in the case of kinetic symmetry, the momental ellipsoid is a figure of revolution and Poinsot's construction reproduces the results obtained in § 5:2. [See also examples (iv), (v) pp. 217-220 which follow.]

§ 5: 6]


211

It is possible to obtain a formal integration of Euler's Dynamical Equations for free rotation, but in general this requires the use of elliptic functions. Starting from eqns. (5.30) with T(0) = 0 we obtain

Ad), — (B — w2co3 =0, Bth, — (C — A) w,w, = 0,

(5.35)

Cth, — (A — B) ahco, = 0. Multiplication respectively by w1, w2, a), followed by addition gives

A coi thi Bwath,

(0363 =0

which can be integrated to give the energy equation

Act)? Bo4

Co4 = 2T = D ,u2 =constant.

(5.36)

Similarly, on multiplication respectively by A w1, B w 2 , Cco, we obtain

A 2a),th1

B2co 2 th 2

C2a)363 = 0,

which can be integrated to A2col

B2c4 C2a4 = h2 = D2 p2 = constant,

(5.37)

expressing the constancy of the angular momentum. The constants of integration D, it are related to T, h by

D = h212T , u = 2T/h,

(5.38)

and, since T > 0, h > 0, we deduce that D > 0 and ,u > 0. Elimination of co, and co, in turn from (5.36) and (5.37) gives co, and co, in terms of w2. We substitute these expressions into the second of eqns. (5.35) and obtain A B 6 2+ IOC (D — ,u2— BC(B — C)

x 11[A D (A — D) ,u2— AB(A — B) 04] = 0,

(5.39 a)

ABGth, 11[C(h2— 2CT) — BC(B — C)4 x [A (2 A T — h2) — AB(A — B) = 0.

(5.39b)

OF

The integration of these equations normally requires the use of elliptic functions; we could also obtain similar differential equations in col and w3. We give further discussion of the free rotation without integrating eqns. (5.39).

212


1. Permanent axes of rotation Suppose the angular velocity is constant, then

dw

°=

dt

at.)

Ow =

.9t +

t •

x

= 0 , 62 = 0, 63 =0.

From eqns. (5.35) we obtain

(B — C) w2co3 =0,

(C — A) co,co, = 0,

(A — B) coico, = 0.

Since, in general, none of the factors B — C, C — A, A —B is zero, we deduce that at least two of the resolutes of w are zero, i.e., the body must be rotating about a principal axis. Such motions are the only possible steady motions of a rigid body which has no kinetic symmetry.

2. The polhode cone We consider the other possible motions by finding the polhode cone in the various cases. From eqn. (5.33) we obtain the length ON, the perpendicular distance from the origin to the invariable plane, as

0N=

1

.M84co

(A2co1 B2c4 + C24) •

Substituting co, = cox/r, co2= coy/r, co3 =co z/r, we obtain

ON2— 2

T/1/s4

(m84)2

h2 A2 x2 + B2 y2 + C2z2 • A2x2 B2y2 C 2z2= m e4h21(2T) = D m 84

(5.40)

The polhode curve is given by the intersection of the quadric (5.40) with the momental ellipsoid

A x2 + By2

Cz2 =.211 e4

(5.41)

The surface with the equation

k (A x2 By2+ Cz2— 1 I e4) + A2 x2 + B2 y2 C 2z2— DM e4 = 0 (5.42) passes through the polhode curve for all values of k. By choosing k = — A, — B, —C respectively we obtain

— B (A — B) y2 — C (A — C) z2 =.111 s4(D — A), A (A — B) x2— C (B — C) z2 = Me4(D — B) , A (A — C) x2+ B (B — C) y2 = M 84(D — C) .

(5.43) (5.44) (5.45)

§ 5: 6]


213

These are the equations of cylinders which each pass through the polhode curve having generators parallel to the respective coordinate axes (since one coordinate is absent from each equation). We can regard each equation as the projection of the polhode curve on the respective coordinate planes x = 0, y = 0, z = 0. If we choose k = — D in eqn. (5.42) we obtain A (A — D) x2 B (B — D) y2 C(C — D) z2 =0, (5.46) which must be the equation of the polhode cone. Eqn. (5.46) represents a real surface if, and only if, all three projections are also real curves. In general, (5.43) is an ellipse, (5.44) is a hyperbola, and (5.45) is an ellipse. 3. The possible motions Since the constant D has the dimensions of moment of inertia and the constant y those of angular velocity, we consider the possible motions according to the value of D. There are 4 cases when D lies in a range of values (a) D> A, (c)A >D >B,

(e) B >D > C,

(g) C > D,

and three transitional cases (b) D = A,

(d) D = B,

(f) D = C.

(a) D > A (i.e., h2 > 2A T). In this case the ellipse (5.43) is imaginary and the equation of the polhode cone can be satisfied only by x = 0 = y = z. Therefore no motion is possible in this case. (b) D = A (i.e., h2 = 2 A T). In this case the polhode cone equation is B (A — B) y2 C (A — C) z2 = 0. Since A > B > C, this equation can be satisfied only by y = 0 = z, i.e., the polhode cone is the axis Ox of Fig. 62, and the motion is one of the permanent steady rotations. (c) A > D > B (i.e., 2AT > h2 > 2BT). In this case all the curves (5.43-45) are real and the polhode cone

A (A — D) x2— B (D — B) y2 — C (D — C) z2 =0 cuts the plane x = 0 in imaginary lines, but cuts the planes y = 0 and z = 0 in real lines. Hence the polhode curve is one of the closed curves of Fig. 62 encircling the axis Ox.

214


(d) D = B (i.e., h2= 2B T). In this case the polhode cone degenerates into a pair of planes A (A — B) x2 = C (B C) z2 which intersect in Oy. It can be shown that in a motion in which J starts at a position such as that shown in Fig. 62 it moves towards one end or the other of Oy but cannot reach the end in a finite time. The rotation is ultimately about the "axis of mean moment" Oy. (e) B > D > C. This is very similar to case (c), the polhode curve enclosing Oz. (f) D = C. This is a permanent rotation about Oz. (g) C > D. As in case (a), no motion is possible with C > D. Examples. (i) We can complete the solution in terms of elementary functions for case (d). We eliminate cooand co, between eqns. (5.37) and (5.38) and obtain

4=

B (B — C) B (A — B) Cu2 04), (4 = CF - 04) • A (A — C) C (A — C)

We substitute from these equations into the second of Euler's equations and find

(B A

th

—

C /(A- B)(1,0

62 n

0

2

0,4)

,

C

where n2 = ( B — C) (A — B)/(CA) .

.• co2 = f µ tanh Lan (t — t0)],

(1)

where to is a constant of integration.

•° . col = WI= ±

/( A B2— B C 2— A C

I

11 (fr: A 1 j:2)2 2

sech [.t9t(t — to)],

(2)

sech Lan (t — to)]•

(3)

In order to satisfy Euler's equations we must choose the ambiguous signs in (1), (2) and (3) so that two are positive and one negative, or vice versa; this ambiguity is normally settled by the initial conditions. Eqns. (1), (2) and (3) show that, as t —> co, CO/ 0, W2 -3,- ± kt , CO3 0, showing that 0 y is the ultimate axis of rotation.

§ 5: 6]


215

(ii) A uniform rectangular plate is such that the acute angle between its diagonals is 2oc. Show that, if it is suddenly set spinning about one diagonal with angular velocity co, then it will be spinning entirely about the other diagonal after a time go 2 d9 col/ (cos2a) 1/ (1 — sin2a sin2 q2) • 0

The principal axes at 0 G) are shown in Fig. 63, the moments of inertia being A= sin2a, B = Int/2 cos2a, C = im12, where 0 D = 1.

FIG. 63. Euler's equations become th1 sin2a — co2co3(cos2a — 1) = 0,

the cos2a — cosco, (1 — sin2a) = 0,

(b3— co, co2(sin2 oc — cos2a) = 0, 61

+ 02(03 = 0, ci.)2— co,(01= 0, cbs+ col co2 cos2a = O.

(1)

First we note that co, (;), + 0)2 u,2 = 0. = constant = 0)2.

.•. co!

The constant of integration 0)2 is• obtained from the initial conditions, viz. = {co cosy co sing 0). Hence we can write co, = w cosA, cos= w sinA, where initially A = a, cos= 0. Eqns. (1) now become —0) A sinA + co,co sinA = 0, 0) A cosA — cosco cosA = 0)3

0)2

0,

cosA sinA cos2a = 0,

i.e., co, = A,

+ co2cos 2a cosA sinA = 0.

We integrate the latter of these, remembering that initially co, = A = 0, to obtain A2 = CO2cos2a

(sin2 — sin2 A).

The plate will be spinning entirely about the other diagonal when co, = A = 0, i.e., sin2 2 = sin2 a, or A = a, — a. Initially A = — 0)2cos 2a cosa sing < 0. The

-

216

-


variation in A is therefore a libration between the values a and — a. The time required for A to change from a to — a is a

T=

r" J 1

1

dA

co (cos 2 0,)v f v(sin2a — sin2A)

—a

The integral is transformed by the substitution sinA = sin a sing), cosA dA = sing cos 99 cico. The limits of integration become — n/2 and n/2, so that 7r/2

T=

1

ni2

( slim cos q) dq)

co (cos 2 oc.)'/ j cos A sin a cos co

2

d7

co (cos 2 ocPzof

- sin2 oc sin2 T)1/2

(iii) Two particles, each of mass m, are fixed at the ends of a diameter 1 of a uniform circular disc of mass 4m and radius a, which is free to rotate about its centre. The disc is set in motion with an angular velocity the components of which are S2 j/2 along 1 and .(2 normal to the disc. If co is the component of the angular velocity along the diameter perpendicular to 1, show that at any subsequent time 1 =— V 2 (20 — co2). Deduce the components of the angular velocity along the principal axes of inertia at any subsequent time, and prove that the disc ultimately rotates with angular velocity S212 about the diameter perpendicular to 1. Find the ultimate direction of the axis of rotation in relation to the initial position of the disc. The moments of inertia are A= 4m . ict2 = ma2,

B = ma2 + 2ma2 = 3ma2 , C = 4ma2.

(See Fig. 64). Euler's equations become, after some cancelling, w2 w3 =

362 — 30301 = 0, 463 + 2co1w2 = 0. (1) (2) (3)

The initial value of co is {,Q y2 0 S2), and the energy and angular momentum equations are 4 + 34 + 44 = 6522, wi + 94 + 164 = 18522. Hence, in the notation of eqn. (5.38), D,a2 = 6522, D2/42= 18S22.

(4) (5)

.• . D = 3, ,a = S2 V2. This is an example of case (d) of p. 214. Following the lines of example (i) above, with co, = co, we obtain co?.= 2522— co2, 4 = ‘22— ia>2. .•.

=2w3, i.e., col. =

co3 V2.

(6)

§ 5: 6]


217

Because of the initial conditions we choose the plus sign, and eqn. (2) becomes — V 2 (S22— 1(1)2) = 0 , equivalent to the stated result. .•. co = V 2S2 tanhS2 t, there being no constant of integration since w = 0 when t = 0. .•. co, = S2 sech S2t, wl = 2(03= V 2S2sech Slt on using eqns. (6) and choosing signs of square roots to conform to the initial conditions. As t—> 00, co {0 Q V2 0).

FIG. 64. Since ultimately the disc rotates about 0 the angular momentum vector lies along this direction and so 0 E3ultimately coincides with the invariable line. Using the initial values we see that the invariable line has the direction of A = 0 iy2). This vector makes an angle a. with the initial direction of w, so that cos a —

1 3

.

V2 3

+

2V2 3

.

1 3

—

V2 3

.

In Fig. 64 OJ, OH are the directions of the initial angular velocity and of the angular momentum respectively and both lie in the initial position of the plane and the direction OH is that of the ultimate position of 0 2 (iv) A solid consists of two equal uniform right circular cones, each having height b and vertical angle a right angle, rigidly joined at the vertex 0 so that their axes are in the same straight line. If 0 is fixed and the solid is set rotating about a common generator of the cones with angular velocity Q under no forces except gravity and the reaction at 0, show that the solid will be rotating about the same generator after a time (107ty2/3S2). After what time will the axis of the solid have its original position ? We include this (and the next) example to illustrate the application of Euler's equations to the case of kinetic symmetry. We leave the reader, as an exercise, to 8 MIES

-▪

• 218


show that for this cone A = B = 4mb2, C = io m b.. We suppose that the initial angular velocity is in the plane 0$1 $3, Fig. 65, and has resolutes {Q/1/2 0 D/V2). Euler's equations reduce to 561—

3co2 co3 = 0, 562 + 3co3co1 = 0, cb, = 0.

Hence co, = n = 12/1/2 = constant. Using the device of the complex combination of the other equations we obtain 5(cb1 +ith2) + 3in (co, -I- ico2)1112 =

3ipt 51/2 ]'

S2 exp [ =— 1/2

wl +

0,

where we have used the initial values of co, and co,. Hence

col= (12/1/2) cospt, co2 = — (0/1/2) sinpt, co, = (3

where p = 30/(5 j/2). Hence the resolutes of a time 27tIp = (10n )12)1(3 S2). The angular momentum vector is

h(0) = 3mb2 I-cospt 4 1/ 2

which is in the direction e =

co return to their initial values after

4V 2

sinp t

5

5

2 y 9 cospt — 729 sin pt

Q 10 V 2 I 2

. This is the

direction of the invariable line about which the precession takes place. The angle y ( = oc-0', see Fig. 45) between the invariable line and 0 3is given by cosy =

k.e= (0 0 1).

5

V 29

cospt

5 snip t V 29

21 1/ 29

f

V 29

§5:6]


219

If zj, is the velocity of precession, then the velocity of the point P (position k) on is, because of the precession, 5

V 29

5 V29 sinp t

cospt

5 = {— V29 sing

2

x {0 0 1}

V29

5 —— 29 cosp t

01.

V

But the velocity of P, regarded as a point of the body, is xk

S2

=

S2

{— cospt — — sinpt — x (0 0 1) V2 V2 v2

S2

= — — sinp t V2 5//5

S2

V29

V2

,f2 — V2 cosp t 0} .

S2 i( 29 \ 5 V

V

2)•

Hence the axis of the solid completes one revolution about the invariable line in time 27r 107r V ( 2 \ ,52

29 ) •

(v) The variation of latitude This example is included here because the use of Euler's equations and axes fixed in the body gives the information in the most natural form, for this particular problem. The shape of the earth, although almost a sphere, is in fact flatter at the poles than at the equator. We assume it to be a spheroid with moments of inertia A, A, C (> A) at the centre 0. If we also assume that the resultant of the gravitational forces exerted by celestial bodies acts through 0, then the earth's motion relative to 0 is one of free rotation. Euler's equations become A + (C — A) co2co3 = 0, A 62— (C — A) co,co, = 0 , C = 0 . Hence, w3= n = constant, and w1cih + co2 w2 = 0. .•.wi + co = A2 = constant. We write w1= 2 cow, w2= A sing9 and obtain from (1)

—A2 + (C — A) An = 0. .*. ci9— 8*

C— A A n P,

=Pt,

(1)

220


where we assume cp = 0 when t = 0. .•. co, = A cosp t,

co, = A sinp t, w3 = n.

The plane containing the axis of rotation and the axis of symmetry 03rotates with angular velocity p relative to the earth in a period 2n/p = 27t Al {(C — A) n). (The angular velocity cp = p which we have determined here is the same as the angular velocity — 4 = p determined in § 5:2.) The period 2atAl{(C — A) n) has a value of about 307 days. If a fixed star or the sun is observed from the earth, the elevation of the star when it crosses the meridian determines the latitude. Observations of the latitude do show a variation which must correspond to a movement of the axis of rotation in the earth but the period differs significantly from 307 days. Consequently, our assumptions either about the rigidity of the earth, or about the forces acting on it, or both, cannot be correct. (vi) A solid uniform cube is in motion about one of its corners 0, which is fixed, and the constraint is smooth. If the resolutes of angular velocity of the cube along its three edges through 0 are col , co„ 0)3and gravity is neglected, prove that co, + co, + 0)3 and avo, + co2co3+ a),coiare each constant throughout the motion. The resolutes of the angular velocity are taken, in this problem, along axes which are not principal axes. The coefficients of inertia are

A = B = C = IM(a2 + 2a2) = ;1Ma2, D = E = = Ma2, where 2a is the length of an edge of the cube. Since gravity can be neglected the kinetic energy is constant and the angular momentum vector is constant. At any instant the resolutes of angular momentum are hl

= Ma2 (1 col — (02 — 0)3), h2 = Ma2 ( — col + !0)2 — (03), = Ma2 (— col— co, +

.•. h2

+

+

= M2a4CV (col. + (02 + 0)3)2—

(0)2(03 + °Vol + coic02)]• (1)

Also the kinetic energy is T = M a2 {

2

8

22

— (col + co2 + (03)2 —— 3 3

(0)2 0)3 + 0)3 W1 + W1 W2)] •

(2)

Since h2, T are constant and the determinant 99. 9

242 9

* 0,

3

the given result follows, on solving eqns. (1) and (2) for (w1+ co, + 0)3) and (co2c03 + 0)30)l + w2)•

§ 5: 6]


221

Exercises 5:6 1. A plane lamina, under the action of no external forces, is set rotating with angular velocity Q about an axis through its centre of mass G, whose angular coordinates are 0 = = 13, referred to the principal axes at G. Prove that at time t the component angular velocities about these axes are col= S2 sing cos q), w2= S2 sin oc sings, A—B cos = Q[cos2oc sin2 oc (sin2 i3 — sin2 92)] , A+B where A, B are the moments of inertia of the lamina about the principal axes in its plane. 2. Coordinate axes 01, 02, 03 are taken with origin 0 at the centre of a uniform rigid rectangular plate so that 01, 02 are parallel to the edges and 03 is normal to the plate. The moments of inertia about 01, 02, 03 are A, B, A + B, respectively, with A > B. The plate can rotate freely about 0 and is set spinning about an axis of spin through 0 having direction-cosines (0, cos a, sin g) with respect to the A—B coordinate axes. If tan2 oc > A2— , prove that the axis of spin rotates conA+B tinually in the same direction relative to the plate. If tang a = A2, prove that the plate will ultimately spin about the axis 01. 3. A body whose principal moments of inertia at its centre of mass G are A1, A2, A 3 spins about a fixed axis through G which has constant direction cosines (/„ /2 , /3) with respect to the principal axes at G, being constrained by smooth bearings. Prove that the angular velocity is constant and show that the body exerts on the bearings a couple of moment {(A2— A3)2 + (A3— A1)2 l3li + (A1— A2 )2/T/V/2, where co is the angular velocity. 4. A rigid body whose principal moments of inertia at the centre of mass are A, B, C (A < B < C) is rotating freely about its centre of mass under no external forces. It is rotating with angular velocity S2 about its axis of greatest moment of inertia when a small angular momentum A w is given to it about the axis of least moment of inertia. Show that the instantaneous axis moves in the body with a period 27clp where AB p2= (C — B) (C — A) 522. Find also the maximum and minimum angular displacements of the instantaneous axis from the axis of greatest moment. 5. A uniform solid right circular cone of height b, mass M and semi-vertical angle 44 moves freely about its vertex 0, which is fixed, under no forces other than the reaction at 0. If the cone is set spinning with angular velocity S2 about a generator 01, find the components of the angular velocity at any subsequent time t in the right-handed orthogonal frame Ox1, Ox2, Ox3fixed in the cone, where Ox3is the axis of the cone and 01 bisects the angle between Ox1and Ox3. Find the reaction at 0 and show that its magnitude is 3 1/29 .111b122.

40

222

A C01711SE OF MATHEMATICS

6. A plane lamina moves under no forces with a point 0 fixed. The moments of inertia about the principal axes at 0 lying in the plane are A and B (A > B). Show that 0, the component of the angular velocity in the plane of the lamina, is a constant of the motion, and that if the motion is started in any manner that makes the total kinetic energy A122 then the lamina will ultimately rotate about the principal axis associated with A. 7. A thin elliptic disc of eccentricity 1/y2 is free to rotate about its centre. The disc is set rotating with angular velocity Q about an axis through its centre inclined at 30° to the rigid axis and 60° to the normal to the plane of the disc. Discuss the motion.

5:7 More general motions We conclude this chapter with some examples of the use of the preceding techniques in cases of a slightly more general character than those already considered. The problems are solved either by the use of rotating axes or by the use of Euler's equations, whichever is the more suitable method. The first example concerns the rotation of a rigid body subject to a resistance. Example. A rigid body possessing axial symmetry and with principal moments of inertia A, A, C at its centre of mass is given a general motion of rotation. The body is surrounded by a medium that produces a retarding couple that at any time is a constant multiple, — k, of the instantaneous angular velocity. Show that, if principal axes of inertia at G are taken as axes of reference fixed in the body, and are such that the initial angular velocity is {Q10 123}, the angular velocity at any subsequent time is tf-21 e- k t/A cos0 e — k tiA sin 0 123,- k t/c) where

k, AO = (C — A) C S2,(1 — e- ktiC).

Hence show that a motion of spin about the axis of symmetry is stable or unstable under the action of a disturbing couple of this kind according as C is greater or less than A. Here

r (0) = — k co and eqns. (5.30) become (C — A) co2co3= — koh,

(1)

A(62— (C — A) co,coi= — 7cco2,

(2)

C 63 = —kw,.

(3)

A(i)1

Eqn. (3) integrates at once to give (03 = s23,—ktIC

(4)

§ 5: 7]

FREE ROTATION AND STEADY

MOTION

223

Also from (1) and (2) we obtain d

2k (coi+ (14) = — — (coi + 04) • A 12?,2 kt IA = • co? + w2

dt

If we write co, =

k

IA Cos(),

(5)

s? e- k th4 sin 0 , co, =i

(6)

thereby satisfying eqn. (5) identically, substitution in (1) or (2) [and use of (4)] gives AO = (C — A) co, = (C — A) S2, e- k tic Ale° = (C — A) C.23(1 — e- kt/C) as required. The angle made by the instantaneous axis of rotation with 0$3 (0z), the axis of symmetry, is tan-1

1'04 + co,

— tan-,

Diek" Die-k tic

— tan-1 {9'- ekt(A - C)I(AC) S23

and this angle tends to zero or in according as C > A or C < A i.e., if a rotation starts about an axis near 0$, the axis remains near Desif and only if C > A. (If C = A, w ne-ktIA and the direction of w remains fixed, i.e., the motion is stable.)

We can also investigate by these methods the motions of certain bodies with kinetic symmetry when they roll on a rough horizontal plane. Such a motion displays some of the characteristics of gyroscopic motion because the angular momentum vector of a rotating body changes direction. Examples of such motions, and the forces required to maintain them, are found in the motion of an aircraft's propeller when the aircraft circles, or the motion of the wheels of a bicycle which is following a curved path. Examples. (i) A uniform circular hoop of radius a rolls uniformly on a rough horizontal plane so that the point of contact describes a circle of radius r in time 2n/co, the plane of the hoop being inclined inwards at an angle 0 to the horizontal. Show that g cos0 = co2sine (2r — ia cos0). Find also the least value of the coefficient of friction if the hoop is not to sideslip. This is an example of steady motion which we investigate by using rotating axes with origin at the centre 0 of the hoop, with 0 e, as the axis of symmetry and 0 along the radius of contact OD (Fig. 66).

224


The motion of the frame The angular velocity of the frame is 0 = {— co sin0 0 co cos 0) ; the velocity of the origin is To = V. The motion of the body The angular velocity of the hoop is co = {— co sin0 0 n) . Since the hoop is rolling, the particle of the hoop at D is stationary. .•. V = {— co sin0 0 n) x {— a 0 0) = {0 — an 0).

We are given that the point of contact D moves around a circle of radius r with angular velocity co. Hence the velocity of the point {a 0 0), considered as a point on the frame, is {0 cor 0). {0 co r 0) = vo + 0 x OD = v, +

act) cos 0 0).

.•. vo = V = {0 co (r — a cos0) 0} = {0 — an 0). .*. n = co(cos0 — r/a), and

p=mV=m {0 co (r — a cos0) 0).

The moments of inertia are A = ma2, C = ma2. h (0) = fma2{— co sin° 0 2n).

The forces At 0, {0 0 0), the weight mg {sin 0 0 — cos 0); at D, {a 0 0), the reaction {./P cos0 — R sin0 Y F sin0 + R cos0}. (The second component, Y, is not indicated in Fig. 66.) .•. F = (F cos° — R sine + mg sin° Y F sin° + R cos 0 — my cos() and

F(0) = {0 — a (./7 sin0 + R cos0) aY}.

(1)

§ 5: 7]


225

The equations of motion The equation of linear motion is dp Ot

F

having resolutes — mco2(r — a cose) cose = F cos° — R sine + mg sine,

(2)

0 = Y,

(3)

— ma)2 (r — a cos0) sine = F sine + R cose — mg cos0.

(4)

The equation of angular motion is

WO) + 0 x h(0) = at

r (0)

having resolutes (the first resolute is identically satisfied): ma2 co2cos@ sine + ma2a) n sine = — a(F sine + R cose) , 0=aY.

(5) (6)

Substitution in (5) from (4) gives ima2co2cos() sine — ma2 con sine = mga cos 0 — ma co2(r — a cos 0) sine,

g cose = co2sine (2r — 2a cose), on using (1). (This equation is the condition for steady precession.) From eqns. (2) and (4) we deduce

F = — mco2 (r — a cos0), R = mg. Hence

F R

w2(r — a cose) g

cos0 (r — a cos0) sine (2r — 2a cose)

2 cos @ (r — a cos 0) sin0(4r — 3a cose) (ii) An aeroplane is turning in a horizontal plane with uniform angular velocity the axis of the propeller being horizontal. Prove that the forces on the propeller (apart from that exerted by the engine on the driving shaft) reduce to a force at the centre of gravity together with a couple Cn,V, in an axial vertical plane and a couple (A — B)nlp in an axial plane which revolves with twice the angular velocity n of the rotation of the propeller. It is assumed that the axis of rotation is a principal axis, and that A, B, C are the moments of inertia of the propeller, C being about the axis. S a CMES

226


The frame O 1 2 e3 in Fig. 67 is fixed in the propeller, W, being the axis of rotation. The circle VCHWD, centre 0, is the plane in which the propeller rotates, 0 V being the vertical. The plane of VC WD makes an angle s with a fixed vertical plane, shown in the figure as V F W. The position of the propeller is given by the angle V 0 C = 0, in the plane of rotation. The propeller has two angular velocities, viz, y about 0 V, and 6 = n about 0 3. Hence = {y cos —i sin 0 6).

Fm. 67. We reduce the forces to the base 0 and assume they are equivalent to F through 0 and a couple T(0). We write down Euler's equations of motion with

F (o) =

A cb, — (B — C) co,w3 = A — dt

L2

L,) ,

cos0) (B — C) ny sine' = L,,

d

Bth2— (C — A) co ,co„ = —B —sine) — (C — A) mb cos0 = L2, dt C63—

(A — B) co,co2= C6 + (A — B) 1/./2sine cos0 = L3.

Remembering that tp and 6 (= n) are constant we deduce

L1 = — (A — B) L2 =

sin0 — Cny sine,

(A — B) cos0 — Cm? cos0

(1)

(2) (3)

§ 5 : 7]


227

[In addition L3= (A — B) sine cos0, but this is the torque which must be provided by the engine to keep the propeller rotating uniformly.] We now resolve L, and L2in the vertical and horizontal directions OV, OH. Lv = Licos° — LZ sin0 = — (A — B) n/j) sin20, LH = L, sine + L2 COS° = A — B) (

cos20 — CnIp.

The axis of the couple C n V, is in the direction HO, i.e., the couple is in the vertical axial plane. The axis of the couple (A — B)nip makes an angle 20 with OH below the horizontal; hence its plane is an axial plane which rotates with angular velocity 2n, as stated.

We consider in Chapter VII one or two more examples of this and other types of motion but there we investigate the stability of the motions. Exercises 5:7 1. A rigid body has principal moments of inertia A, A, C at a point about which it moves under no forces except a resisting couple equal to k times the resultant angular velocity. If at time t = 0 the instantaneous axis makes angle oc with the axis of symmetry, prove that at time t it makes angle co with this axis, where tang) = tancx . exp {kt(C-1— A-1)). Also show that the plane containing the instantaneous axis and the axis of symmetry has rotated about the latter through an angle (C A)

Co) cos oc Ak

kt 1 exp (— 7)1 ,

where w is the magnitude of the initial angular velocity. 2. A rigid body with an axis of symmetry GS moves freely about its centre of mass G, which is fixed, and the principal moments of inertia at G are (A, A, 2A). The body is given an initial angular velocity co about an axis through G equally inclined to the chosen principal axes of inertia at G, and in the ensuing motion, a constant retarding couple of amount 2k2A acts about the axis of symmetry and persists until the angular velocity about this axis vanishes. If all other forces have zero moment about G, prove that when the retarding couple ceases to act the direction of the angular momentum vector has turned through an angle 1 cos c°2 (-)1 cos-1 {1/3 6k2 relative to a frame fixed in the body. 3. A gyroscope, pivoted so that the centre of mass remains fixed, spins about its axis with constant angular speed n and is acted on by an external couple. Show that in steady motion the axis of the gyroscope moves in a plane at right angles to the plane of the couple. 8a*

228


4. A uniform sphere, of mass m and radius c, rolls and spins without slipping on the edge of a circular hole of radius a cut in a fixed horizontal plane. In steady motion the centre of the sphere remains at a fixed height, c sin 0, above the plane, and describes a circle of radius b = a — c cos 0 with constant angular velocity co. Show that the angular velocity of the sphere about the radius drawn to the point of contact is (7bco/2c) tan0 — (5 gf2cco) 5. A sphere of radius a and mass M is loaded so that, while its centre of mass remains coincident with its centre of figure, its principal moments of inertia are A, A, C. Show that a steady motion is possible in which the sphere rolls on a rough horizontal plane with the axis of C inclined to the vertical at a constant angle cc and describes a cone with constant precessional angular velocity co about a vertical axis which is at a distance c from the centre of the sphere. If n is the component of the angular velocity of the sphere along the axis of C, show that in such a steady motion n is given by a(C — A) n sin = ca )(A M a2). 6. Show that if the momental ellipsoid at the vertex of a uniform solid right circular cone is a sphere, then the radius of the base of the cone is double the altitude h. If such a cone rolls in steady motion on a rough table, prove that the resultant of the reactions along the generator of contact cuts the generator at a distance h{j/ 5 + 4ho.)2/g} from the vertex, co being the spin of the plane containing the axis of the cone and the generator of contact. Show that 3co2 h/(4g y5) must be less than the lesser of the values 17/16 or it being the coefficient of friction between the cone and table.

Miscellaneous Exercises V 1. A rigid body is free in space under no external forces and its motion consists of a pure rotation about an axis fixed in the body. The principal moments of inertia A, B, C at the centre of mass are all unequal. Establish that: the angular velocity is constant in magnitude; the axis of rotation is also fixed in space; the centre of mass lies on the axis of rotation; the axis of rotation coincides with a principal axis. Prove further that if co is the scalar angular velocity dco ABCco— +(B — C) (C — A) (A — B)wi a)2 w3 =0, dt

Ex. V


229

and hence show that co dco = + v {(A dt

w2) (a, 0) (v — 0)),

where BC 2. = 2 T(B + C) — H2, and bi and v are similarly defined. 2. A rigid straight light rod, of length 5a, is rigidly and normally attached at one end to a point 0 on the surface of a uniform solid sphere of radius 5a. To the other end of the rod is fixed a particle of mass equal to that of the sphere. The system is free to turn about 0 and is given an angular velocity Q about a line OA where A is a specified point on the sphere such that OA = 8a. Prove that at time t later the component of angular velocity about OA is

SI 2 .Q t — 25 {16 + 9 cos3()1 . Show also that the instantaneous axis of the system describes in space a cone of angle 2 tan-1(6/7). 3. A particle of mass m is attached to the surface of a uniform sphere of radius a and mass M. The sphere rotates with a constant angular velocity n about the radius to the particle, which makes a constant acute angle a with the upward vertical. The vertical plane through this diameter rotates with constant angular velocity ,12, and the whole arrangement slides along a smooth horizontal table. Show that

a k (2k + 7) 02cosa — 2ank (k + 1) f2 + 5g (k + 1) = 0, where k = M Irn. 4. A rigid body has unequal principal moments of inertia A, B, C at a fixed point and moves freely about this point under no forces. Show that the acute angle between the angular velocity vector and the angular momentum vector reaches a maximum wherever the angular velocity reaches a maximum, and deduce that this occurs only when the instantaneous axis crosses a plane containing two of the principal axes of inertia. In particular show that, when the instantaneous axis crosses the plane of the principal axes associated with A and B, then in a motion with kinetic energy T and total angular momentum H the value of the angle is

cos-1

1'12 AB H [2T (A + B) — H2 i •

2T

5. The vertex of a uniform solid circular cone of height b and semi-angle /3 is freely pivoted to a fixed support, and the edge of the base is in contact with a smooth horizontal table. The cone is set in motion so that it rolls without slipping on the table, and the axis moves with constant angular velocity Q about the vertical, and makes an angle a with the downward vertical. Show that the reaction R on the table is given by

Rb sin (a — /3) = sina eosin/Mg b + S22 {C sina cot/3 — A cosoc}], where C, A are respectively the moments of inertia of the cone about its axis and about a perpendicular line through the vertex, and M is the mass of the cone.

230


6. A perfectly rough horizontal plane rotates with angular velocity w about a vertical axis which meets the plane in a point 0. A top, whose axial and transverse moments of inertia at its vertex are C and A respectively, spins on the rotating plane with its vertex at a point P distant a from 0. The top is of mass M and the distance of its centre of mass from its vertex is h. Obtain the equations of motion. Show that if the direction of the axis of the top remains fixed relative to the plane, then the plane through OP and the axis is vertical and the angle 0 between the axis of the top and the vertical is given by A we sine cos° — Cnco sine + Mgh sine Mahco2cos0 = 0, where n is the spin of the top. 7. An axially symmetric top of mass M is freely pivoted about a point P on its axis. Its centre of mass G is at a distance / from 0 and its principal moments of inertia at 0 are A, A and C. Initially the top is set in motion with its axis horizontal and with a small angular velocity n about its axis. Show that, when G is at its lowest point in the subsequent path, the axis of the top makes an angle approximately

Cn 1/(2Mg IA) with the downward vertical. 8. A light rigid isosceles triangular frame ABC of altitude h carries a flywheel of mass M on the base BC as axle, at the mid-point G of BC. Show that a steady motion is possible with the vertex A at rest on a perfectly rough table, AG at an angle 0 with the upward vertical, and the plane ABC vertical (with C above B) and rotating with angular velocity D, provided

S2 cos° {(A — .M h2) 12 sin0 — C n} = Mgh sin0 , where C, A are the moments of inertia of the flywheel along and perpendicular to BC, and n is the angular velocity of the flywheel about BC. 9. A top, whose principal moments at the point of support 0 are (A, A, C), is of mass M, and the distance of its centre of mass from 0 along the axis is h. If the cosine of the inclination of the axis to the upward vertical is x, show that

= 2 (g11) (x — a) (x — 13) (x — where 1 = Al M h, and oc, E, y are numbers satisfying the inequalities — 1 < < < 1 < y. [It is assumed that the axis of the top is never vertical during the motion.] Show further that the vertical component of the reaction at 0 exceeds the weight of the top by (Mgh/l) {3x2— 2 (a + p + y) x + (04 + fi'y + yoc)), and that this vertical reaction is least when x = /3 or when x = + + y)/3, according as a + y > or < 2°. 10. A uniform circular disc of radius a moves on a smooth horizontal plane with its centre of mass G at rest, the axis normal to the disc at G making a constant angle oc with the vertical through G, about which it precesses uniformly at a rate D. Find an equation giving the spin n of the disc about its axis.

Ex. V


231

If the same disc now rolls on a rough horizontal plane, with G moving in a circle of centre 0 and radius A a, with its axis at the same constant angle a with the vertical and precessing about the vertical through 0 at the same uniform rate .12, show that the spin of the disc about its axis is n/3, and that A is n/(3f2). 11. An axisymmetric top of mass M and axial moment of inertia C is hinged freely at a point 0 on its axis at a distance Is from the centre of gravity G. The spin of the top is n, and its precessional angular velocity is f2 when the angle between OG and the vertical is a maximum. Prove that, if the path of G is cusped, then CnQ = 2Mgh.

CHAPTER VI

LAGRANGE'S EQUATIONS 6:1 Generalised methods In the preceding chapters we have developed and generalised the Law of Motion and the Law of Action and Reaction, which in their primary form concern the motion of a particle, so that they can be applied to systems of particles and rigid bodies. We have made considerable use of non-Newtonian (e.g. rotating) frames and we have used a variety of means, other than rectangular cartesian coordinates, for specifying the configuration of a system. The equations of motion, of energy, and of momentum all have been expressed in terms of parameters other than cartesian coordinates. In § 2:4 we introduced variables q1, g,, , to denote any set of parameters which can be used to specify the configuration of a mechanical system. The equations of motion, in the form first introduced by J. L. Lagrange about 1760, express the laws of motion in terms of such an arbitrary set of parameters. These parameters q1, g,, , qn are called generalised coordinates, and many of the equations of motion we have already obtained by other methods can be obtained, more directly, by quoting Lagrange's equations. When the configuration of a system is specified by generalised coordinates, it is essential that the position of an arbitrary particle of the system, denoted by the suffix p, can be expressed in terms of q1 , g,, , qn and, possibly, the time t, i.e. rp = rp(q, , g,, or

xp = x„(gi,

q2 , '",qn , t) ,Yp

zp = zp (q,, q2 ,

, g„, t); = Yp(qi, q2, qn , t),

qn ,

(6.1)

where {xp yp zp} are the coordinates in some, arbitrary, Newtonian frame. Unless equations such as (6.1) can, in principle, be written down for every particle of the system, the parameters a,1 a,2 q55 are not suitable as generalised coordinates. 5

232

5 • • • 5

§6:2]

233

LAGRANGE'S EQUATIONS

Examples. (i) A simple pendulum whose bob is at {x y}, where

x = / sine, y = / cosO, is a system whose configuration is specified by one generalised coordinate O. (ii) The bob of the spherical pendulum of § 3:4 has coordinates given by

x = a sine cosy, y = a sine

z = a cose.

This is a system specified by two generalised coordinates 0, v. (iii) A particle constrained to move on a helix (see the example of p. 85) is a system specified by one generalised coordinate 0, for

x = c cosa cosO, y = c cosoc sine, z = cO sing. (iv) The example (ii) of p. 153 is a system specified by two generalised coordinates (v) A top, or any rigid body, constrained to rotate about a fixed point 0 is specified in position by Euler's angles, i.e., three generalised coordinates.

6:2 The dynamical variables In order to make the formulae more compact in the following sections we shall use the summation convention (Vol. IV p. 101) for expressions involving sums over the generalised coordinates: in formulae a repeated dummy suffix i, j, k, ... in any term implies that all such terms are added for which the dummy suffix has the values 1, 2, ..., n. Thus ax P

aq;

uq: =

n n

ax •

1=1

P

vq:, aii4i4;

=

.2: aii 4i 4j J-1

•

We now give expressions for some dynamical variables in terms of generalised coordinates. 1. Kinetic energy The kinetic energy of a system is the sum of the kinetic energies of the constituent particles.

•• • T = z my Vp = 1my (4 ± yv ± 4) •

234


(The sum denoted here and subsequently by particles of the system.) But, from eqn. (6.1), P Xp =

ax p

a qi

2

axp

P

aqi

qi

is taken over all the

ax p

at •

ax, axv at u'

4, + 2 aqi

(exp

at )

with similar expressions for g and 4.

2b11,

T= where au = Z p rap

/ ax,

c),

(6.2)

ax, ± ay, ay, + az, az, aqi eq.; agi

aqi aqi ax, axp bi = mp ( at

ayp ayp aq, at

azp azp) aq, at

7mPfr xtP )2 ± ( °aYtP )2 + ( eazit) )2]

(6.3) (6.4) (6.5)

Although the general expression for T in terms of the generalised velocities is (6.2), the coefficients au , c being functions of the coordinates qt and t only, there is an important case in which T is a homogeneous quadratic function of the velocities. This case occurs when the "equations of constraint" (6.1) do not involve t explicitly. In this case

ax, ay,az, at at at bi =0, c = 0 ; T

0

' (6.6)

2. Momentum The distinction between linear and angular momentum cannot be made with reference to generalised coordinates so a generalised momentum pi , corresponding to qi , is defined by

aT Pt = ai ,

(6.7)

If the coordinates gi are chosen as rectangular cartesian coordinates then the pi defined in (6.7) agree with the elementary expressions.

§ 6: 2]

235

LAGRAN.GE'S EQUATIONS

Examples. (i) We obtain an expression connecting the pi with the momentum components of the constituent particles. Since

T=

ki q;+ c,

aT

pi= —= aikqk + z a.; + bi = aqi

+ bi,

since c/o = aji . .". = Emp

(ax, ax, aqi

I 49X,

R

=

y, y,

oXp oXp

_t_ a Yp .

z,

q.1+ az, a z, aqi at )

a Yp)aY p

49q; qj + at ) aqi

qi qj+ at )

MP

z,

q.,+— —4 aqi aq i J+ aqi ax, ax, ay, ay, aqi at aqi at

azp az,\ az, + (qi 4J+ t aqi = m, P

P

ax, ay, az, + Y„ + aqi aqi P aqi )

(1)

The terms mpt, are the components of momentum of the constituent particles. The formula (1) will occur again in connection with generalised forces below. [The reader should note that the summation convention operates in the relations such as a x„

aq.1

q,+

axP

dx,

at

dt

±p

which are obtained by differentiating eqns. (6.1).] (ii) The simple pendulum has one coordinate 0 and one momentum.

.• T = irra262; po = m126. (iii) The spherical pendulum has two coordinates and momenta:

T = Ima2(02

sin20) ; po m a2

sin20 .

The generalised momentum py, is here the same as the (upward) vertical resolute of the angular momentum. (iv) Two uniform rods AB, AC, each of mass m and length 2a, are smoothly hinged together at A and move on a smooth horizontal plane. At time t the masscentre G of the pair of rods is at the point (x, 9) referred to fixed perpendicular axes Ox, Oy in the plane, and the rods make angles 0 + 92 with Ox. Find the components of generalised momentum.

236


This system was considered in the example on p. 150 of Chapter IV. From the expressions obtained there we find that the kinetic energy of the motion of AB relative to G is sing9)2 + (a0 cos99)2] + ima2 (0 + cl7 ) 2 . Here the first term arises from the motion of the mid-point of AB relative to G, and the second term from the rotation of AB about its mid-point. Similarly, for AC the corresponding kinetic energy is

im[(a6 sinq9)2+ (—a12 cosw)2] + ima2 (6 — q9)2.

The kinetic energy of translation is . 2m

.• T = m [12 p7, = 2m

a2 (1.

+

sin2w) 02

a2

cos20 4,2]

•

= 2m'g, po = 3ma2 O(1 + 3 sin2 g7), = ima2 42(1 + 3 cos2q9).

The momenta pX, RT, are the resolutes of the linear momentum of the whole system but RI, pg, have no immediate interpretation in elementary terms. (v) The particle in the bowl of example (i) of p. 89 (Fig. 23) is specified by one coordinate 0, but, referred to a suitable set of axes, its coordinates are

x=

— a sinO,

y= — a cos0

This is an example of a system subject to a constraint which varies with time. For this system

T = im,(a262— 2a6 ft cos0 f 2t2), 2)0 = ma(a0 ft cos0). (vi) A smooth wire in the form of a circle rotates with constant angular velocity w about a vertical tangent and a bead of mass m slides on the wire (see Vol. III p. 316). This is a system subject to a moving constraint in a conservative field of force, whose configuration is given by one coordinate 0 (see Fig. 68). If we choose the initial position of OA as Ox, with Or along the axis of rotation, and Oy to complete the r.h. set, then

x = a(1 + cos0) coswt, y = a(1 + cos0) sinwt, z = —a sine.

(1)

These are special cases of eqns. (6.1) showing explicit dependence upon t. The kinetic energy, potential energy and momentum are

T = fma2[62+ (1 + cos0)2col, V = — mga sine, pe = ma20. We notice that, despite the occurrence of t in (1), in this case T is independent of t.

§ 6: 3]


237

6:3 Generalised forces In connection with this section the reader is advised to read again § 2:4, on Virtual Work. In order to work entirely with the generalised coordinates q„ we must replace the elementary process of resolution of vectors by a more general method. The Law of Motion of a particle states the equivalence of the resultant force FP on a particle, a line vector passing through the particle, to the line vector mp fp , where fp is the acceleration of the particle of mass mp . For a system consisting of interacting particles we therefore have two sets of line vectors, the force vectors and the vectors mp fp , which are equivalent in the sense

of § 2:3 and we express this equivalence by equating the virtual works of the two sets of vectors in an arbitrary displacement consistent with the constraints. A virtual displacement of the system consists of a displacement 6 = (6 x, 6y, 6z,} of each particle and corresponds to increments (i = 1, 2, ... , n) in the generalised coordinates. The two sets of vectors are equivalent at an arbitrary instant and so the displacement must be made consistent with the constraints prevailing at that instant; at a subsequent time the constraints may be different, a fact expressed in general by the occurrence of t in eqn. (6.1). Consequently the connection between 6 r, and 6q, is given by 0 Yp 6qt, 6zp = aaz 6x = 6 qi , 6y P (5qi, qi qi Pqi P

(6.8)

238


in which there are no terms containing ax piat, aypiat, &p/at, even though the time may occur explicitly in eqn. (6.1). If Fp , {X, Y, Zr), is the resultant force (including both internal and external forces) acting on particle p, the virtual work is

6W =

F, 8ri, =

(X,6xp Ydy„ Zpoz p).

On substituting from (6.8) we obtain

6W = Q j 6go

(6.9)

where (xp a x, + yp ay, + zp az,

Qi P

agi

agi

Ogi

(6.10)

The coefficients Qi in eqn. (6.9) are defined to be the generalised forces (see § 2:4), and are given explicitly in eqn. (6.10). [This equation should be compared with the corresponding equation for momenta in example (i) p. 235.] We include internal forces in eqn. (6.9) if they contribute to the virtual work in an arbitrary displacement of the system, and the final equations of motion obtained later can refer to internal motion of a system if the coordinates gi are suitably chosen. If any forces, e.g., the internal forces in a rigid body, do no work in an arbitrary displacement they contribute nothing to 6 W and so do not occur in the final equations of motion. If the forces in a system are conservative the virtual work 6W is a perfect differential — d V, where V is the potential energy function, and

Qi = —

av aqi •

Examples. (i) The position of a particle is specified by spherical polar coordinates {r 0 s) and it is under the action of a force whose resolutes in the radial, longitudinal and transverse (azimuthal) directions are R, L, T respectively. Find the generalised components of force. In a displacement given by dr, 60, dv the virtual work is

6147 = Rdr + Era@ + Tr sin06v. = R, Q5 = Lr,

= Tr sin0

(ii) Two rods AB, BC are freely jointed at B and hang in a vertical plane from A, which is fixed. If AB = a, BC = b, and the rods make angles 0,99 with the vertical, find the generalised components of a horizontal force F acting at C.

§ 6 : 3]


239

The distance of C from the vertical through A (a fixed line) is x = a sin 0 b sin (p. In an arbitrary virtual displacement 60, 6q),

6x = a cos060 + b cosT699. .• 6W = Fox = Fa cos000 + Fb cosq)692. Qo = Fa cos0, Q,, = Fb cosy). (iii) An external force with resolutes {X Y} along the axes acts at A on the rods of example (iv) p. 235. Find the generalised components. The coordinates of G are given by X = xA+ a cos q) cos 0, Y = YA + a cos q) sine. In a virtual displacement 6X., 69,60, 6q9,

6xA= 6Z + a sing) cos0699 + a cosy) sin000, oy A= 6f, + a sing) sin04) — a cosy) cos000. .•.d W= XoxA + Y6yA

= X 61; Y dg QX

+ cosq)(X sin() — Y cos0) 00.

asinq)(X cos0 + Y sine)

= X, Qi = Y, QQ, = asinq)(X cos0 + Y sine), Qo= acosq)(X sine — Y cos0).

(iv) A rigid body, which we take to be a frame of axes OSjS2~3,canrotate about the origin and is subject to a couple r= {L1 L2 L3}, where the resolutes are taken along the axes 0E1 2 3. If the orientation of the frame is specified by Euler's angles according to the scheme of Fig. 4 (§ 1:4), find the generalised components of force. An infinitesimal rotation is specified by a skew-symmetric matrix (see § 1:6) 0 — h3 h 2 hs

h=

0 —

—h, h1

0

whose non-zero elements constitute a vector h = {h1 h2 h3}. In such a small rotation the work done is

6147—r.h—L1h1+ L2h2 + L2h3.

(1)

If the original orientation corresponds to R and the final orientation to R + (SR, then R + (SR R(I + h). •*. 6.11 = Rh,

or

h=

(2)

240


When the frame 0$13is rotating continuously the angular velocity resolutes along 0$1$23are given by 0 — a), C°2 R'R=S1 = (c03 — co2

0

—

col

co,

0

and in a short interval of time St 0 R' liSt = IV dIt = (

coot

— co2 St

— w3 St 0

co2 6t —

oh&

colOt) • 0

Writing SO = 6 St, by = 4 St, Sy, — y St, we obtain the expressions for h1, h2, h3 by using the results (1.25) for oh, co„ .• 6W = L1(60 sing) — Sy, sine cosg)) + L2 (6 0 cosg) + Sy, sing) sine) + L3 (6q) oy, cog). Q8 = L1sinyo + L2 cosg),

= L3,

= — L, sine cos q1 + L2sine sing) + L3 cos0 .

6:4 Classification of constraints In general, the number of generalised coordinates needed to specify the position of a system is equal to the number of degrees of freedom in the system, but in certain cases it may be convenient to use a larger number of coordinates and to restrict the degrees of freedom by requiring the coordinates to satisfy a number of conditions. For example, a rigid body rotating about a point has 3 degrees of freedom and can have its position specified by Euler's three angles. However, we can also specify its position by the nine elements of the orthogonal matrix R; these elements must satisfy 6 conditions comprised in the relation R R' = I = R' R. In general, this type of situation is represented by using n coordinates qt which satisfy m(< n) relations

F;(qi , q2 ,

q„) = 0

(j = 1, 2, ..., m).

In this situation the system is subject to m holonomic constraints. Another type of constraint occurs chiefly in connection with rolling bodies. When two bodies in contact roll on one another the velocities of the particles, one in each body, instantaneously in contact are equal. Such a condition leads to one or more relations between the generalised velocities; these relations are usually linear in these generalised velocities and are, in effect, differential equations relating the coordinates qi.

§ 6 : 4]

241


When there is more than one generalised coordinate such an equation may not be integrable (see Vol. IV § 4: 4). A system in this situation is specified by the coordinates qi subject to m (< n) conditions

Aki

= 0 (k = 1, 2, ..., m) .

(Note the summation over index j.) When these equations are not integrable the constraints are non-holonomic. (If one or more of these equations are integrable, they lead to holonomic constraints.) The crucial difference between the two cases is that holonomic constraints involve the coordinates only, whereas non-holonomic constraints involve the velocities in relations which cannot be integrated.

Examples. (i) A disc is constrained to roll with its plane vertical so that its point of contact with the ground follows a prescribed curve. The position of the disc can be specified completely by two generalised coordinates, viz., s the arc length OB, and co the angle between the radius of contact and the radius CA fixed in the disc, Fig. 69. Since the centre C traces out a path parallel to that of B with the same speed, the condition for rolling is = ay, i.e., s = a(q9 — sod, where Tois a constant of integration. The system therefore has one degree of freedom, and is a holonomic system. (ii) A similar disc with its plane vertical rolls on the horizontal ground, but is not constrained to follow any particular curve. The position of the disc is given by the coordinates {x y} of the point of contact B, the angle y, and the angle y giving the orientation of the plane of the disc, Fig. 69. [This orientation was fixed in (i) by the direction of the tangent to the curve.] The condition for rolling is = ay cosy,

y= aq)sinp.

We write these, in full, as Px± where

Py#

±

=0, Rx + R,Y +

+ Rop = 0 ,

Px = 1, Py = 0, P = —a cosy,

P, = 0;

Ry =1, R„ = —a siny,

R„ = 0.

Rx =0,

242


The condition that these equations shall be integrable is 0 Pi

013J

0 Ri

0 Ri

0q1

0 qi

0qJ

0 qi

,

(q1, qr = x, Y, 99, V) •

Since OP

= a sine *

P„,

OR

g,

07,p

' ,

— a cosy

092

the conditions for rolling are not integrable. The disc free to roll anywhere on the horizontal ground is subject to a non-holonomic constraint.

Constraints are also clasSified as "rough" or "smooth" according to whether the corresponding forces called into play by the constraint do or do not do work in an arbitrary displacement. Since a convenient method of handling a rough constraint is to remove it and add the corresponding (unknown) force to the applied forces, in general we shall assume that constraints are smooth.

6:5 Application of the Principle of Virtual Work The vectors mp fp are situated at the points rp , {xp yp zp} . Hence the virtual work of these vectors is W = mp (4 oxp

6w = 2' mp

ax p

g,6yp

6 Zp),

, aY p Y1,

aqi

-

-

P

azP)oq. aqi

(6.11)

[Note (i) that the terms a xpiat, etc., do not occur, (ii) that the summation convention over the dummy suffix i is implied.] The manipulations at this point depend upon two identities, viz.,

94 94 9q, ' d axp dt aqi ) with similar identities for y„, zp .

-

Oqi

(6.12) (6.13)

§ 6: 5]

243


Example. We prove these identities as follows. Using the theorem of the total derivative dxp a xP axP ±p 4,+ at dt and regarding qt as an additional set of n variables independent of q„ we see, by differentiation, that (6.12) follows immediately. We apply the same theorem for differentiating w.r. to t to the function Ox plo q„ d dt

oxp

0 (oxp axp ) qk dq , dqk dq, ataq, dq, dqk at 02 xp

(32 xp

qk

-

=

ip dq,

(This implies that the operations d/dt and 010 q, commute.)

Making use of the identities (6.12), (6.13) we obtain xP

d_

dt

Ogi

P

d

ax

P dt \

d dt

d r a 2)1 dt 4i 2XPLI dt

d

qi

ax„ )

a(1-4)

[a(1-41 L ath J

Oq i

•

With the similar identities for yi, and zp, eqn. (6.11) gives

6w _= 1 since T = z .m, (x4

T

d

)•

aT aqi i

(6.14)

Hence the equivalence of the two sets

of line-vectors can now be written d aT Ldt a

aT Qi 6q''

(6.15)

[Note that there is summation over i in both eqn. (6.14) and eqn. (6.15).] Students frequently consider that the manipulations leading from eqn. (6.11) to eqn. (6.15) are the most important part of the proof of Lagrange's equations of motion. Whilst such manipulations are important, it is essential that the student should appreciate the significance of the principles being used and the differences listed below. He must appreciate why, without further qualifications, we cannot equate the coefficients of oq, on each side of eqn. (6.15).

-

244


(a) Holonomic systems with n degrees of freedom In this case any one of the n coordinates qi can be varied by an amount q , keeping the remaining coordinates fixed, without violating the constraints. We can then deduce that the coefficients of d qt on each side of (6.15) are equal: d aT aT )-(6.16) • • dt (aq, aq, =Q, (i = 1, 2, n). The n equations so obtained are Lagrange's equations of motion for a holonomic system with n degrees of freedom. When the forces are conservative Q, = —OVI0g„ V being the potential energy which is a function of the coordinates only. Eqn. (6.16) then takes the form d eL

dt

al,

aL aq,

_

(6.17)

where L = T — V is the Lagrangian function (also called the kinetic

potential). The equations of motion of a top can easily be obtained by using Lagrange's equations; this is done in Chapter VII p. 288. Examples. (i) A particle of mass m moving in space has kinetic energy T = m(i.2 + T 262 + r2,)2 in terms of spherical polar coordinates. The generalised components of force are given by example (i) p. 238. Hence the equations of motion are d (mi.) — (mr02 mresin2 B) = R, (1) dt dt

(mr2 e) — mr2 tp2 sine cos° = r L ,

at(mr2y, sin2 B) = Tr sine.

(2)

(3)

In the special case of motion under a central force R = mf(r), L = 0 = T. .'..r — (rO2

at

sin2 B) = (r),

(r26) — r2V sine cos° = 0,

d — 2y5 sin2 B) = 0. dt (r

(4) (5) (6)

§ 6: 5]


245

If initially 6 = in and rO = 0, i.e., we choose the polar axis so that it is perpendicular to the plane containing the radius vector and the velocity eqn. (5) shows that 6 = 0 at t = 0. Further differentiations show that all the derivatives of 0 are zero when t = 0; hence, by Taylor's theorem 6 = in throughout the motion. The motion therefore takes place in the plane 6 = 1-7r in which the "plane polar coordinates" r, v are given by d

r1:02 = f(r),

dt

(r2

= O.

These are the equations of motion considered in Vol. III § 6:3. (ii) The motion of a particle on a smooth helix under gravity (see the example of p. 85) is given by

T = imc202, V = —mgz = — mg c0 .• L = lmc(cO2 + 2g0 sing). There is one Lagrangian equation of motion

at

sing= 0, i.e., c0 = g sinoc. (mc2 O) — mg c sin •

This is the equation obtained earlier. (iii) We consider the system of rotating rods of example (ii) p. 153. From the expressions obtained there

T = 3ma2[6 2 (1 + 3 sin2 0) + 4.02 cos2 e],

V = —2mga sine.

L = 3ma2 [62 (1 + 3 sin2 0) + 41p2 cos2 0] + 2mga sine. The Lagrangian equations of motion are: d 8 2 ma2 zp2 sine cosh + mat 6(1 + 3 sin2 0)] — [8ma2 02 sine cos° — 3— —dT + 2mga cos/ = 0, 80 (1 + 3 sin2 0) + 8(302+ 41p2) sine cosh — 6 (g la) cos0 = 0, and the second equation

dt

[_ ma2zpcos2 0] = 0, i.e., — at 3

cos2 0) = 0.

These are equivalent to the equations of motion obtained earlier.

246

A COVESE OF MATHEMATICS

(iv) Euler's dynamical equations. When the position of a rigid body with one point fixed is given by Euler's angles, 2T = A(6 sin?' — sine cos g")2 +

cosq, + xj) sine sin02 + C(4) + yi cos 0)2 .

When a couple T (0) = {L1L2 L3} acts on the body the generalised components of force are given by example (iv) p. 239. We obtain the equation of motion corresponding to q2 =co as dt

[C(49 + tpcoso)] — A(6 cosq, + sine sin(p) (0 sin?' — 1 sine cos?) —

— B(— 6 sing" + sine cow) (0 cosq" + tsine since) = L3. The resolutes (.01, w2 , W3of the angular velocity of the body are given by co, = 6 sing,— '4) sine cos (i" , w2 = 6 cos q' + 7i) sin° sing2, w3 =49+zpcose. d co3 d

.C— t —(A — B) col% = L3. This is the third of Euler's equations (5.30). The others follow by cyclic permutation of the quantities, or by writing out the remaining two of Lagrange's equations. (v) Four bars, smoothly jointed together, form a parallelogram ABCD, which is pivoted in a vertical plane about the mid-point 0 of AB. The bars AB, DC are each of length 2a and mass M, and the bars BC, AD are each of length 2 b and mass M'. The mass-centres of the four bars are at their mid-points, and the radii of gyration of AB, DC, BC, AD about axes through their mid-points perpendicular to the bars are K, K, K', K', respectively. If during the motion the bars AB and DC make an angle 0 with the horizontal, and the bars BC and AD make an angle 87 with the vertical, form Lagrange's equations of motion and interpret the results.

FIG. 70. Since the angular velocities of each of AB, DC are 0 and of each of AD, BC are 0, we see that the velocity of H, the mid-point of BC is the sum of two components shown in Fig. 70. The component bq) is perpendicular to BC, and aO is perpendicular to AB. We find the velocities of G, L similarly.

§ 6 : 5]

247

LAGRANGE'S EQUATION'S

The kinetic energies are:

AB; MK202 : BC; AD;

CD; iM(24)2 + iMK202 :

cosg — a6 sin0)2+ (— b g sing + a0 cos0)2] + IM/K1202 : M1 [(bgb cosg + a6 sin0)2 + (— bg sing) — aO cos0)2] +

K'24 2.

T = kM(2K202+ 4b2g2) + M' (2b2g)2+ 2a2 02+ 2K' 202) = (M K2+ M' a2) 62+ (M' K' 2+ 1111 b2+ 2Mb2) 4)2. Also

V= — 2 (M + M') bg cosg. L = (M K2+ M' a2) 62+ (M'K' 2+ .1111 b2+ 2Mb2)(;o2 + 2(M + M')bg cosg.

The equations of motion are

(MK2 + M' a2) 6 = 0 , (M'IC 2 + M' b2 + 2M b2) g + (M + M') bg sin g = . Therefore 0 is constant and the motion in g is synchronous with that of a simple pendulum of length M'K' 2 + M'b2 + 2Mb2

(M + M')b This motion is independent of 0 and 0. (vi) A rhombus A BCD of four uniform pin-jointed rods, each of mass m and length 2a, is placed on a smooth horizontal table on which it moves freely in any manner. If the hinges are lubricated, so that at each joint a couple of resistance to turning is brought into play equal to k times the rate of relative turn, prove that the angle 20 of the rhombus at time t is given by ma2 0

0 = 6o +

3k

(1

e-3k ilina'),

where 0„ 0, are the initial values of 0,0. The kinetic energy of this system is calculated by methods similar to those used for the example of p. 150 (see Fig. 71). [See also example (iv) p. 235.] The result is

T= 4m(±2 + 92) + 2 . ima2 (g) — 6)2 + 2 . ima2 (g) + 6)2 + + 2 . ima2 (g) — 6)2 + 2 . ima2 (4) +6)2= 2m(±2+y2) + gma2(02 + 62). The terms arise as follows: motion of G; the motion of the centres of the rods relative to G; the rotation of the rods about their own centres of mass. In a virtual displacement the couples do work

OW = —4(2k0 .200) = —16k000. Q„ = = Q7, = 0, Qe= —16k0. The equation of motion in 0 is 16 3

.

ma26 = —16k0, i.e., 6 = —

3k 0. mat


248 This integrates to

61-00 expc 3" ), ma2 ,

0=A

—

ma2 60 ( 3kt exp — 97712 ) 3k

where 00 and A are constants of integration. Since 0 = 60 when t = 0, 0 = 00 +

ma260 [ 3k

1 exp /

3 kt —, ma2 /

(vii) Ox, Oy are rectangular axes fixed in a smooth horizontal rigid lamina constrained to rotate about 0 at an angular rate w. An elastic string of natural length 1 passes through a smooth hole in the lamina at the point A, (1, 0), one end being fastened on the underside of the lamina at 0, and the other end attached to a particle P of mass in resting on the lamina. If initially the particle is released from rest at 0 and the modulus of elasticity of the string is mlw2, show that so long as the string is not stretched beyond its elastic limit the path of the particle relative to the lamina is a cycloid with its cusps at the points x = 0, y = — rIn12, r = 0, 1, 2, 3, ..., reached after successive intervals of time r/w. Referred to the rotating axes Oxy in the lamina, Fig. 72, the position of P is r = {x y 0} ; the velocity of P is T=

9

0) +{00 co} x T= m[(t —

y 0} = {t — coy 9 + cox y)2 (g + (0x)2].

0}.

§ 6 : 5]

249


The extension of the elastic string is PA, where PA2 = (x — 1)2 + y2. V = insco2[(x — 1)2 +y2]. L = 19711( — w y)2 + (9 + w x)2— co2 (x — 1)2— w2 y2]. The equations of motion are: d mat (± c") m[w(y + cox) — co2 (x —1)] = 0, i.e., d ma t

+ cox) — 9n[—co(.t— coy) — co2y] = 0, i.e.,

— 2o4 co2 /,

g + 2coX = 0.

FIG. 72. We integrate these equations subject to the initial conditions x = y = 0, t = 9 = 0.

— 2coy = w2 /t,

+ 2cox = O.

Combining these results, with = x iy, we obtain + 2 ico = w2 / t. Further integration leads finally to = 4/[1 — e-2"1 — i.e., x = 11(1 — cos2wt), y = 1/(-2wt + sin2wt). These equations are the parametric equations of the cycloid as stated. Exercises 6:5 (a) 1. One end A of a uniform rod AB of mass m and length 2a is constrained to move on a smooth, straight, horizontal wire; the rod moves in a vertical plane through the wire under its own weight and under a force X acting at B in a direction parallel to the wire. At time t, A is at a distance x from a fixed point of the wire and the rod makes an angle 0 with the vertical. Derive Lagrange's equations for the rod and deduce from them that the rate at which work is done by the force X and the weight is equal to the rate of increase of the kinetic energy of the rod. 2. A framework of four light rods each of length a is in the form of a rhombus ABC D freely jointed so that C can slide along a vertical axis through A which is fixed. Particles of mass m are attached at B and D and one of mass 4m at C. The 9

CDIES

250


whole framework can rotate freely about the vertical axis. When the angular velocity of the plane A BC D is 1p the rods make an angle 0 with the vertical. Prove that the Lagrangian equations of motion are d (V) sin2 0) = 0, dt 6(1 + 8 sin20) + 862sine cos° — ,2 sing cos 0 + 5 (gla) sin0 = 0. Find the angular velocity tj) = Q for which steady motion is possible with 0 = 60°. 3. A hollow circular cylinder of radius b is made to rotate about its axis, which is horizontal, with variable angular velocity w. A uniform circular cylinder of radius a ( < b) rolls without slipping inside the hollow cylinder, the axes of the two cylinders being parallel, and the plane through them making an angle 0 with the vertical. Prove that at any time 3 (b — a) 6 = bth — 2g sin0. Show also that if th is a small constant s, and the smaller cylinder is at rest in its lowest position at time t = 0, then (approximately) 0 = (sblg)sin2ipt, where p2 = 2g/3(b — a). 4. Two rectangular blocks of equal height are placed on a smooth horizontal table with a pair of faces parallel and facing each other at a distance c y2 apart. One block is fixed; the other, of mass M, is free to move on the table in the direction perpendicular to those parallel faces. The movable block is held at rest with a uniform cube, of mass M and edge 2a (> c), placed symmetrically across the gap between the blocks so that two adjacent faces of the cube rest on the edges of the gap. If the system is then released from rest and all contacts are smooth, show that the initial angular acceleration of the cube is g(c — a)/(7a2 — lOac + 5c2)1/ 2. 5. A thin rod moves freely in a vertical plane which is made to rotate with constant angular velocity w about a downward vertical axis Ox; Oy is a horizontal axis in the plane. The coordinates of the centre of mass of the rod are (x, y) and the inclination of the rod to Oy is 0. Initially x = y = 0, x = 0, fj = V, 0 = 0 and 6 = w. Find the subsequent movement of the rod. 6. A uniform hollow circular cylinder of mass m and radius a is free to rotate about its axis which is fixed and horizontal. Inside it rolls, without sliding, a uniform solid circular cylinder of mass m and radius b. Prove that a

1/{

3g 2(a — b)

r t

J

clp v (cos q)

— cos a)

0

where 0 is the angle the plane through the axes of the cylinders makes with the downward vertical at time t, and the system starts from rest in the position 0 = oc.

§ 6: 5]


251

7. A light rod AB of length 1 can move freely in a vertical plane about a horizontal axis through A. The end B is smoothly hinged to a point on the circumference of a heavy uniform disc of radius 1 in such a way that the disc can move freely in the vertical plane of motion of the rod. When the system is allowed to move under gravity, the angle made by AB with the downward vertical is B and the angle which the radius through B makes with the downward vertical is 92. If at any instant during the motion B = 99 prove that the angular acceleration of the disc is then instantaneously zero. 8. A bead of mass m is threaded on a smooth uniform circular wire of mass M and radius a which can rotate freely about a fixed horizontal diameter AB. At time t the bead is at an angular distance B from A and the plane of the wire makes an angle 99 with the horizontal plane through AB; find the Lagrangian equations of motion. If, when t = 0, = , 0 = = cp = 0 , prove that (M + 2m sin2a)aep = 2mg sina when t = 0. 9. For a certain dynamical system the kinetic and potential energy functions are

T = 1[(1 + 2k) 02 +

+ 492],

V= n2[(1 + k) 2 + 9)2] , where 0, co are the coordinates and n, k are positive constants. Write down the Lagrangian equations of motion and deduce that 1+k

(9— el5) + n2

(0 92) = 0.

Hence prove that if B = q), 9 = 99 at t = 0, then 0 = (p for all t. 10. A uniform solid circular cylinder, of radius a and mass m, and a thin hollow circular cylinder, of external radius b (> a) and mass M, are mounted on the same axis with the solid cylinder inside the hollow one and able to turn freely about the common axis. The cylinders are of equal length and are connected by light springs with one end attached to the inside of the hollow cylinder and the other end to the solid cylinder. When the inner cylinder is turned relatively to the outer one from the position in which the springs are unstrained, a couple of total moment K per unit relative angular displacement is exerted by the springs on each cylinder. The outer cylinder can roll without slipping on a horizontal table and the system is released from rest when the relative angular displacement of the cylinders is a. Taking B and 99 to measure the subsequent angular displacements of the inner and outer cylinders respectively, obtain the Lagrangian equations of motion, and prove that, at any instant, ma2 0 + (2m + 4M) b20 = 0. 11. A uniform rod OA, of mass m and length 2a, is pivoted at 0. The rod makes the angle 0 with the downward vertical OZ and the plane A OZ makes the angle fp 9*

252


with a fixed vertical plane. A bead P of mass Ain slides on the rod and is joined to 0 by a light elastic string of natural length a and modulus nmg. Prove that the kinetic energy is given by 2T = maz (62 +

sin20) + m

x20 4. x2492 sin26)

where x = OP. Derive the Lagrangian equations of motion and show that if a x = 4a/3, then 02 3g/(2 a) and n = 6R. steady motion is possible with 0 = 12. A thin uniform smooth wire, in the form of a circle of radius a and of mass M, is free to rotate about a fixed vertical diameter AB. A heavy bead of mass 4M, threaded on the wire, is held at rest relative to the wire by means of a catch at one end of the diameter perpendicular to AB. The system is spun about AB and, when the spin is w, the catch is released, and the system then moves freely. If 9 a co2 = 2g, and 0 is the angle which the radius to the bead makes with the horizontal in the subsequent motion, prove that 8 62 — = 1 + 8 sin0 9 w2

9 9 — 8 sin2 0

Show also that, when the bead first comes to rest relative to the wire, the spin of the latter is 9w.

(b) Systems subject to holonomic constraints We suppose that the n coordinates qi are subject to m (< n) holonomic conditions (1 = 1, 2, ..., m). (6.18) q2) ••• q„, t) = 0 The variable t occurs in (6.18) because such a constraint may, in general, vary with time. As we explained in § 6:3, the virtual displacement must satisfy the constraints prevailing at a given instant.

F (qi , q2 , , q,,, t) = 0, Fi (q,

bqi , q2

42, , q„

t) = 0 ,

i.e.,

aF,

= 0 (1 = 1, 2, ..., m).

(6.19)

There is no term containing aFziat. This implies that the Sq, are not independent but must satisfy the m conditions (6.19). To proceed further we make use of the method of undetermined multipliers (see Vol. II p. 273); we add multiples of eqns. (6.19) to eqn. (6.15), so that

m aF — Qi — Ai,,H 6 qi =O. qi vqi 1=1

aT [ ddt aZi

§ 6: 5]


253

Because of the introduction of the multipliers 2, we may now treat all the bqi as if they were independent and obtain n equations of motion

OT\8T = Qi a qi dt04,)

21 0Fi

d

f= 1

Oqi

(6.20)

Or

d aL

aL

dt

Oqi

(

04i1

OF, aqi •

i

(6.20 a)

These equations, together with the conditions (6.18) are, in principle, sufficient to determine q1 ,5 a, 2 • • • qn and the multipliers 21, 2,„. The form of eqns. (6.20) shows that, effectively, we have to introduce

aF,

unknown "forces of constraint" 2, — into the equations of motion

aqi

for each of the constraints. Examples. (i) A uniform rod of mass m and length 2l is held with one end on a smooth floor and the other end against a smooth wall. Rectangular cartesian axes are chosen so that the wall is the plane x = 0, the floor is the plane z = 0, and the ends of the rod are (2a, — ,8, 0) and (0, jl, 2y). Prove that, if the rod is released from rest in this position, the initial downward acceleration of the centre of mass is 0,2 + 3g 4 2

4162 ± y2 •

Fm. 73. We assume (see Fig. 73) that the coordinates of the ends of the rod are:

A, {2x 2y, 0); B, {0 2y2 2z}. Then G is {x yi +y2 z}, and X2 4-(Y1

— Y2)2 + Z2 = 12

.

(1)

254


Thus, we are using four coordinates subject to one holonomic constraint. The kinetic energy of the rod is [using the formula T = *m(u2 + u. v + v2), see p. 148]

T = lin(t2 + + A +

Y1Y2) •

Also V = mgz, (2)

L = im (2 i2 +2A + 2g + 2i2+ 4192- 39z)• Using one undetermined multiplier the equations of motion reduce to

+ 22(yi— y2) = 0,

+ 2.1.x = 0, lyi +

— 2 A(Y1 — Y2) = 0,

392 +

+ g + 22z = 0,

+ 32x = o, 291+ g2 + 32(y, — y2) = 0,

(3) (4)

g, + 2g2 1— 32(y, — y2) = 0, 4z + 3g + 6Az = O. (5) (6) We are not required to solve these completely but only to find initial values for the accelerations. The initial values of the coordinates are = , Y.to =

-09

Y20 = 1 3,zo

— 7' •

Hence the initial accelerations are Xo =

P10 = 3A 3, Y20 = — 3216, 4 =

+ 9)•

On differentiating the condition (1) twice and substituting the initial values (remembering that initial velocities are zero) we obtain x0 .4 + (yi.o ,•. —

Y2o) (Pio

g2o) + Z0 10 = 0.

Poc2 — 602— iy(24 I g) — 0. Y 2 (x2 + 4i32 + y2)

A Hence the initial value of z is 3

4 = — — (2AY+ g) — 4

3g(oc2+ 4132) 4(00 + 4132 + y2) •

(ii) A particle P, of mass m, restricted to move on a smooth horizontal plane, is attached by an inextensible string of length b to a point A that is made to describe a circle of centre 0 and radius a in the plane with uniform angular velocity w. If 0 is the angle that AP makes with OA produced, show that provided the string remains taut the motion takes place in accordance with the equation d20

at + e

awe

b

sin0 = 0..

If the motion is such that 0 oscillates between — cc and oc, express the tension in the string in terms of 0 and oc, and hence show that whatever the amplitude of oscillation the tension cannot vanish if b > 3a.

-

§6:5]


255

To find F, the tension, we remove the constraint that AP = b and use two coordinates r, 0, where r = AP, Fig. 74. The components of velocity of P along and perpendicular to AP are aco sinO, r (6 + co) + a co cos° . 2 T/m

+ a co sine)2 + {r(6 + w) + aw cos0}2

The generalised components of force are Qr= — F, Qs = 0. Therefore the Lagrangian equations of motion are d

+ aaw sine) — (0 + w) {r(6 + w) + aw cos0) = —

dt dt

(r2 (0 -4- w) + aw r cos0) — [aco(r + a w sine) cos° — aw {r(6 + w) + aw cos0) sine] = 0,

which reduce to

— r(6 -F co)2— awe cos6 = — F Im , d —

{r2(0 + w)} + aw2r sine = 0.

(1) (2)

at

Provided the string remains taut there must be a solution of eqns. (1) and (2) for which r = b, i.e., F = m {b(0 + co)2+ awe cos0}, (3)

b2 0 = —abco2 sine. Equation (4) integrates to give b02 = 2aco2 (cose — cosa) after using the condition 6 = 0 when 0 = +a. Then (3) gives F/(mco2) = a(3 cose — 2 cosa) + b + 2 1/(2ab(cose — cosa)),

(4)

256


where we take the positive sign when 0 is increasing (0 > 0) and the negative sign when 0 decreases, i.e.,

E/(mco2) = 3a(cos0 — cosa) f 21/{2ab(cos0 — cosa)}

b + a cosa.

With x = y (cos0 — cosa),

I 7 (9n, co2) = 3ax2 ± 21/(2ab) x ± (b + a cosa) = f (x).

If the zeros of the equation f (x) = 0 are complex, then 1' > 0 for all values of x, i.e., for all possible values of cos0. But the zeros are complex if tab < 3a(b + a cosa), i.e., if cosa > — b/(3 a). This inequality is always satisfied when b > 3a, since cosa? —1.

(c) Non-holonomic systems We treat non-holonomic systems in a similar way to holonomic systems with constraints. We suppose that the constraints are expressed in the form of m (< n) non-integrable equations, linear in the velocities, Alk

qk — B , (1 = 1, 2, ..., m)

(6.21)

where A,„, B1are functions of the coordinates q1 . The non-holonomic conditions (6.21) imply that if increments 4 qitake place in some motion of the system during a timed t, then

A lk A qk = B14t

.

However, a virtual displacement is a displacement to a position which the system could occupy at the same instant (see § 6:3). Therefore, in order that a virtual displacement given by 6q1, 6q2, ...,6qn shall be "consistent" with the non-holonomic constraints (6.21) the increments must satisfy (6.22) Aik Sqk = 0, (1 = 1, 2, ..., m) the relations obtained by letting LI t —> 0 in the equation for a possible motion. [The reader should compare this with eqn. (6.18) which is equivalent, on differentiation w.r. to t, to

OF,

a qk

qk =

a Ft — at

§ 6 : 5]

257


But the restriction on the 8q1is nevertheless given by (6.19).] Hence the increments 6q.• are not independent of one another. Using undetermined multipliers dui , as before, we obtain d ( 8T) aT — — — — pi dt 4, i=i

c5q, = O.

(6.23)

In this equation we treat all the aq, as if they were independent. We obtain n equations of motion aT d (071 ) — — — = Q • + .11 Aii, (i = 1, 2, dt a Oqi t=1 d dt

or

8L

aL)

n),

A

(6.24)

(6.24a)

which together with the conditions (6.21) can give q1, q2, q„ and the multipliersu2,,da„,. Here again the occurrence of undetermined multipliers in eqn. (6.24) is equivalent to the introduction of unknown "forces of constraint". When constraints, particularly non-holonomic constraints, are present the use of Lagrange's equations loses some of its advantages over the methods given earlier for the solution of specific problems. Example. We derive the equations of motion for example (iii) p. 164. We use Euler's angles to specify the orientation of the sphere: a set of axes 0$1E2$3fixed in the sphere (not shown in Fig. 41) are given in direction by Euler's angles from the fixed directions Oxyz (which are Oxi x2 x3in Fig. 4 p. 13). According to eqn. (1.24) the resolutes of the angular velocity of the sphere along Oxyz are given by col =

—

6 sing + 49 sin° Cosy, cot = 6 comp + sin() singp , co3 = y1 + 49 cost9

(In Chapter I, oh, cot , co3were used to denote resolutes in the directions of 0,..;3 e,: this is not so here.) The conditions for rolling are (see p. 165) —

= fix — 9

9a

OMES

—

aco2 =

—

a() comp +

sin° sinv),

acoi = 9 -I- a (— siny q9 sin° cosy).

(1 ) (2)

58


The kinetic energy is 1 2

T = — m(t2 + 2) + — ma2 (coT + cw2 =

5

+ (4)

2 + 5 2 + 2a262 + 2a2 02 + 2a2y2 + 4a29'y cose). 10 (5t

Hence, with the potential energy V = — mg x sin a,

L = ,16 m(5±2+ 592 +2a2 02 + 2a2 y2 + 2a2y2 + 4a2 0y cose + lOgx sina). (3) We can now write down the equations of motion using undetermined multipliers, 2A/5 for (1), 2p/5 for (2). The equations reduce to = 0,

(4) (5)

a2 — (26) + 2a2 yy sine — 2Aa cosy — 2pa siny = 0, dt

(6)

d a2 a(249 alp coo) - 22ct sine siny + 2pa sine cosy = 0, t

(7)

d a2 -d— (2tp + 20, cose) = 0. t

(8)

—(5X) — 5g sing + 2A = 0, dt

— (5) + dt

Eqn. (8) shows that cb3= 0. We rewrite the eqns. (4-8) as 5X — 5g sina + 2A = 0,

6+

59 + 2p = 0,

(9) (10)

sine = (Ala) cosy + (p/a) siny,

+ iy cose — y,O sine = (A/a) sine siny — Ca/a) sine cosy, + cos° —

6 + (14 sine = (Ala) cosy + (p/a) sing, cose — Oy, = (Ala) sing — (p/a) cosy.

By differentiating the conditions (1) and (2) we obtain X + Di/ = a (6 +

sine) cosy + a (0 sine +

cose — 6y) siny,

9-

sine) siny — a(0 sine +

cos° — Oy) cosy,

= a(0 +

whence

+ ,S4 = A, 9 - (2± = — 4a. Eqns. (9), (10) now become 7X + 209 — 5g sina = 0, 79 -

(12) (13)

sine = 0.

We eliminate Vi from eqns. (12), (13) and obtain, with eqn. (11)

0 sine +

(11)

= 0.

§ 6 : 6]


259

These are the equations of motion obtained on p. 165. The reader can see from this example, which deals with a comparatively simple system, that the elimination of the undetermined multipliers makes this use of Lagrange's equations of little advantage over earlier methods. For complicated systems the manipulations become prohibitively heavy.

6:6 Conservation Laws

(a) Momentum Since the potential energy does not depend upon the velocity,

OT

L

a4i

-

and Lagrange's equations of motion take the form

aL

dp,

dt When L is independent of a coordinate q„,

aL

= 0,

(6.25)

qk

d p, — 0, i.e., dt

p, = constant.

This is the generalised form of the conservation of momentum. The momentum pk is still conserved even when there is no potential energy function (non-conservative forces) provided that a T/eqk = 0 and the corresponding generalised force Qk is zero.

(b) Conservation of energy In elementary treatments we have seen that energy is conserved when the forces acting on the system are conservative, the energy being the sum of kinetic and potential energies. For conservative systems we define a function H which is conserved when L does not depend explicitly upon the time. The definition of the function H is

H (q „ 4,, t) = pkgk - L.

(6.26)

(Note that here there is a sum over the index k.) On differentiating we obtain dH aL 01, aL .. Pkqk qk qk a 4k qk dt — Pk qk at 9 a*

260


When L does not explieitly contain t, aLlat = 0, and from Lagrange's equation and the definition of momentum

aL Pk =

Pk =

agk

aL a lk,

it follows that dH dt

Pk qk

Pkqk

Plc4k

Pk4k = °

i.e.,

H = constant.

(6.27)

In the important special case where T is a quadratic function of the velocities we see that

aT

qk,k =2 T

lkrk

(6.28)

from Euler's theorem on homogeneous functions (Vol. II p. 239). Hence, for such systems H = 2T — (T — V) -- T + V . (6.29) In these cases, therefore, the function H coincides with the total energy. When T has the general form given by (6.2) we write

T=

+ bi qj+ c = T2+ T, + T,.

Then

aT qk

= 2T, + T,

qk

using Euler's theorem. The form (6.29) for H becomes in this case

H=2T 2 +T1 —(T2 +T1 +To —V)=T 2 —T0 + V the terms linear in the velocities being eliminated. Therefore if (6.28) does not hold, H is still constant when a Llat = 0; in such cases the constraints usually vary with time and the total energy T + V is not conserved. The function H is called the Hamiltonian function. Examples. (i) A hollow circular roller, of radius a and mass M, has attached to the centre of its axis a particle of mass m by a light rod of length h. The particle is free to swing about the axis in a plane perpendicular to this axis. The roller is at rest on a horizontal plane, rough enough to prevent slipping, with the rod held at an angle a with the downward vertical. If the rod is then released, prove that the centre of the roller will oscillate through a distance 2m ha2sina/{M (k2 + a2) + m a2) , where Mk' is the moment of inertia of the roller about its axis.

§ 6: 6]

261


In the general position shown in Fig. 75 the kinetic energy is

T=

(k2 +a2) (±/a)2 + im(t2+ 2h±B cos° + h202)

and the potential energy V = — mgh cos°.

.*. L = 4-M (k2 +a9 (t1a)2+ im(±2+ 2h±0 cos0 + h,2 02) + mgh cos0

x Fm. 75. The momentum corresponding to x is conserved, so that k2 a2

Px = M

a2

ll+ m(t + hO cos0) = 0,

since the system is initially at rest. [M(k2 + a2) + ma2] x + ma2hO cos0 = 0.

(1)

On integration, subject to the initial conditions x = 0, 0 = a, we obtain [M(k2 + a2) + ma2] x = ma2 h (sina — sin0).

(2)

The maximum value of x occurs with the minimum value of 0. The energy equation (or conservation of H) gives ma2 (t2 212±O cos0 + h,202) — mga2h cog = — mga2h cos a M (k2 a2) ±2 on using the initial conditions. Eliminating i gives M(k2 + a2) + ma2 sin2 0 02 2 ( (cos° — cosa). m M(k2 a2) ma2 h, This shows that 0 follows a libration motion between the limits 0 = values of x, as shown by (2) vary between the limits 0 and

a. Hence the

2ma2 h sina M(k2 a2) ma2 (ii) A uniform rod of length a and mass m is attached by smooth light rings at its ends to a circular wire of radius a which is constrained to rotate with constant

262


angular velocity w about a fixed vertical diameter as axis. Show that the kinetic energy of the rod is T = Anta2 {1062+ (9 — 8 cos2 0) w2), where 0 is the angle which the radius through the mid-point G of the rod makes with the downward vertical. If initially the rod is vertical and at rest relative to the wire, show that subsequently 562= (6 1/3g/a — 4co2cos e) cos0

The velocity of 0 has components act 1/36 (a y3 sine) a) 0), see Fig. 76. The angular velocity has components {co cos0 6 co sine), and the corresponding (principal) moments of inertia are pi ma2 , a i ma2, 0.

.• T = 1

1 (w2 cos ze + 62) + 2

24

ma2 24

3 a2 (62 + w 2 sin20)

[1062+ (9 — 8 cos2 0) co2].

Since

V — mg L = — = •

II =6

a 1/3

2

cos° ,

a2 [1062+ (9 — 8 cos2 0) 0)2 + 12 1/3 (g ct) cos0]. 24

9/?,

aL m a2 — L = 24 [1062— (9 — 8 cos2 0) a)2— 12 1/3 (g la) cose]. a6

Since the wire is made to rotate, an external agency is doing work on the system to an unknown extent. Hence, energy is not conserved, but the function II is "conserved", i.e., H is constant.

§ 6 : 6]

263


Initially 0 = in, 6 = 0. .'. 1002— (9 — 8 cos2 0) w2— 12 I/3 (g la) cos° = .'. 502 = [6 1/3(g la)

—

—

9 co2 .

4w2cose] cos°.

(iii) A vertical plane is forced to rotate with constant angular velocity w about a vertical axis 0 V in it. A uniform solid circular cone of semi-angle a and generator length / has its vertex attached to 0. The cone is perfectly rough and is in a position of rest relative to the plane when the line of contact is horizontal. Show that the inclination of the line of contact to the vertical at another position of relative rest is given by lOg cos° . lw2 (5 costa — 1)

FIG. 77. This is a system in which the single parameter 0 specifies the position of the system. The chief difficulty in the solution is the incorporation of the condition that the cone rolls on the plane. In order to express this condition and to calculate the kinetic energy we use a set of rectangular axes 0$1$2$3in which 0$1is the generator of contact, 0$2 is perpendicular to the rotating plane and 0 3 lies in this plane as shown in Fig. 77. Since the cone is rolling on the plane its angular

264


velocity consists of that of the plane together with an additional component about The angular velocity of the plane is {— co cos 0 0 eosin 6} ; hence co = {wi 0 co sin0} . The axis of the cone 0 GC is fixed in the cone so that the velocity of an arbitrary point k{cosoc sina 0} of this axis is {w10 cosin0} x k {cosoc sing 0} = k{—wsinOsina eosin° cosa coisina} . This axis is also fixed relative to the frame, so that the velocity of the point on the axis is also 0 x {cos a sina 0} = k {— co cos 0 —6

co sine} x {cos a sin a 0}

= k(— cosin0 cosa eosin° cosa — co cosOsina + cosa} , where 0 is the angular velocity of the frame. —w toss sina + 6 cos a= w1sina i.e.,

,

= 6 cota — co cos°.

Using principal axes of the cone at 0, viz., 0 G, a second in the plane 3 = 0, and the moments of inertia are

Amt2 sine a, 2 31/2 (4 — 3 sine a), Amp(4 — 3 sine a) respectively and the corresponding resolutes of co are = {colcos a — wisin oc co sine} . 3111/2 'T— [2 sine . wi cos2a + (4 — 3 sin2a) co;.sinea + (4 — 3 sin2a) w2 sins]

40

3M12 40

[(1 + 5 cos2a) col sinea + (4 — 3 sin2a) we sin2 0].

Also the potential energy is V = —1111g1 costa cos 0.

L—

3M12 [(1 + 5 cos2a) (6 cota — co cos6)2 sin2a + (4 — 3 sin2a) w2 sin2 0 40

lOg + — cosea cos0]. This does not involve t explicitly so that H is conserved. H

OL = 0

06

— L=

311112 40

[26 cota (0 cota — w toss) sinea (1 + 5 cos2a)

— (6 cota — co cos0)2sinea (1 + 5 cos2 a) — w2sine@ (4 — 3 sin2a) —

lOg

cos2 0 cos 0; j .

§ 6 : 6]

265


From the initial conditions, 0 = gin, 9 = 0,

3 m 12 H=

(4 3 sin2a).

40

(02cotta — w2cos20) sin2cx (1 + 5 costa) = 10 (g 11) costa cos0 — w2 (4 — 3 sin2 a) cos2 0,

02(1 + 5 cost a) = w2 (5 costa — 1) cos° [

lOg

cos0

ho2 (5 costa — 1)

from which the stated condition follows on putting 0 = 0. (iv) A particle of unit mass moves in a field of force whose potential energy is _ jui{z2 r2}'I2 — Zz (p, Z constants) at the point whose cylindrical polar coordinates are z, r, 92. Obtain the Lagrangian equations of motion in terms of •, ri, T, where n are new variables defined by the equations z

r = $71.

-

=

Show that, in addition to the integrals of energy and moment of momentum about Oz, the equations also possess the integral

($2 +

,12)2 i2 =

2E 2 + Z 4— h2g2 + constant,

where E and h are the constant values of the energy and moment of momentum about Oz. Also obtain the similar equation involving i/2. The kinetic and potential energies are 1

T = — (r2 r 2ci9 2 2

i 2) ,

V

(z2 + r2)1/2

Z z .

From the definitions of z, r,

.•

vev+

L = 1 [( 2 1 {(i2 2

, z2 + 2 = ($2 ± ,72)2.

= $j

=

772) (i2

•

71 ,5) 21 +

+ $2,72 4,2] +

(2

1 z(v __

w 4_ n2)

2 kt

1

+ n2)

2

2

Z

—

s

772)

The equation of conservation of energy is (2

+ n2) (i2 + 7,12 )

+272

6,2

4p (e±,2)

Z ($2

)

= 2E

.

.

266


= h., i.e.,

The equation of conservation of (angular) momentum pc,, is h 49 = 9)2)

,12) (2

••• (2

2

h2 en2

712)

— 2E.

z(v —

( 1)

The Lagrangian equation of motion for is d

• =

I/ E2

•

4p$ (2 + n2)2 + z

dt "c We multiply this by i(V + 0 and deduce that d

[ w

i2]

+ n2) (i2

7?2)

di

h2 (2 ±n )

on2

+

+712 +

But from the energy equation 4p

4_ 772

2

•

d 1 dt 2

Ti2)2

(e2

n2)(i2 + 712) ±

= h2 ( 2 e 2 =

h2

h2

n2

Z( 2— 7,2) — 2E.

.1_

+ n2 +

n2) +

2E

+ 2Z$35 +

Integration of this last equation gives

1

n2)2

2

_ 70

equivalent to the required result. When eqn. (1) is multiplied by 2 for nby subtracting twice eqn. (2). •••

n2)

1

2V

Ev

2 Z +n2

constant

(2)

we obtain the corresponding equation

h2

_7 12 — Z24 + 2En2 + constant.

(v) A rectangular frame of reference Oxyz with fixed origin 0 is rotating round Or with constant angular velocity cc, and the position of a dynamical system within

-

§ 6 : 6]

267


it is describable by means of n independent holonomic coordinates q1, q2, ,q„, the time not occurring explicitly. If the kinetic energy relative to the rotating frame is denoted by TR, show that the Lagrangian equations when expressed in terms of TR take the form d dt

OT R \ OT R +ijlj= — a(T7 Oqi alt ) Oqi

—— 1 well l) 2

(i = 1, 2, . ,

a(x, y) , m being a typical particle of the system, V the potential 0 (El, E) energy of the forces corresponding to the qionly, and I the moment of inertia of the system about Oz. Show that these equations possess an integral where i3i;= 2E m m

TR + V - 0)2 1 = constant.

We are given that x = x(q1,q2,

= Y (415425-54n), z = z (qi 4'2

qn) •

When the frame rotates, the velocity components of the particle are

g + cox, w=z.

u = — coy, v = .•. T

fr2 i2 2,0 (x y _ thy)] +

E2n[v

0,2E m (x2 + y2)

= TR + COEM(X9 - Xy) + co2Eni(x2 + y2).

y Ox + coEm(x — — — y), Oqi aqi 041

OT 0 T R 04i

since

[see eqn. (6.12)]. Also d (OT \ dt alt )

0X

ox

0 41

04.i

op = dy alt a qt

5

d (0 TR \ y dt ) alt wpma 04i 02y

02x

0 4iO4k

agiagk

wEm x d - dt

0Ox 4i ) Y) 4k

O TR \ + coL,m ( Ox y

oq, )

+ coEm(x

0 qk aqi 402y

02x

Oqi 0q2

Oqt aqk

(The summation convention implies summing over k.)

Y) 4k

Ox Oqi

°Y )4k + 0A

268


Similarly

a z m c ay ax 0 11 OT=aT R co_ -r OA qk ) qk+a qi 2) Oqi aqi OTR +cozm iox

ay

koqi aqk —

Sgt

02y

+0)Znqx d ( OT\ aT . . dt a It ) OD

02x

aq, k 2

)

(OTR \_ OTR+2cozm idx ay _ Ox ay (21, aqi dqk) a41 1 Sgt kaqk Sgt

d

dt

a ( 1 01_ a 0qV . i 0 qi ±wi3iklk — .'i 2.i

OTR

)

d ( 0 TR dt

y qk+ 0 (1co2, .

OM qk OM qk )

a 11.(02,\ _ d (aTR dt a4, ) OD k 2 )

''

ax 0y) aqi )th`

a TR OD

aq, j

+ cothk 4k_ _

a ( T7 _ — 1 0,21). Oqi

2

)

Since

aT a TR ay ax + co E m (x — — — y) lk, qkOqk = qk aqk aqk aqk OT °41c

1 a TR co2./ T = qk — — + CO m (x" — y) — co E m(xy— 4) — — 2 qk

1 1 = 2TR — TR — —0O21 = TR— —01, 2 2 where we have used the fact that T R is a homogeneous quadratic function of di. But d( OT OT d (OT aT OT qk . ° qk

A, qk — qk— qk

T

up I.,

d

(DT

{— dt — .•.

dt

41,

OT aq

—k =qk (— dV dt

(TR— — u.)2 /) = — 2

qk

T R -I- V — (02 I = constant.

q A k

aV

aA

4k

dV dt •

§ 6: 6]


269

Exercises 6:6 The following exercises from earlier parts of this series can be considered by the method of this section: Vol. III Ex. 9:4 Nos. 1-5; Vol. VI Misc. Ex. V Nos. 3, 8 (in this volume). 1. A dynamical system is in motion relative to a base which rotates about a fixed axis with constant angular velocity. Prove that, if the system is smoothly constrained so as to have only one degree of freedom, we can find the motion of the system relative to the base by superposing the "centrifugal force" on the actual forces. A smooth wire in the form of a parabola of latus rectum 4a with its vertex downwards rotates with constant angular velocity w about its axis, which is vertical. A bead which is free to slide on the wire is in relative equilibrium at any height b above the vertex. If the angular velocity is suddenly reduced to w', show that the bead will oscillate from the height b on one limb to the same height on the other limb in time n/2

0)2

2

(1 + co, 2),/ fi

b

s n'_ 0

__

dd.

0

2. A uniform rod AB of length 2a is forced to move so that the vertical plane through the rod rotates with constant angular velocity w about a vertical axis through a fixed point A. Apart from this the rod can turn freely about A and make any angle with the vertical. If 2 aco2= 3g and the rod is originally horizontal and moving in a horizontal plane, show that, when AB makes an angle 0 with the downward vertical through A, 26 = sin() (2 cos° — 1) co2, and that a time (1/2/co) cosh-1(cot la) will elapse before AB makes an angle a with the downward vertical. 3. A solid uniform circular cylinder of mass m rolls on another such cylinder of mass M which rolls on a rough horizontal plane, the axes of both cylinders being horizontal and their centroids in the same vertical plane. Show that, if the system starts from rest in any position, the coordinates of the centre of the upper cylinder, referred to a fixed origin and axes, are given by the equations (3M + m) x = c {m0 + (3M + m) sind} , y = c cosO,

c being the distance between the axes of the cylinders. 4. A thin tube bent to the form of a circle of radius a is made to rotate with constant angular velocity w about a fixed vertical diameter. A particle of mass m can slide in the tube without friction. Prove that the quantity a (c10/dt)2— aco2 sin20 — 2g cos° is constant throughout the motion, where 0 is the angular displacement of the [see over] particle from the lowest point of the tube.

270


Find a general expression for the applied couple required to maintain the motion, and, if the particle just makes complete revolutions in the tube, show that the couple is 8x(1 — x2) (1 — 2x2) (a g — a z w2 x2)1/2 mac°, where x = sin 10.

6:7 Ignoration of coordinates The procedure followed in this section is a general method which is the counterpart of the procedure in elementary problems of using first integrals of momentum and energy to reduce the number of variables in a problem. An ignorable (or cyclic) coordinate is one which does not occur explicitly in the Lagrangian. L. According to eqn. (6.25) the momentum corresponding to an ignorable coordinate is constant. If a system is described by 1 m coordinates of which x1 , x2, , xi are ignorable and q1, q2, , gm are generalised coordinates as already introduced, the Lagrangian is a function L (q, , q2,•,g„„ 41 , 42 , ...,4m t) i.e.,

DL axk = 0 (k = 1, 2, ..., /). The corresponding momenta are constant, viz.,

aLk = ock (k

■ S

=1, 2, ..., 1) .

(6.30)

We can solve the eqns. (6.30) to find the velocities 4 in terms of 4, (i = 1, 2, ... , m) and a, . We construct a modified Lagrangian function R (gi ,4,, CCk t), called the Routhian, defined by

R = L — ±k k =1

L

(6.31)

OXk

in which all the velocities 4 are expressed in terms of 4, and ock . The variables in R are the coordinates g„ 4„ and the time, and we can derive a set of equations according to Lagrange's formula d ( dt

aR aq,

aR aq i

= 0 (i= 1, 2, ..., m) .

§ 6 : 7]

271

LAGRANGE 5 S EQUATIONS

The Routhian function is used in exactly the same way as the Lagrangian function L to give equations of motion. The use is illustrated in the following example. We do not prove this result here but postpone it until Chapter XI where we discuss more fully the important transformation (6.31), and the similar one (6.26). Example. A uniform rod of length 2a is attached at one end to a particle of equal mass by a light inextensible string of length a, and the system is placed at rest on a smooth horizontal table with the particle at the centre of the rod. If the particle is projected with a speed V at right angles to the rod and in the plane of the table, prove that 3 (1 — cos a) = 2 (4 — 3 cos a) 6 , where 0 is the angle turned through by the rod, a is the angle between the rod and the string, and a = 0, 0 = 0 initially. Show also that, when the string is in line with the rod,

a6 = 3 V/V7.

Fm. 78. Let G, the centre of mass of the rod, have coordinates {x y). see Fig. 78. Then P has coordinates

{x + a cos0 — a cos (a — 0) y + a sin0 + a sin (a — 0)). Tp =

{t

- ccO

g + aO cos° + a (dc — 6) cos (a — O)).

sin0 + a(e< — 0) sin (a —

T = m(i2 + + 2at(et

+ ma262 + lm[t2+ — 2ctO sine + —

6) sin (a — 0) + 2a9O cos0 + 2afj(ec — 0) cos (a — 0)

+ a2 02 + 2a2 6(oc — 6) cosa + a2 (et — 6)2] = n (±2 + + 2 (thla) (dc

+ ima2 [62( —

—

2 cosa)

—

2a0

—

cosa) + de2

6) sin (a — 0) + 2 (g/a) 0 cosO + 2(g/a) (a

—

—

2 (t1a) 6 sine

0) cos (a — 0)].

272


The coordinates x and y are ignorable and the initial conditions are = V.

0 = a = 0, x = y = 0, x= y =0, 6 = 0, OT • per = a= 2m4 — ma[6 sin0 — py =

— 6) sin (a. — 0)] = 0,

T =2my + ma[6 cos0 + (& — 0) cos(a— 0)]= mV. ag

ia [0 sin0 —(a— 6) sin (a — 0)],

p=1-v—ia[O cos0 +(a— 6) cos (a — 0)].

Therefore, since the potential energy remains constant,

OT OT B = T — ± —y 3- — 2 cosa) — 2 6 (1 — cosa) + ci2] m a2[62 (1

+

= — m(t2 +

1 = -T- ma2 [62(-±- — cosa) — ic 6 (1 — cosa) + I ec2] 2 2 3 1 +2 ma V[O cos0 + (a — 0) cos (a — 0)]

—

—mV2. 4

The equation of motion for 0 is

d [6 ( 4 _ dt [ 3

—cosa— I— cos (a — 0)] cosoc) — — 16( (1 — cosoc) ± 17 2a

2a

2

6 sin0 -I (dc — 0) sin (a — 0)] = 0, —— 2a [— 6 (± — cosa) — I ci (1 — d

[

2

3

cosa)]

= 0.

Hence, using the initial conditions, 26(4 — 3 cosa) = 3a (1 — cosa). The equation of motion for a is 1 r 1 d— (1 — cos a) + t [ 2

+

V 2a

ic sina — — r-1 62sina — 2 2

cos c'

f9)1

• •(a — 0) I = 0, V ( ct— 0) sin

—[a — 6(1 — cosa)] = 62sino — de() sinoc. dt

(We include this for completeness although we shall not use it.)

(1)

§ 6 : 7]

273


We now construct the Hamiltonian function

H=6

R —w

R

+a

—

= ma2 [02(-± — COS 3 1 — — & 6 (1 2

-I-

1 2

a 6 (1

—

—

R

OL) — —21 et 0

cosa)

(1 — cosa) + ---V— 6 cos()

2a

1

V

2

2a

+ — et2 +

2a

6 cosa

1 4 et cos (a — 0) — :- 62(— — cosa 2 3

1 V cosa) — — ex2— 6 cos° 4 2a

1, ( 4 = ma2 [— 02 — — cos 2 3

V — —

V —

2a

—

0)

)

(a — 0) cos (cx — 0)]

a — —12 of 6 (1 — cos a) + —14 d121 .

Since H is constant, 262 (4 — 3 cosa) — 6&0(1 — cosa) + 3a2 = 3(V a)2. When cos = — 1, 1402— 12 &el + 3 et2= 3 ( V/a)2, and, from (1), 6 ex = 146, i.e., & = 70/3. 7 1462— 2862 + (49/3) 62= — 62 = 3 (1). 2 3 a V3 V 7a Exercises 6:7 1. A particle of mass m is attached to a fixed point 0 on a smooth horizontal table by a light elastic string of natural length 1 and modulus A. The particle moves on the table in such a way that the string is always taut. Use the method of ignorable coordinates to obtain a second order differential equation for r as a function of time, where r denotes the distance of the particle from 0. Deduce that the polar equation of the path of the particle can be expressed in the form

0

dr r {ar2— 1 — br2(r — 1)2}

where a and b are positive constants. 2. A uniform circular disc of mass 2m and radius a is free to rotate in a vertical plane about its centre, which is fixed. An insect of mass m stands on the rim of the disc at its lowest point, and the system is initially at rest. The insect then proceeds to walk round the rim, the tangential component of his acceleration relative to the rim having the constant value f . Prove that he will not reach the top if / < g sin 20', where 13 is the acute angle defined by the equation tan 0' = 20. 3. A smooth uniform circular wire, of mass 2m and radius a, can turn freely about a fixed vertical diameter and a bead of mass m can slide freely on the wire. Find an equation to give the motion of the bead relative to the wire, and show

274


that, if the system is set in motion by giving the wire an angular velocity rg/a) about its vertical diameter when the bead is at rest at the extremity of the horizontal diameter, the bead next comes to relative rest at the lowest point of the wire. 4. A uniform hollow sphere of radius B and mass M rests on a smooth horizontal plane; a particle of mass m is placed at a point on the lower half of the smooth inner surface of the sphere, and is given an initial velocity V parallel to the horizontal tangent at that point. If, in the subsequent motion, P denotes the position of the particle, 0 the centre of the sphere and A the point of contact of the sphere and the plane, show, assuming that the particle does not leave the sphere, that the angle P OA oscillates between. its original value c and a value of a given by 2g sin2 a. =

M

V2

m+ M R

(cos a. + cos°co) .

6:8 The motion of a charged particle The force acting on a charged particle moving in an electromagnetic field cannot be derived from a potential function V (a a "• qn) because the force depends upon the velocity of the charge. However, if the (generalised) force can be expressed in the form

(au\ au

d

Q — di

04,)

(6.33)

where U is a function of qt and 4• , the equations of motion keep the Lagrangian form by writing L = T — U. (The function U reduces to the potential function V when U does not depend upon 4•.) Using mixed (Gaussian) units the force on a charged particle is F= e [E

7(v x B)] .

(6.34)

In electromagnetic theory the vectors E and B are obtained from a "scalar" and a "vector potential" E=—

aA c at 1

— gradT, B = curl A.

{10A 1 + gradg9 — 7 (v x curl Ad . (6.35) F = —e c

at

If we use cartesian coordinates x, (i = 1, 2, 3) to denote the position of the particle and resolve vectors along the coordinate axes, vi

d xi dt

§ 6: 8]

275


and

aA, "Fi

a99 ax,

—e IC

1

c 81kni

dxk aAs i dt Em 's ax j •

The alternating symbols 8,,,„„ en,„ were introduced in Vol. IV § 1:11 and are there shown to satisfy the relations

= a 99 ax,

eikm ems.

1 aA F, = —e[— c at =—e[

Since

1 aA,

aiq aks

1 c (6ig

dxk a Ad (5ik4s) at axq ]

1 ( dxk aA. _ dock aA a xik )] c dt ax, dt

aq, ax,

c at

aik aqs •

dA, = aA, aA, dxk , (differentiation following the motion), + ax, at dt dt

F, — e

1 aAk dxk ± ..9 991 ax, dt ax, j

[ 1 dA i c dt

By choosing

U = e(—ge

cl A k

d dXtk

(6.36)

we obtain F, in the form F.

d( dt

ari

au ax,

il dA, Lc dt

a 99 axi

1 aAk dxk i c ax, dt j

The kinetic energy of the particle is T = 2 m (4. + 4 + 4).

1 L = m(xi+ 4 + 4) + e (— 99 I Ak±k). This leads to the generalised momentum components

p, = m4,

(e/c) Ai

(6.37)

+ 4 + 4) + eq7.

(6.38)

and a Hamiltonian function

H = p,±, — L =

Example. Starting from rest at the origin at time t = 0, a particle of mass m and charge e moves in uniform, but not parallel, electric and magnetic fields. Prove that the particle describes a cycloid in a plane which is always perpendicular to the magnetic field, and which moves with a constant acceleration; determine this acceleration.

276


We choose the directions of the coordinate axes so that Ox, is parallel to H and E lies in the plane 0x1x3 . H = {0 0 H} , E ={E0 E3} , A = H {— x2 xi 0} , 99 — (Coe, + E3 x,) The Lagrangian function is

L= l m(4 + 2

eH

4 + 4) +

2c

(—ilx2 +±2x1) +e(E1x1+E3x3).

This leads to the following equations of motion

dt d

dt —[972

eH

—

[m

2c

X2

eH

±2

[eH —

— X11 —

2c

2c —

eH 2c

eH

=

±2

±11

— — X2

—

eE, = 0,

eH

= M X2

— it =0.

d

— 3) — eE, = md, — eE, = 0. dt (mt These reduce to

— X2= eEilm,

.t2

+ co =

0 , ;e3

e Edm

where co = eH/(mc). This first two of these equations are solved by writing = x1 + ix, [as in example (vii) p. 248] and have the solution

x1 =

eE, m

o (1

cos co t), x, —

eE„ mc02

(wt — sinwt).

This motion is therefore a cycloid in a plane which moves with constant acceleration (elm) E,(= ;„), in the direction of H. Exercises 6:8 1. A particle of mass m, carrying a charge e, moves in free space in an electric field E = — grad f (r), and a uniform magnetic field

Hr = Ho = 0 , Hi =H (r, 0 and z are cylindrical polar coordinates). At time t = 0, z = 0. Show that 1 eH z, — m (t2+ r2 62) + el (r), mr2 0 + - r2 2

remain constant throughout the motion.

2c

Ex. VI


277

An infinite earthed cylindrical conductor of radius a is enclosed by a concentric infinite cylindrical conductor of radius b whose potential is V; a uniform magnetic field H is applied parallel to the common axis of the conductors. An electron of mass m whose charge is — e(e > 0) leaves the inner conductor with zero velocity. Show that it cannot reach the outer conductor if

V < eH2 ( b2— a2 \ 2 8MC2 b ) 2. Assuming that the Lagrangian function for a particle of mass m and charge e moving in an electromagnetic field is L = imv2— eyo + e(v . A)/c, where co and A are the scalar and vector potentials of the field, find the force on the particle. Prove that the path of a particle moving in a uniform magnetic field Hok is a helix of the form x = (U/w) sinwt, y = (U/w) coscot, z = Wt, where co = eHolmc, and ( U, 0, W) are the components of the velocity of projection. If the field is non-uniform, being given by 11 = Ho (syl + Exj + k), where e is small, prove that the components (u, v, w) of the velocity at time t are, approximately, u = U costa, v = — U(1 — e W/co) sincot, w = W, the circumstances of projection being the same as above. Find the coordinates of the particle at time t to the first order in e. (The unit vectors i, j, k denote the directions of the coordinate axes.)

Miscellaneous Exercises VI 1. A thin, uniform, hollow cylinder of mass M and radius b rolls on a rough horizontal table, and inside it is a rough, uniform, solid cylinder of mass m and radius a. The axes of the cylinders are parallel and their centres of gravity lie in the same cross-section. Show that, when the inclination to the downward vertical of the plane through the axes is 0, the kinetic energy of the system is

Mv2+ m (u + v)2 + m (u2 + 2 uv cos 0 + v2) , where v is the speed of the axis of M and u = (b — a) 0. Prove also that, for small oscillations of the system, the length of the simple equivalent pendulum is 6M(b — a)/(4M + 3m). 2. A solid hemisphere of mass M and radius a can slide freely with its plane face on a smooth horizontal table. A rough uniform sphere of mass m and radius b is placed at rest at the apex of the hemisphere and is then slightly disturbed from

278


this position so that it rolls down the hemisphere, its centre C remaining in the same vertical plane throughout the motion. If after a time t the centre 0 of the hemisphere has moved a distance x, OC has turned through an angle 0, and the sphere has turned through an angle g) in space, show that bg, = (a + b) 0, (M + m) x = m (a + b) sin 0, and that, so long as the sphere is in contact with the hemisphere,

(7M + 2m + 5m sin2 0) (a + b) 02 = 10 (M

m) g (1 — cos0).

3. A uniform wedge of mass M and angle oc is free to slide without friction down a fixed plane of inclination oc, the top face of the wedge being horizontal and perfectly rough. A uniform sphere of mass m is placed on this top face in the vertical plane of symmetry and the system is released from rest. Prove that the speed V of the wedge after it has moved a distance x down the plane is given by

(7 M + 2m + 5m sin2 a) V2 = 14 (M + m)gx sina. 4. A uniform thin hollow circular cylinder of radius a and mass m rolls without slipping inside another such cylinder of radius 2a and mass M. The latter rolls without slipping on a horizontal table. If the larger cylinder has turned through an angle 0 when the plane of the axes makes an angle g9 with the vertical, show that the kinetic energy of the system is given by 4 a2 MO2 + ma2 + 462 + 24(1 + cos)}. T Show that for small motion about the position of equilibrium the equivalent simple pendulum has length 2M

M+m

a.

5. A rectangular rigid frame A B CD is free to rotate about a fixed vertical axis AB about which its moment of inertia is I. A heavy spherical shell of mass M can rotate about a diametral axis coincident with CD under the action of a light clockwork spring attachment to the frame, the potential energy of which is a function only of the angle 0 through which the shell is turned relative to the frame. If at time t the frame makes an angle 89 with its initial position, show that the kinetic energy of the system is

T=

2

Mkt 1(1 +

k2

+

Mk2

) 192+ 2 0 ci9 + 0 2} ,

where BC = AD = a, and k is the radius of gyration of the shell about the axis CD. If the spring is wound and the system released from rest, find the angle co turned backwards by the frame when the shell has turned through the angle 0. 6. Prove that, if a system is not conservative, the equations can still be put into the form containing the Lagrangian L, provided that the generalised forces are obtained from a function U gi) according to the formula

aU Ri=

d (a u dt Ogg

Ex. VI


279

Given

L = — mc2 11 (1 —

VkVk

c2

q99+ — Akvk,

where vk = 4(k = 1, 2, 3) and ?), Axare functions of Cartesian coordinates x1, x2, x3 while m, c, q are constants, set up the equations of motion and reduce them to vector form. 7. The two ends of a rod of length 2a and mass M slide freely on a horizontal wire. The ends of an equal rod are suspended from the ends of the first rod by equal light inextensible strings of length a. A particle of mass m is moved along the lower rod by an internal mechanism, the distance of the particle from the centre of this rod being the prescribed function z (t). The system moves in a vertical plane. Set up equations determining the motion. Supposing now that the angle between the vertical and the strings is small throughout the motion and that at t = 0 the entire system was at rest in its equilibrium position with z = 0, prove that the displacement x of the first rod is given by

m m 2M

f z(r) sin [co (t — -r)] co dr

where w2 = g (m 2M)la M. 8. A heavy circular cylinder can turn about its axis which is vertical and on smooth bearings. A light rod fixed to the cylinder projects horizontally and radially to a distance b from the axis. One end of a uniform heavy rod of mass m and length 2a is smoothly pivoted to the projecting end of the first rod so that the heavy rod can move at right angles to the projecting rod. The system is held at rest with the heavy rod horizontal and is then released. If the moment of inertia of the cylinder about its axis is m (1k2— b2), show that, in the subsequent motion, when the heavy rod makes an angle 99 with the horizontal, the cylinder has turned through an angle

lak(1 — cos 9)) 1 . 3b — tan-1 4k k2 a 2 cow 9. A mechanical system is rotating about a fixed axis, each particle of the system moving in a plane which passes through the fixed axis and rotates about it with constant angular velocity. Prove that the equations of motion, relative to axes rotating with the system, are the same as if the rotation were ignored but the potential energy diminished by the kinetic energy that the system would have when rotating in relative rest in its actual configuration at any moment; and deduce conditions of relative equilibrium and stability. The ends of a uniform rod of mass 3m and length 2l are constrained to move along two smooth wires OA and OB, OA being horizontal and OB vertically downwards. A smooth bead of mass m is free to slide along the rod. The whole system is made to rotate about OB with a constant angular velocity 11(6 gI51). If 0 is the inclination of the rod to the horizontal, prove that it is possible for the rod and bead

280


to remain at rest relative to the rotating wires, with the bead at the middle point of the rod, if sin 0 = 1; that this position of relative equilibrium is unstable; but that it becomes stable if the bead is fixed to the rod at its mid-point. 10. A smooth circular tube of mass M and radius a contains a particle of mass m and is free to rotate in its own plane about a point 0 distant c from the centre. If q, is the angle turned through by the tube, 0 the angle described by the particle relatively to the tube, obtain the kinetic energy in the form

[{M(c2 + a2) + m (c2 + 2ca cos0 + a2))

+ 2ma (a + c cos 0) 643 + ma202].

If no forces act, and if the moment of momentum about 0, the centre of rotation, is zero, prove that the particle must make complete revolutions in the tube. In the case when c = a, show that in a complete revolution of the particle the tube turns through the angle n[l — (M/(M + 2m))1/.] in the opposite direction.

CHAPTER VII

STABILITY OF MOTION 7:1 Introduction In earlier chapters we investigated the possibility of steady motion of various systems, e.g., the spherical pendulum or a spinning top. First we must point out that whether a motion can be called steady depends on the coordinate system used. For example, a particle moving uniformly around a circle can be specified by cartesian coordinates

x = a coswt, y = a sinwt,

(7.1)

or by polar coordinates r = a , d = w.

(7.2)

In the description (7.1) neither x nor y are constant nor are their derivatives. On the other hand, in (7.2) the coordinate r and the velocity 0 are both constant. Hence (7.2) describes a steady motion, whereas (7.1) does not. In general, a steady motion is one for which a suitable coordinate system would have either the coordinate or the corresponding velocity constant for all the coordinates in use. In the case of the spherical pendulum, when specified by the polar angles 0, v, the steady motion is given by 0 = a (coordinate constant), = Q (velocity constant). For the spinning top, specified by three Euler's angles, the steady motion is given by 0 = , //) =52

,

a), = 9.9 + //, cos0 =

+ .S2 cosoc = n

Any motion which persists in practice must, almost certainly, be a steady motion as discussed above; a very important related question concerns the stability of such a motion. If the steady motion is disturbed slightly, will the resulting motion deviate appreciably from the steady motion? Only those motions which are stable, i.e., in which subsequent 10 CMES 281

282


deviations remain small, will persist in practice because of the slight, but inevitable, disturbances which must occur. The normal motion of a train, an aircraft, a piece of rotating machinery, must be a stable motion if it is to be safe. This idea of stability was introduced in Vol. III Chapter IX and developed there for simple systems. In this chapter we consider certain cases of steady motion, and in Chapter IX we consider the question of stability of equilibrium and small oscillations about positions of stable equilibrium.

7:2 Steady motion with two degrees of freedom The general method, which can be used in all investigations of stability, is to use the equations of motion: (a) to give the conditions which must be satisfied by the steady motion; (b) to obtain, correct to the first order, equations governing the variation of the coordinates or velocities from the steady motion. Provided that the variations arising from an arbitrary disturbance remain small, which means that they must be harmonic variations with small amplitude, the steady state is stable. The equations obtained in (b) can then be used to obtain the periods of the oscillations which take place near the steady state. It is important to ensure that the assumed variations from the steady motion are quite general, otherwise we are not investigating the effect of an arbitrary disturbance. In general, the number of quantities necessary to specify the deviation from the steady state is equal to the number of degrees of freedom in the system. Examples. (i) A particle of mass m can slide freely on a light wire bent into the shape of the curve a3y = x4which can rotate about the y-axis, which is vertically upwards. If a steady motion takes place with the particle at x = a, find the angular velocity Q of the wire. The particle is slightly disturbed from its steady motion at x = a. Prove that the period of oscillation about this position is 27E (17 a/8 g)1/2 if the wire is constrained to rotate with the steady angular velocity D. Find the period of the oscillation if the wire is not constrained but moves freely about the y-axis. We need two generalised coordinates to specify this system: the position of the particle on the wire is given by x = q, y = q4/a3; the orientation of the wire is given by an azimuthal angle v.

§ 7: 2]

283

STABILITY OF MOTION

Resolved in two directions in the plane and in a direction perpendicular to the plane of the wire the particle velocity has resolutes {g 4g34/a3 Di)}.

T = 1772,[1 .•, L

16g6/a6] 42 + img2V,2,

16g6 1 977,0 — [(1 + —)+ g2 /P2 =2 a6

V = nigg4/a3 • 2gg4

•

The corresponding equations of motion are 16g6

[ 48g5 42 I. a

c+', [4 (1± a° )

—

at

4gq3] a3

0,

(g2 ) = o.

(1) (2)

When the particle is in a steady state (equilibrium relative to the wire) q = a,

q = 0, ?,b = D. Eqn. (2) is satisfied identically, and eqn. (1) gives aQ2 = 4g, i.e., ‘22 = 4g/a.

(3)

(a) When the wire is constrained to rotate, only one coordinate is required to describe the system and the equation of motion is

6. dt [g

We put q = a +

16,6

a6

[

48,542

a6

+ qQ2

4g q3 1 —. a3

where e is small of the first order. Then 17e— [(a + e) 02 — 4g(1 + 3e/a)] = 0.

The zero-order terms cancel out because of condition (3), and we neglect the 2nd order term 48g642/a4. .•. 17 — (e la) (4 — 12) = 0, i.e., 8ge/(17a) = 0. Hence the period of oscillation is 27r y(17a/8g). (b) When the wire is free to rotate, we need two coordinates q, used. We write q = a e, /i) = + in the equations of motion (1), (2) and obtain

— [(a + e) (.(22+ 2th.)) — 4g (1 + 3 e/a)] = 0 , q2W + 2q0,b = qact + 9j + 2,S2] = 0. 10*

as originally

284


Correct to the first order in $, n, [and after use of (3)] these equations reduce to — [122 $ + 2aQi) — 12g$/a] = 0, aij + 2i52 = 0.

(4) (5)

We integrate (5) to give a~j + 2$S2 =

where 13 is a constant determined by the initial arbitrary disturbance. Eliminating from eqn. (4) we find 17$+ (3Q2 + 12g/a) =

24g$/a = —2452.

Hence the motion in $ is simple harmonic motion, with period 2x y(17a/24g), whose centre is not at $ = 0 but at $ = — cti3Q1(24g). (ii) A conservative dynamical system with two degrees of freedom has kinetic and potential energies given by

T = ia02 + F (0) 4)2 , V = V(0) where a is a positive constant, F and V are functions of 0 only, and F is positive for all values of 0 concerned. If steady motion is possible with 0 = a, = co, where a, ai are constants, prove that a must exist as a solution of the equation

0)2 E' (a) = V' (a). Prove that F (0) cp is constant in any motion of the system and that, if the steady motion is disturbed without altering the value of this constant,

Et

F' 2 { F

F" +

a0+ 2

co2 =0 V ' 10 ,

to the first order in 0 where 0 = a + e. Illustrate these results by considering the motion under gravity of a particle suspended from a fixed point by a light string of length 1 ("spherical pendulum") showing in particular that the period of a small oscillation about the steady motion in which the string makes angle a with the vertical is 2

/ cos a

F/2

(1 + 3 cos2 a)g

The Lagrangian function is

L = -ictO2 + F (0) 492 — V(0). Hence the equations of motion are

a& —

(0) 4)2 + V' (61) = 0, —[I'(0 ) 4)] = 0. dt

(1)

§ 7: 2]

285

STABILITY OF MOTION

For a steady motion — co2 F' (a) + V' (oc) = 0, as required, the second equation of motion being satisfied identically. It also follows from the second of eqns. (1) that F(0) rp = constant. (2) For the disturbed motion we write 0 = a + 0, = co + g. Since the constant value of F(0) 42 is unaltered,

w F (oc) = (co + g)[F (oc) + O F' (a)], .• 0 = F (cc) +

(a).

The first of the equations of motion gives a0 — [F' (cc) + F" (cc)] (w2+ 2cog) + [V' (a) + 0 V" (a)] = 0. The condition co2 F' (cc) = V' (cc) removes the zero-order terms and leaves, after substituting t = — a 0 + 0)20[— F" + 2 F' 2/F + V"/w2] = 0. Substitution of co2 = V'/F+' in the last term of the bracket leads to the stated result. For the spherical pendulum we have (see § 3:4)

T = im12(02+ 422sin20), V = — mgl cos0 We replace "a" of the above by m12and identify

/7 (0) =1-m/2 sine, V(0) = —mgl cos0, = m12 (cos2 oc — sin2oc),

F' = m12sing cosa, V' = mgl sing,

V" = mg1 cosa.

.• co2= g/(1 cosa). The integral F(0) 99 = constant becomes, in this case, (j9 sin2 0 = constant. This is the integral of angular momentum about the vertical, eqn. (3.13). The equation of motion becomes

m12 0 + [4m12 costa— m/2 (cos2 oc — sin2 oc) + m12costa] + (1 + 3 cos2a) g 0

1 coscx This gives the required period.

0.

g 1 cosa

o,

286


Exercises 7:2 1. A light thin wire is bent into a circle of radius a and can rotate about its vertical diameter. A smooth bead can slide freely along the wire. Show that the bead can remain at rest relative to the wire if cos 0 = g/(aw2), where a is the angle between the downward vertical and the radius through the bead, g is the gravitational acceleration, and w > y(g/a) is the angular velocity of the wire. If the bead is slightly displaced from this position, find the period of small oscillations (i) if the circle is constrained to rotate with constant angular velocity w; (ii) if the rotation of the circle is unconstrained. 2. A thin rod OA of length 2a moves freely about the end 0 which is fixed. If 0 is the angle between OA and the downward vertical OZ, and 92 is the angle between the plane AOZ and a fixed plane through 0 Z, write down the velocity of a point P of the rod distant x from 0, and calculate the kinetic energy of the rod. In a steady motion in which 0 = constant = a, show that = constant = w = (3g sec oc/4a)1/2, and that the period of small oscillations about this steady motion is 2 ni[co (1 + 3 cost 4/2] 3. A mechanical system with two degrees of freedom is defined by the coordinates 0 and 1p; the kinetic energy of the system is of the form 2T = A02 + 142, where A and I, as well as the potential energy, V, of the system are functions of 0 only. If for steady motion 0 = 00and 1,i) = w, prove that w is determined by the equation 0/6 = 2 V6, where accents denote differentiation with respect to 0, and the suffix 0 refers to the values of the functions for 0 = 00 . Prove that if the steady motion is slightly disturbed so that 0 = 00 + s, = CO + where 8 and ?) are small, the Lagrangian equations of motion are approximately — CO2 — l'0 00171 Vio' e = 0, /0.n + /0'co = 0. Show that the system is stable when rotating at the speed a) provided rot/r, > (44' — 2r0 2)/(44).

7:3 The stability of free rotation of a rigid body We showed in § 5:6 that the only steady motions possible in the general case (A, B, C all different) of free rotation are rotations about the three principal axes. The diagram in Fig. 62 contains the answer to the question of the stability of these rotations. After a small disturbance the instantaneous axis is close to the principal axis and its intersection with the surface in Fig. 62 follows one of the curves passing near the end of the principal axis. When the axis of steady rotation was an axis of greatest, or of least, moment of inertia these curves are small closed

§ 7 : 3]

STABILITY OF MOTION

287

curves encircling the end of the principal axis. Hence, a steady rotation about either of these principal axes is stable, for the instantaneous axis never moves far from its original position. On the other hand, when the disturbed position of the instantaneous axis is near the principal axis of mean moment then Fig. 62 shows that its path eventually moves far from the original principal axis. However, in order to find the periods of the oscillations about the stable positions we proceed as follows. Suppose first that no two of A, B, C are equal and that the body is given a small disturbance so that w1, the first resolute of the angular velocity, differs only slightly from [2, the original steady angular velocity. We suppose also that w2 , w3are small. Then, correct to the first order, eqns. (5.35) become Ad), = 0, i.e., o), = S2 + s, where e is a small constant, and

B62— (C — A ) Dco3 = 0, Cth, — (A — B) S2co2 = 0 • (Note that we assume variations in all three resolutes of co corresponding to the three degrees of freedom of the body.) Eliminating w3we obtain

iii w+

— C) (A — B) 920)2___. 0 . BC

(7.3)

Hence, iff, (if and only if) (A — C) (A — B) > 0, i.e., iff the axis of the steady rotation is the axis of greatest or of least moment, is the rotation about the principal axis stable. The period of the oscillation is 2n

,f2

BC (A — C)(A — B)1 •

(7.4)

If the body is rotating about an axis of symmetry, which we suppose to be the first coordinate axis, so that B = C, then in place of eqn. (7.3) we have (.62 [ (A — B) i 2 = 0 , (7.5) and the motion is always stable unless A = B. If A = B = C, Euler's equations give 61 = 62 = 63 = 0 ,

so that co/ = constant, co2 = constant, co3= constant. Since the angular velocity vector remains fixed in the body, and in space, close to its original position, the motion is stable.

288


Exercises 7:3 1. If A > B > C and there is no external couple, show that the body can rotate steadily with uniform angular velocity Q about the axis of greatest moment of inertia, and that if slightly disturbed its instantaneous axis will make small oscillations of period 27r "if BC

.S2 II (A — B) (A —

'

2. A uniform circular disc is rotating freely in space with angular speed n about an axis perpendicular to its plane. Show that, if it is slightly disturbed, the period of oscillation about the mean position of its plane is n/n.

7:4 The stability of a top Using the notation of § 5:3 and the expressions obtained there for the kinetic and potential energies of a spinning top we can write down the Lagrangian function in terms of Euler's angles

L = zA(B2

2 sin20 )

2C(0

tp coso)2 — 1V1g1 cos° .

The three equations of motion are

AO — Alp sin0 cos0 d —t [Ay, sin20 d

CnV, sin0

M gl sin0 = 0,

(7.6)

Cn cosO] = A sin2 0 dt

2 .4 Viel sin0 cos° — CnO sin° = 0, (7.7) ()

ii,cosO) = 0.

(7.8)

We have used the integral of (7.8), viz., (i9 cos° = n, to simplify the earlier equations. After division of eqn. (7.7) by sin° these equations are the same as those obtained in § 5:3 eqns. (5.16-18). The condition for a steady precession, 0 = S2 is ,

AD2 cosoc — CnS2 M gl = 0,

(7.9)

eqns. (7.7), (7.8) being satisfied identically. In an arbitrary disturbance the total energy, the angular momentum K = A sin20 Cn cos0 about the vertical, and the spin n all vary slightly from their steady values. Hence we write, for the disturbed motion

0 —

y,

tp = S2 + .

(7.10)

The altered value of n, by eqn. (7.8), does not vary subsequently so it is customary to use the same symbol n, although the system, having

§7: 4]

289

STABILITY OF MOTION

three degrees of freedom, strictly needs three quantities to denote the variations from the steady motion. Substituting the values (7.10) into eqn. (7.6) gives

A g - A (S22 2S2 [sing cosa y (cos2oc — sin2a)] -4- Cn(S2 0 (sin g y cosa) — 1141(sina y cosa) = 0, i.e., correct to the first order,

Ay y AS22sin2a

sincx [Cn — 2AS2 cosa] = 0,

(7.11)

eqn. (7.9) having been used to eliminate the zero order terms and to simplify the coefficient of y. Substituting (7.10) into (7.7) gives, correct to the first order,

Az sin2a

2AS2y sing cosa — Cny sing = 0,

from which we obtain, on integrating w.r. to t, (slim 4= 0),

Az sing = (Cn — 2AS2 cosa) y + j.

(7.12)

The constant of integration p must, of course, be small. The constant is zero only if the disturbance leaves the angular momentum K about the vertical unaltered; this can be achieved by a blow in the vertical plane containing the axis of symmetry. For an arbitrary disturbance, then, p + 0. Substituting from (7.12) into (7.11) gives A2 y + [A2-22 sin2a

(Cn 2AS2 cosoc)9y

= —P(Cn — 2AS2 cosa).

(7.13)

Since the coefficient of y, being the sum of two squares, is positive the steady motions, both the fast and the slow precessions, are stable. The occurrence of the constant term on the r.h. side of (7.13) implies that the oscillations take place, not about y = 0, but about another steady motion very close to 0 = a. The period of the oscillation is T—

V[D2 sin2cx

2n (CnIA — 2S2 cosa)2]

(7.14)

The path traced out on the sphere of Fig. 79, see p. 290, by the axis of the top is a narrow version of Fig. 54(i) or Fig. 54 (ii). Examples. (i) A uniform solid sphere, of mass m and radius a, can move freely about a fixed point 0 on its surface and has a spin 5 y (g la) about the diameter 0 C. ]0 a

CrNIES

290


Prove that the sphere can precess steadily with O C inclined to the upward vertical at an angle cos-1(3/5), and find the two possible precessional angular velocities. Find the periods of small oscillations about these states of steady precession. For this system A = 5ma2, C = 5ma2, and n = 5 y(gla). The condition (7.9) for steady precession with cos a = s gives 7 _ ma2122

3

2 ma2 s^ . 5

7)

nga = 0,

2102— 500 1/(g/a) + 25 (g/a) = 0,

79.

FIG.

whose roots are the two possible precessional angular velocities

Di =

f22 =3 i/(a) •

These values lead to the periods, using (7.14),

Ti

27r

ir 25 16

V [W.-2-5T,— V[ 295

/2 5

2

7

a

1 ]

27r

1 25 6 g + (, 7 2 .5 _ .

)2 1 .j

7n -Ili a \ 2 II 2g 21n 11( 59 % ). 2 V

[The student is advised to calculate these periods from first principles using the methods of § 7:4.1 (ii) We investigate the stability of the vertical top. The discussion given in the text excluded the case sin a = 0. We saw in § 5:4 that a steady motion with 6 = 0 is possible only if C2 n2 > 4AMgl. The equations

§ 7: 4]

STABILITY OF MOTION

291

of motion of the top for small values of 0 are

A0 — Alp20 Cnzp0 — MOO = 0, Aij,0 2A0 — CnO

0,

(1) (2)

d dt (92 + V)

1).

(3)

If we multiply eqn. (2) by 0 we obtain d 1 — (A002— — Cn02) = 0. dt 2

1 Ay) = Cn

s3

2 02

(4)

We have written the constant of integration as s3, where s is of the first order, so that the last term is of the first order, Cn being the zero-order approximation to A 0. This implies that when 0 is increased by a first order increment the angular momentum K is increased (from zero) by a quantity of the third order. However, since the last three terms in (1) are all of the first order because of the factor 0, we use only the zero-order approximations in the other factors, and so, correct to the first order, (1) can be written C2 n2

AO — 0 (—

C2 n2

—

4A 2A

AO — (Mg1

C2 n2

4A

MO) = 0, 10= 0.

Hence, for stability the coefficient of 0 must be positive, i.e., C2n2 > 4AMgl. But this is a necessary condition for the existence of a steady motion with 0 = 0, and so such a motion is always stable, the period of the oscillations being

4nA 11(C2n2— 4AMg1)

Exercises 7:4 1. A peg-top having an axis of symmetry spins with its point 0 on a smooth horizontal plane. M is the mass of the top, A, A, C the principal moments of inertia at its centre of mass G and 0G is of length h. Show that the condition that there may be steady precession of the axis of the top about the vertical in the position in which it is inclined at an angle a to the vertical is C2 n2 > 4A Mgh cosoc, where n is the spin of the top about its axis. 10a*

[see over]

292


Show that the period of small oscillations about such a position of steady precession is

2-1/1

A (A M h2 sins) A2S22 sin2 oc — 4AM gh cosoc

C2 n2

where is the rate of precession. 2. Two circular hoops, each of radius a and made of the same uniform thin wire, are rigidly fastened together at the ends of a common diameter AB so that their planes are perpendicular. The system can turn freely about the point A, which is fixed. Show that, when it spins steadily with B vertically above A, the motion is stable if the angular velocity of spin (n) is greater than 1/(28g/a). If this condition is satisfied, show that the period of a small oscillation about this state of steady motion is 1441/(n2 — 289/a). 3. A symmetrical top of mass M has principal moments of inertia A, A, C at its peg, which is fixed, while the distance of its centre of mass from the peg is h. Show that if properly started it can precess steadily round the vertical with its own axis horizontal at angular rate Mgh/Cn, where n is the spin of the top. Obtain the period of small oscillations about this motion, and show that it will be longest when the spin is equal to (MghA/C2)'/'. 4. A symmetrical top of mass M and centre of mass G moves under gravity about a fixed point 0 on its axis, where OG = h, and the principal moments of inertia at 0 are A, A, C. An additional force is applied at a point on 0 G and perpendicular to the vertical through 0 G, such that this plane has a constant spin w about the upward vertical Oz. Show that the component of spin n of the top about OG is constant, and obtain the equation for the motion in 0, the inclination of OG to Oz. Show that, if steady motion is possible in which 0 is constant and sin * 0, then ICnco — Mghl < Aco2. If the steady motion in which 0 = a is slightly disturbed, so that w and n are unaffected, show that small oscillations about the steady motion will occur, with the period 27c/(a) sin a).

7:5 The gyro-compass When a flywheel is mounted in gimbals as shown in Fig. 80, its axis is free to turn in any direction about its centre of mass, which is a fixed point. Hence the motion of the flywheel exhibits the motion of a freely rotating body with kinetic symmetry (apart from the effect of the inertia of the gimbal rings). If the centre of mass is stationary in a Newtonian frame and the wheel has a large spin about its axle (axis of symmetry), the angular momentum is a vector lying along the axle; by the results of § 7:3, this direction remains constant and is stable.

§ 7: 5]

STABILITY OF MOTION

293

If the gimbals are now made to move slowly, the wheel must receive small disturbances but, because the rotation is stable, remains pointing in the same direction. The system can be made to move relative to a Newtonian frame by mounting the apparatus in a vehicle, ship or aircraft, or even by fixing the support relative to the earth's surface. In such cases the axle should remain pointing in the same direction and consequently can serve as a compass.

In practice the gyro-compass does not consist of a freely mounted flywheel but of one whose axle is constrained to move in a horizontal plane relative to the earth (or in a vertical plane, for certain purposes). We now investigate the stability of this arrangement. If the period of oscillation about the steady position is large the compass will take a long time to settle; hence the machine must be designed to reduce this period as much as possible and to damp out the oscillations. The compass consists of the gimbals of Fig. 80 without the inner ring but with the axle mounted along ER. The apparatus is free to turn about AB. The mass of the frame is assumed to be negligible. In Fig. 81 the frame Oxyz has Ox (not shown in the diagram) due E, Oy due N, and Oz vertically upwards. Relative to this frame the position of the wheel is given by the angle p, in the horizontal plane, between the meridian and the axle W3; the axis Wigoes through a point H fixed in the wheel and makes an angle q2 with the vertical; 0completes the r.h. set and 2

294


is also fixed in the wheel. The apparatus is set up at a point on the earth's surface in latitude A, and the angular velocity of the earth is D. The forces acting on the wheel are reduced to a force at 0 together with a couple T about 0. Using the frame 0 ei e2 e, for resolutions

r = {L1 L, 0}.

There is no resolute L, since the wheel turns freely on its axle. In addition, r can have no resolute along the vertical dircetion k, since the frame is free to rotate about AB. Since k = {cosq9 — sin.q9 0},

I' . k = L1 cos92 — L2singe

= 0.

We now find the generalised components of r corresponding to the generalised coordinates 99, v. In a virtual displacement given by Sy, the wheel undergoes a rotation specified by the resolutes {6vCOS99 -(5v sing) So in the frame 0 ei e,e,. Hence the virtual work is = L16v cos9) — L,Sv simp. = Licoscp —

L, since = 0 , Qro = 0 .

§ 7: 5]

STABILITY OF MOTION

295

Referred to 0 1 2 3 the angular velocity of the earth (i.e., of the frame Oxyz) has resolutes {2 sin). cosv — S2 cos), sinv sirup —

2 sin A sinv — S2 cosA sinv cosp 2 cosA cosv}

Referred to 0 eie2 e3 the wheel has angular velocity relative to Oxyz cos 99

—

sing) VI.

Hence the angular velocity of the wheel is {co, cot oh}, where co, = cosv + 2 (sinA cosv — cosA sinv sinv), cot = — sinv — 2(sinA sinv + cosA sinv cosv), co3 = 97 + 2 cosA cosy). Therefore the kinetic energy is

T = z A (co + + 1 C n2v 2Q, Ip sin A) A (02 + 22 sin22 22 cos2A i iC(cb

Q cos). cosv)2 .

Lagrange's equations of motion are —

dt

[A (ti + 2 sink)] — —

[A22cos2 A sins cosy — C(V + 2 cosA cosy) 2 cosA sin v] = L, sincp — L2 COST = 0,

dt

[C(V + Q cosA cosy)] = 0, i.e., V + 2 cosA cosv = n = constant.

In the first equation of motion we may neglect terms containing 22.

C n S2 cosA sine = 0. (7.15) This equation shows that the positions of equilibrium relative to the earth are given by v = 0, r. By comparing eqn. (7.15) with the equation of motion of a simple pendulum we see that v = 0 is stable and v = is unstable. The period of oscillation is 2

A

nil C n S2 cos A)

296


A gyro-compass in a ship or aircraft is a much more complicated machine than that of the simple example just discussed. In the example we neglected the mass of the frame; also the vehicle usually does not remain fixed relative to the earth but pitches and rolls in addition to moving over the earth's surface. Because of the former movements the flywheel is mounted in complete gimbals and a weight is added so that the centre of mass of the system is below 0. This arrangement keeps the plane in which DC rotates effectively horizontal despite pitching and rolling. The machine has to be designed so that the oscillations induced by pitching and rolling are kept small and are damped out quickly. However, the simple system we have considered demonstrates the essential property that the axis of the flywheel rests in stable equilibrium pointing N. Exercises 7:5 1. A gyroscopic compass is a gyroscope spinning about its axis, which is free to turn in a horizontal plane. Assuming that the centre of the gyroscope is at rest relative to the rotating earth, show that the axis of the gyroscope can remain at rest relative to the earth if pointing north. The spin of the gyroscope about its axis is n, the angular velocity of the earth is co, and n > co. Show that, if the axis is slightly disturbed it will oscillate about the true north with period approximately equal to 27c(AICnco cos 2.)1/2, where A, C are the transverse and axial moments of inertia of the gyroscope, and A. is the latitude. 2. The upper end of the axis of a gyrostat is compelled to describe a horizontal circle of radius c with uniform angular velocity co. Prove that, if oc is the inclination of the axis to the vertical when the motion is steady, (A sin g Mac) co2 =tan a (Cnco

M g a) ,

where a is the distance of the centre of gravity from the end of the axis, M the mass, C the moment of inertia about the axis of symmetry, A the moment about a perpendicular through the end of the axis and n the angular velocity of the fly-wheel of the gyrostat about its axis. Prove that the periods of the small oscillations about the state of steady motion are determined by nip where p satisfies (A sin oc (1)2— co2 sin2 oc) — Mco2ac) (Ap2 sinoc — Mw2ac) = p2 (A co sin2oc — Cm sinoc)2 . 3. A gyroscopic compass, consisting essentially of a heavy symmetrical flywheel whose axis is constrained so that it can move only in a horizontal plane, is mounted in an aeroplane that is flying due north with constant speed. Right-handed axes are chosen at the centre of the flywheel with 01 vertical, 02 horizontal, and 03

§ 7: 6]

STABILITY OF MOTION

297

along the axis; q) is the deviation of 03 to the east of north, A is the latitude, and co the angular velocity of the Earth. Show that the motion of the axis is given by

A45 + Cn(co cosA sing, + A cosyri) = 0, where A, A, C are the principal moments of inertia at 0 of the compass and n is its spin, and squares and products of co and A are neglected. Determine the position of relative equilibrium and the period of small oscillations about it.

7:6 The stability of a rolling wheel Everyone is familiar with the general character of the motion of a rolling wheel, or disc, or hoop. Such a wheel can roll steadily in a straight line with its plane vertical; it can roll in a circle with its plane inclined, either in a large circle, or in a circle very little larger than its own circumference as in the motion of a coin just before it settles flat on a table. Again a disc can spin about a vertical diameter. In this section we investigate the stability of such motions and introduce a geometrical method of representing the stability criterion. In order to specify the position of a disc, of mass m and radius a, we need two angles 0, y and the coordinates of the point of contact D. The situation is shown in Fig. 82. The axis of the disc E e3 makes an angle 0 with the vertical, and the vertical plane containing this axis and the radius of contact ED makes the angle y with the direction of Ox. The frame Oxyz is fixed while the frame 0 ei e2 e3 moves as the wheel

298


moves but is not fixed in the disc. The coefficients of inertia A, A, C have their usual meanings with reference to 0 2a, and we assume that the "disc" may be any figure with the appropriate symmetry, and, while having a circle for its circumference, is not necessarily a plane figure inside the circumference (e.g. a dinner plate, or a discus). When the coordinates {xi, yD 0) of the point of contact D are given, the position of the disc is specified as far as is necessary for most purposes. When the disc rolls on the horizontal plane the constraints are non-holonomic so that the methods of Chapters IV and V using moving axes are just as convenient (if not more so) than the use of Lagrange's equations for obtaining the equations of motion. The derivation of these equations is given in example (ii) below. Here we quote the results:

Ali, sin° + 2,4,00 cos° — C0o)3 = 0, (7.16) w3 (C ma2) — maY sin0 = 0, (7.17) (A + ma2) U — A IP2 sin° cos° (C ma2) co3V, sin° mya cos° = 0. (7.18) In these equations co3stands for the spin of the disc about its axis. First we look for the conditions which must be satisfied for a steady motion. If 0 = a equations (7.16) and (7.17) imply that = S2, co3 = n,

(7.19)

where oc, S2 n are constants. Then eqn. (7.18) gives ,

AS22sing cosy — (C

ma2) Stn sing — mga cosa

0. (7.20)

This corresponds to the condition for a steady precession of a top, showing that, when n, oc are given, there are, in general, two possible values for ,S2, since eqn. (7.20) always has two real roots. In the faster of these two precessions the point of contact D describes a small circle on the horizontal plane. In the slower of these two precessions the point of contact describes a comparatively large circle, this motion being more akin to rolling along a straight line. The investigation of stability follows lines familiar by now (but involves some rather heavy manipulations). We substitute tp = S2 + E, 0 = 04 +

w3 = n +

(7.21)

into the equations of motion (7.16-18), writing them correct to the first order in e, n , and their derivatives. Eqns. (7.16) and (7.17) give equa-

c

§ 7: 6]

299

STABILITY OF MOTION

tions which can be integrated. From (7.16)

A t sing + 1)(2 A S2 cosa — C n) = 0. A t sing = (C n — 2 AS2 cosa) + /3,

(7.22)

where 3is a constant of integration. Eqn. (7.17) leads to

sinoc = 0. (C ma2) — ma2 .*. (C ma2) = nma2S2 sinoc y,

(7.23)

where y is another constant of integration. Eqn. (7.18) becomes

(A + ma2) ij — A (S22+ 2S 2 t) [sinoc cosa mat) (n + 4) (S2 + E) (sing

(C

(costa — sin2oc)]

cosa) + mga (cosoc — sinoc) = 0.

On our using eqns. (7.20), (7.22), (7.23) this reduces to

p2r1 = constant,

(7.24)

where

A (A + mat) p2 = .Q2 [3 A2 costa + A (A + ma2) sin2oc] n2 C (C ma2) — Amg a sinoc. ma2) cosoc - S2 nA (3 C Hence we obtain the condition for stability as s22 [3 A2 cost a + A (A + ma2) sin2oc] —

— S2 nA (3C + ma2 ) cosec n2 C (C ma2) > Amy a sinoc. (7.25) To avoid obscuring the principles with excessive arithmetic and manipulations we now illustrate the further investigation for the case of a ring moving in steady motion with its plane inclined to the horizontal. In this case

A = 2ma2 , C = ma2, eqn. (7.20) becomes .S22 sing cosa — 4S2n sing = 2 (gla) cosa,

(7.26)

and the condition (7.25) for stability becomes 3Q2— 8S2n cosa

8n2 > 2(g I a) sinoc

(7.27)

If we regard {Q n} as the rectangular coordinates of a representative point in a plane, eqn. (7.26) implies that the point must lie on a hyperbola, and eqn. (7.27) implies that this point lies outside the ellipse 3522— 8S2n cosoc

8n2 = 2(g I a) sines.

(7.28)

300


(The two values for the precession Q corresponding to a given n are shown by the points in which a horizontal line intersects the hyperbola of Fig. 83.) If the ellipse is situated in position 1 in Fig. 83, only those values of S2, n lying on the dotted portions of the hyperbola correspond to stable steady motions. If the ellipse is in position 2 in Fig. 83, all possible steady motions are stable. The distinction between these two cases is

that in position 1 there are four real intersections (of the ellipse and hyperbola) and in position 2 there are none. We investigate whether the curves (7.26) and (7.28) intersect in real points by using polar coordinates to represent the point {S2 n} and write S2 = r cos0, n

r sine .

Then the angles 01 , 02 , 03 , 04giving the intersections (only 02is shown in Fig. 83) are obtained by eliminating r from r2sina (cosoc cos2 0 — 4 sin0 cos0) = 2 (g la) cosa , r2(3 cos20 — 8 cosa cos° sin0 + 8 sin2 0) = 2 (g la) sina cos20 cosa (3 costa + 2 sin2a) — 4 cos0 sin° (2 costa — sin2a) + 8 sin20 cosA' = 0. (7.29)

§7 : 6]

STABILITY OF MOTION

301

The ellipse is in position 2 if eqn. (7.29) has no real solutions for 0, i.e.. 4(2 cos2oc — sin2oc)2 < 8 cos2oc(3 costa + 2 sin2 (x) i.e., tan4cx — 8 tan2oc — 2 < 0. Since tan2oc > 0, this gives the condition that tan2 oc < 4 + 3 f2.

(7.30)

If this condition (7.30) is satisfied by cc, the inclination of the axis of the disc to the vertical, then any steady motion with values of Q, n, satisfying eqn. (7.26) is stable. If condition (7.30) is not satisfied (e.g. when the plane of the disc is vertical) a steady motion is stable only if Q, n correspond to points on the dotted portions of the hyperbola in Fig. 83. Examples. (i) We give now an example of one of the two extreme cases in which this stability investigation involves reasonably light manipulations. (We leave the other case to the reader: see Exercises 7:6, Nos. 2, 3). We consider the case of a disc spinning about a vertical diameter. For this motion a = n/2, n = 0, = S2, these values satisfying eqns. (7.16-18) identically. We put = S2 + , n small 0 = n/2 + y, into the equations of motion (7.16)-(7.18) and obtain

As = 0, (C + ma2);?, = ma2S2g. These give on integration

Ae =

(C + m a2) n = y m a212 + y.

Hence eqn. (7.18) becomes

(A + ma2)

— AS-22(— y) +

g

(122

(C + ma2) nS2 — mgay = 0,

mg mga — VS2 Y— A + ma2 A + ma2 )

We see, therefore, that this motion is stable if D2 >

mga A + ma2

302


For the special cases, we have: a ring: 3

A + ma2 = 2 mat,

122 > 2g ; 3a

a uniform disc: 5 4g 2, Q2> — 4 ma 5a

A + ma2 —

In terms of our representative diagram the "steady motion" hyperbola is

Stn = 0, i.e., the degenerate case of the two axes. The limits of stability curve is A(A + ma2) Q2+ C (C + ma2) n2 = Am ga , which is an ellipse. We are concerned only with motions for which n = 0; hence the representative points must lie on the .Q-axis beyond the ends of the appropriate axis of the ellipse, i.e., .Q2>

A raga A(A + ma2)

the result already obtained. (The other section of the steady motion curve Q = 0 refers to the disc rolling in a straight line which we leave to the reader in Exercises 7:6, Nos. 2, 3.) (ii) We now obtain the equations of motion (7.16-18). We refer to Fig. 82 using the axes Eel 23for resolutions.

Motion of the frame Velocity of E: vo = V. Angular velocity: 0 = {— y,sin B 6 l'pcos0}.

Motion of the body The angular velocity of the body is co = {— sing B we). Since D is the point of contact, the condition for rolling gives V = co x DE =

sin° 0 co3) x {— a 0 0} = {0 — aco3aO}.

Hence p = m{0 —aw3 a6}. Also

h(E) = {— A sine AO C co,}, The forces At E, {0 0 0}, the weight {mg sin 0 0 — mg cos 0} ; at D, {a 0 0}, the reaction {X Y Z}. F = {X + mg sin° Y Z —mg cosh),

r(E) = {0 — a Z a Y} .

§ 7: 6]

303

STABILITY OF MOTION

The equations of motion Linear motion: O a+ 0 x p = R. .•. mae2 + mapco, cos° = X + mg sine,

(1)

—maths + malpe sine = Y,

(2)

ma + maco3lp sin° = Z — mg cos°. Angular motion: h (E) + 0 x h(E) = (E).

(3)

at

— A(yi sine + A6— Cco3

cos0) + C co,0 — A'pO cose = 0,

sin0 cos° + C co,lp sin° = — a Z , =

a

Y.

(4) (5) (6)

We eliminate Y between (6) and (2); Z between (5) and (3) and, with (4), obtain the three equations,

AV) sin° + 2.A V,6 cos° — COco3 = 0, (C + ma2) 63

ma2zPO

sine = 0,

(A + ma2) 0 — A)i2 sine cos° + (C + ma2) co3V) sin° + mga cos0 = 0. These are the equations quoted in (7.16-18). Exercises 7:6 1. Obtain a diagram for the stability of steady motion of a disc rolling steadily with its plane at an angle a to the horizontal, and give the limits of a for which all possible steady motions are stable. 2. Prove directly from the equations of motion that the disc of § 7:6 when rolling steadily along a straight line is stable if

>

A mga C (C + ma2)

3. Prove that the periods of oscillation of the plane of a rolling ring, and a uniform disc are respectively when the plane is vertical, \ 2grf(8an 3a 2 — 2g )'

2n1/(

5a ) 12an2 — 4g •

4. A heavy uniform sphere of radius a, that is spinning with angular velocity n about its vertical diameter, is gently placed on the highest point of a perfectly rough fixed sphere of radius c — a. Show that the arrangement is stable if [see over] n2 > 35cg/a2.

304


Show also that when the arrangement is unstable the two spheres remain permanently in contact provided n2 > 945 cg/34a2. Explain why the angular momentum of the system about the vertical diameter of the fixed sphere is not constant. Miscellaneous Exercises VII

1. Oxyz is a rectangular system of axes of which Ox, Oy rotate with constant angular velocity co about Oz. Prove that the resolutes, parallel to the instantaneous positions of Ox, Oy respectively, of the acceleration of a point P (x, y, 0) are — 2cog — co2 x and

g + 2co± — co2 y.

A bead is strung on a smooth wire in the shape of an ellipse of eccentricity e which rotates in its own plane about a focus with constant angular velocity co. Show that the bead can remain in stable relative equilibrium at one end of the major axis and that the period of small oscillations about this position is

27r ( 1 — e e 2. A symmetrical top of mass m has principal moments of inertia A, A, and C at its fixed peg 0. Its centre of gravity is at a distance h from 0. Initially the top is spinning with angular velocity n about its axis of symmetry, which is vertical. Its motion is then disturbed by a small impulsive couple of magnitude G about a horizontal axis through 0. Show that if C2 n2 > 4mghA its axis of symmetry never makes an angle greater than 2G

11(C2n2— 4mghA) with the vertical. [For the effect of the impulsive couple see § 8: 1. 3. A spinning top is free to turn about its peg 0 which is fixed. The distance from 0 to the centre of gravity G is 1, the mass M, the moments of inertia at 0 are A, A, C, and the (constant) spin of the top about its axis OG is n. Prove that the angular momentum of the top about 0 is

Cn u+ Au x

u,

where u is a unit vector along the axis. Deduce the equation of motion when the top moves under the action of gravity, and prove that the upward vertical position of rest is stable if p2 > q, where 2p = CnIA, q = MglIA. If this condition is fulfilled, and if at t = 0 the axis moves through the vertical with a small angular velocity co, determine the motion. Give a sketch of the path of G for the case p2 = 4q/3. 4. A pendulum consists of a light rod with a flywheel attached to the free end, the axis of the flywheel coinciding with the line of the rod. Show that the Lagrangian function of the system can be expressed in the form

(A(62-H where

71.12

sin2 0)

7/: cos0)2}

is measured from the downward vertical.

Mgh cos°,

Ex. VII]

STABILITY OF MOTION

305

By means of Lagrange's equations, or otherwise, obtain the equations of motion. If, initially, p = 0, 0 = 0, = 0, 9 + = n and 6 = co, where co is small, show that at time t co 0 = — sinpt, y = Cnt/2A, and y = — C )t, 2A approximately, where 19 2

Mgli A

C272,2 4A2

5. A uniform circular disc of radius a rolls on a rough horizontal plane. Establish the equations 50 — y,2cose sine + 6co3 W sine + 4 (g./a) cos° = 0, sine + 260 cos° — 2o)30 = 0, 363—

sine = 0,

where 0, y are the polar angles of the axis of the disc referred to fixed rectangular axes, of which Oz is vertical, and co3is the spin of the disc about its axis. Find the least velocity which will ensure stability when the disc rolls, with its plane vertical, in a straight line. 6. A uniform rigid circular disc of mass M and radius a rolls and slides with its edge in contact with a smooth horizontal plane. The axis of the disc OC makes an acute angle 0 with the upward vertical; the plane through OC and the vertical OZ through the centre of the disc, 0, makes an angle y with a fixed vertical plane, and the angle made by a particular radius of the disc with the plane Z 0 C is y. Show that the kinetic and potential energies of the disc are respectively given by 2T = M(±2 + 92) + (A + Ma2cos20) 02 + A02 sin2O + 2 A (Y + cos0)2, V = Mga sine -where A is the moment of inertia of the disc about a diameter, and x, y are the coordinates of the projection of 0 on the horizontal plane referred to axes fixed in this plane. Derive the equations of motion of the disc and show that, if it is set spinning about a vertical diameter with angular velocity co, the motion is stable to small variations in 0 provided that aco2 > 4y.

CHAPTER VIII

IMPULSIVE MOTION 8:1 Elementary discussion When a force F acts for a time interval x = t, — t, the impulse imparted by F is defined to be tz (8.1) dt. I= As was shown in Vol. III § 10:1, the change in momentum imparted to a particle of mass m by the action of the force F is obtained by integrating the equation of motion, i.e.,

m(v

—

v0) = J F dt = I.

(8.2)

An impulsive force is one whose magnitude is very large and which acts for a very short time; by (8.2) such a force imparts a change of momentum I to the particle during which interval the particle moves through a small distance. The distance depends upon the way in which the force varies with time but in all cases the distance tends to zero as -c 0 (see Vol. III § 10:1 p. 332). By impulsive motion, we mean the limiting case of motion under the type of impulsive force described above, the limit being that for which tz -c—> 0, IF' a oo, such that I =fF d t, tz

(8.3)

where I is finite. We keep the term impulse for the limiting case 0, F -± oo, of an impulsive force. Such a force is, of course, also called a "blow". We may describe the present discussion briefly thus: so far in this volume we have considered the behaviour of bodies when they are pushed around, now we consider their behaviour when they are kicked around. The important condition which applies to all cases of impulsive motion is that no change of position takes place during the action of a 306

§ 8: 1]

IMPULSIVE MOTION

307

set of simultaneously applied impulses, but there is a discontinuous change of velocity. Two sets of velocities will be involved in the equations, viz., those of the particles (or parts) of a mechanical system immediately before the impulses act, and the corresponding velocities immediately after; the coordinates giving the position of the system are the same before and after. When a system which consists of two particles joined by a light, rigid rod receives a blow applied to one of the particles, in general the velocities of both particles alter discontinuously and an impulse acts from the rod on the second particle altering its velocity. This impulse is an internal impulse of the system. This impulse can also be said to arise from the constraint that the rod is rigid. Hence, in considering the impulsive motion of a mechanical system we must consider not only the applied or external impulses but also "internal impulses" and "impulses of constraint". Such impulses are classified in exactly the same way as the corresponding forces (see § 4:1). In the following text we usually denote velocities immediately before an impulse by v0 and immediately after by v. (Sometimes, however, we find it convenient to use different letters, and we also use suffixes to distinguish between particles, or between resolutes, or components.) We denote a typical external impulse by I or I and a typical internal impulse by I' or I. (The suffix p denotes the particle of the system on which I or I' acts; we omit this suffix when it causes no confusion to do so.) We assume that the internal impulses obey the Law of Action and Reaction (Vol. III § 1 :3) exactly as do the corresponding forces. A typical particle conforms to the Law of motion ,

nt„ (v, — v„) = I + ri,. ,

(8.4)

When we consider every particle of the system, by the Law of Action and Reaction the impulses Ip of the set cancel out. By addition of equations such as (8.4) for the whole system and by taking moments we obtain, using the fact that 27 1; = 0, = GJ mp (vp

vp0)

= P — Po,

(8.5)

and, using the fact that 27 (ri, — rA ) x Ip = 0, (rp — rA)

x

I , = G Mp (rp

rA) x (v„ — vi3O) = h (A) — ho (A). (8.6)

308


Eqns. (8.5) and (8.6) are the two fundamental equations of impulsive motion; the 1.h. side of (8.5) is the resultant external impulse, and the 1.h. side of (8.6) is the moment of the external impulses, or impulsive moment, about an arbitrary point A. These equations are, of course, equations of linear and angular momentum. When a particle receives an impulse the change in its kinetic energy is given by 2 m(v2 — (8.7) vo) = 1I . (v [see Vol. III, eqn. (10.10)]. When we add equations such as (8.7) for every particle of a system we obtain the relation

T

To = I- 2Ip • (vi

v „) + zLI 1p • (vp + v „).

(8.8)

We interpret the two sums on the r.h. side as giving the energy imparted to the system, or the work done by, the impulses. The term . (vp v„) is the work done by the external impulses, and the other term is that done by the internal impulses. If a particle hanging freely from a fixed point is subjected to a blow J in a direction making an acute angle with the downward vertical, the particle must move in a horizontal direction just after the impulse. In addition to J, another impulse K acts on the particle from the suspension which constrains it to move horizontally. From eqn. (8.8) the final kinetic energy of the particle is z (J K) . v. In this case K . v = 0 and the kinetic energy is iJ . v. Here we may call J the applied impulse and K the impulse of constraint. In more complicated systems we can classify the impulses acting on the system either into external impulses Ip and internal impulses rp , or into applied impulses Jp and impulses of constraint Kp. The impulses Kparise when the system contains rigid bodies, hinges, etc., or when the system is confined by surfaces, rails, strings, etc., to move in a restricted manner. The constraints considered are all smooth or frictionless constraints. This means that the sum Kp . vp= 0, i.e., the impulses of constraint do no work, for any velocity vpspecifying a motion " consistent with the constraints". This condition is the definition of the term smooth or frictionless. This condition is expressible as a linear relation between the velocities of different particles of the system ; when we know that such a linear relation is satisfied, for arbitrary velocities, we can equally well say that the

§ 8:

IMPULSIVE MOTION

309

system is subject to a smooth constraint. Here we do not consider any other type of constraint. The formula (8.8) and the idea of a constraint as a linear relation between velocities play an important part in subsequent sections of this chapter. We give now some examples in which a number of impulsive motions are considered from an " elementary " point of view on the basis of eqns. (8.5) and (8.6). Such problems are effectively exercises in the use of the equations of momentum. Since we consider that no motion takes place during the action of the impulse we need not take account of any motion of the point A in eqn. (8.6) for moments about A. Also, any force F whose magnitude is finite, e.g., weight or other gravitational force, does not contribute to the equations since, in this case, ta lim f F dt = O. r->0

Often the motion of a system consists of a continuous motion, followed by an impulsive motion, and then a second continuous motion. A wheel rolling towards a step, striking the step, and then rolling along the new level, provides an example in which eqns. (8.5) and (8.6) are required to investigate the impulsive change when the wheel strikes the step. In this sort of motion the first continuous motion gives the velocities v0 , and eqns. (8.5) and (8.6) determine the discontinuities v — v0 , where v gives the initial velocities of the subsequent continuous motion. Examples. (i) Two equal uniform square plates OABC, ODE F, each of mass m, are smoothly hinged at 0 and placed at rest on a smooth horizontal plane with the diagonals EO, OB in line. An impulse I is applied at B in the plane of the plates. If w, w' are the angular velocities imparted to OABC, ODEF respectively, and measured in opposite senses, show that w = 3w' and that k.e.I2/4r

712/4m.

This is a problem in plane motion, which we specify in terms of four unknowns, viz., u, v, the resolutes of the velocity of 0, co and w' (see Fig. 84). There is an internal impulse at 0 which need not be found. We apply eqn. (8.5) first (to the whole system), in terms of resolutes: in the direction OB: 2mu = in the direction AC; m(v + 1w) + m(v + lw') =

(1) (2)

where 21 is the length of the diagonal of each square. We use the principle of independence of translation and rotation to find the angular momentum of each square about 0. Since no external impulse having a

310


moment about 0 acts on ODEF, eqn. (8.6), applied to ODEF only, gives

—m/(v +

—

m (T 12)(.1 ) ' = 0.

v=—

(3)

The equation of angular momentum about 0 for OABC gives 2

/2

m/(v + /co)+ —m— 3 co = 2115. 2

(4)

F

FIG. 84. The solution of (2), (3) and (4) is

_ry 9I v = — --, co — 4m1 m

,

co —

3/, 4m 1 '

showing that w = 3w'. The velocity of B has resolutes (u, v + 2/co) so that, using (8.8),

1 T = — {Ixu + y (v + 21w)) 2

( Ix2

742

2 2m 2m J 4m

(costa + 7 sin2a),

where we have written Ix = I cos a, Iy = I sing. The kinetic energy can be written 12 {1 1 7 7 /2 T= — + — cos 2a + — — — cos 2a = — 4m(4 — 3 cos2oc)• 4m 2 2 2 2 Hence I214mT B > C. The body is initially at rest and is set in motion by means of an impulsive couple, the axis of the couple lying in the plane containing the axes A

§ 8 : 1]

311

IMPULSIVE MOTION

and B, and making a small angle a with the A axis. Show that the least angular displacement of the instantaneous axis from the A axis is A2 (A — B)1112 a. IBC (A — C)

f

We use a frame of axes, with origin at the centre of mass, which coincides with the principal axes of inertia. This is a case in which the impulsive motion gives the initial values of the velocities for the subsequent continuous motion. Since the external impulse is a couple, eqn. (8.5) shows that the linear momentum is zero, i.e., the centre of mass remains at rest after the impulse. The impulsive couple [which is given in eqn. (8.6) by E (r5— rA) x Ip with A = 0] has resolutes {G cosa Gsina 01, where a is small. Since h (0) = {AwlB co, Cw3), we deduce from eqn. (8.6) that col = (0 I A) cosa, w2 = (GIB) sing, co, = 0.

(1)

This completes the investigation of the impulse part of the motion. These resolutes are the initial angular velocity resolutes for the subsequent free rotation of the body about its stationary centre of mass To investigate this we use Euler's equations. Briefly, the working is as follows. Since a is small co2 , co, remain small subsequently (rotation about this principal axis being stable, see §7 : 3) and wi 4-(GIA)cosa.* GIA, correct to the first order in a. Euler's equations then give 62 —

C— A G,, co AB

63

(A — B) CA

G we,

(C — A) (A — B) G2 (03. A2 BC The solution satisfying condition (1), is Ga co, = -B— cos n t ,

co, —

Ana sinnt, A—C

n2—

02 (A — C) (A — B) A2 BC

This shows that a point on the instantaneous axis traces out an ellipse about the first principal axis; the major axis of this ellipse lies in the direction of the third principal axis and the maximum angular displacement of the instantaneous axis is, correct to the first order in a, (w

3 /max

Ana Ar A2(A — B) 1'12 A — C G IBC(A — C)

[The result (1) can also be obtained by integrating Euler's Equations (§ 5:5) over the duration of the impulse, remembering that, since no change of position tz takes place, f ahco;dt = 0.]

312


(iii) A uniform rectangular block, of edges 2a, 2b, 2c, stands on a horizontal table, the edges of length 2c being vertical. A string is attached to one of the top corners J and is given a sudden jerk vertically upwards. Prove that the block will at once rise clear of the table if c.4 - 2 c2(a2

b2) — 5,2b2 > 0 .

If this condition is not satisfied, show that there is an impulsive pressure on the table, supposed smooth, at the corner diagonally opposite to J, given by 5a2 b2 + 2c2(a2 b2) _ 7a2 b2 + 4c2 (a2 + b2) + c4

P,

where P is the impulsive tension in the string.

Since the table is smooth and the impulse at J is vertical, eqn. (8.5) shows that the centre o mass G must move vertically after the impulse (Fig. 85). Hence the motion of the block cannot just be one of rotation about either OD or OF. The ensuing motion must therefore be one of two possibilities: either (a) the block rises clear of the table and there is no impulse from the table at 0; or (b) the block remains in contact at 0, but D and F rise clear, and an impulse (of constraint) acts vertically at 0. Since, in this latter case, the point 0 moves along the table the vertical impulse at 0 does no work. We make the following assumptions for the motion after the impulse: the velocity of G is {0 0 w}; the angular velocity is Oh co2 co3}.

p = M{0 0 w}, h (G) = M {(b2 + c2 ) wl (c2 + a2) co2 (a2+ b2)

§ 8 : 1]

313

IMPULSIVE MOTION

(a) In this case the only impulse on the block is {C 0 11 acting at the point {2a 2b 2c). The condition that 0 rises clear of the table is that the third resolute of its velocity is positive. Now vo =

— b — c} .

0 w} + {col co2 w3} x {— a

w — bco, + aco2 > O.

(1)

Eqns. (8.5) and (8.6) give p = {0 0 P}, h (G) = {a b c} x (0 0 P) .

Mw = P, bP = 1111(b2 + c2)co,, —aP = 1.111(c2 + a2) co2, 0 = 3 M (a2 + b2) coo, i.e.,

coi -

w=

3P b M (b2+ c2)

—

3P a M(c2 a2)

a)2

Substitution of these values into (1) leads to the stated condition. (b) If the condition (1) does not hold then an impulse of constraint (0 0 K} acts at 0 and the third resolute of vo is zero.

w = bco, — aw2 ,

(2)

and, from eqns. (8.5) and (8.6), p = {0 0 P + K}, h(G) = {a b c} x {0 0 P} + {— a — b — c} x {0 0 K} .

Mw =P + K, b (P — K) = 3M (b2+ c2) col, — a (P — K) =

(c2 + a2)

w3 0

3b(P — K) P + K = M (bco, — aco2), (01 — M (b2 + c2)

3a(P — K) (02 — m (e2 a2) •

When we eliminate co, and W2from these we obtain

P + K = (P — K)

3b2 b2+ c2

3a2

c2 + a2 )

This leads to the given formula for K. (iv) A homogeneous billiard ball of radius a lies at rest on a fixed inelastic horizontal plane and is struck by a vertically downward impulse J whose line of action passes at a distance b from the centre. Show that slipping occurs if the coefficient p of friction is less than (5 b)/(7 a) and that the ball will then move through a distance

pJ2 2M2 g before it begins to roll. 11 CMES

5b

7a/ )

1

1

314


We assume that the relations and laws concerning the properties of rough surfaces under impulsive forces are exactly the same as for finite forces. In the given case we assume a frictional impulse F (see Fig. 86) and assume that the velocities are as indicated. (Since the plane is inelastic the ball does not leave the table and C moves horizontally; hence the vertical reaction from the table is J as shown.)

F FIG.

86.

(a) If slipping does not occur, a co = u,

F pJ.

The horizontal resolution of eqn. (8.5) gives F =mu. Taking moments about C gives

Jb — Fa = *ma2co = *Fa. i.e., for no slipping p (b) If slipping occurs

5Jb = 7 Fa,

= (5b)/(7a). a co * u,

F = pJ.

Then the equation of linear motion gives ,uJ = mu. The equation of moments about C gives Jb — ,uJa = ema2co. ,. u =

m

aw-

5J(b — pa) 2ma

These velocities are the initial values for the ensuing motion of the ball, initially slipping, on the horizontal plane. Since aw — u = J (5b — 7 pa)1(2ma) > 0 the force of friction acts from left to right in the (continuous) motion which takes place after the blow until slipping ceases (see Vol. III p. 268). The equations of motion for this (continuous) motion are = ,ug, a6 = —5,ug12.

,•,

=

Ita

pgt, au

5J(b — pa)

5pgt

2ma

2

§ 8 : 1]

315

IMPULSIVE MOTION

Slipping ceases when ± = cal after a time t given by J ( 5b

gt =

1) .

m 7,u a

In this time the ball has travelled a distance

t

x= m

1 + — „t2— ka2 1( 5b 2m2 g 2 7,ua

— 11.

(v) N equal uniform rods A0 A1, A1 A2, AN_i AN , each of mass m, are smoothly jointed at A1, A2, AN_iand lie in a straight line on a horizontal plane. The point A, is fixed and an impulse P is applied at AN perpendicular to the line of the rods. The velocity of Anin the direction of the impulse P after application of the impulse is denoted by un . Prove that un

+ 4u. +

+2 = 0

(n = 0 to N — 2).

Prove further that u„ =

6P m

(— 1)N - n

sinh nec 2 sinhNa — sinh (N — 1) a

where a = log (2 + y3).

A0

u,

u2

Ai

A2

UN - 2

UN -I

1

UN

AN _,

(i) n

aZ7a7ffi= 0

(ii) Fro. 87. Fig. 87 (i) shows the velocity of each hinge which is perpendicular to AO AN because A, is fixed. In this system an internal impulse acts at each hinge. This impulse is shown as In corresponding to the hinge An in the "exploded" diagram of Fig. 87 (ii). We keep other internal impulses out of the equations by taking moments. First we note that: 11*

316


"an— un _i 2/ ' the velocity of the centre of mass of a rod is i(un where each rod has length 2/ and n = 1, 2, ..., N. the angular velocity of a rod is

Moments about A,,_1for A.„_,A„, m12 u„ —

3

im(

2/

+ un-') 2 )

211'

Moments about An±1for u„) m12 ( u„.1 21

(u„+1 + un)

2

— —21 In'

where n = 1, 2, ..., N — 1. On subtracting these equations we obtain un _1+ 4un +

(1)

= 0•

This is equivalent to the stated result. Also, by taking moments for the rod AN_i AN about the point AN_i we obtain

3— m/2 ( 3

UN-1 )

2/

+1M (

+ _i)

uN _ + 2uN = 6P/m.

2

- 21P, (2)

The solution of the difference equation (1) (see Vol. II § 2:9) subject to the condition uo = 0 and (2) is as stated. In problems of this sort, though with fewer rods, it is often useful to take moments about each hinge for that part of the system lying to the right of the hinge, so that the internal impulses do not occur in the equations. [See also examples (ii), (iii), (iv) of § 10:4 of Vol. III.]

Exercises 8:1 1. Three equal uniform rods AB, BC, CD, smoothly jointed together at B and C, lie at rest on a horizontal table with the points A, B, C, D in the same line. The system is set in motion by a horizontal blow applied at the point of trisection of BC which is nearer to C and in a direction perpendicular to the line of the rods. Prove that the rods AB, BC, CD begin to move with angular velocities of which the magnitudes are in the ratios 1:2:4. 2. Three equal uniform rods AB, BC, CD, each of mass m and length 2a, are smoothly jointed at B and C, and are at rest in a straight line. An impulse of magnitude I is given to the middle rod, at a point distant c from its centre 0, and in a direction perpendicular to the rod. Show that the initial velocity of 0 is 21/3m. Find the initial angular velocities of the rods, and the kinetic energy of the system.

§8:11

IMPULSIVE MOTION

317

3. Two uniform rods AB, BC, of equal length 2a and equal mass, are smoothly jointed at B and rest on a smooth horizontal table with their directions at right angles to one another. A horizontal impulse is applied to the mid-point of AB and the rods start moving so that their angular velocities are equal and opposite. Find the direction of the impulse and prove that the initial speeds of A and C are as 1/13: 1. 4. A uniform cube rests with its face ABCD on a smooth inelastic horizontal plane. A blow J is applied to the corner A so that the components of the impulse along AB, AD are J sin a, J cos a, and a < n/4. If the initial motion is one in which both A and B move along the plane, find the point of AB at which the resultant impulsive reaction of the table acts upon the cube, and prove that tan a s 3/5. 5. AB and A'B' are parallel diameters of the circular ends of a uniform solid right circular cylinder, of mass M, radius a and height h. The cylinder is suspended by means of a string which is attached to the point A. When it its hanging in equilibrium, an impulse of magnitude P acts at B' in the direction of the normal to the plane AA' B' B. Prove that the initial kinetic energy of the cylinder is 3 P2(3 a2 + 2h2) 2M(3a2 + h2) 6. A uniform square lamina ABCD, of mass m and edge 2a, is free to turn about the corner A, which is fixed, and is struck by an impulse I normal to the lamina at the diagonally opposite corner C. Show that the lamina begins to turn about the line through A parallel to BD with angular velocity 6 1/21/(7ma). If the support breaks immediately before the blow, show that the lamina begins to turn with angular velocity 3 1/2II(ma) about a line parallel to BD dividing AC in the ratio 5 : 7. 7. A uniform rectangular lamina of mass M, with adjacent sides OA, OB of respective lengths 2a, 2 b is free to move and is initially at rest. The lamina is set in motion by an impulse at 0 whose components are X along OA, Y along OB and Z along the normal to the lamina. Find the angular velocity of the lamina and the velocity of 0 just after the impulse, and show that the kinetic energy generated is 3(bX — aY)2 X2 + Y2 + + 7 Z21 . 2M 12 a + b2 If the motion just after the impulse is a pure rotation, show that the line of action of the impulse lies either in the plane of the lamina or in a particular plane perpendicular to the lamina. 8. A uniform lamina in the shape of a parallelogram OPQR, acute-angled at 0, is free to turn about the vertex 0, which is fixed. When it is at rest, it receives a blow at the vertex P perpendicular to its plane. Prove that the instantaneous axis of rotation is the line OS, where S divides RQ in the ratio 3 : 1. 9. A body is moving about its centre of inertia G, which is fixed. When GI is the instantaneous axis of rotation G is suddenly released and a point P of the body, not on GI, immediately becomes fixed. If PI' is the new instantaneous axis,

318


show that the diametral plane of PI' with respect to the momental ellipsoid at P is parallel to that of GI with respect to the momental ellipsoid at G. 10. ABCD and A'B' C' D' are the opposite faces of a uniform cube of edge 2a and mass M, and AA', BB', CC', DD' are the edges perpendicular to these faces. The cube is spinning about the diagonal A C' with spin S2, when the diagonal AC of the face ABCD is suddenly fixed. Show that immediately afterwards the spin w about AC is 2 y212/5 j/3, and prove that the same change of motion can be secured by applying an impulse Maw parallel to DB through a point P on AC such -->that AP = 11 . PC. 11. A uniform cube, of edge 2a and mass M, rests on a smooth horizontal plane. A blow I I/10 is applied at a corner A which is in contact with the plane so that the components of the impulse along the horizontal edges through A are land 37. Show that initially A has velocity components — 21/M and 391/(5M) in the direction of the impulse components I and 3/, respectively. Find also the equation of the axis of the screw for the initial motion of the cube with respect to the edges at A as coordinate axes.

8:2 Generalised methods When a mechanical system consists of a number of parts, e.g., rods, laminas, etc., which are connected by smooth joints or other constraints, the formulation of these constraints in terms of the velocities of the system usually involves the introduction of many unknown velocities and internal impulses which are linearly related. The solution of the equations of motion obtained from (8.5) and (8.6) by elementary methods then becomes very cumbersome. As with continuous motions, so here the use of generalised coordinates and velocities enables many of these constraints to be incorporated in the coordinate system, and there is a corresponding generalised formulation (Lagrangian equations of impulsive motion) which excludes internal impulses and most impulses of constraint. We discuss this formulation now. We assume that the system is specified by n generalised coordinates (q1, q2 .", qi • • qn) in terms of which the cartesian coordinates xp yp zp} of a typical particle, of mass mp , can be expressed. The velocity components of such a particle are therefore

t

ax p

= aqi 41

ayp

= aq4 4i, zp=

aZp 4i.

(8.9)

(For the remainder of this chapter we shall omit the suffix p denoting a particle of the system and we shall use the summation convention for generalised coordinates. A repeated suffix, as in (8.9), is understood to

§ 8 : 2]

319

IMPULSIVE MOTION

be summed over the range 1 to n.) Since we assume no change of position during the impulse we can regard the derivatives a xia qi , etc., simply as coefficients in linear expressions giving x, y, z in terms of 4,. The generalised velocities 4, are the fundamental quantities in discussions of impulsive motion. To obtain the Lagrangian equations we use the Principle of Virtual Work to express the equivalence of two sets of line-vectors ; the first set comprises the changes or discontinuities in the linear momenta of the particles of the system; the second set comprises the impulses, both internal and external. We use a virtual displacement in which the particle of mass m at {x y z} undergoes a displacement Sr, fax by z} where

6x =

ax ay , az 0 , Or = —oq,. a q, 6q,, by = —oqi a q, a qi

The virtual work of the momentum discontinuities is m (v — v0) . or = L'm f(v„ — vx0) Ox (vy — v„)6y

=

ax mi(vz— vz0) -- + (vy — a q, ,

, ay a qi

0)

+ (1)

(vz— vzo) Oz}

, az 6q,. a q,

Vz 0)

(8.10) (The sum denoted by E here is taken over all the particles of the system : a summation over i is also implied.) From eqn. (1) p. 235 the coefficient of 6q, can be expressed thus, 77? (V x

m (vxo

ax

aT = pi =- a aq, ay az aT vy ovz o v) = PL o = a li )0 • Vy

ay

aq q, a ,

ax aqi

+ VZ

az

(8.11)

The virtual work of the external impulses is

a x + , ay + T az . or _ z , a q, a q, a q,

— Ji 6q,, (8.12)

where the coefficient J, is, by definition [see § 6:3 eqn. (6.10)] the generalised component of the impulses acting on the system. In practice, the determination of the Ji often presents difficulties to those unfamiliar with the procedure. The virtual displacements must be made "consistent with any constraints" that are incorporated—" built-in "—in the system. Since we are concerned with velocities and not coordinates,

320


the virtual displacement is taken to be that which the system would undergo in a short time interval 6 t if it had an arbitrary kinematically possible motion. If the generalised velocities in this arbitrary motion are u„ then each 6qi = ui 6t. The generalised impulses then appear as the coefficients of the respective terms ut 6t in the expression for the virtual work of the impulses. The evaluation of generalised components is illustrated in the examples below. The equivalence of the two sets of line vectors is expressed by the equation a T (OT (8.13) (Pi — Pio) 6 qi = [ . . )1 6 qi = 6 qi •

Oqi

aq, 0

If the 6q, are independent we may equate the coefficients of each increment separately and obtain

Pi

Pio =

T

—

(aT o

= J., (i = 1, 2, ..., n).

(8.14)

If the system is subject to constraints then (8.13) holds subject to (< n) linear relations

(k = 1, 2, ..., m).

= 0,

(8.15)

Using the method of undetermined multipliers we obtain

aT

—

(8T

)

,„ ±k =1 Akbkt.

(8.16)

The distinction between holonomic and non-holonomic constraints need not be made here since the coordinates themselves are fixed and either sort of constraint reduces to a relation of the type (8.15). In all discussions of impulsive motion we assume that T = z au4,4i, a homogeneous quadratic function of the velocities. We saw in § 6:2(1) that terms linear in and independent of the velocities only occurred in T when the constraints involved the time. When this is so the state with all qi =0, i.e., stationary w.r. to the "frame", nevertheless has nonzero pi and kinetic energy. Hence the frame must be moving. In the special circumstances of impulsive motion we simply add the velocity of the "frame" on to every velocity of the system, an unnecessary complication. Therefore, we assume T = z auq, qi where ai; = ai [The reader can see that the terms bi in the expression for pi (see p. 235)

j8:• 2]

321

IMPULSIVE MOTION

cancel out in the equations of motion (8.14) and (8.16), this confirming that it is unnecessary to consider them.] From the symmetry of auwe deduce a reciprocal theorem. Suppose that one possible motion of the system is given by velocities q, and another motion through the same configuration by 4/, . The correspond. ing generalised momenta are Pi = aii 4J, 19; = ai.i6 •

(8.17)

P;4; = each side being equal to a1

4, .

Examples. (i) We give first a proof, using generalised coordinates, of the result (8.8) for the kinetic energy generated by impulses. The equations of motion (8.14) give

Pi — Pro •• • Pi

—

Ji •

Pro 4r = Jr 41, Pi 410 -- Pio

(1)

We note first that

T=

414J= iPi 40 To = 1-4io• Pio

Then, on adding the two eqns. (1), we obtain

2(T — To) + Pi 4 io

-

Pro 4r = Jr

J s lio•

Using the reciprocal theorem (8.17) we deduce

T — To = Pi(41 -1-4io)• This is the generalised formulation of (8.8). (ii) Two uniform rods PQ and QR, each of mass m and length 2a, are smoothly hinged at Q. Initially they lie in a straight line on a smooth table. PQ is acted on by an impulse ma at P in a horizontal direction orthogonal to PQ. Use Lagrange's equations to determine the initial velocities of the respective centres of mass and the initial angular velocities of the rods. We use the generalised velocities u, wl , co, to denote an arbitrary motion of the system, Fig. 88. By using the velocity of Q as one of the velocities we incorporate the constraint that the rods are hinged at Q in the coordinate system. To find the generalised components of impulse we suppose that the system receives a displacement u St, a), St, W 2 St in the three coordinates This gives a displacement (u 2aw1) St to the point P. Hence the virtual work of the applied impulse is 1920C (U 2aw1) St = moc(uot) + 2maa(co1 ot). lla

CIVES

322


From the coefficients in this expression we find that the generalised components are: corresponding to w1 w2 u impulse components ma 2macc 0. The kinetic energy of the system is

T = im ((u + acoi)2 + 1,22 4) +

{(u — (10)2)2 + 3a204}

= m(u2 + auco, — auco, + fa2 c4 + *a2 4). The Lagrangian equations of motion are, remembering that u, = w,0 = con = 0, m(2u + aco, — aco2) = ma, m(au + ta2co1) = 2maa, m(—

+ 42 a2co2) = 0.

W2

ma

u+o 03,

U - 0 W2

Fm. 88. The solution of these equations is u = —a, aro, = 9a/4, aco2 = —3a/4. The velocities of the centres of mass are indicated in Fig. 88 underneath the respective points and are for the two rods:

PQ: QR:

vel. of centre

5a/4, ang. vel. —

(9a)/4a); —

(3 oc)1(4a).

(iii) A framework A BCD consisting of four equal uniform rods AB, BC, CD, DA, freely jointed at their ends, lies on a smooth horizontal table in the form of a square. An impulse, J, applied at the corner A, causes the framework to start moving as a rigid body. Prove that the impulse acts in a direction parallel to DB and that, if M be the mass of the framework, the kinetic energy generated is 5 J2/(4 M). To specify the motion of this system we use two linear velocities u, v which are the resolutes of A's velocity and col , co2the angular velocities of opposite pairs of rods, Fig. 89. In an interval St when u, v, col , w 2 have arbitrary values the displacement of A is (u St, v St) so that the virtual work of the impulse is ,

J (u cosa + v sin a) ot = J cos cx (u S t) + J sinoc(v St) .

§ 8: 2]

323

IMPULSIVE MOTION

The generalised components of impulse are: corresponding to impulse components

u J cos a

v co, w2 J sin a 0 0.

The velocity resolutes parallel to AB, AD of the mid-points of the rods are BC: {u — aco2 v + 2aw„};

AB: {u v + acoi};

CD: {u — 2aco2 v + aco„}; DA: {u — aco2 v}. Hence T = m[u2+ (v + aco1)2 + 02 4 + (u — aco2)2 + (v + 2aco1)2 + la2c022 + (u — 2aco2)2 + (v + aw1)2 + -1a2 c4 + (u — aw2)2 + v2 + 02°4] = im[u2 + v2+ 2 (u — a w2)2+ (u — 2 a co2)2+ 2 (v + a coi)2+ (v + 2 a coi)2 + ea2 coT + -1a2 4]. cy,

J sin a

Fro. 89. It is usually unnecessary to simplify further expressions such as this. The Lagrangian equations of motion are: u-equation, m(u ± 2u — 2aco2 + u — 2aw2) = J cosa.; v-equation, m(v + 2v + 2aco1+ v + 2aco1) = J sing ; wrequation, m,{2a(v + awl)

2a(v + 2aco1) + ia2 w1) = 0;

co2-equation, m {-2a(u — aco2) — 2a(u — 2aw2) + la2 co2) = 0. 4(u — awe) = (Jim) cosa, 4(v + awl) = (Jim) sing, 3v + 5acoi = 0, —3u + 5aw2 = O. .• u = 11 a*

5J cosa, v = — sing, awl = 8m 8m 5J

3J 8m

sing, awe =

3J 8m

cosa.

324


If the framework moves as a rigid body c.), = w2and therefore — sin ce = cos a , a = — 7r/4 , or 3 n/4 . This shows that J is parallel to the diagonal not passing through A, the two possible senses giving exactly equivalent motions. v = — 5J/(8 1/2m) and the kinetic energy When a = —x/4, u = 5J/(8 generated is 1 J 5J2 5J 2 T = — J (u cos oc v sin oc) = — (u — v) = 2 2 1/2 16m 4M since M = 4m. (iv) Three uniform rods OA, AB, BC each of mass m, are smoothly jointed together at A, B and hang in equilibrium under gravity from a smooth fixed joint at 0. An impulsive couple G is applied to the rod A B in a vertical plane through 0; prove that the initial kinetic energy of the system is (5702)/(26m12), where 1 is the length of AB.

V

w

Fm. 90. We use the velocities of A, B, C as the generalised velocities to specify the motion of the system, Fig. 90. This automatically includes the constraint that the rods are jointed together at A, B. In a displacement with arbitrary values for u, v, w, AB turns through an angle (v — u) dt/l and the virtual work of G is G(v — so that the generalised components are: corresponding to u impulse components — Gll

v w, 0.

§ 8 : 2]

325

IMPULSIVE MOTION

To calculate the kinetic energy we use the expression *m(u2 u • v v2) for each rod.

T = m (u2 u 2 v 2 ± tv + v2 w 2 v w) = m (2 u2± 2 v2 w 2+ u v v w) The Lagrangian equations are: im(4u v) = — 011, m(4v u w) =

int (2w + v) = 0.

The solution of these equations is u

30 G 13m1

27G 13m/ 1 3m1

150 —

13m1

These velocities correspond to an angular velocity 57 G/(13m /2) for AB. Hence the kinetic energy generated is, using the generalised form of eqn. (8.8) given in example (i) p. 321, 1 1 0 57 G 57 02 2

2

/

13m1

26m12

(v) We give now the Lagrangian solution of example (v), p. 315. Briefly, the problem is that of a chain of N identical rods which receives an impulse P at the point AN, A, being fixed. We use the N velocities u1, u„, oiN as the generalised velocities. It is easy to see that the corresponding generalised impulses are 0, 0, 0, P i.e., only the last component is different from zero. Using the formula for the kinetic energy of a rod in terms of the velocities of its ends we obtain

T = m + + u 2 + 4) + (4 + u 2u3 + 4) + + + (4_ 1+ uN-1 uN + uN)) =

PUT

U1U2 +

+ u2 u, ± 214

2u2N _ I

uN + uN}.

The Lagrangian equations of motion are:

4ui +

= 0,

4u2 ?A,

= 0,

u2 + 4u3 +u = 0,

UN _ 3

+ 426N _ 2 + UN _1 UN _ 2 + 4 UN _ + UN

= 0, = 0,

UN -1 2UN =

6P/m.

These are the same as the final equations obtained by the elementary method. (vi) A smoothly jointed framework of four uniform rods AB, BC, CD, DA, each of length 2a and mass no, moves on a smooth horizontal plane. The intersection 0 of the diagonals CA, DB has coordinates x, y referred to fixed axes to which CA, DB are instantaneously parallel; CA makes an angle co with a fixed line in the plane and the angle DAB is 20. Show that the kinetic energy of the system is 2,07;2 .. 92) I ma2(A2 + 4,2),

326


and that, if impulses X, Y parallel to CA, DB are applied at A, the generalised components of impulse in the coordinates x, y, 0, co are X, Y, — 2a X sine, 2a Y cos 0 respectively. If, when the framework is rotating as a square with uniform angular velocity co, and with its centre of mass at rest, the point A is suddenly fixed, show that the immediately ensuing angular velocity of CA is *w. With the coordinates specified as generalised coordinates the kinetic energy is calculated using the independence of translation and rotation, Fig. 91. The velocity of G contributes I- 4m(t2 + 92) to the kinetic energy. To calculate the kinetic

Fm. 91. energy of the system about G we regard G as at rest, the angular motion being given by 6 and 49. The angular velocities of AB, CD are both CP - A, and of BC, AD both 99 + 0. Relative to G the velocities of the mid-points of AB, CD have magnitudes a(49 + 6), and those of the mid-points of BC, AD have magnitudes a (49 — 6). We now use the independence of translation and rotation to find the kinetic energy of motion relative to G. It is 2 . ima2 (4) + 0)2 + 2 .*ma2 (49 — 6)2 + 2 . m

— 0)2 + 2 .

a2 (49 + 61)2

= 3ma2 (02+ 492).

T = 2 m (x2 + 92) + 3ma2 (62+ 472). When the coordinates are increased by amounts ox, oy, dB, (549 the point G receives a displacement {dx oy}. Relative to G, the point A is displaced {6 (2a cos 0) 2a cos° acp}

§ 8: 2]

327

IMPULSIVE MOTION

since GA is increased by 6(2a cos 0) and rotated through an angle 6q9. The resultant displacement of A is {6x — 2a50 sine by + 2a6cp cos0}. The virtual work of the impulse {X Y} is

X(ox — 2a60 sine) + Y (6y + 2a6q9cos0). From this we identify the generalised components:

x y 0 99 X Y— 2aX sin 0 2a Ycos O.

corresponding to impulse component

We suppose that the impulse {X Y.} is sufficient to bring A to rest and note that before the impulse x = y = 0, b = 0, = a) and that 0 = x/4 at the instant of the impulse. The Lagrangian equations (8.14) are 4m(^"c — 0) = X,

4/n( — 0) = Y ,

16 3

ma2 (O — 0) = —2aX •

1/2

1 2 (W— co) = 2a Y . — , V2 3 ma

16

X Y 9 _ 9 y= 4m m, 4m

3X

— co +

8 1/2m,a

3Y 8 1/2ma

(1)

The general expression for the velocity of A after the impulse is {± — 2a0 sin 0 g + 2a 4) cos 0} . In our case this velocity is zero and so = 1/2a0, g = - V2act . When we substitute these values into eqns. (1) we obtain

X = 0,

4m

= — 1/2a (a) +

3Y 81/2ma)'

i.e., Y—

81/2 maw . 5

Hence = w — =3 w.

Exercises 8:2 1. Two equal uniform straight rods AB, BC are freely jointed at B and lie at rest on a smooth horizontal table; the angle ABC is a right angle. A horizontal impulse is applied at A in a direction that makes an angle oc with the direction BA. Show that the ratio of the initial angular velocity of AB to that of BC is 3 tang.

328


2. Four rods, each of mass m and length 2a, are freely jointed together at their ends to form a rhombus ABCD with the angle ABC equal to 2a. The rods are moving on a horizontal table, without rotation and with velocity V in the direction of the diagonal CA, when the point A strikes an inelastic obstacle. Prove that after the impact the angular velocity of each rod is (3 V cosa)/{2 a (1 + 3 cos2 oc)], and that the loss of kinetic energy caused by the impact is 2m V2/(1 + 3 cos2 a). 3. Four equal uniform rods AB, BC, CD, DA, each of length 2a and mass m, are freely hinged together at A, B, C, D so as to form a square, and they lie on a smooth horizontal plane. A blow m V is delivered at A in the direction AB. If u1 , u2 , u3, u4 are the resolutes of the velocities of AB, BC, CD, DA, respectively, parallel to AB, BC, CD, DA, respectively, show, by the Lagrangian method or otherwise, that = b V, u2 = 0, u3 = I V, '114 = 0. 4. Four equal rods AB, BC, CD, DE smoothly jointed at B, C, D and each of mass m are at rest upon a smooth horizontal plane in the form of a square, with A and E in contact. Simultaneous impulses I act at B and D parallel to AC and CA respectively. Show that in the ensuing motion the velocity of A relative to E is 121/25m along A C. 5. Four equal uniform rods, each of mass m, are freely jointed to form a square. An impulse J is applied to the mid-point of one of the rods along a line which makes an angle a with the rod and lies in the plane of the square. Prove that the kinetic energy developed is J2 (2 + 3 costa) . 16m 6. A chain of 2n equal uniform bars of unit mass, freely jointed at their ends, is laid out in a straight line and set in motion by an impulse J, at right angles to its length at the middle joint A. If denotes the initial speed of the rth joint on either side of A, prove that the initial kinetic energy T of the system is given by 3T =

{ur2 + r =1

'It, _1+ 11).2-1} •

Derive the equations of impulse and show that they are satisfied by = A a" +

r = 0, 1, n,

where oc and p are the roots of the equation x2 + 4x + 1 = 0, and A and B are constants to be determined. 7. A chain of 2n + 1 rods, A0 A1, A5 A2, A2n A2,7+1, freely hinged together, lies on a smooth horizontal table. The rods are uniform and of equal length, and their masses are alternately m, 5m, m, . m. The system is initially at rest

j 8 : 3]

329

IMPULSIVE MOTION

with adjacent rods at right angles and the centres of all rods on a straight line. An impulse J is applied at A, in the direction A0 A1. The component of translational velocity of the rod along itself is denoted by ul . Show that ui = 0 if i is even and that the following relations hold: 6J

5 u2k + 1 -I- 26u2k+3 + 5u2k

16u1 5u 3 = — , 16u2 , +1 + 5u2 „ _ 1 = 0,

m

(0 Hence show that (— 1)k U2k +1 —

2J m

k

= 0, 72, —

2).

k 52n +1—k ± 3 5 52n +2

9

(0

k

n).

8. Four particles each of mass m lie on a table at the four corners A, B, C. D of a rhombus and are connected by strings along the sides of the rhombus. If the particle at A receives a blow J outwards along the diagonal CA find the velocities 7i- and of the individual particles in terms of J, m and the angle CAB = a < -± (

verify that the kinetic energy generated is greatest when the rhombus is a square. 9. Two equal uniform square plates ABCD, BF OH, each of mass m, lie at rest on a smooth horizontal plane, and are connected by light parallel rods DE, CF, freely jointed to the plates, so that ADEH and BCFG are straight lines. If the system is set in motion by a blow I at A along AB, prove that the kinetic energy generated is 7/2/8 m. 10. A uniform circular hoop of mass M and radius a can turn freely about a point 0 of its circumference. When it is at rest, is is struck at right angles to its plane by a blow of impulse J at another point P of its circumference. Show that the angular velocity generated by the blow cannot exceed 3J/(jI2Ma), and find the distance OP for which this greatest angular velocity occurs. 11. Three uniform rods AB, BC, CD each of same length and of the same mass m, are smoothly jointed at B and C and are placed in a straight line. The rods are set in motion from rest by an impulse I applied at A at right angles to A B. Show that the kinetic energy generated is 26/2/(15m) if the rods are free, and is less than this by /2/(390m) if the end D is fixed.

8:3 General theorems There is a number of theorems concerning the behaviour of a mechanical system under various types of impulsive action. These theorems usually state that an expression, a kinetic energy expression, has a maximum, a minimum or a stationary value. In order to understand these theorems it is important that the reader should appreciate what is being varied, and how, so that the appropriate expression displays the stated stationary property.

330


To explain this point more fully we consider three motions of the system: (a) the motion immediately before the impulse, denoted by u, the velocity of a typical particle; (b) the actual motion immediately after the impulse, denoted by a typical velocity v; (c) a comparison motion of the system immediately after the impulse denoted by a typical velocity v'. This comparison motion is a kinematically possible motion of the system (i.e., this motion is consistent with any "built-in" constraints of the system) and is supposed to be one which ensues after the impulse when additional (i.e., additional to the built-in) constraints, or linear relations, have been imposed on the system. Application of these theorems to the determination of the motion of a system usually assumes that the comparison motion v' can be made to differ infinitesimally from the actual motion v by the imposition of suitable constraints. The method of applying the theorems regards the velocities v' as constituting a number of continuously variable parameters (the resolutes of the velocities of different parts of the system); these parameters are varied, subject to any "built-in." constraints and any conditions laid down in the theorem, and when they take values which make v' = v the kinetic energy expression takes its stationary value. The determination of the motion is then given by the investigation of stationary values, instead of by solving the equations of motion (8.5-6), or (8.14), or (8.16). However, as is shown in the examples below, the actual linear equations which must be solved in order to find the stationary values are frequently the same as the equations of motion. Since there is thus little effective difference between the methods, it is more important that the reader should understand and know how to use Lagrange's equations of motion, than that he should remember a number of theorems which apply to certain stated conditions only. We shall give three of these theorems—there are others—and in their proofs we need the following result, and shall use the notation which we introduce now. The kinetic energy of a system when u is a typical velocity, or when v is a typical velocity, is

T.. = zZ7mu2 , T„ = Z mv2 . When the system has the motion in which the typical velocity is v — u, the kinetic energy is denoted by T., and is called the "kinetic energy of relative motion", i.e.,

T., =1 Z m(v — '1)2 = T.. + T„, — Z mv .u.

(8.18)

§ 8 : 3]

IMPULSIVE MOTION

331

When expressed in terms of generalised velocities this result becomes Tuv

= Tau + Tot, —

ui

(8.19)

where Tau

= 2a u ui ui> Ty, = 2 a u vi vj

(1) Kelvin's theorem This can be stated as follows:

When a system is set in motion by means of impulses applied to certain particles so that these particles have given velocities, the kinetic energy of the motion which ensues is less than the kinetic energy of any other motion which would be produced by the imposition of additional smooth constraints. We use formula (8.18) replacing the velocity u by v', the velocity in the comparison motion, i.e., Tv „, = T„„ + T„,,, — my . v'. Subtracting the relation 2 T„ =Emv.v from the above we obtain T y„, = — Tv „

my. (v — v').

Tv . v .

We use the equations of motion to evaluate the sum E m v . (v — y'). For a typical particle the equation of motion is my = I + F. Here I stands for the impulse applied at those particles which have the prescribed velocity, I' is the internal impulse corresponding to the "builtin" constraints of the system. Because these "built-in" constraints do no work in any possible motion of the system we can deduce that / I' . v = 0 , and 7 I' . v' = 0 , since v, v' are both possible motions of the system. Also v — for those particles on which I acts, m v . (v — v') =

=

0

(I + I') . (v — v') = / I' . (v — yr) = O. Tv „, = Tv,„, — Tvv .

Now, since T y ,' is the kinetic energy of a possible motion of the system,

O. T,,,, and the theorem is proved.

Tut,'

332


Examples. (i) If a rod is set in motion by giving one point A a specified velocity, it can be constrained by making it rotate about some other point, B. As B is made to approach A, so the angular velocity of the rod must increase in order that vB shall keep its prescribed value. Hence, by making AB sufficiently small the kinetic energy of the rod can be increased indefinitely. This is an illustration of Kelvin's theorem. (ii) Three similar uniform rods AB, BC, CD, each of mass M, are freely jointed together at B and C, and the rods are initially at rest forming three sides of a square. The end A is jerked into motion with speed u at right angles to the plane of the square. The kinetic energy acquired by the system is T1if the system is free and T2 if D is held fixed. Prove that T2 - Tl= Muz/4680.

We specify the motion of the system by means of the velocity resolutes shown in Fig. 92. Resolutes labelled 1 are perpendicular to the plane of A B C D, resolutes labelled 2 are parallel to BC, and those labelled 3 are parallel to AB. By using the velocities of the joints we incorporate the constraint that the rods are jointed together, and by using, e.g., v2 at both B and C we incorporate the constraint that the rods are fixed in length. Hence this set of velocities incorporates all the "built-in" constraints. (a) When D is free

T = ((u2 + 4

+ 4 + 4 + uv„) + (4+ 4 + 4 + + 4 + 4 + v„w„ + + 4+ v3w3) +(wi+ 4 + 4 + 4 + 4 + 4 + xiwi+ x2v2+ 4)} = m (u2 + 24 + + 24 + 2u4.± 424 + 4 + 4 + uv, + v,w, + I v 3 w3 + x1w1

x2 v2).

(1)

§ 8 : 3]

333

IMP ULSIVE MOTION

Using Kelvin's theorem we regard T as a function of seven independent variables v1, v2 , v3, w1, w3, x1, x2 and choose these so that T has a stationary value. (Since T is a positive definite quadratic form in these variables this value must be a minimum.) The equations are = 0, 10v2 + x2 = 0, 4v3 w,= 0, 4w1+ v„ + x, = 0,

4v1+ u

8w3 + v3 = 0, 2x1 +

(2)

= 0, 2x2 + v2 = O.

The solution of these is = -

7u 26

V2 = 0, V3 = 0, 2V1 =

, W3 = 0, x1 = -

26

X2 = 0,

and the corresponding kinetic energy is T1 = 15Mu2/104. (b) When D is fixed we take x3 = x2 = w3 =0 giving T = *DI {u2 ±

± 524 + 224 ± 224 u

We choose values of the four remaining independent variables v1, v2 , v3, w1, which make this kinetic energy a minimum. The equations are 4v1+ u + w, = 0, 10v2 = 0, 4v3 = 0, 4w1 + = 0,

(3)

whose solution is v1 =-

4u 15

,

v - 2 =0 -

v3 = 0 w1 = - • 15

The corresponding kinetic energy is T2 = 13Mu2/90.

The fact that T2 -T, = Mu2/4680 > 0 is also an illustration of Kelvin's theorem. For comparison we now give the solution of this problem by Lagrange's equations. Alternative solution by Lagrange's method We use eight independent variables by treating u as an independent variable; we also assume an impulse I at A in the direction of u and components of impulse K1, K2 K3 at D, to cover the case when D is fixed. The generalised components of impulse are: ,

corresponding to impulse components

u v1 v2 v3 w1 w3 x1 x2, I 0 0 0 0 K3 K1 K2.

The eight Lagrangian equations are given by using expression (1) for the kinetic energy */1/(2u + v1) = I, *M(4v1+ u + w1) = 0, *M(10v2 + x3) = 0, i-M(4v3 + w3) = 0, iM(4w1 + v1 + x1) = 0, *M(8w3 + v3) = K3, M (2x, + w1) = K1, -4- M(2 x2 + v3) = K2 •

(4)

334


(a) When D is free we put K1 = K2 = K3 = 0. We notice that, apart from the first, eqns. (4) are identical with eqns. (2). When we include the first relation 2u + v1 = 61/M we deduce that I = 15Mu/52, and the remaining velocities follow. We find the energy T1by quoting eqn. (8.8) for this case, viz.,

T1 = -}1u = 15Mu2/104. (b) When D is fixed K1, K2 , K3 are no longer necessarily zero, but xl = x, = w, = 0. Eqns. (4) now give K, = K3 = 0, w1 = 6K1IM. Again, the remaining eqns. (4) apart from the first, are identical with eqns. (3). Hence we find I = 13Mu/45, K1= Mu/90. The work done by the impulse K1is iK,x, = 0, as would be expected for an impulse of constraint. Eqn. (8.8) gives

T2 = jIu = 13Mu2/90. The reader can expect from this illustration that the use of Kelvin's theorem leads to the same equations for solution as Lagrange's equations of motion. (iii) A framework AB CD of four uniform smoothly jointed rods of the same constant mass per unit length in the form of a rectangle of sides AB = CD = a, AD= BC = b, rests on a smooth horizontal plane. The corner A is seized and made to move with speed V y2 along the bisector of the right angle DAB. Show that the opposite corner C begins to move in the opposite sense with velocity resolutes bV1(3a + 2b), aV 1(2a + 3 b) along the rods CD, CB respectively.

If we assume that co, and co, give the angular velocities of opposite pairs of sides, Fig. 93, we have specified the motion of the system, being given the velocity of A. We denote the mass per unit length by e, so that the kinetic energy is

T=

{3 V2 + (V -F acoi)2 + V( V + acoi) + V2 + V2+ 2(V — bco,)2 + + (V + aco1)2 + V( V + acoi) + (V — bco2)2} +

§8:3]

335

IMPULSIVE MOTION

+

{V' + 2(V + aco1)2 + (V — b(02)2+ V (V — bw2) + (V + aco1)2 +

+ 3 V2 + (V — bco2)2+ V (V — bco2) + V2) = iect{5V2 + 2(V + aco1)2+ 3(V — bco2)2 + 2V(V + wl)}+

+ eb {5 V 2+ 3(V + aw1)2 + 2(V — bco2)2+ 2V(V — bcoz )} • We do not simplify this expression further. We now use Kelvin's theorem and find col> (02 to make T stationary: 6 aT a Ow, 6 8T b a we

= a{4(V + awl) + 2 V} + e b {6(V + aco,)} = 0, = ec t {— 6(V — bco2)} + eb {— 4(V — bco2) — 2V) = 0. aco, —

3 V (a ± b) 2a + 3b

bw2

3 V (a ± b) 3a + 2b

•

These angular velocities lead to the stated velocity resolutes for C. (2) Bertrand's theorem

This theorem can be stated as follows:

When a system in motion is subjected to a number of given impulses, the motion which ensues has a greater kinetic energy than that of any other motion which would be produced by the same impulses with the imposition of additional smooth constraints. We give a proof of this theorem formulated in terms of generalised coordinates. (It is quite possible to formulate the proof in terms similar to those used for Kelvin's theorem.) We assume that the generalised coordinates have been so chosen that the given impulses have generalised components I,, 12, , 1k , corresponding to coordinates q1, q2, ..., qk , these impulses being the prescribed impulses. The impulse components in the remaining coordinates are zero for the actual motion. We denote by K (i = 1, 2, ..., n) the impulses of additional constraint which are required to produce the comparison motion. The initial motion, the actual motion and the comparison motion are denoted by generalised velocities uu vi , respectively. Then eqn. (8.19) shows that T,•,• — at; vi = Ty, — + at; v't vj — ai vi v; (p; — p;) vi . = T„ — T,•,• a,;(4 — vi ) vi = T„ — Tv' v•

= T„

,

(8.20)

336


Here 23: , p, are the momenta corresponding to v:, v1respectively. The Lagrangian equations of motion are, for the natural motion:

Pi — Pio = Ii (i = 1, 2, •••5 k), p,. — p„ — 0 (r = k 1, n) ; and for the comparison motion:

Ki, P — Pro = Kr. — Pio = Since the impulses K, arise from smooth constraints to which the comparison motion v', conforms, then Ki = 0 (summed from j = 1 to n). (20; -

p,)v; = 21 (I, + K, — i -1

24 ±

Kr 2j, = K1 24 = 0. r=k+1

Since Tv „, 0 we conclude that

T„,

Tv,„,

.

The direct application of Bertrand's theorem, without assuming Lagrange's equations of motion, usually leads to the determination of a stationary value subject to a condition. This condition expresses the fact that the applied impulses are the same for all the possible comparison motions. To formulate the condition we note that there are two expressions for the final kinetic energy of the comparison motion, viz., and, from eqn. (8.8),

T

= a,i vi v; -1-1;(24 + u f). =To

.*, auv; vi, — 11(4 +u;) — 2 T, = .

(8.21) (8.22) (8.23)

Hence the applications of Bertrand's theorem consist of choosing the parameters v', so that T , in either form (8.21) or (8.22) is stationary (in fact a maximum, since T is positive definite) subject to the condition (8.23). This means that its use is often rather cumbersome. Examples. (i) The rod in example (i) p. 332 is set in motion by a given impulse I at A and is then constrained by making it rotate about B. As B is moved closer to A, so the angular velocity is reduced—reduced to zero when B coincides with A. This is an illustration of Bertrand's theorem. (ii) AB, BC are two uniform rigid rods each of mass m freely jointed at B and freely suspended from a fixed support at A. When they are hanging in equilibrium an impulse I is applied horizontally at C. Prove that the system is set in motion with kinetic energy 12/2/(7m).

X8:3]

IMPULSIVE MOTION

337

We use the two velocities u, v in Fig. 94 to specify the motion. Then 2 uv v2) . T = *m, (u2 u

(1)

Also, since any impulse at A does no work, the kinetic energy is T = iIv. Hence, to determine the motion by Bertrand's theorem we find u, v so that T = lIv

(2)

is a maximum subject to the condition m(2u2 + uv + v2) — 3/v = 0.

(3)

We use the method of undetermined multipliers, noting that, from (2), aT

0u

aT 1 = 2 I. v

The equations for stationary values of T as given by (2) subject to condition (3) are A {4mu + mv} =0, + A {mu + 2mv — 3/) =0. Since I 0 we cannot choose A = 0, 4u + v = 0, /(1 — 32) + Am(u + 2v) = 0, i.e., u = ,uI, v = — 4,u/, it = (1 — 6A)/(14Am). (In effect, we use it instead of A as the undetermined multiplier.) Substitution into the condition (3) gives it (7mu +6)=0.

338

A COURSE OF MATEMATICS

The solution ft = 0 gives no motion, hence tC = — 6/(7m). 6/

24/

7m

7m

(4)

The corresponding (stationary) value of T is 1

12/2

2

7m

T = — Iv —

(5)

Alternative solution by Lagrange's method The generalised components of impulse are: u v corresponding to components of impulse 0 I. Using T = *m(2u2 uv v2), Lagrange's equations are: im(4u v) = 0, im(u ± 2v) = /, solutions of which give (4) and (5).

(3) Robin's theorem This theorem concerns the situation where a system is subject not only to applied impulses but also to the simultaneous application of smooth constraints. The sort of constraint involved here is one such as bringing a point or part of the system suddenly to rest, or causing a point to slide along a smooth surface. The constraint must be one which does no work in any possible subsequent motion of the system. The theorem includes the case in which the system suffers a blow at the same instant as the imposition of the constraint. If the velocity of a typical particle is u in the motion before the impulse, v in the natural motion, and v' in a comparison motion, then Robin's theorem states that the expression

/1 =

m(v' — u)2 —

I . (v' — u)

has a minimum value when v' = v. We give a proof of this theorem using the Principle of Virtual Work. The system is subject to two sets of impulses; I the given external impulses, and K the impulses of constraint suddenly applied. Hence the two sets of line vectors m (v — u) and I + K

§ 8: 3]

339

IMPULSIVE MOTION

giving the natural motion have equal virtual works. Suppose now that we use virtual displacements v t and v'6t; these are both possible motions of the system after the impulse and so are consistent with K.

K . v 6t = 0, and

L'm(v —u).v=

,

K v' St = O.

Zm(v

ZI.vi. (8.24)

We suppose that K' represents a typical additional impulse required to constrain the system to take up the comparison motion v' after the impulse. Hence the sets of line vectors m(v' — u)

and I

K K'

(8.25)

have equal virtual works. The displacement v' at is consistent with K and with K' (whereas v St is not necessarily consistent with K'), i.e., K . v' = 0, K' . v' = 0. Therefore, using v' 6t as a virtual displacement to express the equivalence of the line vectors (8.25), we obtain

m ( v' — u) . v' =

. v'.

(8.26)

From the second of eqns. (8.24) we obtain now L'm (v — u) . v' = Z m(v' — u) . v'

(8.27)

and, from the two eqns. (8.24), m(v — u) . (v' — v) =

I . (v' — v).

(8.28)

The difference between the values of f for the comparison and natural motions is

f'

—

f = =

= =

m {(v' — u)2— (v — u.)2} —

I . (v' — v) m(v'— v) . (v' v — 2u) — m(v — u) . (v' — v) m(v' — v) . (v' v) — m(v' — v) . v Z m(vr — v)2 = T„,,, O.

The two extreme special cases of this theorem are given now, (4) When only the impulses I act on the system, no constraints are imposed, the actual motion is that velocity which makes

m(v' u)2— I . (v' — u) 2 a minimum. (5) When only constraints are imposed on the system, i.e., I =- 0, the actual motion is that motion which makes the kinetic energy of relative motion, i.e., 71„,,, = m.2 (v' — u)2, a minimum. —

340


In this case, in the absence of applied impulses, eqn. (8.26) shows that 2' m(v — u) . y' = 0. Also eqn. (8.18) gives =

T„,,, — 7 my' .0 T,,,, —

my' . y' = Tau — T„,„,.

Hence in this case the minimum value of f is T„ and is equal to Tau — T„ which is the loss of kinetic energy. The theorem may therefore be stated thus: When smooth constraints are suddenly imposed on a moving system the resulting motion is one for which the kinetic energy of relative motion has a minimum value, the minimum value being equal to the kinetic energy lost by the system. Examples. (i) We can use Robin's theorem to solve the problem of example (vi) P. 325. When the system is moving with an arbitrary motion

T

2m(iz + 0)

=

g

(62 + ,p2)

and the velocity of the point A has resolutes + 2aci9 cos0).

4 — 2a0 sin°

We consider the case in which 0 = n/4 and the system is moving in the shape of a square with its centre of mass stationary, i.e., initially = 0, 9 = 0, 0 = 0, 4) = co. To apply Robin's theorem we determine x, ', 0, q) such that

Tuv, _ 2m(±2 ± 0) + ?woo 3ma2 (,

(0)2

is a minimum subject to the two conditions — -112a0 = 0,

1/2acp = 0.

(No impulses act, only constraints are imposed.) Using two undetermined multipliers 2, p the conditions are 4mt + A = 0, 49.74 + p = 0, yma2O — 1/2a2 = 0, 3ma2(cp — co) + 1/2a,a = 0.

_

A 4m

9=

itt —

4m

0 —

32 81/2m a

(i) —

81/2ma

§ 8: 3]

341

IMPULSIVE MOTION

If we compare these equations with eqns. (1) on p. 327 we can see that they are identical if 2 = — X, p = — Y. Hence, the solutions are identical from this point onwards. (ii) A plane non-rigid framework consisting of four equal uniform straight rods AB, BC, CD, DA, each of unit mass, is in general motion in its own plane. At a certain instant, when its form is that of a square, the motion is such that the component velocities of A B, B C, CD, andDA along their lengths are respectively a, b, c, and d. At this instant a constraint is suddenly applied that maintains the framework in the form of a square. Prove that the resulting loss of kinetic energy is (a — b + c — d)2. C

C

D

A

B

a FIG.

95.

A system of four connected rods has 4 degrees of freedom and the given allocation of velocities is sometimes a convenient set of generalised velocities to use, Fig. 95. With these velocities vA has resolutes {a — d} , vB {a b} . etc. Hence the kinetic energy is (3,2 b2 d2 _ bd) in{5 (a2 b2

T=

c2

d2,)

2 (a c + b d)).

(1)

abcd

The condition that the system shall rotate as a square is that the angular velocities of AB (or CD) and BC (or AD) shall be the same. Since co AB

b+d AB

c +a 0BC —

BC

this condition is equivalent to b + d = a + c.

(2)

To solve this problem by Robin's theorem we have to choose a', b', c', d' so that T„„, = int {5(a' — a)2 + 5(b' — b)2 + 5(c' — c)2 -I- 5(d' — d)2 — 2 (a' — a) (c' — c) — 2 (b' — b) (d' — d)}

342


is a minimum, subject to the condition, b' + d' = a' + c'. Using an undetermined multiplier the stationary conditions are 10(a' — a) — 2(c' — c) — 2 = 0, 10 (b' — b) — 2(di — d) + A = 0 ,

(1)

10 (0' — c) — 2(a' — a) — A = 0, 10(d' — d) — 2 (b' — b) + A = 0 . By adding all these equations we obtain a' + b' +cr +di =a+b+c+d. From the condition, we obtain a' — b' + c' — di = 0.

. a' + c' = (a + b + c + d) = b' + di . From the 1.h. pair of (1) a' + c' — a — c = 12, and from the r.h. pair b' + di — b — d = — eA.

;.A=2(—a+b— c+ d), a' = /(3a

b—c+

c' = 1(—a + b + 3c + d),

4(a + 3b + c — d), =1 = i(a — b + c + 3d),

i.e., a — a' = — (b — b') = (c — c') = — (d — d') = (a — b + c — d). Substitution of these values into T.; gives

Tay, = i(a — b + c — d)2 . The theory shows that this minimum value of the kinetic energy of relative motion is equal to the kinetic energy lost.

Alternative solution by Lagrange's eauations We use a', b', c', di as generalised velocities and assume that corresponding generalised impulses K , Kb, K , Kd must be imposed to make the system move as required. Since these constraints do no work

Ka a' + Kb b' + Ka c' + Kd d' = 0 for an arbitrary motion consistent with the constraint. But the velocities in such an arbitrary motion must satisfy a' — b' + c' — di = 0. Hence, since the motion is arbitrary, we must choose

Ka = p,

Kb

—p, Ka — p, Kd

= A,

§ 8 : 3]

IMPULSIVE MOTION

343

where y is unknown. Under the influence of these constraints the system must satisfy the Lagrangian equations obtained from the kinetic energy expression

T = m (5 (a' 2 + b' 2 + c'2

d' 2) - 2 (a' c' + b'd')),

i.e., (10(a' — a) — 2(c' — c)) = ,a, -km (10(b' — b) — 2 (d' — d)} = —y, *in (10(c' — c) — 2(a' — a)) = y,

(10(d' — d) — 2(h' — b)) = —y.

These equations are identical with eqns. (1) if y = — m/1./6. The rest of the working is identical. We determine the change of kinetic energy from the expression (8.8)

T — To= k (Ka(a + a') + Kb(b + b') + Kb(c + c') + Ka (d + d')) • In this expression

= — Kb = K, = — Ka = = im(a — b + c — where we have used the value of A obtained earlier. Since the constraint is smooth Kaa' + Kb b' Kcc' + Kd d' = 0.

T — To= -km(a — b c — d) (a — b + c — d) = *m(a — b + c — d)2. Exercises 8:3

1. Oxi, Ox2, Ox, are rectangular axes at the centre 0 of a uniform cube of mass M and side 2a, the axes being parallel to the sides of the cube. The cube is set in motion from rest by an impulse applied at the point (a, a, a), the impulse being such that the initial velocity of this point is (u1, u2 , u3). Show that the kinetic energy is 22 [5 (24 + 2G2

+

4) + 6 (u2 u3 + usui + u1u2)]

2. A uniform rectangular plate A BCD of mass M, for which AB = 2a, BC = 2 b, is at rest and is then given a blow of impulse J perpendicular to the plate at C. Show that, if A is fixed but the plate is free to turn about it, the kinetic energy generated by the blow is 12J2/(7M). If the plate is completely free find the kinetic energy generated by the blow and the velocities of B and C. 3. A system with n degrees of freedom has generalised velocities and momenta 4, and pr (r = 1, 2, ..., n). Show that it is possible to express the kinetic energy of the system in the form P Q, where P is a quadratic form in pi, p2 , ..., PS and Q is a quadratic form in 4„1 , 48+2 , ln. Hence, or otherwise, show that if a system is started from rest by the application of specified impulses the kinetic energy taken up by the system is greater than if the system were subject to additional smooth constraints and acted on by the same impulses. 4. A rhombus composed of four equal, smoothly-jointed rods AB, BC, CD and DA each of length 2a and mass m, moves in its own plane so that the component velocities of its centre in the direction of the diagonals are u, v, the angular velocity

344


of the diagonals is w and the angular velocities of the rods relative to the diagonals are co'. Show that its kinetic energy is 2 m(u2 v2) m a2 (,02 co(2) If the rhombus is moving as a rigid body in the form of a square with velocity u parallel to CA and the middle point of AD is suddenly brought to rest, prove that the loss of kinetic energy is 7 mu2/5. 5. A rhombus ABCD consisting of four freely jointed uniform rods each of mass m and length 2a is moving in any manner on a smooth horizontal table. Denoting by u and v the component velocities of the centre of mass of the system in the directions AC and BD and by eh and w2 the angular velocities of CD and AD, show that the kinetic energy of the rhombus is 2 m (u2 + v2) + m a2 (4. +

.

The rh( mbus receives a blow I in the direction CA at the point C; calculate the corresponding generalised components of I if the angle BAC is oc. The rhombus, in the form of a square, is moving without rotation with velocity U in the direction AC when C strikes a vertical, inelastic, wall at right angles. Prove that the angular velocities of the rods are all 3 U/(5 a y2) after the impact. Find the impulse I at C. 6. Four equal uniform rods are freely hinged together so as to form a rhombus of side 2a. The system falls from rest in a vertical plane under gravity with one diagonal vertical, and with each rod making a constant angle a with the vertical, until impact occurs against a fixed horizontal plane. Show that immediately after impact the angular velocity of each rod is 3(U + V) sinoc 2a(1 + 3 sin2a) where U is the downward speed of the rhombus immediately before impact, and V is the upward speed of the point of contact immediately after impact.

Miscellaneous Exercises VIII 1. A uniform solid right circular cylinder is of radius a and height 2a. AB is a diameter of one of the ends. The cylinder is held at rest in any position and released. Immediately an impulse is applied at A, so that A has a given velocity V in a direction perpendicular to AB in the plane of the end; show that the initial velocity of the centre of the cylinder is 7 V/33. 2. Six equal uniform rods AB, BC, CD, DE, EF, FG are freely jointed at B, C, D, E, F and are lying at rest on a smooth horizontal table so that ABCD and CDEF are squares, while FG is perpendicular to EF. If a blow is applied at A in the direction ADE, show that immediately afterwards A is moving 782 times as fast as G and in the opposite direction. 3. Three uniform equal square laminae ABCD, DEFG, FHIJ, freely jointed at D and F, rest on a smooth horizontal table with the diagonals BD, DF, FI in

Ex. VIII

345

IMPULSIVE MOTION

the same straight line, the length of each diagonal being 2a. A blow P is given at G to the middle lamina in the direction GE. Show that the angular velocity of the outer squares is (P/2m a) immediately afterwards, and that it is y(1/o) (Pima) when their sides are first parallel to the diagonals of the middle square, m being the mass of each lamina. 4. An infinite number of straight uniform rods A0 B0 , B0 A1, A1 B1, B1 A2, ... are smoothly jointed at B,, A1, B1, ... and lie on a horizontal table in zig-zag formation, so that Ao , A1, A2 , ... are in a straight line and Bo , B1, B2, ... are in a parallel straight line, the angles at Bo , A1, B1, A2, ... all being right angles. Each rod is of the same mass m and of the same length. If Aois suddenly seized and made to move with velocity vo in the direction Bo Ao , show that the longitudinal velocities of the rods Bo A1, B1 A2, B2 A3 , ... all vanish, whilst the longitudinal velocities of the rods A„B„, n = 1, 2, 3, ... are v„ = vo(21/6 — 5)". Show that the kinetic energy generated is 2 1— .1 (3

12

+ 21/6) 4.

5. A uniform cube, of edge 2a and mass M, is rotating freely about a diagonal with angular velocity D. One of the corners not on the diagonal of rotation is suddenly brought to rest by a smooth constraint. Show that the impulse required is 11 )1 ( )Mcif2 and that the new angular velocity of the cube is of magnitude 121/(17) 11

•

6. A uniform square lamina of mass M and side 2a is set in motion by impulses applied at two adjacent corners A and B. A moves with velocity U perpendicular to the plane of the lamina and B moves with velocity V in the direction perpendicular to AB in the plane of the lamina. Prove that the kinetic energy is V2

7 U2 1 .

1 3 + 96 7. Two uniform rods AB, BC, each of length 2a and mass M, are freely jointed at B, and lie on a smooth horizontal table. The midpoint 0 of AB is smoothly hinged to a fixed vertical axis. Initially ABC is a straight line, and a horizontal impulse I is applied at C perpendicular to ABC, resulting in a velocity v for C and a kinetic energy T1. If the joint B is clamped and the point C is again given the velocity v, the resulting kinetic energy is T2 ; while if B is clamped and I is applied at C, the energy is T3 . Show that 49432

Mv2 = T2 T1 > T1 — Ta =

mi

MV2.

8. A uniform circular disc of mass m, initially at rest but free to move in any manner, is suddenly set in motion by an impulse applied to a point A of its circumference. The initial velocity of A has components U along the radius of the circle 12

CMES

346


through A, V along the tangent to the circle at A and W along a third perpendicular; prove that T, the initial kinetic energy of the disc, is given by the equation 2T = m(U2

V2 ± W2).

9. A rigid body with one point 0 of itself fixed is in motion with angular velocity w. Show that, if m, is a typical particle of the body with position vector r relative to 0, the angular momentum of the body is (Emr2)to — Em (r co) r. Hence derive expressions for the components of angular momentum of a rigid body referred to rectangular axes at 0 for which the moments and products of inertia are A, B, C, F, 0, H. If the body is initially at rest and is then acted on by an impulsive couple (L, M, N), give equations to determine the initial angular velocity components w1, w2 , 0)3of the body, and write down their formal solution. 10. A rigid body, whose moments of inertia about its principal axes at 0 are A, B, C, is at rest with 0 fixed. It is set in motion by an impulse I applied in the line L whose equations, referred to the principal axes at 0, are (x — a)/l = (y — b)/m = (z — c)In. Find the initial component angular velocities of the body. If the line L is perpendicular to 0G, where G is the centre of mass of the body, show that the impulsive reaction at 0 is also perpendicular to 0 G. 11. A uniform circular hoop of mass 8m and radius a is pivoted at a point 0 of itself so that it can swing freely in a vertical plane. A smooth bead of mass m can slide on the hoop. If A is the centre of the hoop and B the position of the particle at any time, and the inclinations of OA and AB to the downward vertical are 0 and (p respectively, derive the kinetic and potential energies for general motion. For small oscillations about the stable position of equilibrium, show that the periods of normal modes are x/p and 23r/3p, where p2= g/8 a, and find the normal coordinates (see § 9:2, 3). If the system is set in motion from the stable position by a small impulsive couple Q = 16m a2 co applied to the hoop, find expressions for the subsequent values of 0 and p in terms of the time t. 12. A uniform heavy sphere of centre 0 is free to move on a rough horizontal plane. The coefficient of sliding friction is y and the velocity of the point of the sphere in contact with the plane at P is To at the instant t = 0. Prove that the velocity of the point of contact at time t (< 21)017 pg) is (1 — 7 ,ug ON) vo and that the path of 0 is generally an arc of a parabola but reduces to a straight line if, at time t = 0, the velocity of 0 is in the direction of v0 . If such a sphere, at rest on an inelastic horizontal plane, is struck by a blow whose line of action passes through P, and which has horizontal and vertical (downward) components I and J respectively, show that the sphere is again at rest when 0 has travelled a distance (I — ,uJ)2/211/2yg.

CHAPTER IX

THE OSCILLATIONS OF A DYNAMICAL SYSTEM WITH A FINITE NUMBER OF DEGREES OF FREEDOM—NORMAL MODES 9:1 Introduction In Vol. III § 9:6, we considered some simple examples of dynamical systems with two degrees of freedom executing small oscillations about positions of stable equilibrium. We showed there that the two coordinates specifying the configuration of such a system are each the sum of two periodic terms. Further, by applying a real linear transformation to the coordinates, we expressed the kinetic and potential energies of the system as the sums of squares in the normal coordinates and showed that the motions in these normal coordinates are harmonic motions which are completely independent of one another. In this chapter we extend these ideas to dynamical systems with n degrees of freedom. Where convenient we use the methods and results of matrix algebra (Vol. IV Chapter V) and the Laplace transformation (Vol. V Chapter II). First we consider an example in which is required the displacement of one particle of a system with three degrees of freedom.

Example. A light string PQ, of length 4a, is stretched at tension T on a smooth horizontal table and particles of masses in, 21m/20, m respectively, are attached at the points of quadrisection A, B, C, with the unequal one in the middle. Motion is started at time t = 0 by a transverse blow I on particle A, where //m is small. Prove that, correct to the first order, the motion of each of the particles is transverse. Prove also that the displacement of B at any subsequent time is

10/ ( sinoct 29m

)'

where oc2 = 4T 1(7 ma), ( 2 =10T/(3ma). 12*

sini3t

347

348


Suppose that at time t the transverse displacements of the particles are x, y, z and PA, = a + e, PB, = 2a + Pc, = 3a + Fig. 96. In a displaced position the length of the string must exceed its undisturbed length and so the potential energy of the string is least in the undisturbed position. But the initial kinetic energy of the system is /2/(2m). The forces acting are conservative and therefore the kinetic energy of the system cannot exceed /2/(2m). • [±2 + S2 + g (92 + 7,2) + 4, C2] /2/(2 m) Therefore each of x, y, z, e, i,

c must be of order //m. Since x2) = a + e + 0 (/2/m2) ,

PA = {(a+

B

C

Fm. 96. the tension in AP increases by an amount proportional to e, say kE. We obtain similar results for the other particles. Then the equations of motion of A, B, resolved parallel to PQ are, on our neglecting terms of order /2/m2, — 2e),

=

m = k( — 277 + C), = kfri subject to the initial conditions e = = = 0, = equations clearly have the solution = n= c = 0

= = 0 at t = 0. These

and so, to the first order in //m, the particles do not move parallel to PQ. [This result can be extended to any case in which a purely transverse (small) initial displacement or velocity gives rise to transverse motion only.] Note that the extension of each part of the string is of the second order in //m and so, to our degree of approximation, the tension throughout the string remains equal to T. The equations of transverse motion for the three particles are y— x x a a

A,

—T

B,

21 — mg—T (z—y 20 a

C,

m—

T (y — z z a a

a )'

§ 9: 1]

THE OSCILLATIONS OF A DYNAMICAL SYSTEM

X — n2 (y — 2x)

349

0,

9 - nn2 (x — 2y + z) = 0, — n2 (y — 2z) = 0, where n2 = T/(ma). We solve these equations, subject to the conditions x = y = z = 0, = //m, y = 0 = z at t = 0, for y by means of the Laplace transformation. The transformed equations are (p2 + 2n2) X — n2 Y = //m,, — n2 X + (2p2-1- 2n2) Y — n2 Z = 0, — n2 Y + (p2 2n2) Z = 0, whence Y=

20/n2 m (21p4 + 82p2 n2 + 40 n4) 1 10/( 29m p2 + 4n2/7

29m ( p2 + a2 Y "y

20/n2 m (7p2 + 4n2) (3p2 + 10 n2)

1 p2 + 10n2/3

p2+ 132 10/ sinat 29m a

sin 13 t /3

).

Exercises 9:1 1. A light string of length 3/ is stretched horizontally between two fixed points. Gravity is neglected. Masses 15m, 7m are attached to the points of trisection. The tension in equilibrium is Aml. The particle of mass 15m is drawn aside a distance a, the other remaining undisplaced, and both are simultaneously released. Prove that in the subsequent motion the displacement of the particle of mass 7m is 15 32 — a / cos. V—t— cos 26 35

2

2. A light string of length 9/ is stretched under tension P between two fixed points. Masses 2m and m are attached to the string at distances 1 and 5/ respectively from one end. The particles are drawn aside equal small distances a in opposite directions at right angles to the string, and released simultaneously, so that the motion takes place in one plane. Show that the displacement of the heavier particle at time t is a cos

( 3P )1/= t. 4m/

350


3. A light string of length 4a is stretched to tension T between two fixed points A and B. A particle of mass 3m is attached to its midpoint C and two particles, each of mass 4m, are attached to it at the midpoints of AC and CB. When the system is at rest small transverse velocities IL are suddenly given in the same direction to both the particles of mass 4m. Prove that the displacement of the other particle at time t is given by 4u co t 116 sin — V 6— sin cot) 5 co

where w2 = T I (m a) .

9:2 Systems with two degrees of freedom The characteristics of a system with n degrees of freedom oscillating about a position of stable equilibrium are best illustrated by considering systems with 2 degrees of freedom only. In this section we consider systems of this type; the results and methods are generalised by the use of matrix algebra in § 9:3. First we consider a simple example, which illustrates the methods and results without requiring elaborate mathematical techniques or complicated manipulations.

A particle of mass m is slightly disturbed from rest at the point

x = 0, y = 0 of the inner surface of a smooth bowl whose equation is 2z = ax2 2hxy by2

,

where the axis Or is vertical and Ox, Oy are horizontal. We find conditions on a, b and h for the ensuing motion to remain small, and discuss the nature of the oscillations in this case. Suppose that the particle is slightly displaced from the equilibrium position (0, 0, 0), Fig. 97. Then

T _ ?,.m

w

f/ 2 +

,

V=

(ax2+ 2h x

b y2 + ...)

.

§ 9:2]


351

Since .t = 0 (x4 correct to the second order in small quantities we can write T = m(x2 + y2), V = img(ax2 2hxy by2) (9.1) and Lagrange's equations are

g(ax hy) = 0, (9.2) g g(hx ± by) = O. (9.3) The solutions of these equations can be obtained by several methods. Here we indicate two methods of solution. First, we seek solutions of eqns. (9.2) and (9.3) of the form

x = F ei" , y = Ge' Pt ,

(9.4)

where F, 0 are complex constants and by the notation x = Feipt we mean that the real quantity x is the real part of F pt [Phase differences are given by differences in the arguments of F and O.] Then .

(ag — p2) F hgG = 0, hg F (bg — p2) G = .

(9.5)

For a non trivial solution (for F, G) of eqns. (9.5) we must have

ag — p2 hg

hg = 0, bg — p2

(9.6)

which reduces to p2 \ 2

er)2

' (a + b) g

g

ab — h2 = .

(9.7)

For stability, i.e., for)the oscillations to remain, small, eqn. (9.7), considered as a quadratic in p2/g, must have real positive roots. [Otherwise, if this quadratic has negative or complex roots, one of the values of pis of the form cc — i19 where 13 > 0, and then (9.4) implies that x and y do not remain small for large t.] The discriminant of (9.7) is (a — b)2 4h2 and therefore the roots of (9.7) are real since the discriminant is the sum of two squares. Further, since the sum and product of the roots of (9.7) are a + b, ab — h2respectively, these roots are both positive provided a > 0, ab > h2, (implying b > 0 also). This is simply the condition that the quadratic form in (9.1) giving the potential energy should be positive definite, illustrating the fact that a position of equilibrium is stable only when it is a position of (strictly) minimum potential energy. [The case of equal roots arising when (a — b)2 + 4h2 = 0, i.e., a = b, h = 0, ,

352


occurs when the bowl is a surface of revolution.] The amplitudes of x, y in the vibrations corresponding to the two solutions of (9.7) are related by the equation

F— h p2Ig — a ' where p takes in turn the values p1, p2given by the roots of (9.7). These vibrations are therefore in phase. Then

x Alcos (pit

, y =

or

x = A2 cos (p2t + E2),

y

—

pi/h — a Alcos (pi t + 8,) 2 a h— A 2 cos (p2t + 82).

P2ig

The general solution is the sum of these solutions where Al and A2 are arbitrary and is, therefore, an arbitrary combination of two characteristic vibrations. We note that, if {F1G,}, {F2 02} correspond to the roots A. , respectively, then

F1 F2 0 1 02 = k[h2(2971g — a)(Alg — a)] where k is a constant of proportionality. From eqn. (9.7) we have (pi

F1 F2

lag = a + b, 24 .241g2= ab — h2 . = k2 [h2 a 2 — a (a ± b) ab —

= O.

This shows that the "vectors" {F1 Gil and {F2 02} are perpendicular. Since axe 2hxy by2 is a positive definite quadratic form, rotation of the coordinate axes about Or from Oxy through the angle 8 given by h cos 28 + z (b — a) sin 20 = 0, to 0 i where ,

= x cos0 y sin°,

= — x sin° y cos0,

(Vol. I p.143) reduces V to a sum of squares and gives T = 2 m(2

1) 2), V = z m(131 2+ 1477 2),

(9.8)

where pi + p2 = (a + b) g, pi p2 = (ab — h2) g2as defined above. A section, z = constant of the bowl is an ellipse for small values of z and the rotation of axes to give (9.8) is a rotation about Or from Oxy to the principal axes of this ellipse. The equations of motion

=0, +

=0

§ 9: 2]


353

imply that the vibrations along the directions W, On with frequencies p j1(27r), 2o2/(27-t) are independent. These new coordinates n are linear combinations of the old coordinates x, y and are called normal coordinates of the system. The motions in the two normal coordinates are each harmonic and are independent of one another. When the system oscillates so that one normal coordinate remains zero whilst the other normal coordinate varies, the system is said to be oscillating in a normal mode. Any other oscillation is an arbitrary combination of the two normal modes. NOTES

(a) An arbitrary general motion cannot be periodic unless pi/p, is a rational number, i.e., in general no position and motion of the system can be repeated in a finite time. (b) From (9.8) we see that when we use normal coordinates the potential energy and kinetic energy of the system can each be expressed as the sum of squares. Since the forces are conservative and the motions in the normal modes are independent of each other, the energy in each mode is conserved and the total energy of the system is the sum of the energies in the modes separately. (c) We have seen that when eqn. (9.7) has equal roots the bowl is a surface of revolution; in this case any pair of perpendicular axes 0 xy can serve as principal axes and the coordinates x, y are normal coordinates. A particle set oscillating in a plane through Oz remains in that plane; it only remains in the initial plane for a non-spherical bowl if that plane corresponds to = 0 or n = 0. A method whereby the normal coordinates can be obtained directly in the case of systems with two degrees of freedom is given below and illustrated in the subsequent examples. Eqn. (9.2) + A (9.3) can be written d2 (x + Ay) + g{(a + Ah)x + (h Ab) y} = O. d t2

(9.9)

We choose the constant A so that this equation is a differential equation in the dependent variable (x + 2 y) by taking 1 a + 2h

A h Ab ' h A' + (a — b) A — h = O. 12 a

CMES

(9.10) (9.11)

354


[Here, to obtain (9.10), we have equated the ratio of the coefficients of x and y in the first term of (9.9) to the ratio of these coefficients in the second term.] Then, if Al , A2 are the roots of (9.11), we obtain two harmonic equations in the normal coordinates = x y, = x + A2 y. Examples. (i) A light inextensible string ABC has AB = 3a, BC = 2a. It is tied to a fixed support at A, a particle of mass 2m is attached to the string at B and a particle of mass m is attached at C, so that the system hangs in equilibrium in a vertical straight line. Prove that for small oscillations in a plane through the equilibrium position the normal periods are 2n (alg)11 and 47r (alg)'ls. If 0, sp are the (small) angles of inclination of AB, BC to the vertical, find normal coordinates for the system in terms of these angles.

(i) FIG.

98.

In this case, see Fig. 98(i), the kinetic and potential energies for small oscillations:, i.e., correct to the second order, about the equilibrium position 0 = 0 = 0 are

T = im a2(27 62 + 1204, + 44)2), Lagrange's equations gives 276 + 6

V = imga (902 +2 T2) .

+ 9n2 0 = 0,

6 6 + 40 + 2n2

=0,

(1) (2)

where n2 = g la. Then eqn. (1) + A(2) give (27 + 6 A)6 + (6 + 42) 45 + n2 (90 + 22,92) = O.

(3) We choose 2 so that the ratio of the coefficient of 0 to the coefficient of cp equals the ratio of the coefficient of 0 to that of i.e., 27 + 62 6+42

9 22

§ 9:2]


355

The resulting quadratic, 222+ 32 — 9 = 0, has roots 2 = 3/2, — 3. When A = 3/2, eqn. (3) reduces to d2 n2 d 1 (30 + 97) + — (30 + 99) = O. 4 When A = — 3, eqn. (3) reduces to d.2 dt2

(30 —2p) + n2(30 — 299) = 0.

Hence the periods of the normal modes are 4n/n and 291/n and the corresponding normal coordinates are proportional to 30 + 9), 30 — 2p respectively. Therefore we can write + r2, 32= —97, 90 =

T = m a2 (8i2 + 9,j2) = m a2 (3S2 + i12) V = 1m a2 (n2 +n2) imga (e• g e + 17,2) . This corresponds to eqns. (9.8) except that n are here proportional to the normal coordinates in (9.8). NOTES (1) In this case the motion is periodic. (2) When the system is oscillating in the normal coordinate (so that n = 0, i.e., 30 = 2p), the configuration is as shown in Fig. 98 (ii). When oscillating in the normal coordinate 77, (30 + = 0), the configuration is as shown in Fig. 98 (iii). (ii) A bead P of mass m is threaded on a smooth uniform circular wire, of mass M, centre A and radius a; one point 0 of the circumference of the wire is fixed and the wire can rotate freely about 0 in a fixed vertical plane. If the system is slightly disturbed from its configuration of stable equilibrium, find the normal coordinates and the periods of the normal vibrations. Find a condition for the motion to he periodic. Let the general configuration be defined by the (small) angles 0, p made by OA, AP with the downward vertical, Fig. 99. Then the kinetic and potential energies are given (correct to the second order in 0, q9) by

T = a2 {(2M + m) 02 + 2mOp + mp2} ,

V = lga{(M m) 02+ m02}.

Lagrange's equations give (2M + m) 6 + m0 + n2 (M + m) 0 = 0,

(1)

6+q +n2 = 0,

(2)

where n2 = g la. Eqn. (1) + 2 eqn. (2) gives (2M + m + 2) 6 +(m + 2) 0 + n2 (M + m) 12 a*

2 n2 = 0.

(3)

356


We choose A so that the ratio of the coefficients of 0 and ()) in (3) equals the ratio of the coefficients of 0 and 9, i.e., so that m+ A 2M +

+

M + m'

or

A2 + M A — m (M + m) = 0 , whence

A=m or — (M + m). When A = m we find 2 {(M + m)

+ m q5} + n2 {(M + m) 0 + m99} = 0 ;

(4)

when A = — (M + m),

M (6 — q;) 4- n2 (M + m) (0 — 99) = 0 .

( 5)

FIG. 99. Therefore the normal coordinates are proportional to {(M + m) 0 + mq9} and (0 — 99). Clearly from (4) and (5) the periods of the normal oscillations are 27rI /( M

27ry 2

n

n

—+m

The motion is periodic iff (i.e., if and only if), the ratio of these periods is a rational number, i.e., iff

V2

19 {M/(M + m)} — q

or 2(M + m) q2= Mpg,

where p and q are integers. (iii) The kinetic and potential energies of a dynamical system are given by 2T =

02+ 472

2 V = oc02+ 26099 + oc992,

where a and 8 are constants. Prove that the motion of the system is one of vibration provided cc > Is!, and find the periods of the normal modes.

§ 9: 2]


357

Show that there is a possible motion in which 0 = A sinpt cosqt, p and q being constants to be determined, and find the corresponding value of y. Describe also the motion if 8 is small. Lagrange's equations are 6 + cc0 + ecp = 0, q + 80 + aq2 = O.

Then, by the method of examples (i) and (ii),

(6 + 4) + {(a. + Ae) 0 + (8 + Aa)(p) = 0, and the equation giving A reduces to 22 = 1. Therefore d2 do (0+ 50) = (oc ± 6)(0f(p), taking the upper or lower sign throughout. Each of these equations represents a harmonic vibration provided oc f e > 0, i.e., provided oc. > 161. In this case the normal coordinates are (0 ± q2)/ y2 and the periods of oscillation are 27c/y(oc f s). A possible motion is given by 0 +q = Asin {t l/(a+ e)}

0— = A sin {t V(a — a)}, i.e., where

0 = A sinpt cosqt, 99 = A sinpt cospt , 2p = 11(a + e) + (oc — a), 2q = 1,I(a + 8) — 11(a — e)

[We have chosen equal amplitudes for the oscillation in each of the normal coordinates.] When a is small, so that 62can be neglected in comparison to oc, p = y a , q = e y • and = A sin (t cc) cos( et] I a), 92 = A sin (Is tVoc.) cos (tf oc) Near t = 0, the motion (and energy) is primarily one of harmonic variation in 0, with period 29r/yoc, q2 remaining small. As t increases towards 96/(a yo) the motion and energy are gradually transferred to the 92 coordinate. In fact we can regard the motion in each coordinate as a harmonic motion with period 2mlya and slowly varying amplitude A cos (fat Vac) as illustrated in Fig. 100.

358


If e = 0 the expressions for T and V show that the motion consists of two independent, identical oscillations. The term 2e0g) in V when c is small represents a "weak coupling" between these two oscillations. The effect of this coupling is to produce two normal frequencies close to, but on either side of the original frequency yoc/(2n). The ensuing motion shown above is one in which the oscillation having the original frequency passes to and fro between the two original oscillators. This phenomenon of weakly coupled vibrations is very common not only in mechanics but in other types of oscillation. Exercises 9:2 1. A uniform rod AB of length 2a and mass M is suspended from a point A which can move horizontally subject to a restoring force — Ax, where xis its displacement from a fixed point. Give expressions for the kinetic and potential energy of the system in terms of x and the angle 0 made by the rod with the vertical, and write down the equations of motion. Show that if 0 is small the equations of motion reduce to .. A + a 61 + — x = 0,

m

+ ta6 + g0 = 0, and hence find the equation determining the values of n for which the system can perform small simple harmonic oscillations with period 2n/n. What happens when A=0? 2. A heavy uniform circular ring of radius a is suspended from a fixed point 0 by a string OA of length b attached to a point A of the ring. The diameter AB of the ring, and the string OA, make angles 0, and 99, respectively with the downward vertical. Motion takes place in the vertical plane of the ring. Show by means of Lagrange's equations, or otherwise, that, if 0 and g) are always small,

a0 + 2a8 +

+ gg9 = 0, + g0 = 0.

If the motion is started in such a way that both coordinates oscillate harmonically with period 2n/p, show that permissible values of p are determined by the equation 1

2\

g2

0. p4 g (— + --)p2 + a b ab = 3. Two uniform rods AB, BC are freely hinged at B and hang from a smooth pivot at A. The mass and length of AB are 2m and 1, and of BC,m and 21, respectively. If the system performs small oscillations in a vertical plane about its position of equilibrium, show that in one normal mode the inclinations of the rods to the vertical are in the ratio — 1:1, and in the second normal mode in the ratio 2 : 1. Prove that the corresponding periods of oscillation are 2n1/ (//3g) and 2n1/(111/60).

§ 9 : 2]


359

4. A uniform rod AB, of mass 5m and length 2a, turns freely about a fixed horizontal axis through A, perpendicular to the rod. A light rod BC, of length 2a, is freely joined to the rod AB at B, and carries a mass m at the other end C. If in a displaced position AB, BC make angles 0 and 92 respectively with the downward vertical, show that the normal periods for small oscillations in the plane ABC are given by 27r/co,, 27r/w2 where 2a co! = 3g and 20a ct4 = 7g. Find the two corresponding normal modes of vibration and indicate these modes in a diagram. 5. A uniform rod AB, of mass 3m and length 2a, is freely pivoted at its centre C; a light elastic string, of natural length a, is fastened to the end A and to a point vertically below C and distant 4a/3 from it. A particle of mass m is suspended from the end B by a light inextensible string of length b and in the position of equilibrium the rod is horizontal. Prove that the principal periods of small oscillations in a vertical plane are 27r (bIg)112 and V7r(a/g)'1.. 6. A uniform cicular disc of centre 0 and radius a has a point P on its circumference smoothly hinged to one end of a uniform rod of equal mass and length 3a which has its centre C fixed. The system makes small oscillations in a vertical plane. Find the normal coordinates, the periods of the normal modes, and the general expressions for the inclinations 0, p of CP, PO to the vertical at time t. Show that, if CP, PO are initially vertical, it is possible for the system to move (i) so that 0PC remains a straight line, (ii) so that 0 remains in the vertical through C. 7. A thin uniform circular cylindrical shell of radius 2a can swing about one of its generators which is fixed horizontally, the plane through the axis of the cylinder containing the generator making an angle 0 with the vertical. Within the shell is placed a uniform solid circular cylinder of radius a and of the same mass as the shell. The plane through the generator of contact and the axis of the solid cylinder makes an angle 92 with the vertical. If the system is initially in equilibrium, show that if it is slightly disturbed and there is no slipping at the common generator, the normal coordinates in the subsequent motion are 0 — (1 + 4y5) 92, and the lengths of the corresponding equivalent simple pendulums are 1(5 + y5) a. 8. If the kinetic and potential energies of a system are given by the equations 2T = 47+

4; 2V=pi gi+pigs

,

where pi, p, are nearly equal constants, show that, if the coordinates be transformed to ul, u2 , where = qi Pq2, u2 =

igg2,

a, /3 being any constants, then, if initially the motion is entirely in the coordinate u„, it will be so again after an interval 2n7r1(pi— pa), where n is an integer; but after an interval (2n + 1) nl(pi— pa), the motion will be entirely in the coordinate u2 . 9. A car has a wheelbase of length 3a and its mass-centre is at a horizontal distance a forward of its back axle. The stiffness of the back springs is 3s, and that of the front springs 28. Find the frequency equation for the normal modes of

360


oscillation (there being no lateral motion) in terms of the constants already given, the mass m and the radius of gyration k about a horizontal transverse axis through the mass-centre of the car.

9:3 Stability of equilibrium: free oscillations of a system with n degrees of freedom In Vol. III §§ 9:2-3 we discussed the stability and oscillations of a system with one degree of freedom under the action of conservative forces. We now give the corresponding discussion for the more general case. The system we consider is a holonomic system with n degrees of freedom in which the forces are derived from a potential function V (q1,a ••• qn) and the constraints do not depend explicitly on the time. (If the constraints do depend upon the time the question of equilibrium hardly arises; the steady states are states of steady motion already considered.) We suppose, without loss of generality, that in the equilibrium position q, = q2 = = qn =0 and that V = 0 at this position. Now, equilibrium being a special case of motion, the "motion" given by = q2 = = qn = ° ql = 42 = = = , ql = q2 = = qn = °

must satisfy Lagrange's equations of motion d dt

(aT)

0 71 a qi

av

=o

(i= 1, 2, ..., n) .

(9.12)

The asumptions we are making imply that T = z ai k4, 4„ (see § 6:2), so that for the equilibrium position the equations of motion reduce to

av = 0, qi

(i = 1, 2, ..., n).

(9.13)

This shows that V has a stationary value (chosen to be zero) at the equilibrium position. To investigate the stability of this equilibrium we consider the solution of eqns. (9.12) after a small, arbitrary, disturbance from equilibrium has been given to the system. The equilibrium is stable if all the coordinates, and their derivatives, remain small. In these circumstances we can write, correct to the second order,

T — aiko4, q„,

V — z bi , qi qk •

(9.14)

§ 9 : 31


361

The quantities a, ko are the values of ai , at the equilibrium position: since the factors qi 4, make every term of the second order, we need not consider (first order) variations in these coefficients and we treat aiko as constants. (With this understanding we shall omit the suffix 0 in the remainder of this section.) If we expand V by Taylor's theorem about the equilibrium position we obtain (q1 q2, "' qn) = bt k qi qk

° (0),

where

bik —

a2v aqk )o .

These coefficients are, therefore, also constants. The earlier terms of the Taylor expansion vanish because of our choice of zero level for potential energy and because of eqn. (9.13). Because the kinetic energy can never be negative, or zero, for any possible motion the quadratic form for T must be positive definite. Using the forms for T and V in eqn. (9.14) we obtain Lagrange's equations of motion as at k qk

bik

qk =0, (i = 1, 2, ..., n),

(9.15)

(note the summation over k). Eqns. (9.14) and (9.15) can be written in matrix form T = 2q'Aq V = (9.16)

A.4

Bq = 0,

(9.17)

where q = {ql q2 • qn}

A = (aik ), B = (NO

The equilibrium is certainly stable if q (i.e., every element of it), is the sum of quantities which vary harmonically with the time with small amplitude. (C.f. § 9:2 in which the particle remains near 0.) We therefore seek solutions of (9.17) of the form q = aei ", where p is a real number. (We use complex numbers on the understanding that the actual coordinates q1are the real parts of the elements cxjei"). Hence (B p2 A) a = 0.

(9.18)

Since the column matrix a is not zero, the matrix B — p2 A must be singular, i.e., (9.19) det (B — p2 A) = 0.

362


In Vol. IV we discussed the solution of equations such as (9.18) and (9.19). It was there shown that the roots of eqn. (9.19), an equation of the nth degree in the unknown p2, has roots 24, ..., p2„ which are all positive if and only if the quadratic forms of B and A are positive definite. The general solution of eqn. (9.17) is then a linear combination of terms such as ak et Pkt and such terms vary harmonically with t only if all the p, are real. (If some of the p: are negative or complex then there will be at least one value of pk = oc — > 0; this leads to a corresponding term in the solution ock eet el"". Since # > 0, this term cannot remain small for all t.) If one, or both, of the quadratic forms of A and B are not positive definite, then at least one of the roots of eqn. (9.19), considered as an equation in p2, is negative. Hence we deduce that, if the value of V at the equilibrium position is a strict minimum, the equilibrium is stable, i.e., V having a minimum is a sufficient condition for stability. If V has a "saddle point" in the equilibrium position the equilibrium must be unstable, because at least one term in the general solution for q increases indefinitely with time and so invalidates our assumption, on which eqns. (9.14-17) are based, that q remains small. (Note that we have obtained sufficient conditions for stability or instability; there is a variety of critical cases which must be discussed separately; see Vol. III Chapter IX for one-dimensional examples.) Further results given in Vol. IV concerning the solution of eqns. (9.18) and (9.19) enable us to investigate the oscillations which take place about a position of stable equilibrium. It was shown that corresponding to different roots 23: of (9.18) there are "characteristic vectors" aj, ock which satisfy a; A ak = .

It was also shown that if one, or more, of the roots At is repeated f times, it is possible to find f linearly independent vectors amt , (i = 1, 2, ..., f), which satisfy (B

—

p2,„A) a,„1= O, oc: „, A anii =0, (i

j, i, j = 1, 2, ..., f)•

Hence we can say that, whether the roots pi! are distinct or not, we can always find n linearly independent vectors °C2 ,••', eXn which satisfy (B

—

p! A) ock = 0 , a; A ak =6ik .

(9.20)

Since the vectors ack can be multiplied by a scalar and still satisfy (9.18), we have multiplied them by an appropriate normalising factor in order to make them satisfy the second of eqns. (9.20). (If the vectors

§ 9: 3]


363

are not normalised we can say that ar, A ak =0, but that the quadratic form when j = k is not, in general, unity.) The reader should compare these results with those in § 9:2 in which the two characteristic vectors are perpendicular (there A = I), and when eqn. (9.7), corresponding to eqn. (9.19), has equal roots we can find pairs of perpendicular axes to serve as normal coordinate axes. Having determined n characteristic (normalised) vectors ak we can write down the general solution of (9.17) in the form Ak ak e'Pkt,

q

(9.21)

k=1

viz., a linear combination, with complex coefficients Ak , of the individual characteristic solutions. The real part of the r.h. side of (9.21) has the form q=

Bki3kCOS(pkt

ak).

(9.21a)

k

In (9.21a), Bk , sk are arbitrary, replacing the arbitrary real and imaginary parts of Ak. The same phase constant sk refers to each coordinate qt in the matrix q, i.e., all the coordinates oscillate in phase together when all the Ak (or Bk) except one, are zero. Then the system is oscillating in a normal mode. (Cf. § 9:2 where normal modes are introduced, and where the same phase constant, el or 82, occurs in the expressions for x and y.) To find the normal coordinates •" En, we form the matrix R whose columns are the characteristic vectors al , a2 , ..., an , i.e., R = (al a2 ... an).

(9.22)

Then the normal coordinates are the elements of the column matrix given by

g = {e, e, ... en

}

or

q=R

g=

R-1q.

(9.23)

(cf. the rotation of axes in § 9:2.) We give in the following examples the expressions for T, V and the equations of motion in terms of normal coordinates. Examples. (i) We prove that T =

n

E a, V = E pnP

The matrix product AR is made up of columns A ak , i.e.,

AR = (Ace, A ce2 A cf„).

364


The transposed matrix R' is made up of rows cf;„ i.e., n].

'

)

e:e R' AR = ( • (An' A n2 A nn)

nh (

a, Aai c4Aa2 ... al A ak \ c4Aa1 a21 Aa2 .

\

0 1

= I.

0

an Aak

an' A al

0

/1

1 11

2 =1

----“'R'Alt& =1-

f.

Also BR = (Ba1 Bat B an).

But, from (9.18), Bak = AA.ock . an).

.•. BR = (AA cc, AA ci2

.•. R'BR

AAoc2

pTallA a, .24a4A oc2

pnaiAan

pi c4 A al c4A oc2 pin,c Ant

p2n Acen)

pi 0 =.

pn

0

p n2 an' A e„

0

0

= diag (p!, pi, .. pn2 ) V = 1-q'Bq = igifeBR =14' [cliag

,

,

g.,)] =

e• i=1

(This working is an example of a linear transformation q = R which diagonalises A and B simultaneously.)

(ii) When we apply this same transformation to the equations of motion, eqn. (9.17) becomes Alle BRA = 0.

§ 9: 3]


365

We multiply on the left by R', obtaining It'

leB.F4 = 0

i.e.,

, p„2) g = 0.

diag

Expressed in terms of the elements this equation is equivalent to

4; +

= 0 , 4.2 +

± .22,21 =

E2 =

0.

This shows that oscillations in any one of the normal coordinates take place independently of the other normal coordinates.

In the following we apply these methods to some specific examples. The reader should follow through the detailed working in these examples in order to become thoroughly familiar with the procedure. Examples. (i) A uniform rod OA of mass 3m and length 2a swings about a smooth horizontal axis at 0, and a light elastic string AB of natural length a and modulus of elasticity mg is fastened to its lower end A and has a particle of mass m attached at B. If the system oscillates in a vertical plane, the rod and string making angles 0, so respectively with the downward vertical, and the extra extension of the elastic string caused by the small oscillations being x, explain the nature of the normal modes of oscillation in terms of 0, co and x. Show that the motions in the three normal modes are synchronous with those of simple pendulums of lengths 2-a, 2a, and 4a.

FTC. 101.

Using Fig. 101 we find that, correct to the second order,

1 T = — m(8a2 O2 + 12a20 2

+ M2 02 + 2) ,

(g a

0 ) (1O a2 2 + 6a + x2) , V = 2m —

366


where we have written AB = 3a + x. (We choose this form since in equilibrium AB = 3a.) Lagrange's equations can be written in the matrix form (16 12 0 ) a0 12 18 0

0

a 43

0

q2 — 0 .

+ (gla) 0 6

0 0

x

0 2

a0_

10 0 0

x

1

0

The equation giving the period can be written 10(g/a) — 16p2

—12p2

0

6 (gla) — 18p2

—12p2 0

= 0,

0

(g/a) — 2p2

0

This has roots p2 = g/(4a), 5g/(3a), g/(2a) and the corresponding characteristic vectors (see Vol. IV) are proportional to a, = {1 2 0},

ay = {6 —5 0},

ct 3= {0 0 1}.

Instead of normalising these, as described in the text, we can work with the unnormalised vectors using a matrix 1 6 0

S= 2 —5 0 . ( 0 0 1 Then the coordinates x = {x1 x2 x3} given by q = Sx are proportional to the normal coordinates. The kinetic energy is T = imei` Ai 2

0) 16 12 0

= iin,'(1 6 —5 0 0 (40 = iniP 36 0

0

1

0

2 —5 0

0

0

1

0 2

48

0

1

6

0

—18

0

2

—5

0

2)

0

0

1

0

6

12 18 0

= m(136 4 + 3064 + 2 4) .

0

1

= im de

136

0

0

0

306

0

0

0

2

§ 9 : 3]

THE OSCILLATIONS OF

367

A DYNAMICAL SYSTEM

And the potential energy is (1

V= 19ng q' Bq = imgx' 10

= inagx' (60 0

2

0

10

0

0

1

6

0

6

—5

0

0

16

0

2

—5

0

0

0

1)

0

0

1)

0

0

1

12

0

1

6

0

— 30

0

2

—5

0

0

1

0

1) 0

x

34

0

0

0

510

0

0

0

1

x = +mg x'

x

= img (344 + 5104 + 4) = img(136 . /4.+ 306 . M + 2 . i4). This shows that the frequencies are associated as follows:

1 g

2

=

with xi ;

=

a

3

1 g = — — with x3. 2 a

with x2;

Since (5

6

0

—1

0

0

0

17

(5

6

0

a0

0

aq2 ,

S-1= -I-,2

x = S-lq = ik 2 0

—1 0

,

17 )

x

xi = a(50 + 60/17, x2 = a(20 — 9907, x3 = x. The actual normal coordinates are a — 1/136(50 + 6q))

e2

a 1/ 306

(20

q)) x3= x/1/2.

(ii) Two identical uniform rods BC, CD each of mass m and length 2a are freely jointed at C and rest on a smooth horizontal table. Two light elastic strings AB, DE are attached to fixed points at A, E respectively so that ABC D E is a straight line, with AB = DE = 2a. When the system is at rest the tension in AB and DE is F. The system performs small, free oscillations so that C moves in a line at right angles to A.R. Calculate the kinetic energy T of the system correct to the second order in the velocities and show that if x, y, z denote the displacements of B, C, D, respectively, perpendicular to AE, the potential energy of the system is

V = F (x2 + y2 + z2

—

xy — yz)/(2a).

Prove that the frequency of the normal mode of oscillation that is anti-symmetrical about C is (3F/ma)'/./2n. Prove also that the symmetrical modes have frequenciesp,I2n and p2/2n, where map = F(4 + V13) and map = F(4 — V13).

368


Let B1, C1, D, be the feet of the perpendiculars from B, C, D to the line AE, Fig. 102. Then, since C moves perpendicular to AE, AC, = 4a = EC,. —

.• A Bi = 4a — 1/{4a2— (y — x)2} = 2a + (y 4ax)2 + 0

(y a3x)4} —

• (1)

Similarly

1

ED = 2a+ (y z)2 +0 {(Y z)4 4a

t a1

(2)

Therefore the velocities of B1, D1along the line AE are of the second order in x, y, z and, using the formula T = *m(u2 + u . v + v2) of the example on p. 148 for the kinetic energy of a rod, we find

T = im(±2-F jc9 + 292 + 9i + 2) correct to the second order. Using (1) and (2) we find X2 ±(y — x)2

AB =11 (ABT + x2) = 2 a + Similarly

z2

DE = 2a -I

4a

(y

+ fourth order terms.

z )2 + fourth order terms.

4a

The extensions of the strings from the equilibrium state are therefore of the second order and hence we obtain the potential energy of the system (the work done against the tension) by multiplying the extension of each string by F and adding. .'. V =

(x2 + y2 + z2— xy — zy)/(2a).

Lagrange's equations are 2x+

+

+

3F

+ — (2x — y) 0, ma

3F ma

2y — z) = 0,

(3) (4)

3F

9 + 2i + — (2z — y) = 0. ma

(5)

For the anti-symmetrical mode x = — z, y = 0. In this case (4) is satisfied identically and (3), (5) become 3F x 3Fz = , 7. — ma ma

§ 9: 3]


369

and so the anti-symmetrical mode has period 2zi y{ma/(3F)), i.e., the frequency is i{3 I (m a)}1/2/zr as stated. For the symmetrical modes x = z. In this case (5) becomes identical with (3) whereas (4) reduces to 3F X + 2y + — (— x + y) = 0. (6) ma Then (3) + 2(6) gives 3F (2 + A) X + (1 + 2A) 9 + — {(2 ) — 2 x + (— 1 + A) y} =0. ma Choosing A so that 1 + 2A — 1 + 2— 2 + A.

i.e., 3A' — 2A 4 = 0

with roots Al , A, = (1 + y13)/3, we obtain the frequencies of the normal modes as 3F(2 — A) 1} 1 and the corresponding normal coordinates as (2+ A) x + (1 + 22)y 2n II ma(2 + where A takes the values 2,, A,. These give the required frequencies. (iii) The kinetic and potential energies of a mechanical system are given by 2T = ±2 1/2 a 2 2 V = p2 (x2 + y2) + 2aa0(x — y) + q2a2 02 , where a is small compared with p2, q2, and p2— q2 . Prove that the normal frequencies of vibration about equilibrium are a2 P P

p (p2

q2)

q

q (p2 q2)

neglecting terms of order a4. Determine the normal coordinates (correct to the first order in a I p2). Lagrange's equations are equivalent to 1

(

0

(2

0) ( X )

0

1

0

9

0

0

1

a8

+ ,

0

x

y = 0 .

0

p2

—a

a

—a

q2

a0

The period equation (period 270) is p2— 22 0

0 p2

-

22

a (p2

22) [(p2

22) (,2

0

a

— a = 0, q2 22 22)

2,„2] _ 0.

0

370


Since a is small compared with p2, q2, p2— q2, the roots of (p2 _ 22) (q2 — 22) — 2a2 is a root of this quadratic in 22. Then = 0 are close to p2 and q2. Suppose p2 e 2a2

e= p2

q2

2a2

e

8

p2 q2

neglecting terms of 0 (a4). Similarly the root near q2 is q2— e. The values of A are therefore p, il(p2+ e), y(q2— e) which reduce to the given expressions on using the binomial theorem. The corresponding characteristic vectors are, neglecting terms of 0 (a2/p4), 1 1 2 = p2, {1 1 0} normalised vector — — ;

1/2 1/2

22= p + e, {a — 22 = q2

1

e} normalised vector / 112

a a p2

8

V2 a 1/2

q2) normalised vector

pz

1,2-1%2

-

q2

Therefore, with the notation of the text, 1 V2 R=

—

1

V2

q2

p2

1

1

V2

V2

p2— q2

•

1

aV 2

Since A = I, this matrix R is orthogonal and so the normal coordinates are given by 1 1

g = R' q =

V2

V2

1

1

V2 —a

V2

p2

q2 p2

q2

0

al I 2

Y a0 ) •

1

Therefore the normal coordinates are sel

=

x+y

x—y

ea0

V2

a1/2

a (y — x) p2

q2

+ a0 .

(iv) Two particles of equal mass 711, lie on a smooth horizontal table at points P, and P2 which have polar coordinates r1, 01and r2 , 02(01 > 02) referred to the fixed point 0. The particles are joined by a light elastic spring and each is joined to 0 by a further spring. All three springs have the same modulus 2 and the same natural

§ 9 : 3]

371


length a. Express the kinetic energy T and the potential energy V in terms of coordinates defined as follows: = 0, — 02 --

2a$ = ri — r„ 2a77 = ri + r2 — 2a,

, 2y = + 02.

The system is in a state of motion such that are small and the generalised momentum corresponding to tp is zero. Show that to the second order of small quantities T and V are proportional to e+ 972 + i2 and 2V + 3172 + 2 3/37a + 3x2. Hence find the ratios of the periods of the three normal modes. Indicate in a sketch the nature of the three types of oscillation. With the given notation

T = m + 4 + 46! + 4 OD {(i + 71.)2 +

+ 77.)2+ (1 + $ + 77)2 (5c + v.))2 + 02) ,

+ (1

V=

a

[(r1— a)2 + (r2— a)2 + (1/

A = — [(4 + a

+ — 2r1r2cos (01— 02)) — a)2]

— a (ri + r2 ) + 3a2/2 — r1r2cos (01— 02)

— al/{rT +

4—

2r1r2cos (01— 02))]

= A a [2 (1 + 27)2 + 2v — 2(1 + 71) + 3/2— {(1 + 77)2—

— {2(1 + 77)2 + 2v — 2[(1 +

- $2] cos

cos (i n

+ 2x)

+ 2x))].

Since the motion is such that the momentum

T P= — 011) =0, i.e.,

[(1+ 71)2 +

= 0.

tp + 2$(1 + 97)

When $, 77 are small, of the first order, this relation shows that 1/., is small of the 2x. Therefore, in the expression for T we may put = 0 and second order; we find, after some reduction and neglecting terms of higher order than the second, ,

x

T = ma2 (i2+ 77.2 + V = l2a(2$2+ 3772 + 21/3nx + 3z2). Lagrange's equations 1

0

o)

i.

2

0

0

$

mat 0

1

0

:4

+1Aa 0

3

V3

n

(0

0

1

i

0

113

3

x

0 =

o 0

372


give the period equation _ p2

0

ma 0

3 2ma

0

V32 2ma

0

V3 2ma

p2

32 2ma

=

0.

p2

This equation has roots p2 = 2/(ma), 2(3 + 1/3)/(2ma). Hence the periods of the three normal modes are in the ratios 31: V(1/ 3 + 1) :1/(1/ 3 — 1) .

(i)

(ii)

Fla. 103. The characteristic vectors corresponding to the normal modes are {1 0 0) , {0 1 1) , {0 1 — 1) . The normal coordinates are proportional to normal oscillations are: o=x

(ii)e (iii)

o, =0,

,

i.e., ri = a (1 +

=x =

,

— x ,

—x

r2= a (1 —

i.e., r = r = a (1 + , i.e.,

+ x and the corresponding

,

= r2 = a (1 + 71),

01— 02 =

—

02 =

e, —

02 =

,

— 2n,

+ 271.

In (i) the angle P10 P2 remains constant but P1, P2oscillate about their equilibrium positions in antiphase with equal amplitudes. In (ii) and (iii) r1 = r2 and P1, P2oscillate in phase but the angle P,OP, increases and decreases respectively with these lengths. The results are shown in Figs. 103 (i), (ii) and (iii) respectively.

(v) A light string under a large tension P stretched between fixed points x = 0 and x = na carries n — 1 particles of equal mass m at the points x = ra,

§ 9: 3]


373

r = 1, 2, . - 1. Show that the small uniplanar transverse vibrations yr of the particles are given by n-1 sgr cos(ps t + ss), Cs sin Yr = s= STC

where ps = 2 (P/ma)V2 sin 2n , and Cs, esare constants determined from the initial conditions. A,

A,

Ar*2

ir--------r y+

----

t

Yri-2

i

13,1

B,

Br+2

FIG. 104. Let the equilibrium position of the rth particle be B„ Fig. 104, and the displaced position at time t be A„ where Ar Br = y,.. Then following the method of the example of p. 347, the equations of motion are clearly ma 9,.+1= — P (Yrd-i — Yr) + P(Yr+2 — Yr+i), Pr+1

(4)2 (Yr+2

2 Yr+1

where

± Yr)

(r = 1, 2, ..., n — 1),

0

=

(1)

and we take yo = 0 = ys.

cot = P/(ma) Assuming a solution in which Yr

we find A.7.+2

=

(2)

Ar e'vt

- (2 — p2/(02) Ar+, + A, = 0 ,

r = 0, 1,

(n — 2),

(3)

with A, = 0 = A„. The solution of (3) is Ar

=

C2m2 ,

(r = 0, 1, ..., n)

(4)

where m1, m2 are the roots of the equation m2— (2 — p2/co2) The condition A 0 = 0 gives C, = — C2

=

.• A, =

+ 1 = 0.

(5)

C. — m2)•

The condition An = 0 gives in; = m2, i.e., m1 = Z17/2,

where z" = 1. Hence z = 027' isln, s = 0, 1, 2, .. , n — 1.

(6)

374


From eqn. (5) we have m,m2 = 1.

enign, m2

.•

(7)

Also, mi. + m2 = 2 — p2/co2, 2 cos (ns/n) = 2 — p2/w2,

ns p = 2co sin (—),

(8 = 1,2, ..., n — 1).

2n

(8)

The values for m, and m2 in eqn. (7) give

1 c(enisrin

A,

e-nisrin) = iC sin

nsr

9

2 leading to the expression for the sth mode yr = i C sin

7r8r

n

el Ps' , (8 = 1, 2, ..., n — 1).

(9 )

We omit the case s = 0, since eqn. (8) shows that for this value p = 0, i.e., there is no oscillation. Writing i C = C, ei6s and superposing the n — 1 normal oscillations gives the required result. Exercises 9:3 1. A uniform rod, of mass m and length 2a, is suspended by a light elastic string of natural length a/3 attached to one end. If the rod be given a slight displacement in a vertical plane, prove that the lengths of the simple equivalent pendulums are in the ratio 30 : 7 : 2, the length of the string in the equilibrium position being 4a/5. 2. A smooth circular ring of radius a has a particle of equal mass fixed at one point of it and rests in stable equilibrium over two parallel horizontal rails with its centre at a depth c below the rails. Show that, if the string is slightly disturbed so as to remain in contact with both rails, the periods of the normal vibrations are equal to those of simple pendulums of lengths 2a,

3a2 + 4c2 + 4ca 2(2c + a)

a2 c 4(a2— c2)

3. One end of a light elastic string, of natural length a, is attached to a fixed point and the other end of the string is fixed to an end of a uniform rod, of length 2a. When the rod hangs vertically in stable equilibrium the length of the string is b. Prove that the periods of small normal oscillations in a vertical plane about this

b —a

position are 2n and the two values of 2n/p, where p is given by the 11() equation 3g2p2 — g(4a + 3b)p + ab = 0. 4. Three particles of different masses are attached to points of a light string and hang in equilibrium, one end being attached to a fixed point. Prove that if the

§ 9: 3]

375


periods of the three normal modes of oscillation in a vertical plane are the same as the periods of simple pendulums of lengths Al , A2 , A3, then Al + A2 + A3 is equal to the distance of the lowest particle from the point of suspension. 5. A light string ABODE is stretched at tension T between two fixed points A and E at distance 4a apart, and particles each of mass m are attached at B, C, and D, which are equally spaced along the string. The system rests on a smooth horizontal table and the particles can oscillate transversely. Show that the squares of the reciprocals of the periods of the normal oscillations are in the ratios 2— 1/2 :2 :2 + 1/2. Sketch the typical configuration in each of the three normal modes. 6. One end of a uniform rod of mass 3m and length 3 L is pivoted to a fixed point and to the other end is attached a light elastic string of natural length L — L', which carries a mass m at its end. In the equilibrium position, the length of the string is L. Show that the small oscillations of the system in a vertical plane have periods of simple pendulums of lengths L', 3L, 2L/5. Find the normal coordinates. 7. A thin uniform square plate of side 2a is suspended by two light inextensible strings, each of length 4a/3, which are attached to the ends A, B of an edge and to two points C, D at the same level at a distance 2a apart. Describe the normal modes of oscillation and prove that the lengths of the equivalent simple pendulums are 4a/3, 4a/9, 2(2 ± l/3) a/3. 8. A chain of n + 1 equal light springs, each of unstretched length land strength A (i.e. A(L — 1) is the force in the spring when its length is L), connects two fixed points a distance (n + 1)1 apart and in the same vertical line. Each junction of two springs is loaded with a particle of mass m. Show that in equilibrium the distance of the kth particle from the top is

la +

mg k(n + 1 — k). 2A

The system executes small oscillations in the vertical line joining the two fixed points. Show that the circular frequencies co of the normal modes are given by the vanishing of the n, x n determinant 2 cos 0

—1

0

0

.

—1

2 cos 0

—1

0

0

0

—1

2 cos 0

—1

0

D,

0 0 where cos 0 = 1 — mw2/2 A.

0 0

—1 0

2 cos° —1

—1 2 cost)

[see over]

376


By establishing a recurrence relation for D„, or otherwise, prove that

D„= sin (n + 1) 0/sin 0 and thus find the frequencies of the normal modes.

9:4 The oscillations of a linearly constrained system—Rayleigh's principle Suppose that a system of bodies acted upon by a system of conservative forces and specified by the generalised coordinates ql , q2 , qt, is subject to a linear constraint which does not involve the time explicitly, so that, in a small displacement, the changes oqi are not independent. Then, as in § 6:5, Lagrange's equations d

(aT\

1

aT OTT 1 A aqi +agi f'qj

J

=1 dt 1 agiJ

(9.24)

and the equation of constraint

ci 6q, = 0 give the (n d dt

(9.25)

1) equations

(aT )

aT

av

a4, — , ,,-+-w,.+;•eci

i=1

Cith

=

= 0, i= 1, 2, ..., n , 0,

(9.26) (9.27)

for the n coordinates qi and the undetermined multiplier )c. [By a linear constraint we mean that the c's in (9.25) are constants.] Consider now an unconstrained system in which the qi are proportional to the normal coordinates, and the corresponding periods are 27c/pi , where p1 < p2 < < p„. We can write 2T = ai g2 ,

i=1

2 V = gad.

i=1

(9.28)

If now we impose the linear constraint i=1

ci qi =0

(9.27)

the equations of motion become

at qi

pPat qi

xci =0, i = 1, 2, ..., n.

(9.29)

§ 9 : 4]


377

Assuming that qi = Ai ei pi is a solution of (9.29), we find

cli Ai (p2—

= xe1 .

(9.30)

Substitution for Ai in (9.27) gives the period equation Ci2 i=1

0

,

ai (P2 —

which can be written in the form, on clearing of fractions,

III

i=1 s=1 '*i

(9.31)

(p2— 23!)} d/ai = 0.

When p = pk , k = 1, 2, ... , (n — 1), the l.h. side of (9.31) has the PE.

value

Ak =

(d/ak)

II

s =1 k

(23! — 292); when p = pk„ this 1.h. side is

n

2+1Iak+1) 17 (A,

Ak+1 = (c

=1

— 292). But, since T

0 for all 4i, ak > 0 and

therefore Ak differs in sign from Ak+1. Therefore the l.h. side of (9.31) changes sign as p increases from pk to pk,1. It follows that the roots of this equation of the (n — 1)-th degree in p2 separate pi , pz , p„2 , i.e., the roots of the period equation of the constrained system separate those of the free system. Note that the period of the slowest mode is decreased and the period of the fastest mode is increased. Starting now with the original unconstrained system, we apply (n — 1) linear constraints

(i = 1, 2, ..., n),

qi =

(9.32)

so as to leave the system with one degree of freedom specified by the parameter O. Then 2T —(L7 aite)62, r=

217 =( Li n iga,N 2) 02 .

-1

(9.33)

The corresponding period is 2nIP where P2 =

a=

I 1.;

ail4)

(ai,(4)

(9.34)

Clearly P lies between the greatest and least of the p's. Suppose now that the motion of the system approximates closely to motion in the r-th normal mode, so that qt and therefore piis small except 13

CMES

378


for i = r. Then P differs from 29,, by a quantity of the second order in u j t , (i r). It follows that, when linear constraints reduce a system to one degree of freedom, the frequency of oscillation of the constrained system varies as the constraints are varied. The stationary values of this frequency, for first

order variations in the constraints, are the normal frequencies of the unconstrained system. At these frequencies the constraints force the system to oscillate in the appropriate mode. Note that, although we have used normal coordinates in establishing this result, in applications this is unnecessary since we have established a physical property of the system. In practice we impose any set of (n — 1) convenient linear constraints on the system as illustrated in the examples of p. 379. We can obtain an alternative physical interpretation of the above results as follows. When (n — 1) constraints are imposed on the system so that the kinetic and potential energies are given by (9.33) and 0 = cc ei wt, the mean values over a period (see Vol. I p. 241) of T, V are T, V where

T = coo

ai i=1

pz ai IL?

0,2 i=1

Therefore T = V when co = P, as defined by (9.34), and so, if we impose (n — 1) linear constraints on a system with n degrees of freedom, assume a harmonic vibration ei a" and determine co by equating the mean values over a period of the kinetic and potential energies, 27r/co will lie between the greatest and least of the natural periods of vibration. This result is the basis of Rayleigh's principle, for calculating the period of the slowest fundamental mode of oscillation of an elastic system, in the form :

In the fundamental mode of vibration the mean values (over a period) of the kinetic and potential energies are equal. Use of this principle enables us to approximate to the fundamental period of a continuous system (see Chapter X p. 437) if the system can oscillate but does not give a proof that the system can, in fact, oscillate at all. The principle can be applied to a wide variety of problems concerning the vibration of elastic solids. The application of Rayleigh's principle in this way gives an approximation to the slowest mode, but always overestimates the frequency.

§ 9: 4]


379

Examples. We use the theorem of stationary modes to solve two problems. (i) We reconsider the double pendulum of example (i) of p. 354. Impose the constraint so = ,a0 so that 2T = ma2 (27 + 12,u + 4,u2) 02 ; 2 V = mga(9

2,u2) 02.

The corresponding period of oscillations is 27cIP where P2—

g (9 + 40) a(27 12y + 4y2)

obtained by equating the expressions for T and V. Then P2 is stationary when 4,u (27 ± 12,u + 4y2) — (12 + 8,u) (9 + 2y2) = 0. This equation has roots y = — 3, 3/2. Hence the values of P2 when the system is oscillating in a normal mode are g/a, g/(4a) and the corresponding normal modes are proportional to 30 + 99, 30 — 29), agreeing with the results found earlier.

(ii) ABODE is a string of length 4a stretched under tension F between fixed points A, E on a smooth horizontal table where AE = 4a. Three particles, each of mass m, are attached to the points of trisection B, C, D of the string. Find the periods of the normal modes for small symmetric transverse vibrations on the table.

With reference to Fig. 105 and by calculation in the usual manner, the energies can be shown to be given by 2T = m(A + ±

g),

2 V = F{A-F (Y2 — n)2+ (Y3 — Y2)2 + YD/ct-

Imposing the constraints yi = y2 , y3 = yy2for symmetric oscillations we find the period of oscillations 2nIP is given by P2

— F(4,u2 — 4y + 2) ma(2y2 + 1)

The normal periods and modes correspond to 0P2

0,u

4F (2y2— 1) —0 ma (2,u2 + 1)2

which give y = +10, corresponding to the symmetrical modes with periods 2x y {ma (2 ± 1/2)/F} 13*

380


Exercises 9:4 The student is advised to rework, using the theorem of stationary modes, the worked examples of pp. 356, 365, 367, 369 and some of Exercises 9 : 2, 9 : 3. 1. The kinetic and potential energies of a mechanical system are given by 2T =

tik

2V =

vik x; xk ,

(j,k

= 1, 2, . , n)

where tik , vik are constants. Establish Rayleigh's principle, that, if p and q are the least and greatest of the natural frequencies, then

E v./ k Y.; Yk q2 E t jk YJA

p2

for any arbitrary set of real numbers y,, y2 , ..., y„. If the inertial constants tJk are changed to t'kf , in such a way that 2 T' = tjk ±i > Itjk ±j tk = 2T for any set of velocities ±1,x2 , ±„, and if p' and q' are the least and greatest of the natural frequencies of the new system, prove that p' < p and q' < q. 2. A heavy rod AB of length 2a has radius of gyration k about its centre of mass which lies at the mid-point. A light inelastic string BC of length b is attached to the rod at B and to a fixed point at C. The system makes small oscillations in a vertical plane such that the string and rod make angles 0, so respectively with the downward vertical. By use of the stationary property, or otherwise, prove that when a2 7c2 = ab the normal modes have the periods of pendulums of length b f y (ab). Describe the motion in the two modes. ,

9:5 A reciprocal theorem In § 2:4 and § 6:3 we showed that, if the forces in a system are conservative, then the generalised forces Qi can be obtained from the potential energy function by the equation Qi =

—

a

V , (i = 1, 2, ..., n).

(9.35)

qi

In particular the equilibrium position occurs when Qi =0, i.e., at the stationary points of V (a .J1; a2 '•' TO' Further, in § 9:3 we showed that, if V is a strict minimum at (q,),, the system has a position of stable equilibrium at (q•),. We investigate now what additional external (impressed) forces are required to give the system a small disturbance from the equilibrium state. _L

§ 9: 5]


381

Suppose that the system is acted upon by external (impressed) forces with generalised components S. Then Lagrange's equations d dt

/8T\ a 4, )

aT

ay-

eqi

eq,

= Si

,

i =1, 2, ..., n,

indicate that the equilibrium position corresponding to T = 0 is given by av = Si i = 1, 2, ..., n. (9.36)

0q,

,

These equations can be used either (a) to determine the generalised impressed forces required to maintain a specified equilibrium position or (b) to determine the equilibrium positions corresponding to specified impressed forces. [Note the sign difference in (9.35) and (9.36).] When the displacements from a position of statical equilibrium are small and we choose the generalised coordinates so that they are all zero in this equilibrium position, we can write (using the summation convention) (9.37) V — Vo — where bo = (a2viaqi am. The linear terms on the r.h. side of (9.37) vanish since V is stationary at qt =0. It follows from (9.36) and (9.37) that Si = bij q j• (9.38) These equations give the generalised forces required to maintain a small displacement from the equilibrium position. Suppose that primes denote a second disturbance qc from the same equilibrium position caused by impressed forces S,. Then

St

= bi i gi

(9.39)

From (9.38) and (9.39) it follows that Si g: = S;qi .

(9.40)

This is a reciprocal theorem and should be compared with that of p. 321 concerning the momenta generated by two systems of impulses. In particular, if all the forces in the two cases vanish except Si and , we find

qm

Sr — 5,1„

382


(no summation convention). If follows that if q , q,,, are coordinates of the same type, (the displacement in coordinate m due to a "force" in /) is equal to (the displacement in coordinate / due to an equal "force" in coordinate m). If we now let n oo so that the system becomes a continuous system with an infinite number of degrees of freedom, the above result still applies. Examples. (i) ABC are particles attached to a tightly stretched string OA B C D with 0, D fixed on a smooth horizontal table. Then the displacement of A due to a transverse force S (in the plane of the table) applied to C is equal to the displacement of C when an equal force S is applied to A. (ii) If AB is a nearly horizontal light elastic beam supported at P, Q, the deflection at C due to a load W at D is equal to the deflection at D due to an equal load at C. Miscellaneous Exercises IX 1. A light string of length 31 is stretched under tension F between two fixed points. Particles of mass 5m, 8m are attached to the string at the points of trisection, A, B respectively. The system is at rest until a small transverse velocity V is suddenly given, at time t = 0, to the particle at A. Prove that in the subsequent motion the displacement of the particle at B is 20 sin int V( 3 )1 14n a 3 I V 20

V2 sin ( nt 1/2

where n2 = F/(ml). 2. A uniform triangular lamina is supported in a horizontal position by three equal springs at its vertices. Show that, for small oscillations in which the vertices move vertically, the lengths of the equivalent simple pendulums are lo f10 , el„ where 10is the compression of each spring in the equilibrium position. If the system is set in oscillation by a small downward impulse I applied at one vertex, show that the displacement of either of the other vertices is ,

I

mw

sincot(1 — 4 coscot),

where co2 = g/10and m is the mass of the lamina. 3. A smooth circular wire, of mass 8m and radius a, swings in a vertical plane, being suspended by an inextensible string of length a attached to one point of it; a particle P, of mass m, can slide on the wire. If the system is released from rest with the ring in its equilibrium position and P is displaced through a small angle /3 from its equilibrium position, show that at time t the angle which the string makes with the vertical is

nt — {cos (— ) — cosnt} , 21

where n2 = 3g/a.

2 1/2

Ex. IX THE OSCILLATIONS OF A DYNAMICAL SYSTEM

383

4. A light string OA B is tied to a fixed point at 0, and carries a particle of mass 2m at A and a particle of mass m at B. The lengths OA, AB are 11, 11 respectively. The string is free to move in a vertical plane, and the system oscillates about the Position of equilibrium. The inclinations of OA, AB to the vertical are denoted by 0, co respectively. Find the normal coordinates. The system is held with the string straight and inclined at a small angle a to the vertical, and is let go from rest in this position at the instant t = 0. Show that at any subsequent time 0 = *a (2 cosn t + cos 2n t) ,

= 3 a (4 cosnt — cos2nt), where n = y(g//). 5. A particle moves on the inside of the smooth surface x2 y2 2z—— —— a b

=0

where a and b are positive and the z-axis is vertically upward. Prove that it can execute small oscillations about the origin, and find the lengths of the equivalent simple pendulums of the normal modes of oscillations. 6. A uniform disc of mass m and radius a is free to turn about a smooth horizontal axis through its centre 0, perpendicular to the plane of the disc. A particle of mass m is attached to a point A on the circumference of the disc by a light inextensible string AB of length 1. When the system is in the equilibrium position, the disc is given a small angular velocity about the horizontal axis. Show that 3a6 + 24 + 2g0 = 0,

ctO +

+ gq2 = 0,

where 0, co are small angular displacements of the disc and string repectively. If 1 = 5 a/4 show that the periods of the normal modes are x y (a g) and 2ny (5 a/2 g). Describe, with the aid of sketches, the motion in the normal modes. 7. A heavy uniform rod AB, of mass M and length a, is attached to a particle C of mass m by a light rod BC of length a. The end A of the rod AB is freely attached to a fixed support, and the system moves in a vertical plane through A. If at time t the rods AB, BC make angles 0, 9) with the downward vertical, find the kinetic energy and the potential energy of the system. If M = 18m, and the system executes small oscillations about the position of 5 ) n stable equilibrium, prove that the frequencies of the normal modes are —

12

and

where n2 = 2g/a; and find the normal coordinates.

2n

7-c 8. A uniform rod AB of length 2a is fastened at A to a light elastic string AC which is fastened at C to a fixed point. The unstretched length of the string is a and its modulus of elasticity is six times the weight of the rod. Show that the equivalent simple pendulums for the principal oscillations in a vertical plane about the position of equilibrium are of lengths 4a,

384


9. Two points A and B are on the same horizontal level a distance 3a apart. A uniform rod AC of length 2a and mass 4m is suspended from A, and an inextensible string BD of length 2a is suspended from B. A particle of mass m is attached to the string at D and connected with C by a light elastic spring of natural length 3a and modulus 15mg. If the system performs small vibrations in a vertical plane, show that the periods / a of the normal modes of vibration are 2n1/(— , where n = 5 + 119, and that the

ng

system, when released from rest, will swing in a normal mode if, and only if, the angles which the rod and the string initially make with the vertical are in the ratio 15 : (-± 4 1/19 — 2). 10. Two particles of mass m are suspended by light inextensible strings of length a from the ends of a uniform thin rod of length 1 and mass 6m/5. The rod itself is suspended by light inextensible strings of length 3a/2 attached to its ends and to two points a distance 1 apart on a fixed horizontal support. The system makes small oscillations in the vertical plane. Determine the frequencies and forms of the normal modes and show that however the motion is set up it always has period 6n/n, where n2 = g/a. 11. A rod of mass M is supported by two smooth rings, through which it can move freely, on the same level. Two particles .P5 , P2 of the same mass m are attached, one to each of two different points 01, 02of the rod, by two light strings of the same length 1. The system is set in motion from rest by an impulse I applied to P1in a direction parallel to the line of the rod. Find the configuration of the system at any subsequent time, assuming that the inclinations of the strings to the vertical are small throughout the motion. If m/M is small, show that the motion of each particle is approximately a simple harmonic motion with a slowly varying amplitude; show that initially the semiamplitude of the oscillation of P1is approximately //inn, where n2 = g/l, while that of P2is approximately zero, and that these conditions are reversed after a time nM/mn. 12. A body with a vertical axis of symmetry is suspended by a light spiral spring from a fixed support. The downward force F and the torque L (about the axis) that must be applied in order to lower the body through a distance x and to turn it time through an angle 0 are given by

P = ax + bO, L = b' x

ce,

where a, b, b', c are constants. Show that the forces exerted by the spring on the body are conservative if and only if b = b'. Find an expression for the potential energy when this condition is satisfied. Prove that there are two normal modes of vibration of the system in which the body describes a screw motion x = leO. If the mass of the body is M, its moment of inertia about the axis is I, and /Mb2is small compared with (la — Mc)2, show that for one of these vibrations the period is approximately 2n V(M/a) and k is

Ex. IX


385

approximately equal to (Ia — Mc)/(Mb), while for the other the period is approximately 2x Rik) and k is approximately equal to — (Ib)/(Ia — Mc). 13. A dynamical system possesses n degrees of freedom and is oscillating about a position of stable equilibrium. Its configuration at any time is specified by the independent coordinates 01, 02, ..., 6„. If constraints are applied so that

01

a1

02 a2

= =

B„ a,,

where a1, a2 , a„ are independent of time, obtain an expression for the period T of the oscillation of the system. Prove that T always lies between the greatest and least periods of the normal modes of the system. Show further that, regarded as a function of a1, a2 , a„, the stationary values of T occur at the periods of normal modes of the system. Three particles, of equal mass m, are tied to the points of quadrisection of a string, tightly stretched to tension P between two fixed points distance 4a apart. Prove that the periods of the normal modes of small transverse oscillations of the particles are given by 2n/ps(s = 1, 2, 3) where

pi = 2w2 ,

p2 = (2 + 1/2) w2, ps = (2 — 1/2) 0)2 and co2 = Plma

Let the displacement of the central particle be x2and of the others be x1, x3 . Suppose that a constraint x2 = A xi , x3 = is applied, where A is independent of time. Obtain a formula for the period of oscillation of the system and find its stationary values with respect to 2,, commenting on the result.

13a

GMBS

CHAPTER X

THE VIBRATIONS OF STRINGS 10:1 The fundamental concepts of wave motion The transference of energy by wave-motion is a common and important feature of physical systems. For example, during the passage of a water wave the level of the free surface of the water rises and falls and the fluid particles undergo oscillatory motions of various forms. When the wave has passed, the free surface has resumed its equilibrium position and the fluid particles have returned (more or less) to their original positions. Mechanical energy has been transmitted by the waves from the seat of the original disturbance to other regions without convection of the material in which the energy originally resided. This is the distinctive feature of wave-motion. Energy is transmitted without permanent bodily transference of the medium. Similar characteristics are displayed by electromagnetic waves (see Vol. VII). In this chapter we consider waves (of a very simple type) associated with the vibrations of strings and membranes.

The wave equation The general equation of wave-motion (or wave equation) for a scalar variable 99 is 2 g,

e2 V2m

(10.1) t2

I-

where t is the time and c is a quantity with the dimensions of velocity. For the time being we consider only cases in which c is constant. If we use rectangular cartesian coordinates Oxyz, then _ a2 v2 =

a2 +

Other forms of V2 are given in Vol. IV. 386

+

02 z2 •

§ 10: 1]

THE VIBRATIONS OF STRINGS

387

Since eqn. (10.1) is linear and homogeneous in q, we can superpose (add) solutions. This is of particular use in § 10 :3, where we superpose normal modes and employ Fourier series to fit arbitrary initial and boundary conditions.

Waves in one dimension When 99 depends on only one space coordinate x and the time t, eqn. (10.1) reduces to 2 a2„„ T

c

=

5 x2

-r

(10.2)

at2

Then the substitution u = x — c t, v = x + c t reduces (10.2) to the form 52 m

4 with solution p = / (u)

=

au v 0

g (v), i.e.,

g9= f(x — ct)

g (x

ct),

(10.3)

where / (x — ct), g (x ct) are arbitrary functions of their respective arguments x — ct, x ct. Consider now the curve in the x, y plane with equation y = f (x — ct).

(10.4)

If we increase the time t by ZI t and the abscissa x by c 4 t, then the ordinate y remains unaltered. Accordingly, eqn. (10.4) represents a curve, whose equation at time t = 0 is y = / (x), moving from left to right with uniform speed c. In fact (10.4) is the equation of a progressive wave, with wave profile y = 1(x), moving with speed c from left to right (in the direction x-increasing) without change of shape Similarly y = g(x

ct)

(10.5)

is the equation of a progressive wave, with wave profile y = g (x), moving with speed c from right to left (in the direction x-decreasing). Therefore the general solution (10.3) of the one-dimensional wave equation (10.2) represents the superposition of two progressive waves with arbitrary wave profiles, each moving with speed c, but moving in opposite directions along Ox. 13a*


388 Harmonic waves The special solution

99 = a cos{m(x — ct)}

(10.6)

of eqn. (10.2) represents a progressive wave with wave profile a cos mx, i.e., a harmonic progressive wave. Here the constant a is the amplitude and A = 27t/m is the wavelength, i.e., A is the distance (parallel to Ox) at any instant between two successive crests (or troughs) of the wave. The period (the time of one complete oscillation at any point, x remaining constant) is 2ic/(mc) = A/c. The frequency v of the vibrations is the number of complete oscillations per second and is the reciprocal of the period, i.e., v = c/A. NOTES (1) Eqn. (10.6) can be written in the alternative forms 99 = a cos {27t (x A

x ct)} , 99 = a cos 12nv (t — — )1 . c

(10.7)

(2) The expression q, = a cos(mx — nt) = a cos {m x — n t m represents a harmonic progressive wave with amplitude a, wavelength 2rc/m and velocity n/m. (3) The expression 99 = a cos {27-c(kx — vt)} represents a wave with wavelength A = llk and frequency v. Here k is called the wave number and is the number of waves (complete cycles) per unit distance along 0 x. (4) The expressions 99, = a cos {27r(vt — kx)}, 992 = a cos {27r(vt — kx)

e}

represent two waves. The crest of a wave in 992 is at a distance el(2nk) ahead of the corresponding crest of 99,, at any instant (see Fig. 106). Alternatively, the oscillation of a point x = x, in 99, will attain the position and velocity of the same point x = x, in 992 after a time el(27tv). The angle e is the phase dillerence between the two waves; when s > 0, q), is said to lead 99,, or 99, is said to lag behind 992 . When s = 0, 1 2n, ± 4n, ... (even multiples) the two oscillations are in phase; when

§ 10: 1]


389

e = ± r, ±3n, ±5n, ... (odd multiples) the oscillations are out of phase, or in opposite phases. In order to avoid uncertainties of sign it is advisable to arrange that the sign of tin expressions such as (pi , 992, eqn. (10.7), etc. is positive. (5) It is convenient to use a complex representation 99=

A exp {ip(t — xlc)),

where A may be complex, with the convention that the physical quantity co is the rea! part of the r.h. side (or, if so stated, the imaginary part). This is possible because an equation relating complex numbers is equivalent to two equations relating the real and imaginary

951 95,

Fm. 106. parts of each side of the given equation. This notation is of considerable value in discussions of the reflection and transmission of harmonic waves, for, in the application of the boundary conditions, the exponential factors cancel out whereas trigonometric terms may not do so. [Phase differences may, however, arise and show up as differences in the arguments of the complex constants A.] (6) Standing waves. When we superpose two progressive waves, which have equal amplitudes a, wavelengths ,. and velocities c, but travel in opposite directions, so that 99 =a cos { 2n (et

—

27t x)} + a cos{—2— (et + x)} ,

we find 27rx 99 = 212 COS (

COS

( 2 nAct) .

(10.8)

This represents a disturbance which, for fixed x, varies harmonically with t with period A/c; the amplitude of the disturbance, 2a cos (2n x/2), depends on x. We can equally well regard (10.8) as a cosine curve whose

390


amplitude 2a cos (27rctIA) varies with time. The disturbance is shown A A A — — . The points of intersection in Fig. 107 at times t = 0 A ' 12 c ' ' 4c ' 3c of the waves with Ox [where cos (2n x12) = 0 and co = 0 for all t] are called nodes. The points of maximum displacement are called antinodes. This type of wave is called a standing or stationary wave.

FIG. 107.

(7)

= A exp I-27"; 1 (lx I- my + nz — ct)}

satisfies the wave equation in three dimensions a 2 9, e2 q, a29, 1 a299 axe

provided that

ay2

az2

e2

at'

(10.9)

/2 + m2 + n2 = 1.

In this case, since (p remains constant on the plane

lx + my + nz — ct = constant,

(10.10)

co represents a plane progressive harmonic wave with amplitude A I, wavelength A, propagated in the direction normal to the plane (10.10), i.e., in the direction e = {l m n} , with speed c. Equation (10.10) is the equation of the wave-fronts at time t. (8) Since any periodic function of x, y, z and t can be expressed as a Fourier Series [see Vol. V § 1:4], and terms such as sin a x sin tg y sin sin w t COS COS cosy cos can be expressed in the form sin (px qy rz cot), any periodic cos disturbance can be expressed as a sum of plane waves and the definitions of wavelength, period etc. still apply.

§ 10: 1]


391

Separable solutions of the wave equation can be obtained by the methods of Vol. II § 2:6, Vol. IV § 2 :1, Vol. V § 1:5. We give here one example. Example. Find a solution of eqn. (10.9) which is periodic in t and such that 99 =

x = 0 and on x = a,,

0 on

= 0 on y = 0 and on

(1)

y=

(2)

= 0 on z = 0 and on z = a3.

(3)

If the frequency v, is large, so that vac >> 1, deduce that the number of possible solutions with frequency less than v0 is approximately 47ra1a2a34/(3 c3). We try a separable periodic solution with frequency v of the form

cp = X (x) Y (y) Z(z) cos (27rvt). Substitution in (10.9), cancelling cos (27r vt) and dividing by XYZ gives 1 d2 X

X

1 d2 Y Y dye

dx2

1 d2Z

Z

d Z2

4 n2 v2 c2 •

Each of the terms on the 1.h. side of this equation must be constant. Further, the student should show (c. f. Vol. II §2:6) that in order to satisfy conditions (1), (2), (3), Xoc sin (n,nx/cri), etc. In fact the required solution is

q, = A sin ( where

n, 76 X al

lsin

nl

n2" sin n3nz cos (2n v , a, a2 94

92,3

4v2

a2

a3

c2

(4)

and 92,1, n2 , n3are integers. Thus the possible modes of oscillation, although infinite in number form a discrete spectrum, i.e., v cannot take any value but only values defined by (4). The number of possible modes of vibration with frequency less than vo is clearly the number of integral solutions of the inequality n23. n22 n3 4v2, (5) a1 + 7 2 + a3 < c2• Regarding n1, n2 , n3as rectangular cartesian coordinates, we see that (5) is satisfied at each point at which n1, n2 , n3are integers inside the ellipsoid whose equation is given by (4). In fact the number N of solutions of (5) for which n1, n2 , n3 are all positive is approximately the volume in the positive octant enclosed by the ellipsoid (4) and the coordinate planes, i.e.,

N

1 4 8 3

2a1y, 2a2 v, 2a3vo c c c

4na1a2 a3 4 3 C3

392


Exercises 10:1 As exercises in solving the wave equation and allied partial differential equations in their various forms we suggest the following taken from earlier volumes of this series. Vol. II: Ex. 2:6, Nos. 1-5, 8-12: Misc. Ex. II, Nos. 46-48: Misc. Ex. V, No. 36. Vol. IV: Ex. 2:1, Nos. 3, 5, 6, 7, 8, 9, 10, 11; Misc. Ex. III, Nos. 5, 7, 11, 13, 14, 15. Vol. V: Ex. 1:5: Misc. Ex. I, Nos. 10, 12, 14, 17, 18.

10:2 Transverse vibrations

(1) Introduction We first consider small transverse vibrations of tightly stretched (but elastic) strings or wires such as are to be found in a piano or on a violin. We take the undisturbed position of the string to be the axis Ox, where 0

PoF** 8x'1 0o FIG. 108. is a suitably chosen point of the string, Fig. 108. We neglect the effect

of gravity so that, when undisturbed, the string is straight and in tension T. The string undergoes a transverse displacement when every particle of the string moves perpendicularly to the undisturbed position Ox. By small oscillations we mean oscillations in which y and 8y/8x are both small. In a disturbed state the element, originally P0 Q0, of length ox, becomes PQ, of length Os, where

Ss

11{(Sx)2 +(0y)2) =V{1 + (ax)2

V

6x

•

ox

Ox )2

Ox 0{(7 ox}

}

Ss — ox

2 x)

0 1( Oy\21

ax) J •

§ 10 : 2]

393


It follows that the increase in tension is of order (ay/a x)2 and, therefore, provided y is small so that (ay/a x)2 is negligible compared with y and ay/a x, we can suppose that the tension of the string is unaltered by the disturbance and remains equal to T. Later, § 10:6, we consider longitudinal vibrations of strings and elastic rods in which each element of the string moves along the line of the string so that the variations in T are of the first order in the displacement. (2) The equation of motion In the remainder of this section we consider only strings with uniform line density (of mass P per unit length). Retaining only first order

terms, the equation of motion of the element PQ, Fig. 109, is

T sin (y

6v) — T 1

2T cosy sin (-2-

a2 y = 06x at , = 06x

a

e

, te .

On the assumption that the string has no kinks or sharp corners on it so that the gradient ay/ax is everywhere differentiable, and since the oscillations are small, (1 1 A 1 6 ( Oy 1 ( 2 y \6x. cos y 1, 2 'IP) * 2 "11P ax) 2 ax2 ) Then the equation of motion becomes, correct to the first order,

y a2 y — 6x = e6x , T ax2 at

T a2 y ea x2

a2y a t2 •

394


Writing Tie = 62, so that c has the dimensions of a velocity, this equation takes the form of the one dimensional wave equation a'Y c2 8x2

a'Y

(10.11)

8 t2 •

Example. In the above proof we assumed that y is a twice differentiable function of x and t. A more fundamental equation of motion has been given by Jeffreys and Jeffreys. They consider the rate of change of transverse momentum of that part of the string which, in equilibrium, stretches between 0 and P„. The transverse component of the tension at P is, correct to the first order, T (.9 yla x) and this is communicating transverse momentum to the string on the left of P at the rate y/a x). Therefore the momentum equation, expressing the rate at which momentum is communicated by the tensions at 0 and P to that portion of the string originally between 0 and P,, is x

f e aaty dx 1—TM j aX p

t

( aX ax)o '•

0

This is the fundamental equation of the motion. When 0200, 62 yla x2exist, differentiation of this equation by x recovers eqn. (10.11). (3)

The kinetic and potential energies

The kinetic energy 6K, of the element PQ, Fig. 108, is approximately 1 e (41002(3x. Therefore the kinetic energy of that portion of the string originally between 0 and P,, where x = x p is ,

xp

K = -21- f

)2dx .

(10.12)

0

The potential energy, 6V, of the element PQ is equal to the work done in extending the length of PQ from 6x to as and is therefore

ay T (as — Ox) T[1111 +

)2 1— 116x

1

T

a,„ )2 ax

6x.

„„ ...v =1 rT

x

)2 clx

(10.13)

o

It should be appreciated that all energy calculations are made correct to the second order (in y and its derivatives) and that this is the degree

§ 10:2]


395

of approximation in eqns. (10.12), (10.13). In deriving eqn. (10.13), it is therefore unnecessary to consider variations in T. Note that for a single progressive wave y = f (x ct), a y/a t = ± cayiax and therefore, since T = e c2, the kinetic and potential energies in any length are equal. We can regard this energy as distributed along the string with an energy density (energy per unit length) E (x, t) which may vary from point to point of the string as well as with the time. The energy located in the element PQ is E(x,t)Sx, so that xp

K+ V =f E (x,t)dx, where

r E (x, t) =

+

C2 (Z)2} .

(10.14)

The wave carries the energy along the string at the rate which we measure by a function J(x, t). This gives the rate at which, at the given instant t, energy is passing the point x. Energy enters the element PQ at the rate J , t) from the left and leaves at the rate J (x p+ S x, t) on the right: in a short interval of time at the energy present in the element increases by an amount {E(xp , t ot) — E(xp,t)} ox. The conservation of energy therefore requires that

{J(x, t) — J(x

(x, t

x, t)) St =

St) — E (x, t)} Ox,

where we have used x in place of the (arbitrary) x p .In the limit the above relation becomes

sE ax = at •

aJ

We can evaluate the function J (x, t) by considering an imaginary division of the string into two parts at P; the tension in the left hand portion does work on the right hand portion at P. This rate of working gives the rate at which energy is propagated along the string. The transverse component of the tension acting on the right hand portion at P is

—T (a ylOx). J (x,t) = —

ay ay T ax at

where again xi, has been replaced by x.

396


For the wave given by the general solution of the wave equation, viz., y= f(x — ct)

g(x

ct),

0c2{g'(x

ct)}2,

J(x,t) = e c3 {f' (x — ct)}2— qc3{g'(x

ct)}2.

E (x,t) = qc2{f' (x — ct)}2

(10.15)

These results suggest that, for these waves, each wave corresponds to a propagation of energy along the string with velocity c, in the appropriate direction. This result is true only if c is a constant independent of x and t. Examples. (i) For a harmonic wave y = a cos {p (t — x I c)} , E(x, t) = ea2p2 sin2{p (t — x c)} , J(x, t) = o a2p2c sin2{p(t — x I c)} At any point the mean values of these expressions are

E(x) = le ct2p2, J(x) = -i e ct2p2c. (ii) For a stationary wave y = a cos (px/c) cos pt.

E(x, t) = e a2p2 {cost (p x c) sin2pt + sin2(p x I c) cos2pt}, and

E(x) = ea2p2 • px px J(x, t) = —T — — = oa2 p2c sin — cos — sinp t cospt , ay x 0at t

7(x) = 0. This result shows that in a string vibrating in a stationary wave the energy oscillates to and fro along sections of the string but that there is no mean flow in any direction. (4) Solution of the equation for a long string

Suppose that we are given the initial conditions y = q)(x),

ay/at = v(x)

at

t=0

for the initial displacement and velocity of a very long string. In this case 99(x) and v(x) must be known for all x. Then from eqn. (10.3), y = f (x — et) + g (x ay/at =

c I' (x — ct)

ct), cg' (x

ct),

§ 10 : 2]


397

where f, g are arbitrary functions to be determined. Substituting the initial conditions in these equations we find

f(x) — /' (x)

g (x) = cp (x) , g' (x) =

1

(10.16)

v (x) .

(10.17)

Equation (10.17) integrated between 0 and x gives

/ (x)

g (x) = a +

1

(u) du,

(10.18)

0

where a is a constant of integration. Solution of (10.16), (10.18) gives 1 1 1 f(x) = — — a: + — 2 99 (x) — 2 f v(u) du, 2 c (10.19) 1

1

2 4'(x ) -E g(x) =) 2

1

v(u) du. 0

These equations and (10.3) give 1

1

y = -2-{99 (x — ct) 99 (x ct)} + — f v (u) du. 2c

(10.20)

NOTES (a) A general motion can be started with arbitrary velocity Sy/e t and slope awax at any point, but, for a single progressive wave y = 1 (x f ct),a y/at = +cay/ex and so a single progressive wave can only be started with special initial conditions. (b) If the string is released from rest, so that v (x) = 0, y=2 {99(x — ct) ± 99(x + ct)}

(10.21)

and the solution consists of two progressive waves each with the same profile (one half of the height of the original profile) but moving with equal speeds c in opposite directions. We can obtain the form of the string at time t by first drawing the curve y = 99 (x), then drawing two

398


similar curves, one ct to the right, and one ct to the left of this first curve and then combining (adding) the ordinates of the two latter curves. (c) If the original disturbance is confined to a finite length, a, of the string, then eventually, for t a/(2c), the disturbance will consist of two isolated progressive waves moving in opposite directions with speed c.

x

95

X

(iv)

(v)

FIG. 110.

§ 10:2]

399


Example. Suppose that

99 (x) = a (12— x2)2/l4 for Ix I< = 1, (x) = 0 for !xi > /. Then the displacement at time t = 0, //(3c), 21/(3c), //c, 4 //3 c are shown in heavy lines in Figs. 110(i)—(v); the dotted lines have been used to obtain the two progressive waves.

(5) A semi-infinite string with one point fixed

Suppose that the string extends to infinity in the positive x-direction and that the end x = 0 is fixed, i.e., y = 0 there for all values of t. Then using (10.3), we obtain

f(—et)

g(ct) = 0 for t

f ( — z) = — g(z)

i.e.,

0,

for z > 0.

(10.22)

To find the motion resulting from an arbitrary initial disturbance y = co (x), 8 ylOt =v(x) for x 0 at t = 0, we proceed as before obtaining eqns. (10.19) and (10.20). However, in this case 99(x), v(x) are only defined for x 0, and therefore eqn. (10.20) is valid only for x ct. For x < ct we must use (10.22) instead of the first of eqns. (10.19) and this, coupled with the second of eqns. (10.19), leads to C

y = 1 {(p(ct + x) — (7) (c t

—

x)} +

1 —

I+ X

f tp(u)du for

x < ct.

2 c

t- x (10.23) If the string is released from rest at t = 0, so that v(x) =- 0, then y = z {co(x + ct) + y=

09(ct

(x — ct)} for

x) — 99(ct — x)1 for

x

ct,

(10.24)

x < ct.

The second term of the second equation indicates that the reflected wave has a similar profile but opposite sign (i.e. is on the opposite side of the x-axis)to the incident wave represented by the first term.

400


Example. In particular we consider the effect of a disturbance confined to a finite length of string. From our earlier results the motion can be considered as the superposition of two progressive waves moving in opposite directions. The wave moving away from 0 will be unaffected by the fact that 0 is fixed. The reflection of the wave moving towards 0 can be investigated as follows. Suppose that at time t = 0 the profile of a wave moving towards 0 is y = 99 (x), where

q)(x) = 0 for 0 < x < a, (p(x) = 0 (x) * 0 for a < x < a + 1, sp(x) = 0 for

x > a + /,

as shown in Fig. 111(i). Then, for a < x + ct < a + 1, the wave moves, unaltered in form and with speed c, towards 0 and is represented by y = w(ct x).

( )

0

0

(i v)

(v)

_„

/

\

0

FIG 111.

§ 10: 2:1


401

After the wave has reached 0, i.e., for t a/c the solution of the wave equation must be y = Oct x) F (ct — x), where F is determined from the condition y = 0 at x = 0 for all t, i.e.,

F (ct) = —co(ct). .• y = Oct + x) — 97(ct — x). The reflected wave, therefore, has a similar profile to the incident wave but is on the opposite side of the x-axis. (In fact reflection of a progressive wave at a fixed point is accompanied by a change of phase of 180°.) Note that q)(ct x) 0 only for a < ct + x < a + /, a — ct<x 0 is obtained by superposition of these waves.

(6) A string of length 1 with both ends fixed Suppose that the ends x = 0, x = 1 are both fixed. Then y = 0 at x = 0 and at x = 1 for all t. These conditions give, for the arbitrary functions of (10.3), (— ct) + g (ct) = 0 for all t,

g (1 ± ct) = 0 for all t.

f (1 — ct)

The first of these equations implies that

f (— z) = — g(z)

(10.25)

and the second implies that

1) = —

g (z

z

/) = — I { — (z — 1)).

But

f {— (z — l)) = —g (z — l) . g (z ± 1) = g (z — l) or, z replacing z

1, g (z + 21) = g (z)

(10.26)

402


Therefore g is periodic with period 21 and therefore, from (10.25), f is also periodic with period 21. Therefore y is periodic with period (in t), 21/c. Coupling this result with that of the previous example in which it was shown that reflection at a fixed point leaves the amplitude of the displacement unchanged but reverses the sign, we see that the form of a string with fixed ends and given arbitrary initial displacement can be obtained as follows:

FIG. 112.

An arbitrary disturbance (the string being released from rest) can be considered as the superposition of two progressive waves. Figure 112 (i) shows the given disturbance (heavy lines) and the (coincident) profiles of the two progressive waves (each of amplitude one-half of the given disturbance). Because of the reflection at the ends x = 0, x =1 and the periodicity of the displacement we make these progressive waves (1) each of period 21, (2) each anti-symmetrical about x = 0 and about x = 1. After release these waves advance as illustrated in Figs. 112 (ii) to 112 (iii) and the resulting displacement is the sum of the displacements in the separate waves. In working examples with given initial conditions the student x+ct should take care when evaluating integrals of the type f y(u) du x- t to ensure that the correct values of tp (u) are inserted into the integrand.

§ 10 : 2]

403


Examples. (i) If v (u) = /2— u2 for I u < /, v(u) = 0 for kid > /, and

x+ct I= f v(u) du, x—ct

then (a)

for

(b)

for

x+c/ 0 x — ct > /, I= x-ct

= 0;

x + ct > / > x — ct > —/, I= f v(u) du x-ct = 1/3— /2 (x — ct) + g (x — ct)3 ;

(e) for

x + ct > / > —/ > x — ct, / = f y(u) du = 4/3/3;

(d)

1 > x + ct > x — ct > —/, I=

for

x+c/ y(u)du

x—ct

= 2120 — 2x2c1 — ic3t3 ; (e)

x+c/ for I> x+ ct > —/> x — ct, / = f p(u)du 2/

= /2 (x + ct) —

(x + ct)3+ g /3 ;

x+ct

(f)

for —/ > x + ct, I= f 0 du = O. x—ct

The graph of the integrand and the ranges of integration in the various cases are shown in Fig. 113.

Fre,. 113.

404


(ii) An infinite string receives an impulse I at the point x = 0. Find the subsequent motion. The initial profile of the string is co (x) = 0 for all x, but the velocity distribution is a 6-function (see Vol. III § 3:8). The total transverse momentum imparted to the string is I. co

f ey(x)dx = I . Y•

ct

FIG. 114 But since no force, or impulse, has acted at any point save at x = 0, y (x) = 0 everywhere except at x = 0.

.• f (ell) v(x) dx = 1, y(x) = 0, x

0.

- co These conditions are equivalent to the definition of the 6-function. .•. y (x) = (Ile) a(x) = ( Ile) H' (x), where H(x) is the Heaviside unit function. From eqn. (10.20) we obtain x4-ct

1 I

y=— 2c

x-cr

+cr

yi(u) du

/

111(x)dx =

2 ec

2

ec {H(x ct) — H(x — ct)}.

x-ct

This represents "steps" of height //(2 pc) travelling in each direction away from x = 0 with speed c (see Fig. 114). It is clear that the above "solution" is merely a formal solution because we have been using discontinuous functions in a formula which assumes continuity in its derivation. However, it does suggest the sort of behaviour which ensues when a string is set in motion by a blow.

Exercises 10:2 1. With the usual notation, a uniform string, stretched between fixed points x = 0, x = 1, is started in vibration at time t = 0 from its undisturbed position with velocity given by

t = 2cg (x) (0

x

1).

§ 10:3]

THE VIBRATIONS OE

Verify that its configuration at time t

STRINGS

405

0) is given by

ct) — f(x — ct),

y=

where f (u) = f (— u) = f (u + 21) for all values of u, and f' (u) = g(u) (0 < u s 1). If f (u) is the "step function" whose graph from u = 0 to u = / consists of the three linear segments joining the points (0, 0), (V, 0), (31, a), (1, a), draw diagrams showing the configuration of the string at times t = 0, t = 116c, t = 112c. 2. With the usual notation, show that, if the string is infinite in length and initially y = 0, y = (x), then

r

x+ct

= 2c

j tp(x) dx. X - C

2 If T (x) = /lc (x 1 — 12 for — 1 < x b 1 c, there will be a portion of the string of length 2 (ct — b) in which — = 0 ax and which is displaced from its equilibrium position by an amount 1 2c

g (x) dx.

4. Show that if the end of a semi-infinite string is attached to a ring of negligible mass free to move along a smooth wire perpendicular to the length of the string, then the reflected wave due to the incidence of a single progressive wave is unchanged in amplitude.

10:3 Normal modes In Chapter IX we considered the normal modes of small oscillations about stable equilibrium of a system with a finite number, n, of degrees of freedom. There we showed that there exist n distinct modes of oscillation, in each of which every particle of the system moves with simple harmonic motion of the same period. However, a tightly stretched string has an infinite number of degrees of freedom for transverse vibrations. This distinction between a continuous system with an infinite number of degrees of freedom and a discrete system with a finite number of degrees of freedom is shown by the nature of the differential equations satisfied in the two cases.

406


The equations governing the oscillations of a discrete system are a set of ordinary differential equations in qi (t), one equation for each degree of freedom. The equation governing the oscillations of a continuous system is a partial differential equation for a function 99 (x, t). In the transition from the discrete system to a continuous system the set of functions of the time ql (t) q2 (t) qn (t) has been replaced by the single function of two independent variables 99(x, t). The "discrete" label i has become the continuous variable x, and the discrete set of ordinary differential equations has become a single partial differential equation. To extend the idea of normal modes to a continuous system we consider motions in which all particles of the system (string) execute S.H.M. with the same period, 2gtlp, and phase. In such a motion 99 must vary harmonically with the time for all values of x, i.e., the variables must be separated and (in complex notation)

(x, t) = / (x) eon or, (in real terms)

99 (x, t) = (x) cos (pt

E)

There is an infinite number of such normal modes, the periods of which are determined by the boundary conditions. In fact assumption of periodic motion and application of the boundary conditions lead to an eigenvalue problem as discussed in Vol. IV § 2:4. The values of p are the eigenvalues and the corresponding values of y are the eigenfunctions. In many cases the Sturm-Liouville theory can be used. The above remarks refer to the case when c is not necessarily constant. We consider now the case when c is constant and assume a normal oscillation y = 1(x) cos (pt 8) as a solution of eqn. (10.11), so that all the particles of the string are oscillating with the same period and phase. We find c2d2f/dx2 = _ pzi, of which the solution is / = A cos (px/c)

B sin (px1c).

Therefore the solution of (10.11), corresponding to a normal mode with period 27r/p, is

y = (A. cos (px/c)

B sin (px/c)} cos (pt

8) .

(10.27)

§ 10: 3]

407


Imposition of the boundary conditions and the requirement that the arbitrary constants A, B are not both zero lead to the period equation for p. Then by superposition of the eigenfunction solutions we can fit the solution to prescribed initial conditions as illustrated in the following example. Example. The ends x = 0, x = 1 of the string are fixed. Then A = 0 , B sin (plIc) = 0 and so, for a non-trivial solution, sin (pllc) = 0.

p = rac11, r = 1, 2, ....

(1)

[NorE: p must be positive since the period 2n/p is positive.] Therefore there exists an infinity of normal modes with periods 27r/p = 211(rc). The mode with the longest period is the gravest mode or fundamental (or first harmonic), the other modes are the second, third, etc. harmonics, sometimes called overtones. The solution corresponding to the rth harmonic can be written

rnx

(rnct

sr), Yr — Br sill (-1 cos — 1

(2)

where B„ sr are constants. Superposing all such modes (2), we find as the general solution of (10.11) in this case 00 (rct gt rnx cos y = .13, sin + s,.) • (3) If y= 99(x), a yla t= v (x) at t= 0 for 0 sx sl, then co

(x) = 1

B,. cos s,. sin

(7' 7L X )

1

7C C co

, (x) = — T

7Z X

r Br sins, sin H.) .

The (half-range) Fourier sine expansions of 9)(x), y(x), valid for 0 x sl, are 00 r7CX rnx (x) = ft,. sin ( ), OCr sin (--) 97 (X) = 1 where

1

1

2(rorx ocr = — f 99 (x) sin (— / )dx, 13r = — f y(x) sin / / / o o 2

r 7C X

(see Vol. V, Chapter I), and lead to

Br cos sr = ar, ncr.13,siner=

qq

dx,

408


from which B„ e„ can be determined In fact we can write the solution (3) in the form 09 (ract )±113„ sin (rnct sin (rnx ). (4) y= locr cos 1 ncr 1 1 1 It follows from the general theory of Fourier series that the r.h. side of (4) converges. Suppose, for example, that T (x) = ax (1 — x)I12, y (x) = 0, where all is small, so that the string is released from rest when in the form of a parabola. Then (see Vol. V p. 25, Ex. 1:3, Q. 1) 8a 062r = 0, CX2r + 1

8a c° (2rn3(2r + 1)) ±1 rx 1)3 0' fl((2r r- + 1) net y = — Z (2r + 1)-3sin cos 7t3

In many cases we cannot show by term by term differentiation that (4) satisfies the wave equation (10.11) since the differentiated series may not be convergent. In fact, to show that (4) is a solution we express the standing wave sin (r7tx11) cos (mall) as the sum of two progressive waves by the formula sin (\ "rx ) cos rci )— 1

[sin

/

r7 (x — ct)} + sin — (x + ct)}1 . {

Proceeding in this way we can express the r.h. side of (4) in the form f (x — ct) + g(x + ct) which of course satisfies the wave equation. Moreover, the general shape of the profile at any instant is by no means obvious from the Fourier series solution. The energy When the string is vibrating in the rth normal mode [given by (2)], the kinetic and potential energies K„ Vr are given by nc2er2B,2 sine rnet + e„) 1 ()2dx — 41

1 K, = 2 0

I

(5)

i

1

Vr = — T

2

av 2

(=1 dx — ax

nc2Qr2 B,.2cost

(rnict

41

+ sr ) (6)

o

Note that K, + V, = constant, i.e., the energy in each mode remains constant. (This constant of course depends on r.)

§10: 3]

409


This result can be generalised as follows: suppose that a general motion of the string has been analysed into modes as specified by (3). Then K

r

_

2 j

y )2 _

dx -

t

x2c2

er ?±). r_13,. sin cxx

2/2 j r = 1

0

sin

1

2 r?)] dx

r Irct

0

c2 e 212

00 00 a rx sin 1 ) x fLE:ijErsB,B, sin (0

snx

(-1 ) sin (

rnct

7,2 c2 e co

rnct 1

r2 B2 sine

41

+es) sin (

+

snct

-; es)] dx.

sr)

(7)

because of the results

r

f

sin x( "IX sin ( 87rX / /

d

)

= 116„, 2

r, s > 0.

o

[Here we have assumed that the operations of (double) summation and integration commute.] Similarly 1 V =

2

7,2 e c2 CO

Tf

x

dx

4/

,. = 1

r2 B2 cos2 (rnct+ e,) . /

(8)

0

From eqns. (5)—(8) it follows that

00 K=

.=1

00 K„ V =

V, ,=1

(9)

i.e., that the kinetic and potential energies of the string are respectively the sum of the kinetic and potential energies in the normal modes. This result is a direct consequence of Parseval's theorem [see Vol. V § 1:6], and resembles the formulae of the kinetic and potential energies of a system with a finite number of degrees of freedom [see § 9:2, 9:3].

We now consider some examples of the transverse vibrations of a loaded string. The examples chosen illustrate a variety of techniques and results which should be carefully studied. Examples. (i) Prove that, if a string is fixed at x = +1 and is executing small transverse vibrations, the equation for the periods 2x/p of the normal modes is sin (2p11c)= 0. 1 4 emEs

410


If a particle of mass (1/n)-th that of the string is attached to the string at its midpoint, show that the periods 2n/p of the small transverse vibrations are now given by (pile) tan (pllc) = n or pl = rac. Assuming a solution y = f (x) cos (pt + e) of the wave equation, we find, before the particle is attached, as in the example of p. 407,

px B sin (—)} cos (pt + e).

px

y= {A cos (—) Since y = 0 at x =

1, A cos (

pl

\

B sin M = 0.

For a non trivial solution of these equations in A, B, we must have cos (pllc)

sin (pile)

cos (pile) — sin (pile)

= 0,

i.e., sin (2p11c) = 0 as required.

.• p = rnc1(21), (r = 1, 2, ...). When r is odd so that cos (plIc) = 0 and B = 0, the motion is symmetrical about 0 (a symmetrical mode); when r is even so that sin (p //c) = 0 and A = 0, the motion is anti-symmetrical about 0 (an anti-symmetrical mode). When a particle of mass 2 e lln is attached to the mid-point of the string, we find the new periods as follows. (1) For 1 x 0, the solution of the wave equation can be written y = yl cos (pt + e), where

= {,41 cos29 (x

B, sin P x

c )1

(1)

The boundary condition y = 0 at x = 1 gives A1cos ( so that

(1)

pl\

B1sin (.2-1 1 )=0

can be written

y1 = Ai {sill (— Pi )

P1 ) Px ) cos (— P1)1 sin (— Px — sin (— cos (—

unless sin (AO = 0, We consider this case later. Therefore, = Al sin p (1

—

yl

x)lisin (p1) ,

(2)

§ 10: 3]


411

which, since Al is arbitrary, can be written

p (1

y, = D1sin

x) I (3)

c

where D1is arbitrary. Note that the form of (3) could have been anticipated since

Di sin

( — I p(l

x)

cos (p t e)

clearly satisfies the wave equation, the boundary condition at x = 1 and has one disposable parameter D1(to satisfy the boundary conditions at x = 0). (2) Similarly for 0 x 1,

—

y = y2 cos (p t e) = D2 sin

p (1 + x) c

(p t e) . cos J

(4)

Fm. 115. NOTE: We must use the same phase angle 8 in each of the expressions for y, otherwise we cannot satisfy the boundary conditions at x = 0 for all t. (3) The boundary conditions at x = 0 are derived by expressing mathematically the continuity of y and the equation of motion of the particle. The first of these gives lira yl = lira y,.

x

=

Di = D2.

(5)

We suppose that at x = 0 the tangents to the string make (small) angles zp, , 7,02 with 0x for x > 0, x < 0 respectively, Fig. 115. Then the equation of motion of the particle is

2e1 ( 02 y

= T situp, — T sintp2 .

(6)

at2 x_o+

[We could equally well use (a 2y/8 t2)x=0_ on the 1.h. side of this equation. The choice in such cases is usually a matter of convenience.] 14*

412


Since ip„y~2 are small we can replace them by (0 y10 x),„ = (0 yi/d x),, (0 y/8 x)„=o _ = (a y210 x)0respectively. Then (6) gives 2e l

p1 Pi (— p2) Di sin (—) = T {(— 11) D1cos (— — c

D2

p1 cos (— )1 . c

(7)

Since D, — 1)2 we must have, for a non trivial solution, 2e/p2

(pl ) 2T p pl — — cos (—) ,

sin( —

22

which reduces to the given equation on using the relation T = e c2. NOTE. This analysis gives the solution corresponding only to the even (symmetrical)

modes of vibration of the loaded string. In the solution we assumed that sin (p //c) 0. If, however, we take sin (pl/c) = 0, then (2) is satisfied for arbitrary B, provided Al = 0. In this case, using the continuity of y at x = 0, we can take px (px y, = a sin —) cos (pt + s), y2 = a sin —) cos (pt + s). This gives y -=- 0 at x = 0 so that the particle remains at rest. In fact x = 0 is a node for the anti-symmetrical modes of oscillation of the unloaded string and these anti-symmetrical modes for which y = 0 at x = 0 are unaffected by attaching the particle. (ii) If a string consists of two portions of lengths 1,, 12 with different densities

e,, e2 respectively, prove that the periods (27t/p) are determined by k1cot ki l, + k2cot k21, = 0, where

kT

=P2

e1/T, k2 = P2o2IT,

T being the tension. Examine the case e, = 0. The (wave) equations for the displacement y are given by 0 0 2y

02y

t2

—

x2

0 02 y

a2y

,

2 T ci = — ,

'L‘ 11 , 0 0.

(4)

These equations show that the amplitude of the oscillations decreases by a factor 1/e in a distance c/(u) r sing)) and this distance gives a measure of the attenuation (along the string) of the forced vibration. The forced vibration does not decrease with time. Since, by Fourier's theorem, any periodic disturbance can be expressed as the sum of harmonic components, the above analysis is of quite general applicability. Since the attenuation varies with the frequency the wave profile is altered in shape, the lower frequencies penetrating further along the string. When k is so small that k2can be neglected, we can take r = 1, q) = k/w and then eqns. (4) become y = A ekxic cos {w (t xlc)} for x < 0, y = A e-k xic cos (co(t — xlc))

for x > 0.

(iii) We consider now the separable solutions of eqn. (10.28). In the usual way we find that these solutions have the following form, when they are bounded for all values of x: px y = {A cos (—) B sin ( 121}C I" COS {I / (p2— k2)t

,

(1)

provided that p > k. (If p < k, the system is so heavily damped that oscillations are impossible.) The separation constant — p2 can be determined by the boundary conditions, and solutions to fit arbitrary initial conditions can be built up by Fourier analysis as in the case of undamped motion. We note that the effect of friction is to damp out the oscillations through the factor e-kt. The modulus of decay, which is the time in which the amplitude of the oscillations decreases by a factor 1/e, is 1/k. In addition friction decreases the frequency (increases the period) from its value for undamped motion.

§ 10 : 5]

423


10:5 Reflection and transmission at a discontinuity We consider a long, taut string in which there is a discontinuity. This discontinuity may be a sudden change in the density of the string, or the presence of a particle attached to the string. We suppose that a train of simple harmonic waves is travelling along the string toward the discontinuity where the train is modified. In general a modified train of waves is transmitted beyond the discontinuity and another train is reflected back toward the incident train. Strictly, a train of harmonic waves of a single frequency must extend over the whole length of an infinite string: if there is a "beginning" or an "end", or both, to the train it must contain more than one frequency. The state of motion of the string which we consider is one of "steady" oscillation with a single frequency, i.e., a forced oscillation which may be regarded as made up of a combination of the incident, reflected, and transmitted wave trains discussed below. We take Ox along the direction of the undisturbed string and use the complex notation of § 10:1, in which, if we write y = F where F is complex, we understand that the physically significant quantity is the real part of y. A train of S.H. waves is written y = A exp {ip (t — x/c)}. Here A may be complex and, if A = a e", where a and s are real, then a is the amplitude of the wave and s is the phase angle. We shall also discuss energy relations which involve squares and products of harmonically varying quantities. The mean values, over a period, of such quantities can also be obtained from the complex notation. If a particle is oscillating with simple harmonic motion under the action of a restoring force, according to the equation _ 232 x,

then

x = a cos (pt s), T = - mi;2 v _

m p2 x2 .

T = ima2 p2 sin2(pt + e), V = z ma2 p2cost (pt + e), T + V = E = ma2 p2. These results show that the energy E of the system is sometimes potential energy, sometimes kinetic energy, and that it oscillates between the

424


two forms. Since sin2 (pt

e) = 2 {1 -- cos (2pt

2s), cost (pt

e) =

+ cos (2pt

2s)},

the mean values of sin2 (pt s) and cos2 (pt s) over one, or more, complete periods are each 2. Hence we deduce that the mean values (denoted by bars) of T and V are

T = ma2p2 = V, (cf. Rayleigh's principle for normal modes). Using the method of complex numbers this oscillation is given by x = Aeil't , A = aeie. Although x = ipA ei pr, the kinetic energy is not the real part of m (ipAei")2, nor is the potential energy the real part of m p2 w eip This shows that we cannot use the complex number method to give instantaneous values of quantities depending quadratically upon x, or its derivatives. Nevertheless, by the use of complex conjugates (denoted by a star*) we can obtain mean values of such quantities. For example: >1< ip A*eiPt . = ip A e ipt , ±±* = +2 p2A A* = 2a22,2 .

= m(1 ± *) T1 ma2 P2 • Similarly x = Aei", x* = A*e-iPt, -1-xx* = ,4 AA* — A a2 -

xx*) = 4ma2 p2 .

;. V =

In general, if x, y are two harmonically varying quantities given by the complex representation x = AeiPt,

y =Bei"

i.e., the real values are, with A = a eie , B = bei", x= a cos(pt

s) y = b sin (pt + ,

the mean value of the product xy is given by R(ixy*) = R(ix*y) [Hero it denotes the real part.]

§ 10 : 5]

425


The proof of this latter statement runs as follows. The instantaneous value of the product is

ab cos(pt

e) cos(pt

= ab{cos2 pt cos e cosa

sin2 pt sine sina — sinpt cospt (cos e sina

+ sine cosa)} . The mean value of this expression is

xy = 1-ab {cose cosa + sine sina} = lab cos (e — a).

—

But xy* = ab ei (e - a) eiPt e- i Pt

,*. R(Zxy*) = tab cos(e — a) = x —y. The energy of a vibrating string Taking y = a cos {w (t — x/c) , the kinetic and potential energies of the wave over a wavelength (say from x = 0 to x = 2nc/co) are given by n c/co

K

—

2nelw

ay )2 dx = — e a2 (02 sine {w(t — x/c) — at 2

—e 2

e} dx

0

0

na2coec 2 2 nclw

V=

a2 CO2

f (j) 2 dx = T f— sin2{co (t — x/c) 2 2 c2 ex

e}dx

0

ra2 Tw na2 WeC 2c

2

•

. . K = V, K + V = na2wee,

(10.29)

so that the kinetic energy and potential energy per wavelength are equal and the total energy per wavelength varies as the square of the amplitude. Note also that the total energy varies as e c [= ]/(TO)] and this fact must be taken into account at an abrupt change of density (see example (ii) below). The techniques for solving problems are illustrated in the following three examples.

426


Examples. (i) A uniform string extending to infinity in both directions of the x-axis has tension T and to a point 0 of the string is attached a particle of mass m. An infinite train of transverse S.H. waves, of period 2x/w and amplitude a, travels with speed c towards 0 along that part of the string for which x < 0. Find the amplitudes and phase lags of the reflected and transmitted waves and verify that the energy of the incident wave is equal to the sum of the energies of the reflected and transmitted waves. We take the displacement in the incident wave, Fig. 119, to be yi• = a exp {i co (t — x/c)}, Incident

a exp ilw(t—x/c)}

Transmitted

(1) A, exp lica(t—x/c)}

0

Reflected

ARexp tw(t+x/c)}

FIG. 119. where the r.h. side stands of course for a cos {w (t — x/c)}. We take the transmitted and reflected waves, yr , y, respectively as yT = AT exp {i w (t — x/c)},

(2)

y, = AR exp {i w (t + x/c)},

(3)

where AT, ARmay be complex. Then we can take y = yl y, for x -5. 0,

(4)

for x O. (5) Y = YT We choose y, yr , y, m the above forms for the following reasons. (a) Since yTis moving in the direction x-increasing, it must involve x and t via the term t — x/c indicating a progressive wave moving from left to right with speed c. Similarly the reflected wave, moving along the direction x-decreasing, must involve x and t via the term t x/c, indicating a progressive wave travelling from right to left with speed c. (b) The wave equation is linear (and homogeneous) in y. Therefore we can superpose solutions and this explains how we derive (4), (5) from (2), (3). (c) The factor which depends on t must be the same in all three expressions yr' YR, yrotherwise we cannot satisfy the boundary conditions at 0. In fact, when we apply boundary conditions which hold for all t, the time factor, exp (iwt) in this case, must cancel throughout the boundary conditions, when expressed mathematically. Note that exp (iwt) gives the correct period 2n/w. The boundary conditions at 0 are derived from (i) continuity of the displacement at 0, (ii) the equation of motion of m at 0. These give T = AR+ a,

m (42k)A

—1 [( d a x)o+

(6)

( 2 )o-I'

§ 10: 5]


—

MCO2 AT=

T i0)

[ AT

+a—

427 (7)

[Here, for simplicity, we choose (yT)0 rather than (yr +yR)0 as the displacement of m.] Writing (6) and (7) in the forms AR — A2, + a 0,

AR+ (1 + cmpilT) AT — a = 0, and solving by cross multiplication we find

AR — cmwilT Or

where

AT

a

2

2 + criziwIT

AR

AT

a

— i sins

cos s

coss + i sin s

ae- is,

tans = cmw/(2T) = nicol(2ec).

AR = a sinse-"E +42), AT = a cosec's.

(8) (9)

Therefore the displacements associated with the reflected and transmitted waves are y R= a sins cos {co (t + x c) — n — s), yT = a cos e cos {co (t — x I c) — s} . The respective amplitudes are a sins, a cos s and the phases lag behind that of the incident wave by In + s, s. From (10.15), the mean energies carried by the incident, reflected and transmitted waves are la2 w2 pc, 1 20.)2 ec, 2I AT12 CO 2 ec respectively and therefore, from (9), the energy of the incident wave is the sum of the energies carried away by the reflected and transmitted waves. NOTES

(1) If m = 0, e = 0, AR = 0, AT = a as we should expect, since in this case there is no particle on the string and the propagation of the waves is unaffected. (2) If m co we obtain the solution corresponding to a fixed end x = 0. In this case s = 1.7r, AT = 0, AR= — a, i.e., the reflected and incident waves have equal amplitudes but are 180° out of phase, as shown by the results of § 10:2. (ii) Two very long uniform strings are connected together and stretched in a straight line with tension T; they carry a particle of mass m at their junction. A train of simple harmonic transverse waves of frequency v travels along one of the strings, and is partially reflected and partially transmitted at the junction. Find the amplitude of the transmitted wave, and prove that its phase lags behind that of the incident wave by an amount tan-1

27cvmci.c21 T (el + c2)

where c, and c2 are the velocities of propagation in the two strings.

428


Verify that the mean energy of the incident wave is equal to the sum of the mean energies of the reflected and transmitted waves. We choose the wave-trains as indicated in Fig. 120, the expressions incorporating the fact that the velocities are different in the two sections x < 0, x > 0. We assume that the particle is at x = 0. The frequency v is given by v = co/(27r) and is the same in each direction. Then, for x S 0, y, = a exp {icy (t — x/c1)) + A exp {i a)(t + x/c„)), and, for x

0,

y, = B exp {ico (t — x/c2)).

Note that, whereas for x < 0 the wavelength Al = 2nc1/a) = c1/v, for x > 0 the wavelength is 2.2 = 2nc2/a) = c2/v = c2 /11lc1. Incident

■ —►a exp

{iw(t—x/ci)}

Transmitted

B exp {iro(t— x/c2)}

(7) Reflected

A exp {ico(t+x/ci)}

FIG. 120. The boundary conditions to be satisfied at 0 are : (i) continuity of y at x = 0, i.e., y„ = y2 there, .•. a + A = B ; (ii) the equation of motion of the particle 02Y in(0 (2

i.e.,

= ( ax

1..0

ax ix-o'

B a A mw2B icoT (— — C2

c1

C1

These are equivalent to a

A — B + a = 0 , A + ( 1 + imal Cl c2 T ••

1

A 1

C1

C2

i ma)

B 2

T

c1

c1

—0.

a 1 imw + + c, c2 T 1

Hence B = ,a e-,6, where Cl +

2 a c, c2 + ima)c1c2/T

and tan 6 =

2a c2 {(c1 + c2)2 + 4n2 m2 v2 44/T2)./.

M CO Ci C2

T (c, + c2)

2nmvc1c2 T (c, + c2) •

(1)

§10: 5]

429


Because of the minus sign in the exponent of B = u e—ja, the phase of the transmitted wave lags behind that of the incident wave. From eqns. (1) we see that BB*

AA* 1V

( 1

m2 w2

a2 ( 1 + 1V + M2(02

4

62

T2

a2— AA* 4

T2

Ci C2

a2 AA* BB* ..C • c2 l Cl

(2)

But from (10.29) the mean energies carried by the respective wave trains are, since sac = T /c,

EI = in co T a21c,, ER = incoT A A.* Ici , ET = xwTBB*/c2 . Hence, from (2)

(iii) A uniform string is attached to a fixed point 0 and stretches to infinity in the positive direction of the x-axis. A particle of mass m is attached to the string at a point P distant a from 0 and a train of transverse waves y = b cos {w (t xlc)} comes from infinity. Show that in the steady wave-motion which results in the string there is a reflected wave y = b cos {w (t — x/c) e} for x > a, where 2 a) a

e= c

Transmitted

2 tan- 1

I ma) c cot z (an)1 . T

exp fiw(t+x/c)}

I

Incident

b exp ficolt-Fx/c4

Reflected

B exp {4.0(t —x/c)1

P

o Reflected --I.- B, exp {colt —x/0}

FIG. 121. The complex notation expressions for the various waves are indicated in Fig.121. The deflections are then y, = b exp (ico (t + x/c)) + B exp (ico (t — x/c)) y2 = Al exp (i (t + x/c))

B1exp (ico (t — x/c))

for

x

for 0

a, x -< a .

For zero displacement at 0, a fixed point, A1 + B1 = .

(1)

Continuity of the displacement at P gives Aie""Ic Bi e-iw ale = beiwalc Be-""le •

(2)

430


The equation of motion of the particle is

leading to — mco2[A,eia' a/c + B,e-iwalc]

= (icolc) T [beicor,I c

B e- i co alc

Aietwalc

Bo— icoalc].

(3)

To handle the manipulations of these equations we write

q = mwc/T, a1= Ae""I`, 131 = Bo-

b1= beiw°1`, b2 = B

Eqns. (2), (3) then become

a, +

= bi + b„ ig(a, + pi) = bi — b, — +

2b1 =a1(2 + iq) + iq )31, 2b2= — iqa, + (2 — iq),(3. b2 • • b1

2—iq — ig(a,I13,) iq + (2 + iq) (a,11,)

Since A1 + B/ = ale-waie + Piet a)* = 0, OGA. = — e2 a/c. b2= B

b

• • b1

2— iq + i gezi wale iq — (2 + iq) e2 "lc

2e-iwak

ig (e iwoic — e- twalc)

_2eicoalc

ig(et co alc

ctwalc)

q sin (co a/c) — cos (co a/c) + i sin (co a/c) — q sin (w a/c) + cos (coa/c) + i sin (co a/c) 1 — i[q — cot ((oak)]

1 + i[q — cot (coalc)] This gives the ratio

_ B = e2 wa/c ( 1 i[q — cot ((oak)] \ b 1 + i[q — cot (walc)] j —

Hence B = beie, where e= 2coalc — 2 tan-1[g — cot (coalc)] 2 co a

l

mco c co 2 tan-1 [— cot (--)1 , ca )

the stated value.

Exercises 10:5 1. A string extending to infinity in both directions has tension T. The part of the string on one side of a point 0 of the string has line-density ei and the part on the other side of 0 has line-density Q.A train of transverse simple-harmonic waves, ,

§ 10 : 6]


431

of length 2 and amplitude a, travels along the part of the string of line-density (21 towards 0. Prove that the waves transmitted beyond 0 have length e2 21c1and that the amplitudes of the incident, reflected and transmitted waves are in the ratios 1/0. + 1/e2

1/e2 2 fey•

Verify that the energy of the incident wave is equal to the sum of the energies of the reflected and transmitted waves. 2. A very long taut string has mass eper unit length and its tension is e c2. One end of the string is tied to a small ring of mass m which can slide on a smooth fixed wire perpendicular to the string. The equilibrium position of the string is along the positive half of the x-axis and the wire is along the y-axis. A wave train in which the displacement of the string is y = a cos {a) (t x/c)} is travelling towards the ring. Find the amplitude and phase of the reflected wave. 3. A semi-infinite string of uniform line density n2 eis fastened to one end of a finite string of length 1 and line density e, the other end being fixed, and the strings being straight under a large tension Q c2. A train of transverse waves of frequency p12n approaching from the distant end of the strings impinges upon the junction and is being reflected from it with a phase difference s. Show that s P1 cot — = n tan — . 2 4. A uniform string is of indefinite length, stretching from x = — oo to x = 0, and is at tension T ; at its end (x = 0) it is tied to two strings of similar make to the first, each at tension T, which stretch from x = 0 to x = oo nearly parallel to each other. A harmonic train of waves of transverse vibrations perpendicular to the plane of the string, is continually advancing on the first string along the axis of x towards the junction; its amplitude is k. Prove that the amplitude of the transmitted trains and that of the reflected train are 2 (1/2 — 1) k and (y2 — 1)2 k respectively, where the mass of the knot is neglected.

10:6 Longitudinal vibrations We first consider the longitudinal vibrations of a thin rod or heavy elastic spring under the action of no external forces and suppose that, in equilibrium, it is straight, lies along Ox and is uniform and just unstrained. For an introduction to the ideas employed in this section we refer the student to Vol. III § 3:3. Let 130 ,(20 be points of the rod distant x, x ax from 0 in the unstrained state, Fig. 122 (i), and suppose the rod undergoes a longitudinal displacement (i.e., along Ox) so that P0 , Q, are displaced to P, Q where OP = x e, PQ = ax + be, Fig. 122 (ii). Here $, the displacement of P from its unstrained position Pe , will be taken as the dependent variable and we use t and the unstrained length x as independent variables. If eis the (unstrained) line density of the rod and T is

432


the tension at P when the rod is disturbed, the equation of motion of PQ is

pax

a2$ te = ST + 0 {(6x)2). a

aT ate e at2 = ax • Hooke's law applied to the element PQ gives 1(6x + 6 4: — 6x) + 0(6x), T 0(6T) —

(10.30)

6x

where A is the modulus of elasticity of the rod. (10.31)

1-.8x•

Sx+S(H--

F OF

x+e-

Qo (i)

-HP

Q

(ii)

FIG. 122.

Equations (10.30), (10.31) give

a2

at2

c2

=

a2E X2

(10.32)

where 02 = Ale. The boundary conditions in this case are illustrated by the following: at a free end, T = 0, i.e., n• lax = 0; at a fixed end, $ = 0; at a discontinuity of density or of A, and T are continuous. The kinetic and potential energies c5K, O V of the element 6x of the rod are, correct to the second order in $, 2

6K =

ox(---a , 6V = 1 A(6$)2 = 1 2( a c3x. at 2 ax 2 ax 2

Therefore the kinetic and potential energies of the rod lying between 0 and x are

K

fe

0

(— :t )dx, V = 2

(-a--$ 2 dx ax )

(10.33)

§ 10 : 6]

433


Examples. (i) In a normal mode of oscillation of a rod which has the end x = 0 fixed and the end x =1 free, the appropriate solution of (10.32) is (px $ = A sin —) cos (pt + s) leading to the eigenvalue equation sin (pllc) = 0 so that p = r7tc11, r = 1, 2, 3, .... (ii) A stretched string, infinite in both directions, is of density g, when undisturbed, and has attached to it a single particle of mass m. The velocity of waves of longitudinal displacement in the string is c. An infinite harmonic train of such waves, such that the period of the displacement of each point of the string is 2n/co, impinges on the particle. Prove that the train is partly transmitted and partly reflected: that the energies per wave length of the incident, the reflected and transmitted trains are as m2w2 + 4 g2 c2 to m2w2 to 4e2 cz ; and that the change of phase of the transmitted train is tan-1{mw/(2 e c)}.

Incident

Reflected

r

o exp{ ico(1— x/c

Transmitted

- A exp {twit —x/c)

+x /c)

exp

FIG.

123.

We choose 0 to be the point at which the particle is attached and let the incident wave approach 0 from the negative side. Then the incident, transmitted and reflected waves are as shown in Fig. 123. Continuity of the displacement at 0 gives a + B = A.

(1)

The equation of motion of the particle is m ( 2$ )

a o ix=0

= To+— To- =

kax/0-1

— rna)2A = weci[— A + a — B].

(2)

A 2gc

(3)

Solution of (1), (2) gives B

a

—mwi

2gc + mwi

Using (10.33) we see that the energy carried by a wave varies as the square of its amplitude and so the required ratios of the energies per wavelength are a2 1B12

12 = 4 e2 c2 1A

m2,02

m2,02 4(22 c2

434


The change of phase (the lag) of the transmitted wave is

— arg A = — arg

/

2ec

arg (2ec + mcoi) — tan-1 'nu) 2e c

2e c + m i

We now briefly consider the case of longitudinal waves when the rod is initially (uniformly) stressed so that its length is L. In this case the density of the rod is e llL and so the equation of motion is 1 .92 .

aT

e L ate = ax where x is the displacement from the initial (strained) state and the equilibrium tension is

A(L -1) 1

T, —

(10.34)

Then in place of (10.31) we have I 6x

T = lim x ->01

+ oe — /

ox

AL 0$ j =" 11°

(10.35)

Elimination of T and T, leads to at2

=

Cl a2$ ax2 '

(10.36)

where 4.= AL2i(e12). We conclude with an illustrative example involving the action of external forces. Example. A particle of mass M is suspended by a string of mass m, natural length 1 and modulus of elasticity A. Show that, if the particle be slightly displaced in a vertical direction, the periods of the vibration are the values of (2x/z) 1/(m //).), where z is given by the equation Mz tan z = m. We use the technique of Vol. III § 3:3 so that Fig. 124(i) refers to the unstretched state and Fig. 124 (ii) to the stretched state of the string and the element PoQ, of length Sso at a distance so from A in the unstretched state becomes the element PQ of length 68 distant s from A in the stretched state. The relation between stretched lengths and corresponding unstretched lengths of the string is obtained by applying Hooke's law to a typical element. The tension, T, in the element PQ is A(Ss — 880) T aso

§ 10 : 6]


435

whence, in the limit as Os -÷ 0,

as

(1)

so

The equation of (downward) motion of the element Os is, correct to the first order in 6 so, Os° 028 mcSso 0T g+ oso . 1 OP 1 aso 02s

Al 02s m4,

0t2

(2)

A

Po 00

Sso

Bo

M

(i)

(ii) Fro. 124.

Note that we are using so , t as independent variables. The boundary conditions are s = 0 at so = 0 for all t, (3) 2, Mate — Mg — A (--La 9 l) at so = / for all t, (4) a so the latter being the equation of motion of the particle. The equilibrium state can be found by putting aiat. 0. However, we complete the problem, without finding the equilibrium state, by writing s = X + y, (5) where X is the equilibrium value of s and y the disturbance from the equilibrium state. Then the equations satisfied by X, the equilibrium equations obtained by

436


putting 0/0 t = 0 in (2), (3), (4), are

21 02 X 2 a so

0,

g+ m

X = 0 at so = 0 for all t, 0— g

1) at so = / for all t.

(aX M so

(6) (7) (8)

Hence the equations satisfied by y, subtracting (6), (7), (8) from (2), (3), (4) respectively and using (5), are 02y Al 02 y (9)

0t 2

m 04,

y = 0 at so = 0 for all 1, azy =

A ay

ate

M Os,

at .90 = 1 for all t.

(10)

(11)

Assuming a solution y = Aso) cos (pt + s) of (9) and applying the boundary conditions (10), (11) leads to the required result.

Exercises 10:6 1. If a stretched elastic string is of great length and its end A is fastened to one end of an elastic string of different material, whose other end B is fixed, show that if a train of longitudinal waves of period 2x/p advances upon A, the reflected train is of equal amplitude. Show also that each portion of the string forms stationary waves, the amplitudes of the waves in AB and in the rest of the string being

pl

in the ratio sin a : sin — , where m', c' are the line mass and wave velocity for the ' portion AB, m, c are the corresponding quantities for the rest of the string, 1 is the length AB and tang =

mac m' c'

pl

tan — . c'

2. An elastic string AB is of natural length a, modulus 2, and mass m per unit length. In the unstrained state, when AB is straight, a point P of the string is distant x from A. The string is then hung vertically from the fixed point A, and a particle of mass M is attached at the end B. The string oscillates longitudinally and (x, t) is the displacement of P from its equilibrium position under gravity. Show that no920 t2 A0271/0 x2 , and that the circular frequency co of small oscillations is given by

ka tan ka = ma/M, where

k2

w2le.

§ 1. 0: 7]

437


If malM is small, show that the smallest frequency co, is given approximately by (M + ma) a4.= A/a. 3. A uniform bar of length 2/ is compressed by forces applied at its ends so that its length is 21(1 — s). At t = 0 the forces are released. Show that the subsequent displacement of the point x is 8 el ()° (— 1)" + 3312

n0

(2n + 1)2 sin

(2n + 1) nx} 21 cos

(2n + 1) act} , 2/

where the origin is at the middle point of the bar and c is the velocity of longitudinal waves in the bar. 4. Longitudinal waves, of period 27r/n, come from infinity along a string, of density elin which the wave velocity is c1, are transmitted through a string, of length( and density ein which the wave velocity is c and proceed to infinity along a string of density c2 in which the wave velocity is c2 . Show that the amplitude is lessened in the ratio

1(

1+

)2cos2(nllc) + Pica /

+ Cc

sin2 (n//c)}112 :2. Nei

10:7 Application of Rayleigh's principle In § 9:3 we showed that, for a dynamical system with n degrees of freedom acted upon by conservative forces, any general motion following a small disturbance from a position of stable equilibrium can be analysed into n normal modes of oscillation. By a suitable linear transformation of coordinates we can take the normal coordinates as the n generalised q, and express the kinetic and potential energies, T, V respectively, of the system in the forms

T=

1

a, ,

V=

1

c iqq, ?

(10.37)

where ai > 0, i = 1, 2, ... , n, and coi /(2n) is the frequency of the ith mode. Since the solutions of Lagrange's equations give

q, = A, cos (co , t

8,),

it follows that, when the system is oscillating in the ith normal mode, the mean values of T and V, taken over a period 2a/co„ are equal. Hence, for a general oscillation, the mean values of T and V taken over a time interval which is large compared with the period of the longest mode are equal also.

438


We showed also in § 9:4 that, when we impose n — 1 linear constraints on the system so that

qi = ,a,0 , i = 1,2, ..., n, T =

) 02 ,

V = (2' aiwW) 02 ,

the frequency a/(2n), where

a2 =

ai co?itP fil aituP

is stationary when the system is oscillating in a normal mode. In fact (see § 9 :4) the imposition of the constraints makes the value of a2 lie con2 between the greatest and least of 4, 0)22 Rayleigh's principle is used to extend the above results to a continuous system and in essence can be stated for an elastic system, in which the relation between stress and strain is linear, as follows : The distribution

of energy as kinetic or potential energy in the fundamental mode of oscillation of an elastic system makes the frequency of this fundamental mode a minimum. We apply the principle to calculate approximately the period of the fundamental mode as follows. (1) Assume that the system oscillates so that every particle moves with the same period 2n/p, i.e., if the column matrix 4 denotes the disturbance of the system from the equilibrium state, we take 4 4 cos pt, where go is a column matrix whose elements are functions of the space variables only. The elements of 4 must be chosen so as to satisfy the boundary conditions of the problem. (2) The mean values over one period, 2nlp, of the total kinetic energy and total potential energy are equated to one another. This gives an approximate value of p2. Rayleigh's principle implies that this calculated value of p2 (for the constrained system) cannot be less than (02, where a)/(27c) is the frequency of the fundamental mode of the unconstrained system. (3) The trial value of p may be improved by allowing the trial solution to contain a number of arbitrary constants A1, A2 , ..., and minimising the trial value of p2 w.r. to A1, A2 , ..., so that 8(p2)/8 A1 = 0, a (p2)1aA2 =0, , giving a sufficient number of equations to determine A1 , A 2 , ... [see Ex. 10:8 No. 8.]

§ 10: 7]


439

Examples. (i) We calculate approximately the fundamental mode of the transverse vibrations of a uniform string stretched between x = 0 and x = 1. [See the example of p. 407.] We assume a trial solution [the simplest polynomial (parabolic) form satisfying the boundary conditions] y = A x(1 — x) cosp t. Then from (10.12), (10.13)

K = lep2A2 sin2 p t f x2 (/ — x)2dx = (e2p2A215 sin2p 0/60, V = TA2cost pt f (1 — 2x)2 dx = (TA213cos2pt)/6. K = e p2 A2/5/120, V = TA213/12. Equating these values of K, F we find p2 101We/2) = 10 c2/12. The exact value of p2 for the fundamental mode is n2c2I12[4- 9.87c2/12] and so the above method gives an excellent approximation. (ii) We solve the second part of example (i), p. 409, obtaining an approximate value for the period of the gravest symmetrical mode. We assume the trial solution

y = a(1 + 131x111) (1 — !xI/l) cospt. Then

V = 2 . lec2 f (8y/Ox)2dx = = (ec2 a2 (p

c2a2/l4)

f {kp — 1) 1— 2/3x}2 cos2ptdx

+ 3) cos2pt)/(3/),

K = 2 . e a2 f (1 + /3x//)2(1 — x/l)2 p2sin2p tdx 1(4110 a2p2 sin2p t

=

la2 (j92+ 513 + 10 + 30/n) p2 sin2pt}/30.

.•.V = e c2 a2 (32 + 3)/(6l), K = ela2 (132 + 5fl + 10 + 30/n) p2/60 .*. p2

100132 + 3) c2/{(92 + 5/3 + 10 + 30/n) /2).

(1)

We choose /9 so that p2is a minimum, i e , to satisfy the equation 2/3032+ 5/3 + 10 + 30/n) — (2/3 + 5) 032 + 3) = 0,

5 132 + (14

60

) /3 — 15 = O.

(2)

440


To simplify the manipulations, we consider a particular value of n; for convenience we take n = 1. Then (2) is satisfied by (3 = 1/5, which value makes p2 as defined by (1) a minimum; then (1) gives p2 * (20 c2)/(27 /2) so that pile 0.86. The exact value of plIc corresponding to the gravest symmetrical mode is the first positive root of the equation x tan x = 1. But (reading directly from the tables) we see that this root is approximately 0.86, showing that the agreement is good. NOTE that in this example we choose a trial solution for which the string does work on the particle during the motion. A trial solution of the form y oc (12— x2) cosp t would be unsuitable in this case since 0 y10 x then vanishes at the particle and changes in its kinetic energy cannot be due to the work done by the tension of the string.

Exercises 10:7 Find approximate values of the fundamental mode in the following problems using Rayleigh's principle. 1. The example (iii) of p. 414 (first part) when M = 2 el. [Try y = ax (1 + 13x/l) cos pt and find /3 so that p2 is a minimum. (pllc =:= 0.65).] 2. Ex. 10:3 No. 4. 3. The example of p. 434.

10:8 Miscellaneous problems We now discuss, by means of illustrative examples, a number of miscellaneous problems associated with the propagation of waves along a string. Examples. (i). Variable wave velocity; transverse oscillations If p = Po/x2, where eo is a constant, and the fixed end-points are x = a, and x = a + 1, show that the normal functions q,, and normal coordinates 1/),, are given by q)„ = x'1' sin (A„ log (xla)) = An sin (P./ where 4 + = Pop/T, the period equation being 2„ log (1 + 11a) = na, n = 1, 2, 3, ... and A„ and 6„ being constants depending upon initial conditions. The equations of motion are the same as for a uniform string and so the equation for y is

T 02y

1 02y

eo 8x2

x2 0 t 2 •

Assuming a solution for the nth mode of the form y = f „(x) sin (p„t + 6„) we have 2 d2f. P2n n - - ,eo =o. x d x2 T

§ 10: 8]

441


This equation is of homogeneous type [Vol. II § 2:3] with solution

1„ = x'12 [A cos {A,, log x} + B sin {A„ log x}],

(1)

where An = gAIT — I. The boundary conditions, y = 0 at x = a and at x = a + 1, give A cos {A„ log a} + B sin {A„ log a} = 0, (2)

A cos {2„ log (a -F 1)) + B sin (2„ log (a + 1)) = 0 .

(3)

The condition for a non-trivial solution of (2) and (3) reduces to sin {A„ log (a -F 1) —

log a} = sin {A„ log (1 + 11a)) = 0 .

The period equation is, therefore, 2„ log (1 + 11a) = nn, n =1, 2, 3, .... Eliminating A from (1) and (2) and writing B/cos (2,, log a) = A„ we find f„ = A„ x'1' sin {A„ log (x1a)}. Therefore, in the nth mode y = A„x'b sin {2„ log (x/a)} sin (p„t + 6,,), giving the stated results. (ii) The small oscillations of a uniform chain hanging under gravity from a fixed

point. We consider the small oscillations in a vertical plane of a uniform chain OA, of length 1 and line density e, hanging under gravity from a fixed point 0, Fig. 125. Let y be the small horizontal displacement of a point P of the chain from its equilibrium position P,. Since the oscillations are small, the vertical displacement of P is of the second order in y and therefore PP° can be taken as horizontal and y, the inclination of the tangent at P to the downward vertical, is small and we can take tan = y = a y/a x. The tension at P is therefore (vertical resolving for PA),

T = e (l— x)g.

(1)

The (horizontal resolute of the) equation of motion of the element PQ, of length a x, is

az y

(T + T) sin (y + 6y) — T siny = e i.e.,

ate

a x,

az y o(T siny) =

ot2 x.

Letting ox -4- 0 we find

a ax 15

OMER

( T 8Y ) = ax,

82Y

ax2

(2)

442


From (1) and (2) we obtain the partial differential equation satisfied by y, viz.,

02y ay (1 — x)--Tx=— 3 ax

-

1 02y g ate

(3)

To find the periods of the normal modes we write y = f (x) cos (pt e) and obtain

(1 — x)

d2f dx2

df

p2 f

dx

g

——

= 0.

(4)

FIG. 125. The substitution 1 — x = u transforms this equation into d2 f du2

df p2f + — — 0, g du

the solution of which can be expressed in terms of Bessel functions [see Vol. IV § 3:1, p. 177] in the form / = CA (2p 1/(u/g)) + C2 Yo{2.23 1/(u/g)} • ••• y = [C1,10 {2p

V(

1

+ C2 Yo {2p 1 /(1—

X

9

)1

V

g

)11 cos (p t + e) .

§ 10: 8]

443


Since y remains finite when x =1 and Y0 (x) — oo as x 0, we must have C2 0. Then the boundary condition at the fixed end, i.e., y = 0 at x = 0, gives the period equation

( )1g

J {2

=

0.

(iii) An infinite uniform taut string, of line density Q, in tension T, and stretched along Ox, has a particle of mass m attached to it at 0. The system is at rest for t < 0 and at time t = 0 a transverse impulse P is given to the particle. Investigate the subsequent motion of the particle. Y1

-c t

Fm. 126. By symmetry the disturbance is symmetrical about 0, Fig. 126. Progressive waves will start from 0 at time t = 0, each advancing with speed c = 1/(T/e) and so we can take the displacement of the string as y=

— x/c)

for 0

x ct, y = 0 for x > ct,

x/c) for — ct x

y=

0, y = 0 for

x < —ct.

Since the string is at right angles to the impulse P, the velocity of the particle immediately after the impulse is Pim in the transverse direction. The only forces acting on the particle in its subsequent motion are the tensions in the two halves of the string. Hence the equation of motion of the particle is azy m

0t

,=0

= —2T sin (n — v) = —2T I ax ..r=0

The particle commences this motion with

/ a y\ 0t

)

'9 at time t = 0. Since y„=0 = f (t) and (-1

P ) .i3 = rn

(

at ,=0

=— — 1f (t), the equation of motion is c

mf"(t) = —2ecf(t), 15*

(1)

444


subject to the initial conditions f (0) = 0, f' (0) = P Im. Hence = (p/m) e-zect/m, f(t) = (1 e-2ect/m)/(2420. f'(t)

P

.•. = — (1 — e-2Q (c t- x)Im) for 0 2ec y=

2e c

(2)

x

(1 _ e- 2e t + xym) for — ct x

0,

y = 0 for Ix' > ct. As t co, the displacement of the particle tends to P/(2gc). In fact, this is true also for the displacement of any point of the string. (iv) The use of the Laplace transformation, In Vol. V Chapter II p. 74 we used the Laplace transformation to obtain, in closed form, the solution of a problem concerning a string one point of which is given a prescribed oscillation. We give here in illustration the solution of a simple problem by this method. The general results from which the solution can be obtained are given as an exercise in Miscellaneous Exercises X, No. 12. A uniform string, of line density eis stretched to tension.ec2 between the fixed points x = 0, x = 1. At time t = 0, the string is plucked and released from rest with the point x = displaced a small distance a from its equilibrium position. Show that, when 0 < ct 3 ax 1

for Osx5_

1 3

— ct,

3a 1 y = — (x + / — 3ct) for — 3 4/ and find the value of y in this case forll We have to solve the equation ozy oz y c2 — = --a xe ate

1 — ct < x < —1 + ct , 3

ct < x < 1.

0

x

/, t

0,

(1)

subject to the boundary conditions for all t, y = 0 at

x = 0 and at

x = /,

and the initial conditions y = 99(x)

ay at

at t = 0 for 0

x 51,

= 0 at t = 0 for 0< x s /,

(2)

§ 10:8]

445


where

yo(x) = 3ax11 for 0 so(x) = 3a (1 - x)/(21)

x s */,

for it < x 0 is a sinpt. Use the Laplace transformation to show with the usual notation that, for 31 < ct < 5/, the displacement of the end

x = 1 is 2a [sin

p (ct — 1) I c

s

(ct — 31)}1 .

10. A uniform string, of line density Q, is stretched to tension e c2 between the fixed points x = 0, x = 1. To the point x = l is attached a particle P of mass m. At time t = 0 a transverse impulse mV is given to P. Show, by using the Laplace transformation or otherwise, that, for 1 < ct < 21, the displacement of the particle is V 2n(ct — 1) - 2n(0t-0/1 2V -2nct11)

__

e

—

1

e-2,Ket-0//

e

where nm = el.

Miscellaneous Exercises X 1. Prove that if a heavy particle is fixed to the mid-point of a uniform string, the normal modes for transverse vibrations of even order are unchanged but the frequencies of the normal modes of odd order are decreased. 2. A string, of uniform line-density e, is stretched to tension e c2 along the x-axis. Verify that px px y = A cos — sinp t + B sin — cosp t represents a possible transverse motion of the string, A, B, p being any constants such that Ap, Bp are small compared with c. Prove that in this motion the average rate at which energy flows along the string is ABecp2. [see over]

15a*

452


One end of the string, at x = 0, is fastened to a light ring sliding on a transverse wire and subject to a frictional force ky, where k is constant. The other end, at x = 1, is made to oscillate transversely so that its displacement is of the form a cos (pt + a), where a, a are constants. Show that a motion of the above form is possible for this system and find A, B in terms of a. Prove that the average rate of dissipation of energy at x = 0 is

a2 p2 cost (pile) + (kle c)2 sin2 (plIc) . 3. A uniform string of length 2(1 + 1') and line density eis stretched between two fixed points; a length 2l' in the middle is uniformly wrapped with wire so that its line density becomes e'. Prove that, if the tension T = c2 e = c'2 e, the periods of the notes which can be sounded are 2n/p, where p satisfies either of the equations c' tan (pl'

-I- c tan(p //c) = 0 and tan (p11/c1) tan (plIc) = c' lc.

4. With the usual notation, the ends of a string of length 1 undergo equal small transverse motions given by y = a sin xct. Show that the distance x of the nodes from an end is given by 2xx = xl + (2n + 1) gr, where n is an integer, assuming the natural modes of oscillation of the string are not excited. 5. The ends of a taut string of length 1, line density e and tension T are attached to light rings which can slide freely one on each of two parallel wires, distant 1 apart, fixed at right angles to the string. A particle of mass m, attached to the midpoint of the string, is subjected to a periodic force F sin pt parallel to the wires. Show that the maximum displacement of the particle due to the forced vibration is ,

F Ol[p2(pl tang + m )] , where 0 = ip 1 (eiT)hIz. 6. A long stretched string has a portion (of length 1) in the centre whose density is ne, the density of the rest of the string being e. A train of simple harmonic transverse waves approaches this portion from one end; prove that the energy in the reflected wave is to the energy in the transmitted wave in the ratio (It — 1) sin2 T : 4n, where co = 27r12, the wave-length in the central part being equal to A. Show also that the sum of these energies is equal to the energy in the incident wave. 7. A uniform stretched string, of great length, lies along the axis Ox from x = —1 to x = + oo; the end at x = —1 is attached to a fixed point, and a particle of mass m is attached to the string at x = 0. A train of transverse waves in which the displacement is x y = a coss (t — c

travels along the string from x = + oo and is reflected. Show that stationary waves are set up in each part of the string, and that in particular the displacement for

Ex. X


— 1 < x < 0 is z — 2a where a =

ax

cos ft

sin a sin — c

and

+ a) cos (at — (3),

am

tan fl =

453

— cot N .

co

8. If a string of length 1 and tension Tostretched between two fixed points is not uniform but of line density eo/(1 + k x)2, where x is the distance from one end, show that the transverse vibrations are of period 27t/n, when 1/(4n2— x2 c2) log (1 + kl) = 2r cxr r, where c2— Tole, and r is a positive integer. Examine the case of r = 0. 9. A uniformly stretched string, of which the extremities are fixed, starts from rest in the form y = A sin (mxx/l), where m is an integer and 1 the distance between the fixed extremities. Prove that, if the resistance of the air be taken into account and be assumed to be 2k times the momentum per unit length, the displacement after any time t is

y = A e-kt (cosne t — sin m' t ) sin

97L 7C X

m'

where m'2 = (m2 n2 c2//2) — 162and c is the velocity of waves of transverse vibration. 10. A uniform membrane S of mass a per unit area is stretched to a constant tension c2cr, and attached to a fixed frame bounded by the rectangle x = la, y = ± b. If a small impulse J, in a direction perpendicular to the plane of the membrane, is applied to it at the point (a, 9), show that the normal functions 99„,„ (x, y) (normalised so that f [99,,,,]2 dS =1) for the resulting small vibrations are 1 2

tp,„„(x, y) = — (a b)" sin

72 Tr X

sin

a

Trory

; n., rn = 1, 2, 3, ...

and the normal coordinates are

(J1130 99mn(oc, where

m2

sinP n2 )

p2 = c2

b2

a2

•

11. A string of constant density eper unit length, under tension T, is subject to a transverse force F (x, t) per unit length. Show that the transverse displacement y (x, t) satisfies

e

2y

,92y

— T — = F (x, t). ate axe

If the ends x = 0 and x = 1 are kept fixed, assume that there is an expansion of the form co rn,rrx y (x, t) = q'„, (t) sin m-

454


If this series may be differentiated term-by-term twice with respect to x and to t, show that d2

dt2

+ T Mi n)2 m

= 2

F (x t) sin

17141x

dx.

12. With the usual notation, the ends x = 0, x = 1 of a uniform string are fixed and the string is released from rest at time t = 0 with profile y = 99(x). Show that 00 the Laplace transform Y = e-Pty(x, t) dt of y satisfies the differential equation c2

d2 Y d x2

P2Y = — P 9)(x)

subject to the boundary conditions Y = 0 at x = 0 and at x = 1. Assuming a solution of this differential equation of the form Y= A (x) cosh13x (-

B (x) sinh1)x (-),

use the method of variation of parameters, Vol. II § 2:5, to show that dA 1 dx

= — (x) sinh19x (-), c

dB1 px — (29(x) cosh (—) . dx Hence, using the boundary conditions for Y, show that x

pu A (x) = — 1 p(u) sinh (—) du, c

0

1 B (x) = f

pu

(u) cosh p u ) du —

coth n( 1 ) f (p(u) si h (±±) 19 du.

(

0 Deduce that x

Y=

1

c sinh (pile)

sinh P (1

c

x) f99 (U)Sillh(-) 124 du — 0

px

— sinh (—) f q (u) sinh P

c

u) du .

Ex. X


455

13. A uniform string of line density O is stretched at tension T between two points A and B, distant 1 apart, on a smooth horizontal table. At the end A the string is attached to a particle of mass m, which can move on the table only in a direction perpendicular to AB and is subject to a restoring force in this direction of magnitude m02times its displacement from A. At the end B the string is fixed. Show that the system can execute small oscillations, in the plane of the table, at angular frequencies co which satisfy the equation co2= S22 — co cot (0)1) c where c = y(Tie). Find an approximation to the value of co nearest to Q in the case when mD >> Tic and D //c is neither very small nor very close to an integral multiple of 7c.

CHAPTER XI

ANALYTICAL DYNAMICS 11:1 Introduction The methods and theorems which we discuss in this chapter form part of a general, comprehensive theory of classical mechanics which was developed by mathematicians in the nineteenth century. Whilst we do solve a few problems, as illustrations, we must point out that, in general, the methods of this chapter are not suitable for solving specific problems ; those which we do solve are quite simple and solutions can usually be derived more easily by elementary methods. Nevertheless, the results of Analytical Dynamics are important because of their wide generality; they give formal solutions, or demonstrate the existence of solutions, of almost any problem. They are also important because they show that the different equations of motion of continuous systems and of electromagnetic systems can take the same formal shape as the equations of motion of discrete systems. Finally, the general results of Analytical Dynamics form the basis of the analogy between quantum mechanics and classical mechanics. Many of the methods used are generalisations of methods used in abstract pure mathematics, chiefly methods of solving sets of differential equations. The reader can find a full and authoritative account of these topics in Analytical Dynamics by E. T. Whittaker (C.U.P., 1937, or Dover, 1944). Throughout this chapter we shall use the summation convention over generalised coordinates and momenta.

11:2 Hamilton's principle We consider a holonomic system specified by n generalised coordinates

q1 , q2, ..., q,,. For our present purposes it is useful to represent the configuration of the system by a point in an n-dimensional "configuration space" in which q1 , q2 , q„ are the coordinates of the point. All pos456

§ 11 : 2]

457

ANALYTICAL DYNAMICS

sible states of the system are represented by a set of points and as the system evolves with time the representative point traces out a curve, called a "trajectory", in the space. Since, in general, the coordinates qi are continuous functions of the time this curve is continuous and, if we rule out impulsive changes of velocity, the curve is also smooth since the velocities q i are continuous. A given system, then, in its motion between the instants t1 and t, follows some curve MN in the configuration space, M representing the initial configuration and N the final configuration. We can join the points M and N by curves other than the natural trajectory; this corresponds to varying the motion of the system e.g., by the imposition of suitable constraints. (The reader will recall the similar situation with "comparison motions" in Chapter VIII.) Hamilton's principle states that the integral

S = Ldt,

(11.1)

where L is the Lagrangian function L(a1' -‘ 2 • • qn ql • • • , 4„, t) of —a the given system, has a stationary value if the trajectory is varied slightly from the natural trajectory. The variations considered must all correspond to motions leaving M at time t1, and reaching N at time t2. This principle is written concisely

68 = 6

f Ldt = 0,

(11.2)

the symbol 6 standing for the variations of the trajectory. When we use Hamilton's principle to find the motion of the system we must find which of all the trajectories joining M and N gives S a stationary value. The mathematical problem is that of determining n functions qi (t) which, when substituted into L, give S a stationary value. This determination is the subject matter of the Calculus of Variations. We proceed as follows. We suppose that the functions qi (t) give the natural trajectory and we consider a varied trajectory in which, at time t, the coordinates of the representative point are

qi

oqi =q,(t)+ En, (t),

(i = 1,2, . .

n)

(11.3)

where E is a small parameter independent of t and ni(t) are a set of functions of t. In order that the varied trajectory shall start from. M at time

458


t, and end at N at time t2the functions ni (t) must satisfy

ni (ti) = 0 = rl c (t,), (i = 1, 2, ..., n)

(11.4)

but are otherwise arbitrary. The value of S for the varied trajectory is t,

S(s) =

f L (qi

th +

t,

t.

(11.5)

If S is to have a stationary value for the natural trajectory, dS de = 0 when E= 0 for arbitrary functions m (t). But to

dS _ = de

aL

aL

' Oq,

04,

ut

t,

(Here a summation over i is implied.) We integrate the second part of the integrand by parts to obtain to

aL

aL . r

rid` a4ut =

d aL d t a qi

J1,„

t,

) dt.

(11.6)

Because of the conditions (11.4) the first term on the r.h. side is zero. to

dS r aL de = nt jaqi

d( dt

aL

)f

dt.

t,

Since (dS/de),=0must vanish for arbitrary functions ni ,the other factor of the integrand must vanish for each value of i.

aL • • aq,

d dt

L )]

aq• c=o

0.

(i = 1, 2, ..., n).

(11.7)

When e = 0 the function L is the Lagrangian function of the system evaluated on the natural trajectory. Hence we conclude that Lagrange's equations (11.7), (with e = 0), are necessary conditions for S to be stationary. (They are also sufficient.) We shall not investigate whether this stationary value is a maximum or a minimum. Works on the Calculus of Variations use a 6-notation rather than the notation we have used above. The condition for S to be stationary is

§ 11: 2]

459

ANALYTICAL DYNAMICS

that the variation 62 is zero for arbitrary variations 6q, in the functions specifying the trajectory. Consequently, the variation e n;(t) used above is denoted by 6q, and tz

62 = f 6Ldt. t„ (No variation is made in the limits of integration.) In this equation

, aL L 61, = ----- uqi + qi Oqi

(11.8)

uqi•

[There is no variation in the time in eqn. (11.3).] Since e is independent of t we use the function velocity in the varied trajectory and so

qi(t)

e (t) for the (11.9)

(t) = 71t (8M) = at (4)

=

i.e., we can commute the operators 6 and did t. Substitution of (11.8) and (11.9) into the integral for 6S gives

L L d (oq,)}dt. - tog, + qi aqi dt

SS =

Integration by parts, as in (11.6), gives 6s

18 f 6qi {a Li ag

[6q`

cdt (OalLi )idt.

(11.6 a)

The variations 6q, are zero at each end of the trajectory so that f

68 =i 'qi

aL

d

aL

-

t,

Since 68 must vanish for arbitrary variations 6q, we are led again to eqns. (11.7), Lagrange's equations of motion, as the conditions for S to be stationary. Examples. (i) The above method of obtaining eqns. (11.7) is quite independent of the interpretation of the functions of the variables used in the derivation. They are, in fact, Euler's equations for S to be stationary. We give now an alternative form for the conditions that S is stationary.

460


First, we state, in abstract, the result incorporated in (11.7). If G (y„ x) is a function of any number N of functions yi (x) and their first derivatives yj' (x) then the condition that the integral f G(yi , y;, x) dx, where the limits a and b are a

fixed, shall be stationary is that the functions y, satisfy Euler's equations d aG dx Oy; )

G

0

Oyi

(i = 1, 2, ..., N).

*

[This result is the fundamental result of the Calculus of Variations, which, we suggest, the reader should prove for himself by the method used to obtain eqns. (11.7).] We now introduce the canonical momenta and the Hamiltonian function by pi =

0L

H (qi , pi , t) — pt gr — L

a gt

(summation over j on the r.h. side of the second equation). We have assumed that the pt have been eliminated from H by using the equations for pi (see § 11:4). Then the function S is given by tz tz t) dt. S = (P.qi — H) dt = f Gi (qt, P., We regard the integrand Gihere as a function of 2n variables qi , pi , and 2n derivatives 4,, pi corresponding to the N functions yi and the N derivatives yi in 0 above. (The fact that pt do not occur in this special choice of function is a "lucky accident".) Hence Euler's equations for a stationary value for S are 1

d dt \ Ogr i

00

9G1 ('

49./3 d (.1),) + — = 0, dt 0 q, d H dt — (0)— = (329,

—

d (.90, O pt

dt

)

dpi dt

aH a qt

dqi dt

aH pi

(i = 1, 2, ..., n).

These equations are equivalent to eqns. (11.7). They are known as Hamilton's equations of motion. (ii) We consider a particle moving under uniform gravity only. The Lagrangian function is L = ini(±2 + 92) — mgy. to

.•.

S

f (2 y — 2gy)d.t. r,

§ 11 : 2]

461

ANALYTICAL DYNAMICS

We make variations Ox, oy in the functions x and y so that to

— goy) dt.

6S = m f (t6 t, But

to

to

to

f xax dt = [± Ove] — f xaxdt = — f Xoxdt, 13 tg f ysy dt = [9 sy];1 — f gay dt = — f

p 6y dt,

since both 6x, oy must vanish at t = t, and at t = t2.

— (g + g) Oy} dt

.• OS = m f {— Since both Ox, 6y are arbitrary we deduce

= —g

= 0, as the equations of motion of a projectile.

(iii) We consider transverse vibrations of a uniform string using Hamilton's principle. If F is the tension and 1 is the length of the string,

1

2T =

f

1

ay

(.)2 dx, 2V= F f (N 2 dx.

ax

Hamilton's principle therefore requires that /2 I 6 f dt f le (—Y O —) 2

o

rl

(N 2Iclx = O. 0x

at

We write c2 = Fie, bring e outside the integral and make an arbitrary variation which vanishes when t = tl and t = t2as well as where x = 0, 1, in the function y (x, t). Then y 0 ay a OS = 2e f dt f at at (ay) — c2 (— — (6y)} dx = 0. Now

t2 t,

ax) x

ti I

to

ay 0 ay a (6y) dt dt — — (6y) dx = f dx f at at at at 0

0

1 =

12 dx [— 6y — ay at

1,

t,

1 . o

f

t,

o2y

dx ay

dt.

462


Since Sy vanishes for t = t, and t = t2, we obtain t, r, 1 02_ ay a 1dt f — (Sy) dx = — f dt f Sy Y x . at — at 0 t2 d 0 r, h o Similarly t, I to r, I ozy a y 1 I dt f " (Sy) = f dt [-6y] — f dt i 6y— dx dx 0x a x axe , Lax 0 r, t, o o h where Sy vanishes for x = 0 and for x = 1. .• SS =-2o f dt Sy o

az y ,92 y at2— c2 x2

dx.

Since the variation Sy is arbitrary we conclude that the condition for SS = 0, i.e., S is stationary, is a2y

a te

02 y

= c2

axe

Here we have assumed that the functions y and Sy are such that the order of integration can be changed and that Sy satisfies the boundary conditions. This latter assumption is the equivalent, in this problem, of the system "starting from configuration M" and "ending in configuration N" of the general theory. (iv) A particle of unit mass moves along Ox under a constant force f, starting fronirest at the origin at time t = 0. If T and V are the kinetic and potential energies of the particle, calculate to f (T — V) dt. Evaluate this integral for the varied motion (in the Hamiltonian sense) in which the position of the particle is given by x=

t2

s t(t — t0),

where s is a constant, and show that the result is in agreement with Hamilton's principle. What are the essential features of the varied motion that ensure this agreement? With the usual notation X = ft, x = f 12, V = — x. .•. T = if 2t 2, V = —112 t] 2, T — V = f 2t 2. to f (T — V) dt =

§ 11 : 3]

ANALYTICAL DYNAMICS

463

When x = if t2 + e t(t — to), Li; = ft + ef (2t — to),

T = {f t (1 + 2e) -- 6/4) 2, V = — if 2t2— el2t(t — to), T — V = f 2t2(1 + 3e + 2e2) — 2f2tto(€ 4, 0) + I_ IT,. to .• f (T — V) dt = (2 + s2) /2 4/6. This shows that the variation of the integral for an adjacent path is 0(s2), the variation being positive. This is in agreement with Hamilton's principle and in fact we have proved (for this type of variation) that the integral in the actual motion is a minimum. The essential features in the varied motion that ensure this agreement are that the variations vanish at the end points (limits) of the region of integration.

11:3 The principle of least action This theorem is usually ascribed to Maupertuis but subsequently it was stated clearly in a more satisfactory form by Euler and Lagrange. Like Hamilton's principle it concerns the stationary value of an integral, but the variations contemplated here differ from those in Hamilton's principle. Moreover, the motions to which the present principle applies are those in which the Hamiltonian is conserved. The Hamiltonian function was defined in § 6 : 6 eqn. (6.26) by

Pi —

aT

=

aL

H(qi,pi,t) = pi4i — L

(11.10)

and is taken to represent the total energy, although it only coincides with

T + V in the special case in which T = 2 al;4,4i, a homogeneous function of the velocities. The condition for conservation of H is that L, and therefore H, does not depend explicitly on the time, i.e., aL aH .

at

—

at _

The proof of Hamilton's principle used as the comparison motion a trajectory joining M and N in which the coordinates of the particle at time t were qi oqi ,where o qi = i (t), and the variations o qi vanished at the end points. The variations were made at constant time. For the Principle of Least Action we consider a more general variation (but restrict its application to systems with constant H). In the adjacent comparison motion the point R2 corresponds to the point 1? of the natural motion (see Fig. 129) while R, would correspond to R in the comparison motion for Hamilton's principle. The system passes through R2 in the

464


varied motion at time ZI t later than it passes through R in the natural motion. (11.11) qi = 6 qi + gi 4t. Since the 4 qi are arbitrary both 6q, and ZI t are arbitrary. Because the system starts at M and finishes at N we impose the conditions

qi)m = 0,

qt)N = 0.

FIG. 129

But, in general, neither 6q, nor 4 t are zero at M, N. For any function , q„, t) the change corresponding to LI qi is

/ (q1, q2,

a

A

— — LI qi

at

Ot

(11.13)

Li t ;

(a summation over i is implied) i.e.,

al 4ff

64 +

aqi + —) at Zit

di

6q1

(11.14)

The action corresponding to the motion of the system from M to N is defined to be the line-integral N

A =f pi dq,

(11.15)

taken along the trajectory followed by the system. The Principle of Least Action states that A has a stationary (usually a minimum) value for the variations just described when the motion is such that H is conserved, i.e., ZI A = f pi dq, = O. na

§ 11:3]

465

ANALYTICAL DYNAMICS

We write A in the form ra

A = f pi q,dt, ra

where now the limits of integration are altered in the variation. From the definition of H

A= f (H + L)dt = f Ldt k H(t2 -- 4), H being a constant.

t,.

ra

;AA = ZIfLdt t,

t2 Z1 t1) =.61iLdt

The integral S = L dt is a function of q„

t,

qn , t evaluated at the

end points of the trajectory so that, according to eqn. (11.13),

A S = 68(t2) — 62(4) + S(t2)At2— A.9(4)44. Since the variations 6q, in this case, do not vanish at the end points, t2 we cannot write 6S = 6 f L dt = 0. Hence rl ta

8L aL iv,.6 qi + Sqi } dt + [L Llt]::. S=f agi Using Lagrange's equations of motion we write this result as

AS

d ( aL d ) uqi

j

rd

aL

d i)} dt dt (aq

t,

j dt 1 ={

6qi

[ r qi

aL

a 4iL + LLIt

d

r ath aL 1 aL qi O't

[LIM"

[LAW

466


Since the variations A qi vanish at the end points,

S = RL

— 4, v °-6.)Z1 t1:: ,

A A = [( H

L — pi th ) ZI = 0, H L — pi 4, vanishes because of the definition

since the expression of H. This proves the Principle of Least Action. We can express this principle in a different form. So far we have used the configuration space simply to represent the states of the system; we have ascribed no other properties to it. Suppose now that we consider systems for which T = z au4, 4.1 (i.e., the constraints do not involve the time) and ascribe a metric to the configuration space such that the distance between two neighbouring points of the space is given by ds2 = a,„ dq, dq„. ds )2 .• (d — a,, _ 2T .

t

This choice of metric, therefore, defines the distance in the space in such a way that the speed of the representative point in the space is V(2 T).

t, t, A = f p, 4, dt =f 2Tdt, t,

(11.18)

t,

since T = z a,, 4, = p, 4, . We now change the variable of integration to arc length in the configuration space so that dt = ds/V(2T). Then

A = fl/(2T) ds. But

H = T + V, where H is the constant E, A=l/2 f (E — V) ds,

and the Principle of Least Action takes the form Zl iV(E — V) ds = 0. This is Jacobi's form of the. Principle.

§ 11 : 3]

ANALYTICAL DYNAMICS

467

Examples. (i) If we consider a particle in motion in three-dimensional space, T = m(±2 + 92 + 12). Therefore the metric of the configuration space, in which the coordinates are {x y z} , is such that it is identical with that of the physical space in which the particle moves. Hence, a particle with total energy E moves in space between the points M and N in such a way that the integral

f 11 (.g — V) ds is stationary. (ii) If there are no forces acting on a system, then V = 0 and H = T = constant. The Principle of Least Action therefore gives

A f 11T d8=11TA f C1.8=0, M

M

i.e., the system moves so that the distance between the representative points M, N is stationary. Curves having this property are called geodesics. A special case of this result is found in the motion of a free particle. The geodesics of Euclidean threedimensional space are straight lines, which are also the paths of free particles. (iii) If a system moves under no forces, e.g., the free rotation of a rigid body, eqn. (11.18) shows that, since V = 0 and T is constant, tg A = 2 f T dt = 2T(t 2— t1). Since T is constant the condition AA = 0 is equivalent to A (t2— t1) = 0, i.e., the system moves so that the time taken for the motion is stationary (usually a minimum). (iv) State and prove the Principle of Stationary Action for a conservative holonomic dynamical system. (Lagrange's equations may be assumed.) Show that for the motion of a particle of mass m between points M and N under given forces with potential energy V, the principle is equivalent to the requirement that the integral

f 11 {2M(E — V)} d shall have a stationary value, E being the total energy, and ds the element of length along the path. Hence or otherwise use the principle to obtain the path of a particle under uniform gravity near the Earth's surface. (The first two parts of this problem have been covered in the text. Here we solve the illustrative problem only.) In this case V = mg y and, omitting the constant factor y(2m), N

A— f (E — gy) ds.

ai

468


We use y as the independent variable and ds =/{(w dx 2) + 11 dy.

A = f 11{E'— g y) 11 {x' 2+ 1} dy M

Euler's equation for the stationary value of A is d ( 0 gy) (

a ) x'2 + 1)11'1 —a7 {[(E — gy) (x' 2 + 1)Y11 = O.

Since the second term is zero we obtain the first integral at once

a x' RE — gy) (x' 2 + 1)]'!' = C,

1/(E — g y) .

•

11 (x' 2+ 1)

1 1 +— x' 2

dy 2 B — gy dx ) Cz

—C.

—gy

C2 1 — a by, (b > 0)

This can be integrated to give 4(a — by) = b2(x + B)2, where B is an arbitrary constant,

i.e.,

a

y=T

b

. —

(x B)2.

This is a parabola with vertex at (— B, alb), latus rectum 4/b, vertex upwards. The path of a projectile is therefore a parabola.

11:4 Hamilton's equations of motion Lagrange's equations for a system with n degrees of freedom consist of n differential equations of the second order which determine, together with 2n initial conditions, the n coordinates qt as functions of the time. The state of the system at any given instant is determined by the values of qi and the velocities 4i ; these 2n, quantities can be given arbitrary values for an arbitrary state of the system. Hamilton's equations of motion take a somewhat different outlook. Instead of qi and 4, the state of a system is determined by the values of qi and the corresponding generalised momenta 2),; the specification of the state of a system with n

§ 11 : 4]

469

ANALYTICAL DYNAMICS

degrees of freedom requires the values of these 2n variables. In the Hamiltonian outlook these canonical variables pi and qi are regarded as the fundamental, independent variables of a system and the equations of motion are 2n first order differential equations whose solution gives these 2n variables as functions of the time. In the Lagrangian outlook we used a configuration space, each position of the system corresponding to a point in this space; the state of motion of the system was given by the motion of the representative point. In the Hamiltonian outlook we use a phase space in which each point is specified by 2n coordinates, pi and qi together; every state of motion of the system now corresponds to a point in this phase space. In the Hamiltonian outlook all the mechanical properties of a system are incorporated in the Hamiltonian function which is written as a function of pi and qt after eliminating li from the equations. (Our previous use of the Hamiltonian containing velocities was not strietly accorate terminology.) Examples. (i) A particle, moving in a potential field V(x, y, z), is described by the functions

T = lm(±2+ 92 + i2), L = T — V, Px

H = pi 4i — L 1 =

2m

Py, = m9, Pz

px + l ` Py +

in

1 M ( 121 +

2

\ m2

+ 1 )+V

m2 m 2j

(p2x + 293,2+ /4) + V.

(ii) A particle in a plane, using plane polar coordinates, is described by the functions

T = m(t2 r 202), L = T — V (r,0),

po = mr2O.

Pr=

PO 1 (24 Pr — —m H = Pr — + Po m2 m mr2 2

pB ) 1 m2 r2

1 ( p, 2 + r)+ V. 2m r2 (iii) The spherical pendulum (of mass on).

T = 1-ma2(62+ qp2 sins), V = — mga cos0 , = ma2,i, sin20 . po moo, 1 . .

2ma2

2

2

+

Ptp

sin2O

) mg a cos0.

V

470


(iv) Taking q1, q2, q, to be the Eulerian angles 0, q,, y which specify the position of a rigid body having one fixed point 0, show that

Awl = p1sing2 + p2cotg„ cos g2— p3cosec g„ cos q2 , and derive corresponding expressions for B co2, Cw3, where co„ wz co3 are the components of angular velocity of the body about its principal axes at 0, and A, B, C its moments of inertia about these axes. From the Hamiltonian equation i9 = —a Hia q2, recover Euler's form of the dynamical equations for the body. ,

The relation between Euler's angles and the angular velocity of the rigid body has resolutes, eqns. (1.25a, b), co, = —y, sine cos q) + 6 sing) = co2 = y sine sing) + 6 cosq) =

sing, cosg2 + ql sing2 ,

q3sing, sing2+ g„ cos g2 ,

co3= q + y cos 0 = q2 + q, cos g . Since

T = (A coT.+ B cc4 + C coD , 0T a ql

awl = =

0 co,

awe

C°1 — B CO2 —

a qi

C

a43.

.9 4i

=

Ao), sing2 + Bco2cos g2

Similarly

p, = Cw3, p3= — A co, sing, cos q, + B co, sing, sing, + Cw3cos q,. Writing these in the form

— A co, cos q, + B co, singe =

— /32 cos q,) cosec g„

A co, sing2 + B co, cosg, = pi, we obtain the solution

A co, = pl sing2 + P2 Ca qi COS q2

p3cosecg, cos q2

BCO2 = picos q, — p2 cotg1sing2 + p, cosecg1sing2 , Cco3 = 232. After some manipulations we obtain

A2CO2

2 H— A

= pi

+

sin2 q2

A

C—c

B2 04

B

+

C24

cos2g2

C

+V

+ (p2cot gi— p3cosec g,)2

sin2 g2 cos2 g2 + A 1

1

2p1sing2cos g, (p2cot q, — p3cosecgi) (— —--) B +2V V.

A

+

§11: 41

ANALYTICAL

471

DYNAMICS

On differentiating this we obtain H aq2

= {pi

—

(p2 cot g,

— T3 1 ) sing, cos g,

(p, cotg, — p3cosecg1)2}

—

1 1 (cos2g, — sine q2) A B

p3cosec g„) (— —

=. (Pi sing, + p, cotg, cos g, —

V a q2

p3cosecq, cos g2 ) x

1 cosec g1sin g2) (— A 1 1) 0V colco2 A B (— - — — A B 0 q2

(p1 cos g, — p, cot g, sin g,

But fi 2 = 063, Hence the equation of motion fi 3 = - a

1\ ± 0V B /I 0 q,

Hog, becomes

V . aq2 Since q2 = 99, — 0 V la q, is the moment N of the applied forces about the axis of moment C, C 63 (A — B) a vo2 = N. i.e.

cb, =

—

(B — A) coi.co2—

—

This is one of Euler's dynamical equations.

To obtain Hamilton's equations of motion we note that Lagrange's equations are aL . d ( aL\ dpi dt dt 1' 2' •." n).

Now H is a function of pi , qi and t so that its differential is OH , dH = — aq•

H d, pi

aildt. at

(11.20)

From the definition of H

dH = 4i dpi + pi dth — dL aL OL OL = qi dpi + pi d qi — dli — dt 0q, , — at aL — at dt. = 4i dpi —

(11.21)

When we compare eqns. (11.20) and (11.21) we obtain dqi dt

dpi dt

aqi

(i = 1, 2, ..., n).

(11.22)

472


These are Hamilton's equations of motion. We see also that

0H

aL

at at • We saw in § 6:6(b) that when aLlat = 0 the Hamiltonian is conserved. This follows immediately from eqns. (11.22). Since aLlat= 0, aHlat= 0. dH 0.H dqi 0H dpi 011 OH aH a H dt

dt

pi dt

Oqi ap,

0pi aqi

=0.

Poisson brackets If f, g are two functions of the canonical variables

f = f(qi , pi , t), g = g(q„ pi , t), the Poisson bracket of f and g is denoted by [ f, g] and is defined by

[f, g] =

al ag

—

af ag

(11.23)

0 pi aqi

(summation over i). The importance of these expressions becomes more evident in more advanced applications. Here we show one important property. Since f is a function of the canonical variables, it alters as the system moves according to the relation df

dt

of

at

dq, aqi dt

dpi api dt

of at

of aH a g, ap,

of aH

of

bpi 0q,

at

of

of

Hence / is a constant of the motion, i.e., dfidt = 0, if

U, 11] = —

Of

— at , (11.24)

[f, H] = 0

or

when f does not depend explicitly on the time. In particular, if we choose f = pr , pr

[pr , H] = T-97

aH

apr aH w7,

—u

aH„ a H — uri

aH =

0q,.

dp,. dt

=

,

§ 11 : 4]

473

ANALYTICAL DYNAMICS

and similarly, if f = qr ,

[q,.,

=

dq,. . dt

Hence, we can write the Hamiltonian equations in the form dp,. —

dt

[Pr,, H ] ,

dt —

11] •

(11.25)

In this form the two canonical variables occur on an exactly equal footing; the difference between a coordinate and a momentum is no longer evident. Examples. (i) Obtain Hamilton's equations for the motion of a particle of mass m under a central force of attraction ,u/r2. Hence obtain a differential equation expressing as a function of V. We have already obtained the Hamiltonian function for this problem, p. 469. viz., (

H=

po Fc +— —— . r2 r

0

2m

The equations of motion are:

dqi

OH

dt

op,

dpi

Pr

811

at

Ps m r2

qi ;

PO, P = mr3 -r2

pip

= 0.

Hence pp = mh = constant. Eliminating Pr gives m =—

mh2 y r3 +—. m2

The remainder of the solution is now the same as in the elementary treatment. (ii) Use Hamilton's equations to obtain the equations of motion of a particle moving in a plane, referred to orthogonal axes Ox, Oy which rotate with uniform angular velocity w, the potential energy being V (x, y). Since the velocity resolutes of the particle in the directions Ox, Oy are {i; — coy

T =m Px =

16

CMES

9 + x} ,

— co y)2+ (' + co x)2} —(0y),

p y = m (9 + co x) ,

474


i.e.,

x = p„Im + coy, 9 = py/m, „

=

—

wx.

24 + coy px + Py — coxpy P.

Py I- V 2m 2m

1 — (p + 792) + Y 2m

(YPv — x/9y) + V.

The Hamiltonian equations of motion are: dq, a H —= dt pi dp dt

X=

1°X

+

m

. PY coy, y = — — cox:

In

aH

0V

=— —7x oPy 4.

ax '

0V Py

°1x - •

•

Elimination of px and pyleads to the usual equation

av

97(x — 079 — thy) = mco(y + cox) —

77* — 20.0 — 0.)2x — thy) = —

ax

av x

.

Similarly

„(9 20)± — co 2 Y + thx)

7

aV y

(iii) Ignoration of coordinates. An ignorable coordinate q, is one which does not occur explicitly in L, i.e., a Lia q,. = 0. When we discussed this case in § 6:7 we introduced a Routhian function R. From its definition the reader can see that R is defined in a manner similar to H. We prove now that R behaves as a Lagrangian function for the nonignorable coordinates, by using Hamilton's principle. In fact, the method introduced for the ignoration of coordinates is half-way to a Hamiltonian method. We suppose that the coordinates q1 q2, qk of a system are not ignorable, but that the coordinates x1, x2 , ..., x1are ignorable. The Lagrangian function is therefore L = L(qi , q2, ±,, t). qk , 42 , ..., 4,, ±,, x2, ,

We denote the momenta conjugate to the x-coordinates by

m

=

0L

a

'

(m = 1, 2, ..., /).

The Routhian function is defined by R(41, 4i,

7r 2 , • • • , r I , t) = L — .L' nn. ±m • m =1

§ 11: 4]

475

ANALYTICAL DYNAMICS

Hence Hamilton's principle states a.

6f

7t„,4,1dt = O.

+ m

r

We regard this integrand as a special case of a function of qt , and t (in which xm , nn, are absent) which has to be stationary i.e.,

i Xm 5 7rm

Z,fl

f G(qi , eh, x„„ nm , ±„„ Am , t} dt = 0. The Euler equations giving the conditions for this are dt d (00)_ O G qi j qt

d (OR \_ 0R

d(OG

00

d

dt

ax„,

dt

d loG\ 00

d

dt

4, )

dt

anm

0

(i = 1, 2, .

0 qi

(7r,„) — 0 = 0,

(m

OR

(0)

an,„

—

°'

k),

1,2, ...,

(m = 1, 2, ..., 1).

(1)

(2)

(3)

Eqns. (1) show that R serves as a Lagrangian function for qi , eqn. (2) shows that the momenta 76 are constant; eqn. (3) gives the expressions for the velocities r„, when R has been expressed in terms of the momenta n„, instead of in terms of the velocities. Exercises 11:4 1. Prove that the canonical variables satisfy the relations [q.,

= 0, [pr, Ps] = 0 ,

[gr, Ps] = Ors, (r, s = 1, 2, ..., n).

2. Prove that, if I (qi , pi , t) is a function of the dynamical variables

[qr, f] =

Of

f 010,

[Pr,

=—

4' ,

•

3. Prove that the Poisson bracket of two constants of the motion is itself a constant of the motion. 4. Show that f = x — ptlm is a constant of the motion for a free particle under no forces. 5. If 2T = qT.+ qi q2 and V = z v2 qi, v being a constant, show that Hamilton's equations give = a= cos= (vt + 6) + b2sin2 (vt + 6) and b

tan (q2 + y) = — tan (vt + 6), a a, b, y, S being arbitrary constants. 16*

476


6. If the Lagrangian of a particle is L = m (e. + 41.42 + g) — T (qi, q2)

,

show that the Hamiltonian is 2

H = 3m (P? — P1 P2 + p) -I-V (qi., q2) and that one equation of motion is 4 0 V 2 0V mq=———— 3 0 qi 3 0q2 7. A particle of mass m moves in a plane under an attractive force u/r3 to the origin. Taking q1., q2 as the polar coordinates (r, 0), write down Hamilton's equations. Prove that p, = h (constant), and 7/2 — pm constant. pi -I 2 8. A particle of mass m moves in a straight line. The Hamiltonian function is in2 c4)1/2 + V (X),

H (p, x) = (p2c2

where c is a constant. Write down Hamilton's equations of motion, and deduce that mx

OV_ d 0x

dt

(1 — eicz)' /2

9. A particle of unit mass moves in space under the action of a constant force F parallel to the axis Oz of a set of rectangular axes 0 (xyz) and a force e/r2 directed towards 0, r being the distance from 0 and e a constant. By introducing new coordinates $, n defined by

, q'

z= prove that (.1$192 B 2

— rl,

x = 2 V($n) cos.?), y = 21/(en) sin99 ,

e ce p$2) ) + 2 np2 B 1 2

97

77

e

pn2) =

2

where B and C are certain constants. 10. A particle of unit mass is moving in a plane under the inverse square law attraction of two fixed particles of equal mass m placed at the points (0, c) and (0, — c), respectively, of a rectangular coordinate system (x, y). If y is the constant of gravitation, show that the Lagrangian for the motion of the particle is 1 L = —(t2 + 92) + yni 2

1

1

1/{x2 + (y c)2}

1){x2 + (y + c)2}

j 11:5]

ANALYTICAL DYNAMICS

477

Using the transformation

x = c sinh4 sinn, y = c cosh cosh, show that the Hamiltonian for the motion is

H=

4? mc cosh$ 1 p + • 2c2 cosh2 6 — cos2 77 —

11:5 Transformation theory: contact transformations The change from the Lagrangian to the Hamiltonian outlook involved a change of independent variables by means of eqns. (11.10). When the Hamiltonian function is independent of a particular coordinate q,. the corresponding momentum A. is constant and that coordinate is said to be cyclic or ignorable [see § 6:7 and example (iii) p. 474]. In this case dp,. dt

0, i.e., 29,. = oc,. = constant. eq =

If all the coordinates qi are cyclic, the Hamiltonian is a function H(291, 232, , p,,, t) and the solution of the equations of motion is trivial, for we have pi = cc, (i = 1, 2, ... , n) where the oci are constants of integration; the remaining equations of motion are dq,

dt

—

aH Op,

= w1 , (i = 1, 2, ..., n)

where the co, are constants since eHlapi is a function of the pi only, and each pi is constant. Hence the solution is

pi = oci , qi = wi t + Pi, where the 13, are other constants of integration. A given system can be described by more than one set of generalised coordinates, and also by more than one set of canonical variables pi , qi . If we are prepared to weaken, or abandon, the distinction between a " coordinate " and a "momentum" and treat both canonical variables on an equal footing we may be able to find a set of canonical variables such that H depends on only one half of them (which we could call "momenta"). Transformation to such a set of variables would make the solution of the problem trivial. Certainly a suitable change of variables may facilitate the solution of a problem. The investigation we give now concerns the possible transformations, i.e., the changes of coordinates of phase space, which are suitable for this purpose. 16 a CMES

478


The variables q, , pi are changed to a new set Q„ Pi which are, in general, mixtures of the original coordinates and momenta. Hence we do not attempt to maintain any distinction between coordinates and momenta although we continue to use two letters P, Q (n of each symbol). A change to a new set of generalised coordinates is a relation such as Qi

(i = 1, 2, ..., n) .

q2, ••• , q„, t)

=

We call this a point transformation; it is the type of change which leaves Lagrange's equations of motion invariant in form. We now wish to find relations such as

Qi

Pi , P2, "'

Pi Pi(qi = q2, • • • , qn , pi ,

Pn, t),

P2, •••

t),

(11.26)

where the functions Q, and Pi are chosen so that we can find a function P1, P2 , ..., P, , t) such that the equations of motion K (Q1, Q2, ••• still have the Hamiltonian form dQ, dt

lc

a Pi

dP, dt

aK aQ,

(11.27)

An arbitrary choice of functions in (11.26) will not ensure that this is so. Accordingly we must find a method of choosing suitable functions. We saw in example (i) p. 459 that the Hamiltonian equations of motion are the conditions for the variations of an integral to be zero, i.e., in the original variables; I (p,4, — H) dt = 0,

(11.28)

in the new variables; tro

0. (11.29) (P,Q, — K) dt = The conditions (11.28) and (11.29) will be satisfied if the difference of the integrands is the total time derivative of an arbitrary function, i.e., if dF (11.30) (PA — H) — (1' tO 1 — K) = (.j . • When this is so

r

to

/2

(p,4, — H) dt = f (Pi Qi — K) dt F (t2) — F (ti).

§ 11 • 5]

479

ANALYTICAL DYNAMICS

The variations considered are made at constant time, therefore

6F(t2) = 0 = 6F (t,), and

t, t, 6 f(pi qi — H) dt = 6 f (PiQi — K) dt.

Hence, when the sets of canonical variables related by (11.26) also satisfy (11.30) the new variables Pi , Q, satisfy eqn. (11.27) and viceversa. The function F may involve the 4n + 1 variables gi , pi , Qi , Pi , t but, since we require 2n relations (11.26), F effectively depends only upon 2n + 1 of these 4n + 1 variables. We can therefore choose F in one of four ways, viz.,

Qi , t) ; F2(gi , Pi , t) ;

F3(pi ,

Qi , t) ;

F4 (pi ,

Pi , t). (11.31)

Suppose we choose F in the form F1, then dF1 dt

OF,

OF, n. q`

a Fi + at = Pith — PiQi — (H — K)

on our using (11.30). The variables, qi , Qi , t are all independent so that we identify

pz —

a a F, Pi = — F ,K = H a gi

aF . at

(11.32, 33, 34)

Eqns. (11.32) give the 23, in terms of qt , Qt , t and we suppose these solved for Qi as functions of gi , pi , t. Hence we have the first of the relations (11.26). Now substitution of these relations into eqn. (11.33) gives the Pi in terms of pi , qt , t thus providing the second of the relations (11.26). The arbitrary function F1(g„ Qi , t) in this way generates a suitable transformation which retains the Hamiltonian form of the equations of motion. This transformation is a contact transformation. We can also find the form of functions F2, F2 , F4which generate the same transformation as F1above. When we use F2 (g„ Pi , t) instead of F1we are replacing go Qi by the independent variables gi , Pi where Pi = — a FilaQi . This change is very similar to that made when we changed from the Lagrangian to the Hamiltonian function. In the latter case gi , 4, in L are replaced by g„ pi in H where pi = a Lja gi , H = pi4, — L. On this analogy we choose

P2(qi ,Pi, t) = 16 It*

0 + FiQi

(11.35)

480


(summation over i in the last term) and eliminate Q, by means of the relations Pi = — F110Q1 . With this choice of F2,

d F2 dt

= d F1 +

dt

=

p.0 + p.n. _ "

_

—

"

(H — K)

+ /5,Qi

QiPi — (H — K).

But

F, . aF2 = aq, q, + aF, P- + at

dF,F, dt

from which we identify

a F2

= -

°F2 pi

Qt

Pt = aqi

at

z

(11.36, 37, 38)

When F1and F2 are related by (11.35) they both generate the same transformation. When F2is arbitrary then it generates a contact transformation through the relations (11.36, 37, 38). Examples. (i) The functions F3 (pi , Qi , t), F4 (pi , Pi , t) when related to F1as stated below generate the same transformation as F1through the relations stated. [Proofs are similar to that leading to eqns. (11.36-38).] (a)

Qi, t) — pith; pi = F110

P3 (Pi, Qi, t) =

Then F3 qi

aF3 Pt

O pi

(b) F4 (pi , Pi , t) = F1 (qi ,Q1 ,t)

0(21

a F3

K = H at.

Pi Qi - pi qi ; pi = 0 Filaqi , Pi = - FilaQi .

Then

=-

-F4

a Pi

Qi

a F4 -

0 Pi

,

=H

a P4 at .

(ii) The condition for a given set of relations between pi , qi , Pi , Qi , t to be a contact transformation is that pi dqi - Pi dQiis a perfect differential of a function of qi , Qi .

Proof. Since

dF1

pith -

- IC) =

=H dt

then

,

Pi th - PiVi or

OF1

aq OF1

pidgi

d F1 OFi dt

at

Fi ti I -w 2:

a I',

- Pi dQi = — dqi -I-o-T dQi• qt

§ 11: 5:1

481

ANALYTICAL DYNAMICS

The r.h. side of this is a perfect differential since 0

Fl

0(21

dql

and the result follows. (iii) We find now the transformations generated by certain special forms for F1, F2 etc(a) Fi(qi , Qi , t) = qi Qi • From (11.32, 33, 34) we obtain 0F, Pi = — = Qi,

a qi

0 F„ Pi = —

Qi

= — go K = H •

This interchanges the roles of coordinate and momentum, apart from the change of sign. [It is clear from the form of Hamilton's equations that this leaves the equations unaltered.] (b)

F2 (qi, Pi, t) = qiPi•

From (11.36, 37, 38) we obtain F2

a F2

P1

P"

Q1

=

K = H.

013i=

This transformation is the identity, i.e., no change has been made. (c) A generalisation of the last transformation is F2 (qi Pi , t) = Fj

(4'1,q2,• • • , qn, t)

(summation over j on the r.h. side). .• pi =

fi aif — , Qi = f i (qi , q2 , , q„, t), K = H Pi a t

a qi

.

Since Qi depends only on qi and t, this is a point transformation; the other half relating pi and Pi is the usual relation between generalised momenta corresponding to the relation between the coordinates. (iv) We consider the one-dimensional harmonic oscillator. This is a system described by one pair of canonical variables (q, p) with the Hamiltonian H (q, P)

p2

m n2 q2

2m

2

•

The Hamiltonian equations of motion are q=

p , p = — mn2q, 10,

i.e., q =

482


The contact transformation generated by

F1(q, Q, t) = +mnq2 cotQ gives

If', 1 = F1 = mnq cotQ, P = — — = mnq2 cosec2 Q, K = H. Q 2 q But

K = H = (9 n n2q2 + 132 m) _ m (n2 9,2 + n2 q2 cot2Q) _=mn2q2 cosec2 Q = nP . Hence the Hamiltonian in the new variables is independent of Q, i.e., all (the only) coordinate(s) are cyclic. The solution of the equations of motion P = 0, Q = n is

P = a, Q = nt + a = mnq2cosec2 (nt + P). .• q =

mn

) sin (nt + 13), p = 1/(2conn) cos (nt + fl)

where a, fl are arbitrary. Of course, this solution is like using "a steam hammer to crack a nut", but it illustrates a contact transformation which makes all the coordinates cyclic. (v) A dynamical system with two degrees of freedom has a Hamiltonian function

H given by 2H =pi +

-F4'7 + (q2 — qi)2 + A •

Find the coefficients a1, a2 , in the generating function of a contact transformation W = aT(q, + q2)2cotQ1+ 14(q1— q2)2cotQ2, so that the Hamiltonian becomes

H = P, + 1/3P2 . Hence find the normal frequencies and normal coordinates for the system. In this case the given function W is of the type F1, i.e., W

aW

pi= —, = — — , K = H. qi a Qt = 24(q1 + q2) cotQ1 +

— q2) cotQ2 ,

p2 = 24(q1 + q2) cotQ1—

(qi— q2) cotQ2 ,

pl

P1=

aT(q„ + q2)2 cosec2 Q1, P2 =

(q„ — q2)2 cosec2Q2 •

Also

K = H = (24. +

+ z 4i + (q2

—

qi)2 +

P1+ 1/3 P2= aT (qi + q2)2 cosec2Q1 + 113 4(q1 --- q2)2 cosec2Q2

= 4a 4 (g1-I- q2)2 cot2 Q1-I-

(q, — q2)2cot2Q 2

e

-

qiq2+ q,2,,.

§ 11: 5]

483

ANALYTICAL DYNAMICS

We equate the coefficients of the (independent) variables cot2 Q1, cot2Q2 and obtain 4ai = 44 = 1/34, a1

Hence Pi

1 —, =4

2

V3 . =— 4

= (qi + q2) cotQ1 + I (q.1 q2) cotQ2, q2) cotQ1

P2 = (q1

1/3(4'1

q2) cotQ2,

Pl = I (q, + q2)2cosec2Q1, P2 = i V 3(qi — q2) 2 cosec2 Q2 •

(1)

In terms of Qt , Pi the equations of motion are

= 0, P2 = 0 ;

01 = 1, Q2 =

(i.e., all the " coordinates " are cyclic). These integrate to Q1 = t + a, Q2 =

P, = b2 , P2 = c2.

+ /3,

[We choose b2, c2as the constants of integration because, from (1), both P1and P2 are positive.] Hence

(q1 + q2)2= 4b2sin2(t + a), (q1— q2)2 = (44 3) c2 sin2 (f3 t + i.e., q1 +

q2 = f 2 b sin (t + a), q1 — q2= i (2/3V4) c sin (I/ 3t + fl)

This form of the solution shows that the normal frequencies are 1/(2n), y3/(2n), and the corresponding normal coordinates are proportional to q1 + q2 , q1 — q2.

Exercises 11:5 1. Prove that two contact transformations applied in succession produce another contact transformation. 2. Define a contact transformation and show that the following is a contact transformation:

Q1 = +

Q2 =

+ A + + pn;

P, = 2 tan-1 (t)— 2tan-1 p2

;

P2 = — tan-1 (

P2 If the Hamiltonian of the system is given by

H=

+ + 41+ 0)

use the above contact transformation to solve the equations of motion of the system; compare this solution with the solution obtained from the given Hamiltonian using the original variables.

484

A COURSE O1 MATHEMATICS

3. Prove that the transformation defined by the equations q, = A.711 (2Q1)'1' cosP1-F 2:VI'(2Q2)'/' cosP2

(2(4)1/2 cos P, 411' (2Q2)1/' cosP2

(12 = p1 =

,

(2 Al Q1)112 sinP1-I- (222 Q2)'/= sinP2 , 02,1 112 Salsin P2 2 (22.2 - 2 ..,

(22, Qi)liz sin P1

p2= —

is a contact transformation, which changes the equations dq, = dt where

H=

H

0H

dp„ dt

pr

q, (r

2),

(qi + q2)2,

+ + *AT(g1— q2)2 +

into the equations dP,

dQ, 8K dt OP,

dt

0K =—

OQ,

(r

1, 2),

where K = yl1Q1 + 22 Q2 . Integrate the latter system of equations and hence find a solution of the original system.

11:6 Infinitesimal contact transformations We have seen that the function F2(qi, F1 , t) = q, P. generates the identical transformation; we now determine what function generates an infinitesimal transformation, i.e., one in which the new variables are almost equal to the old variables, Q1 = q1 + 6 qi, P1= pi + aP1-

(11.39)

(Here aq,, 6p, are NOT arbitrary variations but stand for the first order differences Q, — q1 , P, — pi .) Since the transformation is almost the identity we expect it to be generated by a function

P2(q1, Pi, t) = q,P1

E

(11.40)

(go Pot)

which differs only slightly from the generator of the identity. The new variables are given by

, 'i—

99 a F2 113i — gi +8 P ,'

F2 p, — aaqi — P, + e

and K = H +

aii 2 at

a q) a qi

i.e., .e.,

a 99 a P, aq)

, i.e. , Op, = — E i.e.,

'

5 q1 = e

OH — e

aq,

a9) at .

'

§ 11 : 6]

ANALYTICAL

DYNAMICS

485

Since we are working to the first order and e is small, we can write these relations

(5q, = e

99 a , (5p, aPi

s a'T a q,

=8

(11.41)

e199

at

The infinitesimal transformation is said to be generated by the function E99 correct to the first order.) Now we consider the evolution of a system with Hamiltonian H(q,, pi , t). In a short interval of time 5t = s the changes in p, and qi are given, correct to the first order, by

q)(qi , pt, t). (We replace P, by pi since we only need

dqt = 84, = s

(5 H =

aH apt

dpt =ep, =

a H 6q, + a H (5p, + a H apt at aqi

—8

aH

.

qi 8=E

(11.42)

aH at

A comparison of eqns. (11.41) and (11.42) shows that the change produced by the infinitesimal contact transformation (11.40) is identical with the motion of the system if we identify p(q, , p, , t) with H (q, , p,, t). Thus the motion of the system in a short interval is the same as an infinitesimal contact transformation generated by H. The motion in a finite time is the result of a succession of such infinitesimal transformations. The result of applying two contact transformations in succession is another contact transformation (see Exercises 11 :5, Q. 1) so that the motion of the system with Hamiltonian H can be regarded as the continuous unfolding of a finite contact transformation.

Poisson brackets We consider now an infinitesimal contact transformation generated by 99 as in eqn. (11.41). As a result of this transformation the function becomes f 6f, where I (qi, q2, •••; qn , p1, P2, "•

e + o f op, — 8 (a f ac p 6!, = aatq, oqi aq, ap, 0Pi

at a\ = ap, aq, )

[

f,97].

If we substitute H for the function f (in the case for which H is independent of t) then

(5H = 8[H, q11. 1 6b CATES

486


Hence 8 H = 0 for all those functions whose Poisson brackets with H vanish. But such functions, as a result of eqn. (11.24), p. 472, are constants of the motion. Hence we conclude:

The constants of the motion are the generators of those infinitesimal contact transformations which leave the Hamiltonian invariant. This means that we can, in principle, find all the constants of the motion (i.e., solve the problem) by an examination of the symmetry of the Hamiltonian. Example. We illustrate this result with the special case of a rotation. Suppose a system is specified by sets of cartesian coordinates {xi xi zi), i = 1, 2, ..., n and the corresponding momenta {pzi py, pzi). We suppose further that the Hamiltonian is unaltered by any rotation about the z-axis. Consider the infinitesimal rotation — a point transformation which is also a contact transformation — given by

X, = xi — yi SO, 17, = y, + xi SO, Z, = z,. In the notation of eqns. (11.39)

Sx, = —y,SO, Sy; = ;SO, ozi = 0.

(1)

Since this is a point transformation the generating function must have the form

yt, zt)

-Fz =

(2)

(summation over j). Also, since it is an infinitesimal transformation F2 must have the form F2 = xi Pxf YiPri 8 99 (xt, Yi, zi, Pxi, PY PZ 1). (3) The new coordinates are given by Q, = a F2I0 Pi , i.e.,

Xi = x • + s

a co

099 Pxt

Yi — Yi + 8

92

Zi - Zi + 8

t/ Pyi

(4)

(3PZi •

Comparison of (1) and (4) gives e = 60,

'9CP

aPx,

y,,

Op — xi , oPy,

Op

O.

aPz

Bearing in mind the forms (2) and (3) we conclude that

9)(x„ yi, zi, Px I, Pn, Pz1) =xt Fr i

Yi PZi•

As earlier, we replace the Px„ etc., in this by pz, and we obtain the generating function = (x,pyi — yi pzi) (summation over i).

§ 11: 7]

487

ANALYTICAL DYNAMICS

This is the z-component of the angular momentum of the whole system. By our hypothesis this transformation leaves H invariant so that the generating function must be a constant of the motion. This is just one illustration of the close connection between the symmetry of a system and the conservation laws (or constants of the motion) associated with the system.

Poisson brackets provide the bridge which links classical mechanics with quantum mechanics. The Poisson brackets of the latter theory satisfy equations which are formally the same as the corresponding equations of classical mechanics, and have analogous interpretations. In this way we obtain the quantum formulation of a problem from its version in classical theory in terms of Poisson brackets. 11:7 The Hamilton-Jacobi equation We saw in the previous section that the initial values of the canonical variables, which we denote for the present by Qi , Pi and the values qi , pi at a finite, arbitrary time later can be related by a contact transformation. If we can find the function which generates this transformation, we have a solution of the equations of motion. Regarding this transformation as one from qi , pi to Q1 , Pi , we require a transformation to constant (initial) values for the coordinates and momenta. We can achieve this certainly if the transformed Hamiltonian K is zero, for then the equations of motion are Pi =0, = 0. The generating function must satisfy, since K = 0 [see eqn. (11.34)],

aF

H (qi , pi , t)

=0.

Using a generating function of the form F,(q,,Q,, t), where pi = a Filaq,, we see that F1satisfies OF, , t)

F, 8F1 H ( ql , q2,

'

Tq,

'

aq.

)

aF, = t

0.

(11.43)

This differential equation is the Hamilton-Jacobi equation (H-J Eqn.) or Hamilton's partial differential equation. A first order partial differential equation has a general integral which depends upon an arbitrary function. A solution which depends upon n 1 arbitrary constants (n 1 being the number of independent variables) is a comp!ete integral. Such a solution, denoted by F1 = S, 16b*

488


is called Hamilton's principal function (H—PF). It is not, of course, unique. But S only appears in the equation through its first derivatives so that F1 = S a is also a solution, i.e., one of the constants of integration is additive. Hence, ignoring such additive constants, we can write S as S (qi , q2 , qn , al , a2 , ..., an , t). We regard S as the generating function of type F1in which we choose the new variables Qi to be constants ( = oci). Since S satisfies the H—J equation the remaining variables must also be constants. The equations 13,

PL

as aQ,

as (. = 1, 2, ..., n) acx,

(11.44)

can be regarded as furnishing the values of the coordinates qi in terms of t and the constants of integration (initial values) at , 9i . The other equations of the transformation 08 (11.45) Pi =

aqi

now give the momenta pi as functions of qt and at , and therefore as functions of t and (xi , Thus we have in principle a complete solution to the problem — provided we can solve the H—J equation! We used the letter S to denote H—PF advisedly, because we have t, already introduced it as the integral of Hamilton's principle, S= f Ldt. t; Now we alter the notation slightly, regarding t1as the time of the initial state and t (instead of t2) as the arbitrary time of the final state (qi , pi ), i.e.,

S = J L (ix tr (using z as the dummy variable of integration). First we note that dS = L(qi , 41 , t)• dt

(11.46)

But S must also depend upon qi , pi , Qi , Pi specifying the final and initial states. Eqn. (11.6a) p. 459, shows that the increment 6S is given, for the actual motion which satisfies Lagrange's equations, by

SS

4, — pi oqi — Pi(5(21 , =[ oqi h

where the variations are made at constant time.

(11.47)

Sg 11 •• 7]

ANALYTICAL DYNAMICS

as

• • avi — Pi 2

as

489

—P,,

.99,

[cf. eqns. (11.45), (11.44)]. But

as as as av, 4i + Qi

dS

L—— dt at

and since the Q, are constants this gives

L=

as

as at = L

+ 2914,, — H (4i, Pi, t)

after substitution for 4, from pi =aLia4,. This is the H—J equation again, and we have verified that Hamilton's principal function as defined on p. 488 is also given by the integral. Example. We give the discussion by the H—J method of the motion of a particle under gravity. The Hamiltonian is H (q,p) =

1 (14 + p),2) + mg y 2m

1 j( OS \2 (as

OS

at

2m ox )

\21 f m" °•

dy

Since ö t and 0 S 10 x are separated from the functions of the remaining variables, we try a solution S= —ht + ax + G(y). Then 1 I (10 —h + + mgy = 0, dy f 2m i.e., dy G= f 1/ {2m (h — mgy) — and S --ht ax f 11 {2m(h, -- mg y) — a2} dy ht

1 3m2g

{2m(h — 'mg y) — ,x2) 312.

490


(We omit the additive constant of the indefinite integral.) Here the constants h, a play the parts of Qi in the theory. Corresponding to (11.44) we have the equations which give x, y in terms of t as follows:

aS

a

= x + m2g{2m(h — mgy) — a2}/ = #,

(1)

0S 1 t —— m {2m(h — rngy) — a2 } 1I' = y. = — oh g

(2)

Eqn. (1), after squaring, shows that the path of the particle is a parabola with vertex upwards. Eqn. (2), after simplification, gives y = A + Bt — It gt2, showing the downward vertical acceleration, g. The remaining equations corresponding to (11.45) give the momenta 0S =

Ox

OS p), = — — {2m (h — mgy) —

=

ay

The first of these equations displays the constant horizontal momentum, p, = a, characteristic of the projectile. The remaining equation enables us to identify the constant h as the total energy. On squaring and writing a = p, we obtain 2m (h — mgy) =

h=

2m

+ 14, .

(14 + 19,2) + mgY

•

When the Hamiltonian is conserved, H(qt , pi ) = E is an integral of the equations of motion, i.e., H itself is a constant of the motion. In this case the H—J equation is

as at

H (qi,

as aqi

I = 0.

Since S depends upon the time only through time variation by writing

(11.48)

as/1H, we separate out the

S (qi , a,,t) = W (qi , oci ) — Et,

(11.49)

(cf. the example above) where E is a constant which may be equal to one of the a, or may be a function of them. The H—J equation becomes H

qi , aw ) — E = 0. (11.50) a qi

(

The function W is called Hamilton's characteristic function. Eqn. (11.50) is also called the Hamilton—Jacobi equation. We can use the function W

§ 11 : 7]

ANALYTICAL DYNAMICS

491

to generate a contact transformation which is different from that generated by S. We take W as the generator of the form F2 , viz.,

F2 (qi , Pi , t) = W (qi , Pi ). This means that we have chosen the new Pi to be the constants of integration oci (and F2is independent of t). The transformation is given by

pi

aw = aqi

aw aw = a Pi — Ooci

and, because of eqn. (11.50), K is

aW

K=H

= H(qi , pi ) = E .

Since .E is, in general, a function of oci , this shows that

K = K (Pi , P2 , ..., Thus the transformation generated by W is one giving new variables in which the Hamiltonian depends only on the momenta, i.e., all the new coordinates are cyclic. This is the case we considered at the beginning of the section, for which the remainder of the solution is trivial. To obtain the solution in terms of the variables qi , pi we use eqns. (11.51) and the solution for Qi , Pi , Qi =

aK

P, = 0,

— wit +13i,

Pi =

The second set of equations (11.51) is solved to give the qi in terms of Qi and oci ; the first set gives the momenta pi immediately in terms of qi and cci . Remembering that the Principle of Least Action applies to systems in which H does not depend explicitly on the time, we expect the appropriate Hamilton's characteristic function to be connected with the Action integral. This is easily shown to be so. Since W is independent of the time, dW aw dt =aqi q i=Pi4i• .*. W = f pi 4i which is equivalent to the Action integral.

492


Example. In illustration of the above we consider the harmonic oscillator. Since p2 mn2 q2 11 (qi, Pi) =+ 2 2m the H—J equation is 1 d.w )2 mn2 q.2 — E, 2m dq 2 of which the solution is, apart from the additive constant [see Vol. I, p. 81 ex. (v)] 2E m nq 2) 2 V(mn 2 — q

- ' m n2 q2 W = — sin-1 I/ 2E n

We choose the transformed momentum P to be P = Eln,, so that W= P sin-1 -1/( 7"-2) 12P

m no

2P

2

—

q2)

(q, P).

This is the generating function which makes all the coordinates cyclic. The function F1(q, Q) which generates the same transformation is given by [see eqn. (11.35)] (1)

(q, Q) = P2 (q, 13) — PQ, where Q = a F2/0 P = OW 10P. OW

Q= a P =sin -111 2P )2P 2 V— — q2 ) mn

2 1/(2P q 2) mn

1/( -272) .• P = imne cosec2 Q.

(2)

We now eliminate P from eqn. (1) by substitution from (2). .• Fi (q,Q) = iQmnq2cosec2 Q

mnq2 cot Q — Qmnq2cosec2 Q

= Imnq2 cotQ. This is the generating function we used in example (iv) p. 481, where we discussed the harmonic oscillator. This makes the coordinate Q cyclic and the new Hamiltonian function is K = nP. Exercises 11:7 1. Define illaupertuis' action A and prove that if the initial and final configurations a• and /9 of a system are given by q„ = a, and q„ = br respectively (r = 1, 2, .., n), then 0A aA 0A — to— t =; (pr)o; Oa, 0 b, H

§ 11.•7]

493

ANALYTICAL DYNAMICS

where pr is the component of generalised momentum in the coordinate q,. and H= V + Epr q,. — T. A heavy bead of mass m is freely movable on a smooth circular wire of radius a, which is made to rotate about a vertical diameter with spin w, 0 being the angle made by the radius through the bead at any time with the downwards vertical. Prove that 112 2g 2H A = f mat{— + — cos° + w2 sin2 0} dO, a mat 01 and verify directly the above general theorem in this case. 2. Establish Hamilton's principle of variable action 61 = — Hot -F EPtOqi where

tl

=

r Ldt,

Pi =

aL

r

H=—L+

pk lk,

and deduce Hamilton's partial differential equation for I. Show that, for the motion of a projectile in the (x, y)-plane, I

1 2

m

(x2—

x0)2 + (y1 —

—

yo)2

to

g(y, + yo) (t1— to) —

1

12 g 2 (t t 0)31

Verify Hamilton's partial differential equation in this case. 3. For a certain system with two degrees of freedom the integral of energy is H (q,, q2, Pi, 232) = h, a constant, and another integral not involving the time is given in the form

V (qi, q2, pig p2) = c, a constant. Show that there exists a function y (q„, q2 , h, c) such that pi

= OV/aqi, P2 = OVMq2,

and that the remaining integrals are 0 c = constant,

a via h = t + constant.

[Hint: show that a function tp exists such that Mi. + P2 q2 = dp/dt.] If the Hamiltonian function is

q1P2 q2P1 + a — a show that there is an integral pigs

and complete the solution.

+ p2 q2— 2 api p2 = constant,

494


4. A particle of mass m moves in the x, y-plane in a conservative field of force of potential U, its total energy being E. If w = q9 + iy is an analytic function of the complex variable z = x + iy such that I dw/dzI2 = 2m(E — U), show that a complete integral of the Hamilton—Jacobi equation is

S = (19 cosoci + sum, + a2 where (xi. , oc2are constants of integration. Hence, or otherwise, determine the trajectories of particles which are projected from infinity with zero total energy in a field of force of potential

U = — A2 0 P2I(A P2. B P2), where A, B, 0, P are the points (a, 0), (— a, 0), (0, 0), (x, y), and A is a constant. Miscellaneous Exercises XI 1. Apply Hamilton's equations to the plane motion of a particle of unit mass in a field of force for which the potential in polar coordinates is x cos 0/r2, and prove that, if at time t = 0 it is given that r = a, r = 0, then

t = {(r2 — a2)/2E)'/2, where E is the constant total energy. 2. The polar coordinates of a particle of unit mass, moving in a plane under a potential energy 2 Ar — /22/2 r2, are (r, 0). Find the Hamiltonian equations of motion, and show that, in the orbit for which the angular momentum is itz and the total energy H is 2a A, there is an apse at r = a, and after a time (a/2 Ariz the particle is at a distance 2a from the origin. 3. A particle of unit mass moves in a plane so that its potential energy is 1n2 (x2 + y2). If x + iy = c cosh -I- in), find the generalised momentapf pn, corresponding to the generalised coordinates 92, and obtain Hamilton's equations in terms of these coordinates and momenta. Show that they admit of solutions = constant, i) = In, and interpret these solutions. 4. Prove that when a dynamical system whose Lagrangian is L passes from a configuration Clto a configuration 02 in time T, the actual path is that which makes ,

f Ldt stationary. A particle of unit mass moves in a plane about a centre of repulsion producing an acceleration n2 x (distance). If the centre of repulsion is taken as the origin of rectangular coordinates and the particle passes from the point (a, 0) to the point (b, c) in time T, prove that T

f Ldt =

2 sinhn T

{(a2 + b2 + c2) coshnT — 2a 1)).

ANALYTICAL DYNAMICS

EX. XI

495

5. The state of a conservative holonomic dynamical system at time t is defined by n coordinates q1, q2 , , q,,, and n momenta pi , p„ ..., p„. At time to the system lies on a closed curve To in the hyperplane t = to in (q, p, t) space. To is defined in terms of a parameter A, where 2,0 s A < Ai , i.e.

qr = qr(A, to ) , 21r = Pr (A, to) and the parameters A, and A. correspond to the same point on To . At time t the coordinates and momenta of the system are q (A, t), p (A, t). By considering the derivative with respect to A of

J (A) = f L {q, (A, t) ; 4, (A, t)) dt to where L is the Lagrangian and integration is along an actual trajectory, prove that

r

d

A, qr(A, t)1

d = 0. 0 A i A

dt ,.2:1 119' (11.' t) — A,

6. Find the generating function for the transformation which corresponds to an infinitesimal translation in one coordinate, i.e.,

Sqi = 66,„ 6 Pi

= O.

7. A dynamical problem of n degrees of freedom with specifying coordinates q,,, possesses the Hamiltonian function H given by n+I

2H =

+Z

(q, - qr _ 1)2

r=1

r=1

in which qo = q,„ = 0. Show that the contact transformation whose generating function is 2 Wr2 cotQ,.

W= r=1

E a„

,

s=1

in which

a„—

2 [(n + 1)

sin

srgr n+1

leads to equations of transformation

Ps =

wr asr

cosQr, qs = E asrkrl' sinQ, ; r =1

r=1

and that, if the constants wr are given by

w,. = 2 sin

rgi 2 (n + 1)

496


the Hamiltonian function in the new coordinates is

K= X W,Pr . r=

1

Hence (i) solve for q„ (ii) show that normal coordinates for the problem arc 2 —

[(n + 1) tv, 'i2 ]

[The results ti

E cost.° — 1

may be assumed.]

sin (n + 1) 0 2 sin /0

1

2

PV2 sin Q r

n

E cos r 1

2

)0 —

sinn0

ANSWERS TO THE EXERCISES Exercises 1:4 (p. 17) 5. cos a = (3 — 1/2)/(4 1/2), axis {V2 — 1 1/3 — 1/6 — 1). 6 ( _9_ —12 4 25 4

25

12 25

11 25

7. My =

5

a5 0

(

-

3 5

0

0

1

0

1

0

1

0

0

1

0

1

0

0

0

0

1

(0 ,

My=

-

,

My = MyMir 1M37 1 . The interpretation of /1/M),M., = Myis that rotation through x/2 about the axes 0 z, Oy, Ox in that order gives the same displacement as the single rotation about Oy. The matrices Mi-, /1/„T1represent rotations through — n/2 (i.e., x/2 in a clockwise sense) about the respective axes Ox, Oy. The interpretation of .11t = My M;13/,71is that rotation about Oy, followed by rotations through —x/2 about Ox and then about Oy again gives the same displacement as the single rotation about Oz. Exercises 1:5 (p. 25)

2. Via; — (V sin fl)/(2a) = — ( VIct) co-O. 3. (i)

w

cos oc

(ii)

w tancx. sin (a. + /3)

; (iii)

co sin fl cosa sin (cx + /3) —

The answers are the same in all cases. Exercises 1:8 (p. 35) 2. ,c T 4. In spherical polars; radial — (v2 sine azimuthal (v2sin/3 cos/3)/r, longitudinal — (v2 sin2,3 cot a)/r. 6. {—w cos2, 0 to sin A). 497

498


Exercises 1:9 (p. 41) 1. (W2 02 N2 — WI Oi N1)/(2a) about a horizontal axis. The centre moves in a straight line. r 2. — a

L3 —12

1

01 , a {- 0 2

}. —

2v 2v

Miscellaneous Exercises I (p. 45) 2. (a) —S2 seca tan a ; (b) S2 secoc. 5. (co'r' — wr)/(r' the cylinders.

—

r); axis distant rr' (co' — w)/(w'r' — wr) from the axis of

6. On an axis parallel to the given axes and passing through the centroid of masses w, w' on the respective axes. 10. Velocity, {0 b0

;

(a + b sine)

Acceleration, {— (a + b sine) 2sine — b Oz b0 — (a + b sine) ,p2 cos0 (a + b sine)

2bh cos 0). Exercises 2:3 (p. 65)

7. G . R' + G' . R = (R . R') 10. (ia, 4 b, 0),

v(a2 +

R2

+

R'2

b2).

Exercises 3:3 (p. 94) 3. x =

g sine( co2 cos2a

(1 cosh S2 t) + a sinlif2t, where S2 = w cos a ;

comes to rest where x = {g sina — f(g2 sjn2 a a2D4))/(0 c0,20 ) Exercises 3:6 (p. 117) 1. West. 3. {4 V3S2(sina cos y cos 0 — 3 cos a sin A) sin2a}/(3g2) ; 60° . Exercises 3:7 (p. 123) 2. 2:1.

ANSWERS TO THE EXERCISES

499

Miscellaneous Exercises III (p. 125) 1. mg -II {1 + 1 2. — 2

f 1- /f

54cos2oc} cosa. g2 a2 (Z ± a) yo2

dz z

1 (y , — z) (z — y,)

5. M (g +

where (M + m) x = md2V 2/x3— (M + m cosoc) g and x is the posi-

tive root of the given equation. Exercises 4:5 (p. 146) 1. The diagonal through the corner, the line through the corner perpendicular to this diagonal and the perpendicular to the lamina through the corner. 2. I Ma2(4 — 3 sine cose). 3. 5Ma2/3,

Mae (7 1/2 + 6)/(6 -1/2).

4. Mae/2, 9Ma2/4, 9Ma2/4. m(az 712 + C2) — m 9. (

—MC$ + V)

—.111-$71

— m$n ). m(0 ± + n2)

Exercises 4:6 (p. 158) 5. m'v0/1/{m(in + m')}, — mvo/1/ (m (m + m')}. Miscellaneous Exercises IV (p. 174) 2. If A is moment of inertia about Gz and G is origin, at the points (0 0 + i/[(A — B)/M]}; cx = sin-1(2/1/5). 3. The tangent at 0 in the plane of the base, the line obtained by rotating the diameter of the base (through 0) through n/8 about this tangent and the third line making a r.h. set. (2n — 3w cos cc) sin g 4. A cone, axis vertical, semi-vertical angle tan-- ' 2n cosoc + w(2 + 3 sin2a )

a cone, axis vertical, semi-vertical angle tan-1

n sing co + n cosoc

7. The equation of motion is 0 + p20 = (cdtp4Ig) cos pt with solution cilip4 t sin pt and the angle through which the lamina swings increases in— 2g definitely with the time.

500


Exercises 5:2 (p. 183) Exercises 5:6 (p. 221) co 4. —

S2'

winA(C — A)1 D 111B (C — B)c •

S2 1 2 2 33/b22 cos p t — — sinpt 5. 5 5 8

—11, where p = 30/(51/2).

7. Referred to axes along the major and minor axes of the ellipse and the normal to the plane of the disc the angular velocity is 1 21 1Dt)}. {y3 sech (- Dt) 1/3 tanh ( Dt) sech (2 2 The motion is ultimately one of rotation with angular velocity 10 I/3 about the minor axis. Miscellaneous Exercises V (p. 228) 10. n = iflcosa — (2g cota)/a. Exercises 6:5(a) (p. 249) 1.

+ a 6 cos0 — a b2 sin0 = X/m, cost) + 4a0/3 + g sine = (2X cost))/m.

2. Q = 1/ (10g/a). 5. x = *gt 2, y = (Vico) sinhwt, 0 = sec-1(coshwt). 8. 6 — (1,2sine cos° — (g la) cos° sing) = 0,

(M + 29n sin20) + 4mel9sin0 cost) — (2mg la) sine cos0 = 0. 9. (1 + 2k) + + n2 (1 + k)0 = 0 , 6 + + n20 = 0 .

10. (2M + m) b2-F K (0 — 0) = 0, 1-ma26 — K (0 — 0) = O. 11.

8 8 (-a2 6 + 2Ax2 0)) — (- 2+ 2/1.x2 ) (P2 sine cose + g (a + Ax) sine = 0, 3a ddt 3

di d AT —

a2 + 22x2) sin2 0} = 0.

Ax (62+ q2 sine) + (g la) {n(x — a) — A a cosO} = 0. Exercises 6:6 (p. 269)

4. 29na2 coe sine cos°.

501


Exercises 6:7 (p. 273) h2 1.r = r3

2(r — I)

ml Exercises 6:8 (p. 276)

2. {(1 ) sin co t co

U

EW

co I_ co

+

EW\

cosco

co

Miscellaneous Exercises VI (p. 277) 5. 20/11 + ak: I - mIk2 1.

q OA, c at

d vi dt V(i vk vk )1 c2

6. m

d v 1 dt 11/(1 — T2/c2) J

q

x,

q OA + q grad cr, — c at

7. — {(2.11/ + m)± + (M + m)ctO cost) +

dt

(M + m) (1

q aAk vk c x,

q

vk

c

0A,

a xk

x curl A = 0.

= 0,

M

+ m cos 0) ctO I (M + m)g sin0 + 2M + 2M + m 2M M m

cos0 = 0.

Exercises 7:2 (p. 286) 1. (i) 27r/(co sing) where cosa = g/(aco2); (ii)

22r co 1/(1 + 3 costa)

2ncoa V(a2w4 + 3g2) Exercises 7:4 (p. 291)

" )21 11/1( mcg: )2+ (-7

3. 27t

4. AO — A co2sin0 cos0 + Cnco sine — Mgt sin() = 0. Exercises 7:5 (p. 296) 3. = a where tang = — Al(0) cos A) ;

=0;

A co cos

27r1/{ On ( • -1 ( 2cos22) A2

1

502


Exercises 7:6 (p. 303) 1. tan2 oc < 7 + 2 1/15. 4. The frictional force on the moving sphere has a moment about this vertical diameter. Miscellaneous Exercises VII (p. 304) 3. The projection of the path of G on a horizontal plane is given by

x = (1/3co/p) cos (pt1113) sinpt, y = (1/3w/p) sin (ptl 113) sinpt or, in polar coordinates, r = (j/3w/p) sin (1/30). 5.1/(1ga). 6. x = 0, y =

at

d

o,

at

+ (i9 cosej = 0,

d

at

[(i) sine + 2n cos0] = 0,

[(A + M a2cos20) 0] + Ma202cos° sine — A (p2 sine cos' + 2A n(p sine + Mga cose = 0,

where n = + cos°. In the variations a constraint is imposed which keeps q, constant. Exercises 8:1 (p. 316) 2. AB, 1(9c + 5a)/(10ma2); BC, 6c//(5ma2); CD, I(9c — 5a)/(10ma2);

T = 12(25a2 + 9c2)/(75ma2). 3. Parallel to A C. 4. 5a(3 — 5 tanoc) from A, where 2a is the length of an edge of the cube. 7. Referred to axes along OA, 0 B and the normal at 0, M1To =

Mw = 3( — 65

11. x— — a 34 3

Y Z),

Z/a (b X — aY)/(a2 + b2)) • y——a

25 51

39 z — —a 34

0

—5 Exercises 8:2 (p. 327)

6. u,+ , 4ur

= 0 for 1 r n — 1,

1= 3J/2, u„_1 2u„ = 0; A and B satisfy the equations 2u0 u 2(A + B) + A (a + B 13) = 3J/2,

A ocn-1+ B/P-1+ 2(oc" A + 13"B) = 0 .


503

8. Referred to axes parallel to CA, DB the velocities are:

A (u 0), B (u + aco sina aco cosa), Gifu + 2aco sing 0}, D{u + aw sina — aco cosa}, where u = J(1 + 2 sin2a)/(4m), w = — (J sinoc)/(2ma) ; T = J2(1 + 2 sin2a)/(8m). 10. 3a/2. Exercises 8:3 (p. 343) 2. T = 7 J2I(2 M); JIM, 7 JIM both parallel to J.

5. I, J, 1 cosa + J sing, — (/ cosa + J sing) ; —14m U/5. Miscellaneous Exercises VIII (p. 344)

9. h

=

—F

—E

B

—D

—E —D

C

/, co where I = (A —F

Io.) = r, co = I-1r where r = {L M N). 10. (I (bn — cm)/A I (cl — an)/B I(am — b1)/C). 11. T = ima2(1702 + 02 + 260 cos (co — 0)), V = — mga (9 cos 0 + cosp); Normal coordinates are proportional to 810 + co, 0 — co; initially 6 = w = cp, and so —

p0 = co(*sin2pt + A sin3pt), pp = co a sin2pt — i sin3pt). Exercises 9:2 (p. 358) 1. T = M(i2+ 2atO cos0 + a262), V = Px2— Mga cos0 ;

at (t + a6 cos0) + 2x/M = 0,

dt (tcos0 + (Tv- — n2) (g

4

o0) + g sin° = 0 ; 4

a n2) — an4 = O.

When 2 = 0, the centre of AB is stationary or has a constant horizontal component of velocity and in addition the rod executes small oscillations about the vertical with period 27c1/(a/(3g)). 4. 30 + 2T, 70 — 399.

504


6. 30 --I- 2T,

0 - 99; 2n 1/(3a/g), 2n 1/{a/(2g)} ;

0 = A cos (n t/1/3 + el) + 2B cos (1/2nt + es), = A cos (nt/1/3 + e,) — 3B cos (1/2nt + es), where n2 =

5k2 )

k2

82

9. — — p2 (11 + — — + 54 — m2 = 0. a2 d2 m Exercises 9:3 (p. 374) 6. If 0, 99 respectively are the inclinations of the rod and string to the vertical and L + x is the length of the elastic string, the normal coordinates are proportional to x, 50 + 99, 30 — 299. 7. One, the plate swinging in its own vertical plane, 443; bifilar suspension, plane of plate vertical, 4a/9; two, the plate swinging at right angles to its own plane in the manner of a double pendulum, -1(2 j/3) a.

8. D„ — (2 cos0) D„ + pH _ 2 = 0 ;

A

rn

2V--) sin m 2(n + 1)1

=

r= 1, 2, ..., n. Miscellaneous Exercises IX (p. 382)

5. a, b. 6. The normal coordinates are 0 + 99, 40 — 599. In the first mode 40 = 5q9, in the second 0 = — 99.

7. T = .41-Ma2 02 + 1-ma2{02+ 2049 cos (0 — q)

4)1,

V = —4-Mgacos0 — mga(cos0 + cos99) ; normal coordinates are 20 + 99, 50 — T. 10. Frequency 2n/n when 0 = 0, 99 = —y; 3n/n when 60 = 599 = 5y; nIn when 20=—cy=— y. 11. If 0, 99 respectively are the inclinations of 0P1, 0.132to the vertical and x is the displacement of the rod (all in the sense of the impulse I), then =

2mnl

1 (mint + sin nAt), 49—

where 22 = (M + 2m)/M; x = 12. .-1- (axe + 2 x0 -I- c02).

( 2mnl

I

Mn22

1 sinn,t — — sinnAt) ,

(nt — sinnt).

A

505


13. Period is

1/( 2 — 22 + 22 , stationary when d = co 2 the modes given by p2 , p3which are symmetric modes.

j/2 corresponding to

Exercises 10:3 (p. 417) in

1. T sin J19(a + b) 1 mpcs c 1

ipa) sin (pb\ . c) c)

a I. j coy — ) 1 3. [b sin f co (/ c — x) 1 ens co t] c J 1 4. n11(4 sin-1 1/(2/3)). Exercises 10:5 (p. 430) 2. Reflected wave is y = a cos (co (t — x/c) — 2s), where tans = mco/(qc); amplitude a, phase lag is 2s. Exercises 10:8 (p. 449) V 5. — (1 — e-ku -xic)) for x < ct, 0 for x > ct. 7. z = A sinpt sin (mg.txla) sin (wry lb), where m, n are positive integers and m2/a2 + n2/b2 = p2/(7,20).

8. f = nPA2a8 (3 +

+ /32)/6,

1.7 = ne A2 p2al° (10 + 5(3 + /32)/120,

2,2 20P(3 + 2 i3 + I32)

a2 (10 + 5fi. + fi2) ; period ÷ 2.6a/c, where c2 = le. Miscellaneous Exercises X (p. 451) 2. A = aincos2 (-2 ) + (— k ) 2 sin2 -1 21—)} ,

ec

B = kAl(Qc).

8. n = ick, period 4n1(ck). 13.

w=+

1 T cot (1 2m c c • Exercises 11:5 (p. 483)

2. qi = a sin (t — oc), q2 = b sin (t — fl),

= a cos (t — a),

p2 = b cos (t — 13), where a, b, a /3 are constants. ,

506 3.


= a, Q2 = b,

=a—Al t, P2 = - A2 t, where a, b, a, ,3

are constants; 2a

q, =11 (—) cos (a — yli t) +11

2b

) cos (13 — A2t),

A2

2b 2a cos (a — 2,t) +1(—) cos (13 — 2.2t). Al Az

q2= —

Exercises 11:7 (p. 492) 3. The two remaining integrals are: p? + p2 constant, tan-'(.2)21.23,) = t I constant. (The roles of pi and q1in the theory of the first part should be interchanged to obtain the integrals given here.) -

--

4. — sina + y cosa = [3 where a, 13 are constants, and = (2m)V2 A (log AP + log BP), y = (2m)'I2 A (A .13x + BPx). The angles Aix, BPx are angles between AP, BP and the x-axis. Miscellaneous Exercises XI (p. 494) 3. Motion in one of a family of confocal ellipses with eccentric angle changing at constant rate n. 6. g) = pr .

7. q,. = Ears as sin (ws t + /3s), where as , j3, are constants of integration. s=i

INDEX Acceleration 29 normal 34 tangential 34 Accelerated axes 86ff. Action 463, 491 and reaction, Law of 130, 307 Amplitude (of a wave) 388 Angular momentum 82, 98, 118, 131, 149, 208, 209, 308 Angular veocity 18ff. 21 matrix 19ff. vector 19 ff. Antinode 390 Apparent gravity 111 Attenuation 422 Axis of rotation 9

Constraints 69, 84, 160, 237, 240ff., 318, 330, 360, 376, 457 frictionless, 308 impulse of 307 Contact transformation 477 ff., 484 ff. Coriolis force 89, 112 Couple 57 Curvature 33 Cyclic-co-ordinate 270, 474, 477 Cyclotron frequency 120

Bertrand's theorem 335 Binormal 32

Eigenvalue 6 Eigenvector 6 Electromagnetism 118 Ellipsoid central 141, 209 of gyration 141 Energy equation 84, 187, 211 Equilibrium 48ff., 68ff., 130 Equimomental systems 142 Equivalent sets of forces 56ff., 71 Euler's angles 12ff., 21, 178 Euler's dynamical equations 207, 246 Euler's equations (for stationary integral) 459 Euler's theorem (on axis of rotation) 5

Damping 418 Determinant 361 Discrete spectrum 391

Calculus of variations 457-458 Canonical momentum 460 Canonical variables 469, 472, 477 Central axis 38, 45, 58 Centrifugal force 89 Central ellipsoid 141 Chain under gravity 441 Charged particle 118 ff. Column vector 1 Complex numbers, use of 89, 92, 389,424 Composition of angular velocities 26 ff., 41 ff. Configuration space 456 Confocal quadrics 141 Conservation of energy 83, 135, 187, 209, 259 of angular momentum 82, 98, 104, 135, 187, 209, 211, of momentum 82, 259

507

Force 48ff. applied 129 external 129 internal 130 of constraint 130

508

INDEX

Forced vibration 418ff., 418 Foucault's pendulum 116 Fourier series 387, 390, 407 Fundamental mode 407 Free rotation 178ff., 209 fE., 286 Frenet (Serret-Frenet) formulae 33ff. Galilean transformation 88 Generalised coordinates 70, 232, 237, 318, 376, 456, 477 Generalised forces 70, 237ff., 380 Generalised impulse 319ff. Generalised momentum 234 Generalised velocity 234, 318 Gimbals 292 Gyro-compass 292 ff. Gyroscope 184ff., 288 Hamiltonian function 260, 469 Hamilton-Jacobi equation 487, 490 Hamilton's characteristic function 490 Hamilton's equations of motion 460, 468, 471 Hamilton's partial differential equation 487 Hamilton's principle 456ff., 463 Hamilton's theorem (on finite rotations) 14 Harmonic (of a vibrating string) 407 Harmonic wave 388 Herpolhode 21, 181 Holonomic 240, 244, 252, 320, 360, 456 Hooke's joint 22

Ignorable coordinates 270, 474, 477 Impulse 306ff. Incident wave 4231f. Independence of translation and rotation 132, 133, 150, 160, 178 Inertia matrix 136ff. Infinitesimal contact transformation 484ff. Infinitesimal rotation 27 Instantaneous axis 38, 178 Invariable line 178, 209

Jacobi's form of Principle of Least Action 466 Kelvin's theorem 331 Kinetic energy 83, 97, 104, 131, 148, 178, 233, 329, 394 Kinetic potential 244 Kinetic symmetry 142, 178ff. Lagrange, J. L. 232 Lagrangian function 244 Laplace transformation 444 Least Action, Principle of 463 ff. Linear momentum 82, 131, 148, 308 Line-integral 70 Line vector 41, 56, 78 Localised vector 56 Longitudinal vibrations 393, 431 ff. Lorentz force 119 Matrix 1, 3 orthogonal 3, 5, 12ff. skew symmetric 3, 4, 7, 8, 19 Maupertius' principle 463 Mean value 424-425 Membrane, vibrations of 447 Modulus of decay 422 Moment of a couple 57 of a force 49, 56 of inertia 136ff. of momentum 132 Momenta' ellipsoid 139, 209 Momentum 82, 131, 234 Motor 56 Moving frame of reference 28 Newtonian frame 88, 110, 232 Node 390 Non-holonomic constraint 241, 256, 298, 320 Normal, principal (to a curve) 32 Normal modes 347 ff., 405 ff. Normal coordinates 353, 360ff. Osculating plane 32

INDEX

Parallel angular velocities 43 Parallel axes theorem 137 Perfect differential 70, 135, 238 Permanent axis of rotation 212 Phase 388 space 469, 477 Pitch of screw motion 38 of wrench 58 Poinsot's construction 210 Point transformation 478 Poisson brackets 472, 485 Polhode 21, 181, 210, 212 Precession 182 steady 187, 195 Principal axes of inertia 139ff. Principal moments of inertia 140 Principle of virtual work 67ff., 242ff., 319ff., 338 Product of inertia 136ff. Progressive wave 387 Rayleigh's principle 376ff., 437 Reflected wave 423ff. Resonance 419 Rigid body 4, 18, 36ff., 136ff., 286ff. Robin's theorem 338 Rolling hoop 223, 241, 297 Rotating axes 86ff., 96, 110, 473 Rotating earth 110ff. Rotation 4 finite 4ff. free 178ff. infinitesimal 27 matrix 7 Rotations, successive 12 ff. Routhian function 270, 474 Row vector 1

Scalar product 3 Screw motion 38 Skew angular velocities 44 Sleeping top 199 Sphere, motion of 160ff. Spherical pendulum 95 Spherical polar resolutes 29ff.

509

Stability 281ff., 360ff. Standing waves 389 Stationary waves 390 Steady motion of a top 187, 195 Summation convention 233, 242, 360, 456 Tangent (to a curve) 32 Top general motion of 191ff., 288 sleeping 199 spinning 184ff. Torque 57 Torsion (of a curve) 33 Trace 6 Transformation of resolutes of a vector 1, 2ff. Transient 419, 420 Translation (of a rigid body) 38, 44 Transmitted wave 423ff. Transverse vibrations 392ff. Undetermined multipliers 252 Variable wave velocity 440 vector 1 column 1 moment 49, 56 product 3, 19 representation of finite rotation 9 row 1 Virtual work 67 ff. Wave harmonic 388 number 388 profile 387 progressive 387 standing 389 Wave equation 386 one dimensional 387, 394 separable solution of 391 Wave-front 390 Wavelength 388 Wrench 58

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