Mechanics, tensors and virtual works

MECHANICS, TENSORS & VIRTUAL WORKS

MECHANICS, TENSORS & VIRTUAL WORKS

Yves R Talpaert Faculties of Science and Engineering at Algiers University, Algeria Brussels University, Belgium Bujumbura University, Burundi Libreville University, Gabon Lomé University, Togo Lubumbashi University, Zaire and Ouagadougou University, Burkina Faso

CAMBRIDGE INTERNATIONAL SCIENCE PUBLISHING

Published by Cambridge International Science Publishing 7 Meadow Walk, Great Abington, Cambridge CB1 6AZ, UK http://www.demon.co.uk/cambsci First published 2003 © Yves Talpaert © Cambridge International Science Publishing

Conditions of sale All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the copyright holder. British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library

ISBN 1 898326 11 8

Production Irina Stupak Printed by Antony Rowe Ltd, Chippenham, England

About the author

Yves R Talpaert received his MS (1968), “Agrégation” (1968), and PhD (1974) degrees in applied mathematics from Brussels University, where he taught mathematics for a few years. A past Professor at Algiers University, Algeria and other universities and engineering institutes in Africa, he is a prolific French author of books on mechanics and differential geometry. He has written papers on dynamics applied to astronomy where he expounded an original fluid-dynamical approach, statistical mechanics models, a variational principle and so on.

PREFACE

This book is intended for third year students in mathematics, physics and engineering within the context of a first one-semester course in mechanics. Most of the text comes from this level courses I taught at several universities and engineering schools. The various chapters connect the notions of mechanics of first and second years with the ones which are developed in more specialized subjects as quantum physics, continuum mechanics, fluid-dynamics, special relativity, general relativity, cosmology, meteorology, electromagnetism, stellar dynamics, celestial mechanics, applied differential geometry and so on. This course of Analytical Mechanics synthesizes the basic notions of first level mechanics and leads to above-mentioned disciplines by introducing various mathematical concepts as tensor and virtual work methods. A measured and logical progression towards notions of mathematics and mechanics give this book its originality. First, the notion of dynam is introduced in Chapter 0, in particular the one of velocity dynam. In Chapter 1, the study of Statics is divided into two parts. After recalling the classic method, the one of virtual work is developed with numerous exercises of classical mechanics. Tensor theory is very expanded and illustrated in Chapter 2. Tensors are intrinsic mathematical beings and are suitable for the expression of laws of mechanics (and physics) regardless of the choice of coordinate system. This property by oneself justifies the extent of this study of tensors. The reader will better view the notion of inertia tensor within this widen context. Devoted to this particular tensor, Chapter 3 especially prepares for the study of continuum mechanics. Inertia ellipsoid and principal axes are examples among others. Chapter 4 shows kinetics and dynamics of systems with the help of dynams. But tensor calculus is also very helpful to write theorems deduced from postulates. Lagrangian dynamics and variational principles are at the root of analytical mechanics introduced in Chapter 5. Lagrangian dynamics shows another formulation of motion equations from the notion of virtual displacements and the profitable use of scalar functions leads to Lagrange’s equations. Euler’s equation and Hamilton’s principle are considered in the context of variational calculus. Euler-Noether theorem concludes this essential chapter. Hamiltonian mechanics, which constitutes the last chapter, is dealt with canonical equations, Lagrange and Poisson brackets, canonical transformations and Hamilton-Jacobi equation finally. This chapter especially prepares for the formalism of quantum mechanics and for celestial mechanics notably.

v

vi

Preface

Many books relating to the developments of tensor theory are either too abstract since aimed at algebraists only, or too quickly applied to physicists and engineers. I have striven for bringing closer these points of view; so the various chapters are intended for mathematicians (which will find an illustrated presentation of mathematical concepts and solved problems) and for physics and engineering students too, since the mathematical foundations are basically introduced in a practical way. As well as tensors, the virtual work concept is systematically used. Being clearly introduced from virtual displacements (and virtual velocities), this notion plays an essential role in continuum mechanics. The two previous mathematical “tools” give the considered mechanics subjects a great unity and must be known by every mathematician, physicist and engineer. And after all, they make the reading of my previous book treating of Differential Geometry easier, which is even better! The present book lets overcome the following difficulty. There is often a gap between academic cycles. Two essential reasons among others are responsible for that. The one follows from the diversity of teaching establishments of first and second years. The second is due to the different nature of academic cycles, namely: general courses at first and second years, specialized courses later. One of my goals has been to reduce this gap, the intensive use of tensor calculus contributes to that; this realization facilitates the writing of a second volume which will deal with rigid bodies, perturbations and continua. All the proofs and 78 solved exercises are detailed. The important propositions and the formulae to be framed are shown by
and . In writing this new book, I had the following assertion fresh in my mind: Pedagogy contributes to Rigor.

Acknowledgements. I am grateful to Professor Michel N. Boyom (Montpellier University) for a critical reading of chapters. Many thanks to my former students who let me expound on the material that resulted in this book.

Yves Talpaert

CONTENTS PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Chapter 0.

Chapter 1.

REQUIREMENTS

1

1.

POINT SPACE AND VECTOR SPACE . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1

Point space (or affine space) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2

Frame of reference and basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.

DYNAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1

Dynam definition and reduction elements . . . . . . . . . . . . . . . . . . . . . . . . . 4 Dynam definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Representation of dynams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2

Properties and operations on dynams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Equality of dynams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Operations on dynams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Equiprojective fields of moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Reduction of a vector system and dynam . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3

Dynam of velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Velocity field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Dynam of velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4

Acceleration vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.5

Sliding velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27

STATICS

33

1.

CLASSIC METHOD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . … 34

1.1

Mechanical actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Moment of a force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Dynam of a mechanical action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.2

Classification of forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 External forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Internal forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1.3

Equilibrium conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Definitions and conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Particular collections of forces applied to a rigid body . . . . . . . . . . . . . . . . . 43

vii

viii

Contents 1.4

Types of equilibrium of rigid bodies and structures . . . . . . . . . . . . . . . . 45

1.5

Stress and contact dynam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contact dynam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dry friction and Coulomb laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.6

Types of constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Punctual constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Rectilinear constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Annular-linear constraint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 Ball-and-socket joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Plane support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 Sliding pivot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Sliding guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Screw joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Pivot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Embedding or welded joint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

1.7

Free-body diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

2.

METHOD OF VIRTUAL WORK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .70

2.1

Number of degrees of freedom and generalized coordinates . . . . . . . . . Number of degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalized coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Types of constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2

Virtual displacements and virtual velocities . . . . . . . . . . . . . . . . . . . . . . . 78 Generalized coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 Definition and expression of virtual displacements . . . . . . . . . . . . . . . . . . . 79 Virtual velocity and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Virtual fields and dynams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . 91

2.3

Virtual work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Definitions, rigid body and ideal constraint . . . . . . . . . . . . . . . . . . . . . . . . . 94 Principle of virtual work (First expression) . . . . . . . . . . . . . . . . . . . . . . . . . 98 Principle of virtual work (Second expression) . . . . . . . . . . . . . . . . . . . . . . 106

3.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

Chapter 2.

TENSORS

47 47 48 50

71 71 72 75

135

1.

FIRST STEPS WITH TENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

1.1

Multilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Linear mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 Multilinear form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

1.2

Dual space, vectors and covectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Dual space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Expression of a covector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Einstein summation convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Change of basis and cobasis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 140

1.3

Tensors and tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

Contents

ix

Tensor product of multilinear forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Tensor of type (10 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

(10 ) Tensor of type ( 02 ) Tensor of type ( 02 ) Tensor of type (11 ) Tensor of type

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

Tensor of type (qp ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Symmetric and antisymmetric tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 2.

OPERATIONS ON TENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

2.1

Tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Addition of tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Multiplication by a scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Tensor multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

2.2

Contraction and tensor criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 Tensor criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

3.

EUCLIDEAN VECTOR SPACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

3.1

Pre-Euclidean vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 Scalar multiplication and pre-Euclidean space . . . . . . . . . . . . . . . . . . . . . . 164 Fundamental tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

3.2

Canonical isomorphism and conjugate tensor . . . . . . . . . . . . . . . . . . . . 166 Canonical isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Conjugate tensor and reciprocal basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Covariant and contravariant representations of vectors . . . . . . . . . . . . . . . 170 Representation of tensors of order 2 and contracted products . . . . . . . . . . 172

3.3

Euclidean vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

4.

EXTERIOR ALGEBRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

4.1

p-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 Definition of a p-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 Exterior product of 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Expression of a p-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Exterior product of p-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Exterior algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

4.2

q-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

5.

POINT SPACES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

5.1

Point space and natural frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 Point space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 Coordinate system and frame of reference . . . . . . . . . . . . . . . . . . . . . . . . . 192 Natural frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

5.2

Tensor fields and metric element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Transformations of curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . 197

x

Contents Tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Metric element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 5.3

Christoffel symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 Definition of Christoffel symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 Ricci identities and Christoffel formulae . . . . . . . . . . . . . . . . . . . . . . . . . 206

5.4

Absolute differential, Covariant derivative, Geodesic . . . . . . . . . . . . . Absolute differential of a vector and covariant derivatives . . . . . . . . . . . . Absolute differential of a tensor and covariant derivatives . . . . . . . . . . . . Geodesic and Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallel transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Absolute derivative of a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5.5

Volume form and adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Volume form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

5.6

Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

230

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

232

6.

Chapter 3

MASS GEOMETRY AND INERTIA TENSOR

207 207 209 211 213 214

220 220 225 228

263

1.

MASS DISTRIBUTION AND INTEGRALS . . . . . . . . . . . . . . . . . . . . 263

1.1

Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2

Integrals of real-valued functions and vector functions . . . . . . . . . . . 265

2.

CENTER OF MASS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

2.1

Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2

Subdivision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

2.3

Theorems of Guldin (and Pappus) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

3.

INERTIA TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

3.1

Moments and products of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

3.2

Inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

4.

INERTIA ELLIPSOID . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

4.1

Moment of inertia about an axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.2

Equation of the quadric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

4.3

Nature of the quadric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

4.4

Radius of gyration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

5.

PRINCIPAL AXES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

263

267

276

Contents

Chapter 4

Chapter 5

xi

5.1

Fundamental theorem about a symmetric tensor . . . . . . . . . . . . . . . . . 281

5.2

Equal eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

5.3

Inertia ellipsoid and principal axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

5.4

Material symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

6.

THEOREM OF STEINER (and HUYGENS) . . . . . . . . . . . . . . . . . . . . 288

7.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

KINETICS AND DYNAMICS OF SYSTEMS

299

1.

NEWTON’S POSTULATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

1.1

Experimental laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300

1.2

Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

1.3

Galilean relativity and inertial frames . . . . . . . . . . . . . . . . . . . . . . . . . . 303

2.

KINETICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

2.1

Kinetic dynam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

2.2

Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

3.

THEOREMS OF MECHANICS OF SYSTEMS . . . . . . . . . . . . . . . . . . 309

3.1

First integrals of a system of particles . . . . . . . . . . . . . . . . . . . . . . . . . . 309

3.2

Linear momentum theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 Linear momentum theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310 Theorem of conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 Theorem of motion of mass center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 Special case of rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

3.3

Angular momentum theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Angular momentum theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Relation between kinetic dynam and dynam of forces . . . . . . . . . . . . . . . . 317 Conservation of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Special case of rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

3.4

Kinetic energy theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 Kinetic energy theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 Special case of rigid bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326

4.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

LAGRANGIAN DYNAMICS AND VARIATIONAL PRINCIPLES 339 1.

LAGRANGIAN DYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339

1.1

Holonomic and scleronomic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340

1.2

D’Alembert-Lagrange principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

xii

Chapter 6

Contents

1.3

Lagrange’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lagrange’s equations in the general case . . . . . . . . . . . . . . . . . . . . . . . . . Lagrange’s equations for conservative forces . . . . . . . . . . . . . . . . . . . . . . Lagrange’s equations with undetermined multipliers . . . . . . . . . . . . . . . .

1.4

Configuration space and Lagrange’s equations . . . . . . . . . . . . . . . . . . 354

1.5

Adjoint Lagrangian and first integrals . . . . . . . . . . . . . . . . . . . . . . . . . 358

2.

VARIATIONAL CALCULUS AND PRINCIPLES . . . . . . . . . . . . . . . 360

2.1

Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 A variational problem and variations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364

2.2

Hamilton’s variational principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 Hamilton’s postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 Hamilton’s principle and motion equations . . . . . . . . . . . . . . . . . . . . . . . . 369

2.3

Jacobi’s form of the principle of least action of Maupertuis . . . . . . . . 371

3.

EULER-NOETHER THEOREM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374

3.1

One-parameter group of diffeomorphisms . . . . . . . . . . . . . . . . . . . . . . 374

3.2

Euler-Noether theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376

3.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

HAMILTONIAN MECHANICS

344 344 348 350

393

1.

N-BODY PROBLEM AND CANONICAL EQUATIONS . . . . . . . . . 393

2.

CANONICAL EQUATIONS AND HAMILTONIAN . . . . . . . . . . . . . 397

2.1

Legendre transformation and Hamiltonian . . . . . . . . . . . . . . . . . . . . . 397

2.2

Canonical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

2.3

First integrals and cyclic coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 404

2.4

Liouville’s theorem in statistical mechanics . . . . . . . . . . . . . . . . . . . . . . 406

3.

CANONICAL TRANSFORMATIONS . . . . . . . . . . . . . . . . . . . . . . . . . 409

3.1

Lagrange and Poisson brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 Preservation of canonical form and Poisson bracket . . . . . . . . . . . . . . . . . . 409 Poisson bracket and symplectic matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 Lagrange and Poisson brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

3.2

Canonical transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 Canonical transformations and brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 Canonical transformations and generating functions . . . . . . . . . . . . . . . . . 419

4.

HAMILTON-JACOBI EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

4.1

Hamilton-Jacobi equation and Jacobi theorem . . . . . . . . . . . . . . . . . . 425

4.2

Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

5.

EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 439

Contents

xiii

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455

CHAPTER 0

REQUIREMENTS

First, let us recall the mathematical notion of point space (also called affine space) from the one of vector space.

1. POINT SPACE AND VECTOR SPACE 1.1

POINT SPACE (OR AFFINE SPACE)

In mechanics, besides vector spaces it is necessary to consider spaces consisting of points. They can be customary as the classic 3-dimensional Euclidean space or very helpful to mechanics as phase spaces. Let E be a (real) vector space. With this vector space, we can associate a point space denoted by E. How? There exists a mapping

φ :E×E → E such that: (i)

∀r ∈ E , ∀a ∈ E , ∃!b ∈ E : φ (a, b) = r ,

(ii)

∀a, b, c ∈ E : φ (a, b) + φ (b, c) = φ (a, c) .

Any vector φ (a, b) is denoted by ab. Now, given an arbitrary point o ∈ E , we define the following mapping:

φ o : E → E : x
φ o ( x) = φ (o, x) = ox . Property (i) implies that, given an arbitrary point o ∈ E , the mapping φ o is a bijection that establishes the connection between the vector space E and the point space E. 1

2

Chapter 0

So, we can clearly define the notion of point space: D

A set E is said to be a point space or an affine space if there is a mapping which connects pairs of elements of E with vectors of E; namely:

φ : E × E → E : (a, b)
ab such that: (i) ab = −ba , (ii) ab + bc = ac , (iii) Given arbitrary o ∈ E , with any r ∈ E is associated only one x ∈ E such that ox = r . Each element of E is called a point. D

The dimension of a point space is the dimension of the corresponding vector space.

1.1

FRAME OF REFERENCE AND BASIS We assume E of finite dimension.

Let B = (e1 ,..., e n ) be a basis of E, o be a point of E. The bijection φ o−1 permits to associate a frame of reference of E with the basis B of E. How? We consider the different points of E defined as follows:

ai = φ o−1 (e i )

i = 1,..., n

or similarly from:

φ o ( ai ) = e i .

Fig. 1 D

A frame of reference of point space E is the set of n + 1 (affinely) independent points o, a1 ,..., a n .

So, a frame of reference of a point space E is defined from a reference point o and a basis (e i ) of the associated vector space E.

3

Requirements

It is denoted by { o; oa1 ,..., oa n }

or simply in general:

R ≡ { o; e ,..., e }. 1

n

Now, let us recall the notion of position vector of any point. Let r be a vector of E, x be the point of E such that x = φ o−1 (r ) . The vector r = ox is expressed with respect to B as n

r = ∑ x i e i = ox i =1

where the various x i are called the components of r relative to B or the coordinates of x with respect to R. It is the well-known position vector of x with respect to R. To conclude this section, we recall: D

A vector space is said to be pre-Euclidean if it is provided with a scalar multiplication. It is said to be Euclidean if the scalar multiplication is positive definite; that is, given any nonzero x ∈ E , the product scalar x . x = x

2

is strictly positive.

A point space is said to be pre-Euclidean (resp. Euclidean) if the associated vector space is pre-Euclidean (resp. Euclidean).

2. DYNAMS The theory of dynams has been notably developed by German researchers in mechanics and the term dynam corresponds to the so-called “torseur” in French. However, the reader will be careful: the dynam terminology is not unique and has not a universal acceptation. The dynams let unify the presentation of the reduction of systems of vectors (as systems of forces in statics), simplify the expression of theorems of dynamics, express the notion of velocity fields in kinematics of rigid bodies and so on. In mechanics, it is interesting to create mathematical beings defined from the resultant and the total moment (or total torque) of any system of (sliding or bound) vectors by “forgetting” this system of vectors and by referring to only previous resultant and moment. These beings are called dynams.1 First, we are going to give the definition and properties of dynams and finally we will introduce the notion of velocity dynam. 1

In this section, systems of vectors will be assumed made up of sliding or bound vectors, by knowing that forces are considered as bound vectors, but as sliding vectors only in the context of unstretchable material systems.

4

2.1

Chapter 0 DYNAM DEFINITION AND REDUCTION ELEMENTS

2.1.1 Dynam Definition

The equivalence of systems of vectors is a well-known notion developed in first courses of mechanics; this is essentially recalled for notation. Let us denote by R(F) the resultant of a system F of (sliding or bound) vectors and by M a ( F ) the moment of F about a point a. The equivalence of two systems of vectors is expressed as follows: R( F ) = R(G ) ,

⇔

F ~G

∀a ∈ E : M a ( F ) = M a (G ) .

In this definition it is not necessary to state the second condition at every point a ∈ E . It is sufficient to specify this condition at only one point, since if this condition is fulfilled at b ∈ E for instance, then it is verified at every point a because M a ( F ) = M b ( F ) + ab ∧ R( F ) . We have thus: R( F ) = R(G ) ,

⇔

F ~G

∃ b ∈ E : M b ( F ) = M b (G ) .

So, we say: D

Two systems of (sliding or bound) vectors are equivalent if they have the same resultant and the same total moment at an arbitrary point.

The equivalence relation which follows from this definition allows the partition of systems of vectors into equivalence classes. Indeed, this definition leads to an equivalence relation on the set C of systems of vectors; that is: ∀F , G , H ∈ C :

F~F , F~G

⇒

G~F ,

[ F ~ G and G ~ H ]

⇒ F~H .

The equivalence class containing the system of vectors F is defined by D = {S ∈C

S ~ F }.

Now, we can set: D
A dynam is an equivalence class of systems of vectors. So, every system of vectors of the equivalence class represents the dynam. The quotient C / ~ is the set of equivalence classes (dynams).

Requirements

5

Since all systems of vectors in an equivalence class are characterized by the same resultant and the same (total) moment at an arbitrary point of E, the definition of dynam can be expressed as follows: D
A dynam D is defined by a set of two vector fields: The one is E → E : a
R( D) ,

which is a mapping such that the image of any point a is the same free vector R. The other is E → E × E : a
(a, M ( D)) = M a ( D) , which is a mapping such that the image of any point a fulfills the condition: ∀a, b ∈ E : M a ( D) = M b ( D) + ab ∧ R( D) . D

(0-1)

The vector R(D) is called the resultant of the dynam D (or vector of D) and M a (D) is the moment of the dynam D at a. These “characteristics” of the dynam D are said to be the elements of reduction of D at the point a. There are also called the components of the dynam.

The dynam is denoted by specifying its elements of reduction as follows: R  D=  . M  a

Remarks. (i)

The elements of reduction are obviously of different nature.

(ii)

If the elements of reduction of a dynam are determined at a point, then there are know at every point. The vector field defined by the resultant R( D) is uniform while the vector field of moments is, of course, not of this type.

(iii) There is a bijection between the set of dynams and the set of pairs made up of the resultant and moment. Later we will introduce operations on the set of dynams so that this bijection will be an isomorphism between vector spaces. Example. Given an orthonormal basis (i , j , k ) of a vector space E, let us consider the vector field E → E : a
w a = (2 + y + l 2 z ) i + (l z − x) j − (4 x + l y ) k

where l is a real parameter and a the point of coordinates ( x, y, z ) . Determine l in order that this vector field is the field of moments of a dynam and find the resultant of the corresponding dynam. Answer. Let b( x + X , y + Y , z + Z ) be any point of E. From w b = [2 + ( y + Y ) + l 2 ( z + Z )] i + [l ( z + Z ) − ( x + X )] j − [4( x + X ) + l ( y + Y )] k ,

6

Chapter 0

we deduce: w b − w a = (Y + l 2 Z ) i + (l Z − X ) j − (4 X + lY ) k . The problem is to find a (uniform) vector field R = ri + s j + uk

such that w b − w a = ba ∧ R , that is: i

j

k

(Y + l Z ) i + (l Z − X ) j − (4 X + lY ) k = − X − Y − Z r s u 2

⇔

Y + l 2 Z = −uY + sZ ,  l Z − X = uX − rZ , − 4 X − lY = − sX + rY 

⇔

[ u = −1, l 2 = s = 4 , r = −l ].

In conclusion, a dynam D is defined for l = 2 and l = −2 ; its resultant is for l = 2 : for l = −2 :

R(D) = −2 i + 4 j − k , R(D) = 2 i + 4 j − k .

2.1.2 Representation of Dynams

We are going to introduce the representation of any dynam from the following: PR1

Every system of vectors defines a dynam, the converse is false.

Proof. First, given a system of vectors, say F, there exists the resultant R(F ) and the moment M a (F ) of the system F at an arbitrary point a. These elements define a dynam D such that: R( F ) = R( D) , ∀a ∈ E : M a ( F ) = M a ( D) . Conversely, a dynam does not (completely) define a system of vectors. This is obvious because equivalent systems of vectors, but which are different, generate the same dynam. D

The system of vectors F is said to be a representation of the dynam D.

2.2

PROPERTIES AND OPERATIONS ON DYNAMS

2.2.1

Equality of Dynams

D

Two dynams D1 and D2 are equal if their elements of reduction are equal at every point.

7

Requirements

The definition is denoted by

R( D1 ) = R( D2 ) , D1 = D2

⇔ ∀a ∈ E : M a ( D1 ) = M a ( D2 ) .

By using (0-1), we have immediately: R( D1 ) = R( D2 ) D1 = D2

⇔ ∃ p ∈ E : M p ( D1 ) = M p ( D2 ) .

Remark. Two dynams are equal iff the moments of each dynam about every point are equal:

D1 = D2

⇔

∀a ∈ E : M a ( D1 ) = M a ( D2 ) .

The necessary condition is obvious. The sufficient condition: ∀a ∈ E : M a ( D1 ) = M a ( D2 )

⇒

R( D1 ) = R( D2 )

is easily proved. Indeed, ∀a, b ∈ E , the equality of the two following equations M a ( D1 ) = M b ( D1 ) + ab ∧ R( D1 ) , M a ( D2 ) = M b ( D2 ) + ab ∧ R( D2 ) and the equality M b ( D1 ) = M b ( D2 ) imply:

ab ∧ [ R( D1 ) − R( D2 )] = 0 that is: R( D1 ) = R( D2 ) . D

The dynam zero, denoted D = 0 , is defined by R( D) = M a ( D) = 0 .

We note that every system of vectors which is equivalent to zero is a representation of dynam zero. 2.2.2

Operations on Dynams

2.2.2a Addition D

R  R  The dynam sum of dynams D1 =  1  and D2 =  2  is the dynam  M1  a M 2  a  R + R2  D1 + D2 =  1  .  M1 + M 2  a

8

Chapter 0

This is well defined because we say: PR2

If the systems of vectors { w1 ,..., w n } and { w n +1 ,..., w n+ s } represent the dynams D1 and D2 respectively, then the system of vectors { w1 ,..., w n+ s } represents the dynam D1 + D2 .

Proof. We have: n

R( D1 ) = ∑ w i ,

R ( D2 ) =

n+ s

∑ wi

i = n +1

i =1

and thus n+ s

R( D1 ) + R( D2 ) = ∑ w i . i =1

In the same manner, given an arbitrary q ∈ E and any ai ∈ E belonging to the line of action of each w i , we have: n

M q ( D1 ) = ∑ qa i ∧ w i , i =1

M q ( D2 ) =

n+ s

∑ qa i ∧ wi

i = n +1

and thus n+s

M q ( D1 ) + M q ( D2 ) = ∑ qa i ∧ w i . i =1

Let us denote the set of dynams by D. Therefore, we define: D

The addition of dynams is the inner law D × D → D : ( D1 , D2 )
D1 + D2 where D1 + D2 is the dynam sum.

2.2.2b Multiplication by a Scalar D

R  The product of a dynam D =   by a scalar k is the dynam denoted kD such that: M  a kR  kD =   . kM  a

This is well defined because we say: PR3

If the system of vectors { w1 ,..., w n } represents the dynam D, then the system of vectors { kw1 ,..., kw n } represents the dynam k D .

Proof. We have:

Requirements

9

n

R( D) = ∑ w i i =1

and thus n

k R( D) = ∑ kw i . i =1

In the same manner, given an arbitrary q ∈ E and any ai ∈ E belonging to the line of action of each w i , we have: n

M q ( D) = ∑ qa i ∧ w i i =1

and thus n

k M q ( D) = ∑ qa i ∧ k w i . i =1

Therefore, we define: D

The multiplication of a dynam by a scalar is the (external) law R × D : (k , D)
k D

where k D is the product of the dynam D by the scalar k. Of course, we note that the definitions do not depend on the choice of q ∈ E . Remark. The set D provided with the two previous operations is a 6-dimensional vector space.

2.2.2c Multiplication of Dynams D

The product of two dynams, denoted by D1 .D2 , is the real

D1 . D2 = R( D1 ) . M a ( D2 ) + R( D2 ) . M a ( D1 ) , where a is an arbitrary point. We note this definition is actually independent of the choice of a ∈ E . Indeed, ∀a, b ∈ E : R( D1 ) . M a ( D2 ) + R( D2 ) . M a ( D1 ) = R( D1 ) . M b ( D2 ) + R( D1 ) . ab ∧ R( D2 ) + R( D2 ) . M b ( D1 ) + R( D2 ) . ab ∧ R( D1 ) = R( D1 ) . M b ( D2 ) + R( D2 ) . M b ( D1 ) ,

this last equality following from the following R( D1 ) . ab ∧ R( D2 ) = − R( D2 ) . ab ∧ R( D1 ) . Therefore, we define:

10

D

Chapter 0

The multiplication of dynams is the law D × D → R : ( D1 , D2 )
D1 . D2 where D1 .D2 is the product of dynams D1 and D2 .

Now, let us establish the expression of the product of two dynams in an orthonormal basis of a vector space E. From R( D1 ) = X 1i + Y1 j + Z1k ,

R( D2 ) = X 2 i + Y2 j + Z 2 k ,

M a ( D1 ) = L1i + M 1 j + N1k ,

M a ( D 2 ) = L2 i + M 2 j + N 2 k ,

we deduce the expression of the product of dynams: D1 . D2 = X 1 L2 + Y1 M 2 + Z1 N 2 + X 2 L1 + Y2 M 1 + Z 2 N1 . Remark. The reader will verify that the properties of the product of dynams are those of the inner product of vectors except for one of them, namely:

PR4

The square of a nonzero dynam is not necessarily positive.

Proof. The square ( D1 ) 2 = D1 . D1 = 2( X 1 L1 + Y1 M 1 + Z1 N1 )

can possibly be negative. We conclude this section by the following D

Any two dynams are said to be orthogonal if their product is zero.

2.2.3 Equiprojective Fields of Moments

PR5

The field of moments of any dynam is equiprojective.

Proof. Given arbitrary points a and b of E, we have immediately: ab . M a ( D) = ab . M b ( D) and so, the projections of M a (D) and M b (D) onto ab being equal, the field of moments is said to be equiprojective. Conversely, let us prove the following: PR6

Every equiprojective vector field is the field of moments of a dynam.

Proof. By the hypothesis, there are vectors M a and M b such that ∀a, b ∈ E : ab . M a = ab . M b

11

Requirements that is ∀a, b ∈ E : ab . ( M a − M b ) = 0 .

- The proof of the proposition is obvious if M a = M b . In this case the resultant of the dynam is zero. - Let us consider the general case where the equiprojective vector field is not uniform as previously. Let m a , m b and m c be vectors of the equiprojective field at three non collinear points. We have: m a . ab = m b . ab , ⇒

(m a − m b ) . ab = 0

which means that m a − m b is orthogonal to ab. Similarly, we deduce that m c − m b is orthogonal to bc and m c − m a is orthogonal to ac. For obtaining a field of moments, we must find a free vector R such that: m b − m a = ba ∧ R ,

m c − m b = cb ∧ R ,

the third equality m c − m a = ca ∧ R being automatically verified. Let us note that R must be orthogonal to the plane determined by the vectors m b − m a and m c − m b ; we have thus: R = k (m b − m a ) ∧ (m c − m b ) = k (ba ∧ R) ∧ (m c − m b ) = k [ba .(m c − m b )] R − [ R .(m c − m b )] ba = k [ba . (m c − m b )] R which implies: k=

1

ba . (m c − m b )

⋅

The vector R is so determined. Since it is immediately proved that an equiprojective field is determined by three of its vectors at three non collinear points, the proposition is thus proved, the dynam being, ∀q ∈ E : R  m  .  q

2.2.4

Invariants

D

An invariant of a dynam is a quantity independent of the point where it is calculated.

(i) The resultant R(D) of a dynam D is a vector invariant, it is a uniform vector field.

12

Chapter 0

(ii) The scalar product R( D) . M q ( D)

is a scalar invariant. Indeed, it is independent of the choice of q: ∀a, b ∈ E :

M a ( D) = M b ( D) + ab ∧ R( D)

and thus: M a ( D) . R( D) = M b . R( D) . (iii) Another invariant has already been encountered; namely: the product of dynams.

2.2.5

Reduction of a Vector System and Dynam

An essential problem consists in determining the simplest system of the set of equivalent systems of (bound and sliding) vectors which define a given dynam; it is the wellknown problem of the “reduction of a vector system.” Let R(D) and M q (D) be the elements of the dynam D at q ∈ E . We are going to classify the dynams associated with the different simplest systems of vectors following from the reduction. The classification will be made from the scalar invariant R( D) . M q ( D) . 2.2.5a First Case: R( D) . M q ( D) = 0 . This scalar product can vanish in three cases: (i) R( D) = M q ( D) = 0 . These conditions define a dynam associated with a vector system equivalent to zero. It is the zero dynam denoted by 0  0  .  q (ii) R(D) = 0 , M q (D) ≠ 0 . These conditions define a dynam associated with a couple of vectors. First, we recall the following: PR7

A system of vectors, say F, defining a dynam D is equivalent to a couple iff [ R( D) = 0 and ∀q ∈ E : M q ( D ) ≠ 0 ] .

Proof. The necessary condition is obvious. Let us prove the sufficient condition, namely: If [ R( D) = 0 and ∀q ∈ E : M q ( D) ≠ 0 ], then the system F is equivalent to a couple.

13

Requirements

Indeed, let M be a nonzero free vector. Let us denote a nonzero (bound) vector located at q by (q, M ) . Let us search the simplest system of vectors defining a dynam D such that: R(D) = 0 ,

M q ( D ) = ( q, M ) .

We locate, at a ∈ E , the (bound) vector (a, v ) of F such that its moment about q is (q, M ) . Both vectors (a, v ) and (q,−v ) form a system of vectors such that: R( D) = 0 ,

M q ( D ) = ( q, M ) ;

it is really a couple of vectors. D

The dynam defined by a couple of vectors is called a couple.

It is denoted: 0  C =  . M  It is a dynam such that the vector invariant (the resultant) is zero and the moment is uniform (free vector). (iii) R( D) ≠ 0 (with R( D ) . M q ( D ) = 0 ). These conditions define a dynam of which we are going to express the simplest associated system of vectors. First, we recall the following PR8

A system of vectors defining a dynam D is equivalent to a simple nonzero vector iff [ ∀q ∈ E : R( D ) . M q ( D ) = 0 and R( D) ≠ 0 ].

Proof. The necessary condition F ~ {v} ⇒

[ ∀q ∈ E : R( D ) . M q ( D ) = 0 and R( D) ≠ 0 ]

is immediate since {v}.{v} = 2 R ({v}) . M q ({v}) = 0 = 2 R( D) . M q ( D)

and R( D) = R ({v}) = 0 .

Let us prove the sufficient condition, namely: [ ∀q ∈ E : R( D) . M q ( D ) and R( D) ≠ 0 ]

Let us consider a ∈ E such that qa =

R( D ) ∧ M q ( D ) R 2 ( D)

⋅

This vector is orthogonal to R(D) and M q (D) .

⇒

F ~ {v}.

14

Chapter 0

We have: qa ∧ R( D) =

1 2

[ R 2 ( D) M q ( D) − ( R( D) . M q ( D)) R( D)]

R ( D) = M q ( D)

So, the system made up of the only vector v = (a, R( D)) has really R(D) as resultant and M q (D) as moment about q. In conclusion, the simplest system of vectors associated with D is equivalent to one (bound or sliding) vector such that the line of action is through q ∈ E if M q (D) = 0 or through a ∈ E if qa ∧ R( D) = M q ( D) .

D

A sliding dynam is a dynam such that the resultant is different from zero and the scalar invariant vanishes.1

It is a dynam of which the moment vanishes at least at a point. We denote any sliding dynam by S. PR9

There is a straight line of points about which the moments of a sliding dynam vanish.

Proof. Let a ∈ E such that M a (S ) = 0 . There exists a point q such that qa is parallel to R(S ) . We have thus M q (S ) = 0 and so there is a straight line of points about which the moments vanish; elsewhere, the moment is orthogonal to the resultant. D

The straight line of points about which the moments vanish is called the axis of the sliding dynam.

We sum up the previous results as follows: PR10 Any nonzero dynam such that the scalar invariant vanishes is either a couple or a sliding dynam. It is a sliding dynam if the resultant is different from zero. 2.2.5b General Case: R( D) . M q ( D) ≠ 0 . We know the Poinsot theorem of reduction: PR11 Every system of (bound or sliding) vectors is equivalent to a single vector through an arbitrary point q ∈ E , plus a couple. The point q is called the center of reduction. It is the general representation of a dynam D of which the single vector is the resultant R(D) and the moment of the couple is the moment of dynam M q (D) . 1

In French, a sliding dynam is called « glisseur. »

15

Requirements

Remark. The reduced system of vectors is a system of bound or sliding vectors. So, the moment of the couple is not considered as a free vector in this study; likewise, the resultant is not viewed as a free vector but as a (bound) vector located at q or a sliding vector through q. We are going to show that the set of three vectors of the previous reduced system, which represents a dynam D, can be expressed in a more interesting form by choosing the center of reduction suitably. We say: D

The central axis of a system of vectors is the set of points where the resultant and the (total) moment of the system are parallel.

First, we note that: ∀a, b ∈ E : M a ( D) = M b ( D)

⇔

ab ∧ R(D ) = 0

and in this case, ab is parallel to R(D) for any points a and b. From this remark, if we find a point where the (total) moment is parallel to the resultant, then the central axis is immediately determined. We recall that the central axis equation is easily obtained. Let s be a point of the central axis. The hypothesis R( D) ∧ M s ( D) = 0 is written ∀q ∈ E : R( D ) ∧ M q ( D ) + R( D ) ∧ ( sq ∧ R( D)) = 0

⇔ R( D) ∧ M q ( D) + R 2 ( D) sq − ( R( D). sq) R( D) = 0 . We simplify this equation by particularizing the point s so as to cancel the third term; it is sufficient to choose s in the plane orthogonal to R(D) through q. Therefore the equation becomes: qs =

R( D) ∧ M q ( D) R 2 ( D)

⋅

Any point a of the central axis is located by the vector: qa = qs + sa ;

that is, by choosing q at the origin o of the frame of reference:

r=

R( D) ∧ M o ( D) R 2 ( D)

+ k R( D) .

It is the well-known equation of the central axis of a vector system. Now, we say the Poinsot theorem of reduction as follows:

(0-2)

16

Chapter 0

PR12 Every dynam D such that the scalar invariant is different from zero is the sum of a couple C and a sliding dynam S. At every point of the central axis, the moment of the couple and the resultant of the sliding dynam are parallel. We note that the moment M q (D) is the sum of M q′′ (D) parallel to R(D) and M q⊥ (D) perpendicular to R(D) . Therefore, the dynam R  R  0  D =   =   +  ⊥  M  q  M ′′ q  M  q

is the sum of a couple the moment of which is parallel to the resultant and a sliding dynam (since R . M ⊥ = 0 ). The sliding dynam is represented by the only vector R along the central axis (for any point a of the central axis, we have M a⊥ = 0 ). PR13 The moment of the couple is minimum at every point of the central axis of the system of vectors. Proof. Let a be any point of the central axis, q be an arbitrary point. The vector invariant R(D) and the scalar invariant such that: M a′′ R( D) = M a ( D) . R( D) = M q ( D ) . R( D )

really imply: M a′′ ≤ M q (D) .

In addition, from M a′′ ( D) =

M q ( D) . R( D) R( D )

,

we deduce for every point a of the central axis the following relation:

M a ( D) =

M q ( D) . R( D ) R 2 ( D)

R( D) .

To conclude this paragraph, we sum up the classification in the following manner: (i) R(D) ≠ 0 : ∀q ∈ E , if R( D) . M q ( D) ≠ 0 : general (nondegenerate) dynam, R( D) . M q ( D) = 0 : sliding dynam.

(ii) R(D) = 0 : ∀q ∈ E , if M q (D) ≠ 0 : couple if M q (D) = 0 : zero dynam.

(0-3)

17

Requirements 2.3

DYNAM OF VELOCITIES

We consider a rigid body B at rest with respect to a frame. There is an infinity of such frames. D

This set of frames, in which B is at rest, defines the space connected with the rigid body as opposed to the frame of reference which is considered as fixed most of the time.

Let R e = { o; e1 , e 2 , e3 } be an orthonormal frame fixed in E, R E = { O(t ); E1 (t ), E 2 (t ), E 3 (t ) } be an orthonormal frame “fixed” in the rigid body B. We note that the point O does not necessarily belong to B.

2.3.1

Velocity Field

We recall that a vector field exists if one vector is associated with every point of E (or part of E).

Fig. 2 Let q be a “fixed” point in the moving body B; it is fixed with respect to R E . The position vector of q is expressed relative to the basis BE = ( E1 (t ), E 2 (t ), E 3 (t )) as following: Oq (t ) = X E1 (t ) + Y E 2 (t ) + Z E 3 (t ) where X, Y and Z are constants. We obviously have the following derivative with respect to R e : (

dE dE dE dOq ) e (t ) = X ( 1 ) e (t ) + Y ( 2 ) e (t ) + Z ( 3 ) e (t ) dt dt dt dt

and with respect to R E it is obviously: (

dOq ) E (t ) = 0 . dt

(0-4)

18

Chapter 0

Let us introduce the vector angular velocity ωEe of R E relative to R e . We recall that this vector does not refer to the only bases BE and Be but refers to the set of fixed bases with respect to Be . So, since ωEe does not depend on the choice of a particular basis of R E and of R e respectively, we denote the angular velocity by ω(t ) or simply by ω. Its well-known expression is

ω=(

dE dE 2 dE ⋅ E 3 ) E1 + ( 3 ⋅ E1 ) E 2 + ( 1 ⋅ E 2 ) E 3 dt dt dt

(0-5)

with dE i (t ) = ω(t ) ∧ E i (t ) . dt

( See e.g. Talpaert [1987] ). Therefore, for every constant vector Op in R E , Eq. (0-4) is written:
D

(

dOq ) e (t ) = ω(t ) ∧ Oq (t ) . dt

This vector is sometimes called the velocity vector of q relative to O.

It is the derivative of every vector “fixed” in the rigid body B. Remark. The angular velocity ω does not depend on the choice of reference point O. Indeed, given (

dOq ) e = ω ∧ Oq , dt

we are going to prove that, for another point of reference O1 : (

dO1q ) e = ω ∧ O1q . dt

(

dO1q ) e = ω1 ∧ O1q dt

A priori, we have:

but (

dO1q dOO1 dOq )e = ( )e − ( ) e = ω ∧ Oq − ω ∧ OO1 dt dt dt = ω ∧ (OO1 + O1q) − ω ∧ OO1 = ω ∧ O1q .

Since for any O1q , we have obtained

ω1 ∧ O1q = ω ∧ O1q , we deduce:

ω1 = ω .

(0-6)

19

Requirements

So, we really have: (

dO1q ) e = ω ∧ O1q . dt

By taking Eq. (0-6) into account, the following equation (

doq doO dOq )e = ( )e + ( )e dt dt dt

becomes:


(v q ) e = (v O ) e + qO ∧ ω .

(0-7)

It is the fundamental formula of rigid body kinematics. It expresses that the velocity of every point of a rigid body B is determined if the velocity of any point of B is known as well as the angular velocity ω of R E relative to R e are known. In other words: PR14 The velocity of any point q in R e is the sum of the velocity of O relative to R e and the velocity vector of q relative to O. In a natural manner, we say: D

The velocity field of the rigid body is the mapping (v ) e : I × E → E : (t , q ) a (v ) e (t , q ) = (v q ) e (t ).

This field is non uniform since it verifies the fundamental formula of rigid body kinematics. The fundamental formula (0-7) can be found again from the formula of the vector derivative, well-known in kinematics: (

du du ) e = ( ) E + ωEe ∧ u dt dt

also denoted by d eu d E u = + ωEe ∧ u dt dt

where u is any derivable vector function: u : I → E : t a u(t ) .

Indeed, for any points a and b of B, that is “fixed in R E , we have: (

doa doa )e = ( ) E + ωEe ∧ oa dt dt

(

dob dob )e = ( ) E + ωEe ∧ ob , dt dt

and

20

Chapter 0

which implies that d (ob − oa )) E + ωEe ∧ (ob − oa ) dt = ωEe ∧ ab.

(v b ) e − ( v a ) e = (

Thus we have obtained: (v b ) e = (v a ) e + ba ∧ ω , that is Eq. (0-7). Remark. Instead of a fixed frame R e , we can consider a second moving frame:

R F = { o(t ); F1 (t ), F2 (t ), F3 (t ) } and we say: PR15 Given moving frames R F and R E of E, there is an angular velocity ωEF of R E with respect to R F such that, for every constant vector u in R E , we have: (

du ) F = ωEF ∧ u . dt

(

du )E = 0 . dt

(0-8)

We have obviously:

In the same manner, Eq. (0-7) is generalized as it follows: PR16 Given moving frames R E and R F of E, the velocities of fixed points a and b in R E are connected as it follows: (v b ) F = (v a ) F + ba ∧ ω EF .

(0-9)

Of course this equation as well as Eq. (0-7) recalls of course the notion of moment of a dynam. Let us introduce this last one. 2.3.2 Dynam of Velocities PR17 The field of velocities of points of a rigid body B is equiprojective. Proof. Let us prove that the projections of velocities at two points of B onto the straight line through these points are equal. Indeed, for every pair (a, b) of points of B, we have:

ab ⇔

2

=c

( c ∈ R+ )

dab d ⋅ ab = (ob − oa ) . ab = 0 dt dt

Requirements

⇔

21

v b . ab = v a . ab .

This equiprojective property means that the velocity field is really the field of moments of a dynam. D
The dynam of velocities or kinematic dynam of B (or R E ) with respect to R e is defined by the following elements of reduction at q ∈ E : - the resultant (or vector) of the dynam, which is independent of the reference point and is denoted by ω(t ) , - the moment about q, that is the velocity v q (t ) of q such that: v q (t ) = v O (t ) + qo ∧ ω(t )

where q and O are “fixed” in B. We denote this dynam of velocities of B by its elements of reduction at q:

ω  v  .  q We can generalize: D
The dynam of velocities or kinematic dynam of a frame R E with respect to a frame R F is defined by the following elements of reduction at a ∈ E : - the resultant (or vector) of the dynam ωEF , - the moment about a, that is the velocity v a (t ) of a such that: (v a ) F = (v b ) F + ab ∧ ωEF where a and b are “fixed” in R E . We denote this dynam by its elements of reduction at a:

ωEF  v ,  a  or also simply by [VEF ] , the moment of which is M a [V EF ] = (v a ) F .

2.4

ACCELERATION VECTORS From (v q ) e = (v p ) e + qp ∧ ω ,

we deduce the acceleration of q: (a q ) e =

de d eω d e qp (v p ) e + qp ∧ + ∧ω , dt dt dt

22

Chapter 0

but d e qp d E qp = + ω ∧ qp dt dt = ω ∧ qp

and thus the fundamental formula of acceleration in rigid body kinematics is d eω (a q ) e = (a p ) e + qp ∧ + (ω ∧ qp) ∧ ω dt

(0-10)

where the points p and q are “fixed” in R E . There is another method which leads to the fields of velocities and accelerations of points of any rigid body. Let us consider a “fixed” point q of a rigid body B in motion relative to a frame R E = { E1 (t ), E 2 (t ), E 3 (t ) }, this last one moving with respect to a frame of reference R e . From oq = oO + Oq ,

we deduce the absolute velocity of q, namely: (v q ) e =

d e oO d e Oq + dt dt

= (v O ) e +

d E Oq + ω ∧ Oq dt

that is (v q ) e = (v q ) E + (v O ) e + ω ∧ Oq

(0-11a)

where (v q ) E is the relative velocity of q. Since we know that the transport velocity 1 of q is (v q ) T = (v O ) e + ω ∧ Oq ,

(0-11b)

we say: PR18 The absolute velocity is the sum of the relative velocity and the transport velocity. In the same manner, the absolute acceleration of q is de de (a q ) e = (v q ) e = [(v q ) E + (v O ) e + ω ∧ Oq ] dt dt

such that the terms of the sum are successively: de dE (v q ) E = (v q ) E + ω ∧ (v q ) E dt dt 1

Called Vitesse d’entraînement in French.

23

Requirements de (v O ) e = ( a O ) e , dt de d eω d E Oq (ω ∧ Oq ) = ∧ Oq + ω ∧ ( + ω ∧ Oq ) , dt dt dt

and by denoting the relative acceleration as follows: dE (a q ) E = (v q ) E , dt

we obtain the following expression of the absolute acceleration of q: d eω (a q ) e = (a q ) E + [(a O ) e + ∧ Oq + ω ∧ (ω ∧ Oq )] + 2ω ∧ (v q ) E . dt

(0-12)

Therefore, we interpret this result as the following: PR19 The absolute acceleration is the sum of the relative acceleration, the transport acceleration and the Coriolis acceleration. Example. The Euler pendulum. Let R 0 = { o;1x0 ,1 y0 ,1z0 } be an orthonormal frame fixed in E,

R 1 = { o' ;1x1 ,1 y1 ,1z1 } be an orthonormal frame in rectilinear translation along 1x0 . We consider a simple pendulum consisting of a mass point p suspended from the moving point o′ and which is swinging in the vertical plane { o' ;1x1 ,1 y1 }. We are going to express the velocity and acceleration

Fig. 3 Let θ be the angle between o'p and the downward vertical. Let R 2 = { o' ;1x2 ,1 y2 ,1z2 ] be the orthonormal frame connected to the pendulum such that 1x2 =

o' p ⋅ o' p

24

Chapter 0

Let (v p ) i denote the velocity of p relative to R i ,

ω ji denote the angular velocity of R j relative to R i . We let oo' = X (t ) 1x0 = X (t ) 1x1 , o ' p = l 1x 2 .

Let us make explicit the velocity (v p ) 0 = (v o ' ) 0 + po'∧ ω20

where p and o′ are “fixed” in R 2 . The formula of vector derivative implies: doo' doo' doo' )0 = ( )1 + ω10 ∧ oo ′ = ( )1 dt dt dt = X& 1x0 = X& 1x1 .

(v o ' ) 0 = (

In addition, the following expressions o' p = l 1x2 = l sin θ 1x1 − l cos θ 1 y1 , ω = θ& 1 20

z1

imply that 1x1

ω20 ∧ o' p = 0 l sin θ

1 y1 0 − l cos θ

1z1

θ& 0

= l θ& cos θ 1x1 + lθ& 1 y1 . Therefore, we obtain: (v p ) 0 = ( X& + lθ& cosθ ) 1x1 + l θ& sin θ 1 y1 = ( X& + l θ& cosθ ) 1x0 + l θ& sin θ 1y0 . We note that this result is found again directly as following: op = oo' + o' p = ( X& + lθ& cos θ ) 1x0 + l θ& sin θ 1 y0 , but (

dop dop dop )0 = ( )1 + ω10 ∧ op = ( )1 dt dt dt

(

dop ) 0 = ( X& + lθ& cos θ ) 1x1 + lθ& sin θ 1 y1 . dt

and thus

Now let us make explicit the acceleration.

Requirements

The fundamental formula of acceleration, Eq. (0-10), in rigid body kinematics is (a p ) 0 = (a o′ ) 0 + po'∧(

d ω20 ) + (ω20 ∧ po′) ∧ ω20 dt

where p and o' are fixed in R 2 . We have: (a o ' ) 0 = (

d v o ' ) 0 = X&& 1x0 = X&& 1x1 . dt

From (

d d ω20 ) 0 = ( ω20 )1 + ω10 ∧ ω20 = θ&& 1z1 dt dt

we deduce: po'∧(

d ω20 ) 0 = lθ&& cosθ 1x1 + lθ&& sin θ 1 y1 . dt

Furthermore, we have: (ω20 ∧ po' ) ∧ ω20 = −l θ& 2 sin θ 1x1 + l θ& 2 cos θ 1 y1 . We conclude: (a p ) 0 = ( X&& + lθ&& cos θ − lθ& 2 sin θ )1x1 + (lθ&& sin θ + lθ& 2 cos θ )1y1 . We note that we can use the method of relative motions. Indeed, let us consider: (v p ) 0 = ( v p )1 + (v p ) T . But the velocity relative to R 1 is

(v p )1 = lθ& 1y2 = lθ& cos θ 1x1 + lθ& sin θ 1y1 and the transport velocity is (v p ) T = (v o ' ) 0 = X& 1x0 = X& 1x1 .

( since ω10 ∧ o' p = 0 )

Therefore, the absolute velocity of p is (v p ) 0 = ( X& + l θ& cosθ ) 1x1 + lθ& sin θ 1 y1 . As regards the acceleration vectors, we have: (a p ) 0 = (a p )1 + (a p ) T + 2ω10 ∧ (v p )1 = (a p )1 + (a p ) T .

25

26

Chapter 0

But d1 (v p )1 dt = (lθ&& cos θ − lθ& 2 sin θ ) 1x1 + (lθ&& sin θ + lθ& 2 cos θ ) 1 y1

(a p )1 =

and

(a p ) T = X&& 1x1 ,

thus the absolute acceleration of p is (a p ) 0 = ( X&& + lθ&& cos θ − lθ& 2 sin θ ) 1x1 + (lθ&&sin θ + lθ& 2 cos θ ) 1 y1 .

2.5

SLIDING VELOCITY

The sliding velocity of a rigid body with respect to another one is introduced in first courses of mechanics. Let B1 and B2 be two rigid bodies moving in a frame of reference R. We assume there is a point of B1 in contact with a point of B2 at each instant. Let v (i ∈ B1 ) denote the velocity of the contact point in the motion of B1 with respect to R at time t. Let v (i ∈ B2 ) denote the velocity of the contact point in the motion of B2 with respect to R at time t.

Let us recall the following D

The sliding velocity of B2 with respect to B1 , at point i, is v s = v (i ∈ B2 ) − v (i ∈ B1 ) .

Since v (i ∈ B1 ) and v (i ∈ B2 ) have directions of the tangent plane to the bodies, at point i, we deduce that the sliding velocity v s belongs to this tangent plane. To conclude this introductory matter, we say: D

Rigid bodies B1 and B2 are rolling without slipping (but pivoting) if the sliding velocity vanishes.

27

Requirements

3. EXERCISES Exercise 1.

Given a plane Π of E, we consider the set D of dynams of which the moments are orthogonal to Π. Prove that the set D ⊥ of dynams orthogonal to the dynams of D is D . Answer. Let { o; i, j } be an orthonormal frame of Π, { o; i , j , k } be an orthonormal frame of E. Every dynam D of D is such that:

∀D ∈ D , ∀q ∈ Π : M q ( D) ∧ k = 0 ⇔

∀o, q ∈ Π : ( M o ( D) + qo ∧ R( D)) ∧ k = 0

⇔

∀o, q ∈ Π : M o ( D) ∧ k + (qo.k ) R( D) − ( R( D).k )qo = 0

⇔

∀o, q ∈ Π : M o ( D) ∧ k − ( R( D).k ) qo = 0

⇔

∀o ∈ Π : [ M o ( D) // k and R( D).k = 0 ]

⇔

∀o ∈ Π : [ M o ( D) ⊥ Π and R( D) // Π ] .

So, every dynam D ∈ D is of the following type: 0  a i + b j  i   j  ck  = a 0  + b 0  + c  k     o  o  o

(reals a, b, c) .

Let us consider the set D ⊥ of dynams α i + β j + γ k  D′ =    l i + m j + nk 

(reals α , β , γ , l , m, n )

which are orthogonal to the ones of D ; that is such that: 0 = D. D ′ = R( D) . M o ( D' ) + R( D' ) . M o ( D) = (a i + b j ) . (l i + m j + n k ) + (α i + β j + γ k ) . ck = al + bm + cγ . Since this expression vanishes for every a, b and c we necessarily have l = m = γ = 0 . Therefore, every dynam D ' of D ⊥ is of type: α i + β j  D' =   .  nk  o

We conclude that the set D ⊥ of dynams of type D ′ is really the set D of dynams of type D whose moments are orthogonal to Π.

28

Chapter 0

Exercise 2. In a frame of reference { o; i , j , k }, we consider the point a ( 2,1,−1) and the sliding vector w = i − 2 j + 2k through a which defines a sliding dynam S. (i) Determine the elements of reduction of S at o. (ii) Find the real l such that the dynam D defined by (l + 2) i + (l − 1) j − 2l k   (l − 4) j + (l − 4) k   o

is equal to S. (iii) Prove there is another value l for which D is also a sliding dynam. Determine the axis of the sliding dynam. (iv) Reduce D in the case where l = 1. Answer. (i) Since

i

j k

M o ( S ) = M a ( S ) + oa ∧ R( S ) = 0 + 2 1 − 1 = −5 j − 5k , 1 −2 2

the sliding dynam S represented by w  0   a

has elements of reduction at point o such that: i − 2 j + 2k   − 5 j − 5k  .  o

(ii) Since D must be equal to S, from the comparison between elements of reduction of D and S, at o, we deduce the value l = −1. (iii) A dynam is a sliding dynam if the scalar invariant is zero while the resultant is nonzero; that is: R( D) . M o ( D) = (l − 1)(l − 4) − 2l (l − 4) = 0 The solutions are: { l = 4 , l = −1 }. The “other value” of the question is thus l = 4. In this case, the dynam is such that: 6 i + 3 j − 8 k    . 0  0

The axis of the sliding dynam is the set of points q ( x, y, z ) such that M q = 0 , namely:

29

Requirements i

j

k

M q = M o + qo ∧ R = 0 + − x − y − z = 0 6

⇔

3

−8

8 y + 3 z = 0  8 x + 6 z = 0 − 3 x + 6 y = 0, 

these three equations defining the axis of the sliding dynam. We note that this axis could be immediately found since it is the straight line through o ( M o = 0 ) and directed along R = 6 i + 3 j − 8k . These equations are immediately: x 6

=

y 3

=

z −8

⋅

(iv) For l = 1 , the dynam D is such that:  3i − 2k  − 3 j − 3 k   o

and the moment at every point a of the central axis is:

M a ( D) =

M o ( D) . R( D) 2

R ( D)

= 18 i − 12 k. 13 13

The dynam D is the sum T = C + S of a couple and a sliding dynam such that: 0   C =  18 12   13 i − 13 k 

and  3i − 2 k   0   3i − 2k  S= . −  18 12  =  18  27  − 3 j − 3 k  o  13 i − 13 k  o  − 13 i − 3 j − 13 k  o

Exercise 3.

A gear, represented as a disk D of fixed center o and radius R, rotates in the (x,y)-plane of a “fixed” frame R e = { o;1x ,1 y ,1z } with an angular velocity ω. (i) A disk D1 in the (x,y)-plane of fixed center c and radius R / 3 rolls without slipping (tooth wheel) on the side of D and on the inside of a hoop C. Calculate the angular velocity ω1 of D1 and the angular velocity ω 2 of C.

(ii)

If the center of D1 is moving when D1 rolls without slipping on the side of D and on the inside of the fixed hoop C, calculate the angular velocity ω1 of D1 and the angular velocity Ω of oc.

30

Chapter 0

Fig. 4

Let 1u =

oc , oc

1v = 1z ∧ 1u .

The dynams of velocities of D, D1 and C with respect to R e are respectively: ω 1  VD e =  z  ,  0 o

ω 1  VD1 e =  1 z  ,  0 c

ω 1  VC e =  2 z  .  0 o

We specify that ω is positive, ω1 and ω 2 are negative. (i) The composition law of dynams of velocities leads to M i (VD1 e ) = M i (VD1 D ) + M i (VD e ) .

From the end of Section 2.3, the point c being fixed when D1 moves, we have: M i (VD1 e ) = (v c ) e + ic ∧ ω1 1z = ic ∧ ω1 1z .

In the same manner, the point o being fixed when D moves, we have: M i (VD e ) = (v o ) e + io ∧ ω 1z = io ∧ ω 1z .

From these results, we deduce that the sliding velocity of D1 with respect to D, at i, is: R

M i (VD1 D ) = − ω1 1v − Rω 1v . 3

The condition of rolling without slipping is thus:

ω1 = −3ω . Now, let us determine the angular velocity of C. The composition law of dynams of velocities leads to: M j (V D1 e ) = M j (VD1 C ) + M j (VC e ) .

31

Requirements

From the end of Section 2.3, the point c being fixed when D1 moves, we have: M j (VD1e ) = (v c ) e + jc ∧ ω1 1z = jc ∧ ω1 1z .

In the same manner, the point o being fixed when C moves, we have: M j (VC e ) = (v o ) e + jo ∧ ω 2 1z = jo ∧ ω 2 1z .

From these results, we deduce that the sliding velocity of D1 with respect to C, at j, is: M j (V D1C ) =

R 5 ω1 1v − ω 2 1v . 3 3

The condition of rolling without slipping is thus: 1

3

5

5

ω 2 = ω1 = − ω . (ii) In the second situation, oc is moving and the hoop C is fixed. Let us consider the moving frame R E = { o;1u ,1v ,1z }. Let us calculate the sliding velocity of D1 with respect to D, at point i: M i (VD1 D ) = M i (VD1 e ) − M i (VD e ).

We have: ω 1  VD1 e =  1 z  , ( v c ) e  c

but (v c ) e = ( =

d oc ) e = ωEe ∧ oc dt

4 3

(since oc is constant in R E )

R Ω 1v ,

thus we have:  ω 1 1z  VD1 e =  4  .  3 R Ω 1v  c

From the end of Section 2.3, the point c being fixed when D1 moves, we have: M i (V D1e ) = (v c ) e + ic ∧ ωD1 e = 43 RΩ 1v − R3 ω1 1v . In the same manner, the point o being fixed when D moves, we have: M i (VD e ) = (v o ) e + io ∧ ω = Rω 1v . From these results, we deduce the following sliding velocity: M i (VD1 D ) = ( 34 R Ω − R3 ω1 − Rω ) 1v .

The condition of rolling without slipping is thus:

32

Chapter 0 4 Ω − 1ω 3 3 1

− ω = 0.

Now, let us calculate the sliding of D1 with respect to C, at j: M j (VD1C ) = (v c ) C + jc ∧ ω D1C = 43 RΩ 1v + R3 ω1 1v .

The condition of rolling without slipping implies: Ω = − 14 ω1

and finally:

ω1 = − 32 ω ,

Ω = 83 ω .

(since c is fixed in D1 )

CHAPTER 1

STATICS

Statics is the subdiscipline of mechanics which studies the equilibrium of material systems under the action of forces.1 Statics is an essential branch of engineering mechanics. First, it is introduced in a classic way by considering the resultant of forces acting on the system and the sum of moments of forces called the total moment or simply the moment. We will quickly deal with this basic method for which many excellent books show numerous engineering applications. On the other hand, the method of virtual work is the subject of our full attention. This analytical mechanics method brings into play virtual displacements or virtual velocities (which are arbitrary vectors a priori). In short, statics is concerned with three types of problems, namely: -

searching for equilibrium positions of mechanical systems in presence of known forces, looking for forces which insure the equilibrium of system, studying the stability of equilibrium.

The classic and “virtual” methods have application fields which are somewhat complementary. So, the classic method is especially used for trying to find forces which maintain material systems in equilibrium. For instance, this method effectively deals with obtaining necessary and sufficient conditions to maintain a rigid body in equilibrium under the action of forces. The method of virtual work is better suited to obtain equilibrium positions of sophisticated mechanisms . It easily allows finding again equilibrium conditions of the classical method and also permits the study of the stability of systems in equilibrium, and so on. Note that the two mentioned methods can be used together in order to solve problems of statics more easily.

1

A material system or mechanical system is a system of particles or a (finite) collection of rigid bodies including possible isolated particles. For example, a mechanical system is a particle, a rigid body, connected bodies,…

33

34

Chapter 1

Modeling In theoretical mechanics, material systems as rigid bodies, mechanisms, etc. are modeled. A modeling of a material system is not absolute but is relative to the study of the considered system. So, various mechanical studies of a same system can lead to model this system in different ways. For example, models for a same rigid body can have 1,2 or 3 dimensions according to the study and we say that the shapes are modeled.

1. CLASSIC METHOD This paragraph recalls notions of first courses of mechanics. 1.1

MECHANICAL ACTIONS

1.1.1

Definitions

D

A cause capable of either altering the motion of a material system, maintaining it at rest or deforming it, is called a mechanical action. If a cause originates outside the system it is said to be an external mechanical action. If a cause is due to an element of the material system it is said to be an internal mechanical action.

As we shall see later, this classification can be arbitrary because of the choice of the mechanical system isolation.

1.1.2 Forces Forces (which are mechanical actions) exerted on a material system can be either given forces or constraint forces, these last being generally unknown. Given applied forces can act on a rigid body, for instance: weight force, spring force, tensile force exerted by a flexible cable, etc. In this last case, if it is a large force compared with the cable weight, then the action line of the tension force is the straight line of the cable; if the weight is not negligible compared with the tension, then this last changes (in direction and norm) along the cable. The cable exerts a force tangent to itself at the attachment point. Besides the given forces applied to a system, there are constraint forces which constrain the motion and also called reactions, this designation emphasizes the manner the material (to which the body is attached) reacts. These constraint forces will later be analyzed. Note that gravitational forces (weight force) are remote mechanical actions unlike contact reactive forces.

1.1.3 Moment of a Force The notion of mechanical action is generally merged with the one of force, but mixing up these notions is justified if actions only correspond to forces.

35

Statics

However, all the mechanical actions cannot only be represented by sliding or fixed vectors that are the forces. The reader will be persuaded by considering the classic example of a welded connection between a rigid bar and a wall (built-in or fixed support).

Fig. 5 A force F is applied to the bar at a point x. It is intuitively obvious that the mechanical action exerted by the bar on the wall depends on the position of x. Therefore, in addition to F it is necessary to take into account of the moment of F about a point a of the weld: M a ( F ) = ax ∧ F . Let us generalize the previous observation to collections of forces.

1.1.4 Dynam of Mechanical Action

S ′ be mechanical systems (solid or not). We suppose that S ′ exerts a mechanical action on S due to several forces Let S and

The mechanical action of

f h , h ∈ {1,…,n}.

S ′ on S is characterized by the resultant of forces:

f = ∑ fh h

and the (total) moment about some point a: M a = ∑ ax h ∧ f h

[ ∀x h ∈ supp ( f h )] .1

h

D

The dynam of mechanical actions of S ′ on S is the dynam defined by the resultant and the (total) moment of forces exerted by S ′ on S about a. The resultant and moment are called the elements of reduction of the dynam at a.

1

supp ( f h ) is the support or line of action of

fh .

36

Chapter 1

It is denoted by  fS ′→S  M  .  S ′→S  a To sum up, we say: PR1

Every mechanical action is well-defined by the dynam of the mechanical action.

In conclusion, the theory of dynams can be used in this context. D

The dynam of mechanical actions is called the dynam of forces and is simply denoted: f  F=  M  a

(if the context is sufficiently clear).

1.2

CLASSIFICATION OF FORCES

It is necessary to specify the various types of forces which may be applied to a material system, so we are going to give a review of these forces acting on every particle of the system. The important thing is to know if no external force has been forgotten or if some of them can be neglected. In other words, the choice of external forces taken into account results from the description of external causes acting upon the considered system. For example, for a body at the earth’s surface, it is logical to neglect the influence of the moon; on the contrary, the same does not apply to the study of tides. The comparison between theoretical model and experience lets generally decide on choosing the forces to consider. If the theoretical results are significantly different from observations or expected conclusions, then supplementary forces must be eventually introduced. A famous example was the discovery of the planet Neptune. By a careful mathematical study, Le Verrier proved that the unexplained deviations from the predicted orbit of Uranus followed from an unknown planet, so Neptune was located. 1.2.1 External Forces The external forces are classified into two categories: (i) Given external forces They are the external forces which a priori are known at every point xh of the material system and at each instant. We recall that the earth’s gravitational force is distributed over the volume of a body or at each point of a system. This force field exerted at each point of a rigid body may be replaced by only one force at the center of gravity: the weight of the body. The resultant of the given external forces acting on a mass point p h is denoted by

Fh(e) .

Statics

37

(ii) External forces of constraint

For a material system and in particular for a particle, we say: D

A material system is subject to external constraints if there are forces originated outside the system which restrict its motion freedom and called the external forces of constraint.

The string of a pendulum, the walls of a container, etc. materialize external constraints. Usually, the external constraint forces (reactions) act on isolated points of the system and these connection forces are generally unknown and thus annoying. The method of virtual work will allow the elimination of their effects. In the classic example of the simple pendulum, the constraint materialized by the weightless string prevents the suspended particle from freely moving. The string exerts a reactive force of connection on the particle which cannot freely follow a parabola. This connection force varies with the position and the velocity of the particle. In the case of a frictionless connection (or smooth connection), the direction of the force is specified. So, the contact between a particle and a smooth surface or a smooth curve shows respectively a connection force normal to the surface or orthogonal to the curve at the contact point. If two bodies are in smooth contact, then the force exerted by a body on the other one is normal to the surfaces. We denote the resultant of the forces of constraint acting on p h by L(eh) . The resultant of all the external forces acting on p h is denoted by f h( e ) = Fh(e ) + L(he) .

1.2.2 Internal Forces

The internal forces are the forces exerted between the different particles or elements of a material system. There are two types of internal forces. (i) Given internal forces

These forces also called the interaction forces are known forces acting between elements of the system. At point xh , they are denoted by Fh(i ) . For example, they are the forces acting between the planets of the solar system (external forces being the ones exerted by the stars of the galaxy). A spring connecting two bodies shows another example of internal force between bodies. (ii) Internal forces of constraint Internal forces of constraint are (unknown) forces which make different elements of a material system being constrained among.

38

Chapter 1

At point xh , they are denoted by L(ih ) . For example, an internal force of constraint is caused by a joint which constraints two bodies to remain in contact. The resultant of all the internal forces acting on p h is denoted: f h(i ) = Fh(i ) + L(hi ) . We recall that, in noninertial frames of reference, it is necessary to take supplementary forces into account.

1.3

EQUILIBRIUM CONDITIONS

1.3.1

Definitions and Conditions

1.3.1a Definition of the Equilibrium of a Material System D

A material system is in equilibrium relative to a frame of reference if each of its points remains at rest relative to the frame.

This implies in particular: D

A rigid body is in equilibrium relative to a frame of reference if its position coordinates1 remain constant.

We can express the following (equivalent) proposition, it being understood that the velocities of the particles are zero at a given (initial) instant. PR2

A material system S is in equilibrium with respect to an inertial frame iff the resultant of all the forces acting on every particle p h of S is zero:2 ∀p h ∈ S : f h = Fh( e ) + Fh(i ) + L(he ) + L(hi ) = 0 .

(1.1)

In particular, this necessary and sufficient condition applies to rigid bodies, but obviously this is not used in practice. In the case of a particle, the above necessary and sufficient condition is equivalent to the following: D

A particle is in equilibrium (or at rest) relative to an inertial frame of reference if the resultant of forces acting on the particle is zero. We denote:

1 2

F +L=0.

We will see that the position coordinates will be called the generalized coordinates (distances, angles,…). The writing iff means if and only if.

39

Statics

Remark. If the frame of reference is not inertial, then it is necessary to take the forces of transport1 and of Coriolis into account. But if we study an equilibrium with respect to the moving coordinate system, then the Coriolis force is not to be considered since: 2ω ∧ v h

(r )

=0

where v h(r ) denotes the relative velocity of p h .

1.3.1b Dynam of Internal Forces We recall the “principle of action and reaction.” PR3

The internal forces of a material system are equal in norm, opposite in direction2 and collinear.

In other words, to every action force exerted by a particle p1 on a particle p 2 , there is a reaction force equal in norm and opposite in a direction along the line joining the particles, which is denoted by

f12 = − f 21 . This principle (or third law of Newton) being valid in classical mechanics, we say: PR4

The dynam of internal forces of any mechanical system is zero.

Proof. The principle of action and reaction implies that the resultant of internal forces of a material system S is zero: f (i ) = ∑ f h(i ) = 0 h

because this principle is valid for every pair of particles of S. In addition, the sum of moments of all internal forces about some point o is zero. Indeed, successive pairs of terms such that:

ox h ∧ f h(i ) + ox k ∧ f k(i ) = (ox h − ox k ) ∧ f h(i ) = xk xh ∧ K xh xk = 0

K∈R

imply that the (total) moment of internal forces is zero. So, if we denote the moment of internal forces acting on any particle p h by M o ( f h(i ) ) , then the (total) moment of internal forces of S is M o(i ) = ∑ M o ( f h(i ) ) = 0 . h

In conclusion, the dynam of internal forces is

1 2

Called “forces d’entraînement” in French. Opposite in direction is said to be de sens opposés in French.

40

Chapter 1

[F ] =  f (i )

  = 0. (i )  M  o (i )

(1-2)

1.3.1c Equilibrium Conditions of a Material System

First undergraduate courses in mechanics set out the reduction, at some point, of any collection of vectors defining a dynam D (see Requirements). Reducing a collection of (force) vectors consists in replacing this collection with a more simple equivalent collection and it is easily proved the following: PR5

Every collection of (force) vectors is equivalent to a collection composed of - a single (force) vector through an arbitrary point, - a couple of (force) vectors.

The single vector is the resultant R(D) of the dynam and the moment of the couple is the moment M (D) of the dynam. The following example describes this reduction. Example. Let us show that a collection of three forces f1 , f 2 , f 3 applied to a material system, for instance a rigid body, is equivalent to a collection composed of the resultant of the collection and the moment of a couple.

Let us note that here the resultant and the couple moment are not free vectors, but are bound vectors at an arbitrary point a. We recall that if we add two opposite vectors to a collection of vectors, then we obtain an equivalent collection. So, every collection consisting of a vector f is equivalent to a new collection composed of a couple and a vector f // that is the representative, at a, of the equivalence class defined by f. The preceding is so summed as: { f }∼ { f , - f //, f // }∼ { M, f // } where M represents the moment of the couple. The previous process is used with f1 , f 2 and f 3 as shown in the following figure.

Fig. 6

41

Statics

So, at a, we obtain the resultant F of three forces and the sum M a of three moments. The collection { f1 , f 2 , f 3 } is equivalent to the collection { F , M a }. Remark. Unlike the resultant, the moment is dependent on the choice of a. In particular, we may choose a in order that the resultant and moment be collinear. The set of such points defines a straight line called the central axis.

The previous proposition about the reduction of a collection of (force) vectors leads to an interesting condition of equilibrium. Indeed, the dynam of internal forces vanishing, we can state the following necessary condition of equilibrium: PR6

If a material system is in equilibrium, then the dynam of external forces is necessary zero. This is denoted: equilibrium where

f ( e ) = ∑ f h(e ) , h

⇒

[F ] (e)

 f (e)  =  = 0. (e)  M  a

(1-3)

M a( e ) = ∑ ax h ∧ f h(e ) . h

Remark 1. The previous condition is not sufficient to ensure the equilibrium of a material system. For example, we consider a plane system composed of two jointed rods. A force F is exerted on one rod and the opposite force is exerted on the other rod.

Fig. 7

The dynam of all exterior forces is − F  F  − F  F  0  +  0  = 0  +  0   p  q   p  p

since F and pq are collinear. So, this dynam is zero, but it is not sufficient for the system of rods to be in equilibrium.

42

Chapter 1

Remark 2. When studying linear momentum and angular momentum theorems, we will see that the condition of zero exterior dynam can lead to motions of the material system (see conservation theorems). So, in particular, we will see that the condition F (e ) = 0 must be fulfilled for a rigid body to be in equilibrium. But it is not sufficient to assure that the body is at rest (unless the body is initially at rest).

[ ]

Example. Find the condition for equilibrium of a rigid body with a fixed point o, in particular in the case of a lever.

The reaction force of constraint at o being denoted by Lo , a first condition of equilibrium (of translation) is F ( e) + Lo = 0 where F (e) is the resultant of given external forces. A second condition of equilibrium (of rotation) is M o ( F (e ) ) = 0 ,

since the constraint force does not contribute to the (total) moment at point o. This equilibrium condition is simply denoted by M o(e) = 0 .

Now, we consider a lever (extremities a and b, fulcrum o) which is in equilibrium under the action of two given forces Fa and Fb .

Fig. 8

The condition of equilibrium of translation lets obtain the reaction force of constraint: Lo = − Fa − Fb .

The condition of equilibrium of rotation is reduced to the following M o( e) = oa ∧ Fa + ob ∧ Fb = 0 ⇔

Fa oa sin α − Fb ob sin β = 0

⇔

M o ( Fa ) = M o ( Fb ) .

These results are well-known.

43

Statics 1.3.2

Particular Collections of Forces Applied to a Rigid Body

A rigid body remains at rest iff the resultant and (total) moment of applied external forces are such that f (e ) = 0 ,

M ( e) = 0 .

The point about which the (total) moment is zero is not indicated because this moment vanishes about any point (since M p = M q + pq ∧ f (e ) ). Given a coordinate system oxyz, the previous necessary and sufficient conditions of equilibrium of a rigid body are written: f x( e) = 0 ,

f y( e) = 0 ,

f z( e) = 0

M x(e ) = 0 ,

M y(e ) = 0 ,

M z(e ) = 0 .

Now, we are going to give a review of conditions of equilibrium for particular force collections. (i)

Collinear forces

A rigid body is in equilibrium under collinear forces iff the resultant of forces vanishes; that is, by choosing the x-direction of forces: f x( e) = 0 .

In particular, if there are only two forces, these must be equal in norm, opposed in direction and collinear. (ii)

Coplanar forces concurrent at a point

If coplanar forces applied to a rigid body are concurrent at a point o then their moment about o is necessarily zero. Let oxy be the plane of forces. The (total) moment of forces about o is M 0( e) = 0 .

In addition, the z-components of forces are zero and thus the equilibrium conditions are f x( e) = 0 ,

f y( e) = 0 .

In particular, a rigid body is in equilibrium under the action of three forces if the polygon of forces is a triangle. So, three forces which ensure this equilibrium are necessarily coplanar. Additionally, the forces must be concurrent. Indeed, if the lines of action were not concurrent, then one of the forces would show a moment about the point of intersection of the other two, but this would be absurd since the (total) moment of the collection of forces is zero (at any point). Collection of three forces plays a very important role in statics because collections of many forces may often be reduced to this simple type.

44

(iii)

Chapter 1

Spatial forces concurrent at a point

A rigid body is in equilibrium under spatial collinear forces iff f x( e) = 0 ,

f y( e) = 0 ,

f z( e) = 0 .

The three moment equations of equilibrium are automatically verified because the moment about any axis through the point of concurrency vanishes. (iv)

Coplanar and parallel forces

We consider for instance a collection of forces in a plane oxy and in x-direction. A rigid body is in equilibrium under these forces iff f x( e) = 0 , M z(e ) = 0 .

Indeed, the forces have the same x-direction and all the moments the same z-direction (normal to the plane). (v)

Spatial parallel forces

For example, we consider a collection of parallel forces in the x-direction. A rigid body is in equilibrium under these forces iff f x( e) = 0 , M y(e ) = 0 ,

M z(e ) = 0 .

Indeed, the force components in y and z are zero and the (total) moment of forces has no component in x. (vi)

Forces concurrent with a line

For example, we choose this line as x-axis. The moment of any (force) vector about an axis is zero if the line of action and axis are concurrent. So, the collection of forces automatically verifies the condition M x(e ) = 0 and therefore the conditions of equilibrium are

(vii)

f x( e) = 0 ,

f y( e) = 0 ,

M y(e ) = 0 ,

M z(e ) = 0 .

f z( e) = 0

Coplanar forces

For example, the forces are assumed in the plane oxy. The conditions of equilibrium of a collection of such forces are immediately: f x( e) = 0 , M z(e ) = 0 .

f y( e) = 0 ,

Statics 1.4

45

TYPES OF EQUILIBRIUM

The necessary and sufficient conditions of equilibrium of a rigid body under a collection of forces let calculate (at equilibrium) the unknown forces of reaction, but these conditions may not be sufficient to evaluate all of them. This is function of types of constraints. D

A collection of forces is hyperstatic if the number of unknown forces is larger than the number of equations of equilibrium condition. A collection of forces is hypostatic if the number of unknown forces is smaller than the number of equations of equilibrium condition. A collection of forces is isostatic if the number of unknown forces is equal to the one of equations of equilibrium condition.

Remark. In the hypostatic case, where the body is called partially constrained, the equilibrium is instable since the unknown forces cannot satisfy all the equilibrium equations. More explicitly, for each unsatisfied scalar equation of equilibrium, there is a corresponding possible motion, either of translation (for an equation of force), or of rotation (for an equation of moment).

In engineering mechanics, it is essential to know the forces acting at different points of a solid structure. These forces can deform and break the structures; but often and in first approximation, these structures can be considered as rigid bodies. Courses of strength of materials detail these notions. D

A structure is statically determinate if there is the minimum number of constraints necessary to ensure the equilibrium.

In this case, from given external forces, all the unknown forces and moments acting on each part of the structure can be determined. D

A structure is statically indeterminate if there are more constraints than necessary to maintain an equilibrium. In other words: if there are more unknown components of reaction forces than scalar equilibrium equations.

So, a statically indeterminate structure is such that the forces acting on its parts are not completely determined by the given (external) forces, owing to stresses within the structure. This type of problem is more difficult to solve than the one of statically determinate structures. D

Constraints are said to be redundant if they can be removed without destroying the equilibrium.

Example 1. A rigid body fixed at two points o1 and o2 is generally a matter for hyperstatic equilibrium.

46

Chapter 1

Let d be the distance between these points, { o1 ; i , j , k } be an orthonormal frame of reference where k is collinear to o1o 2 and has the direction1 of o1o2 , L1 ( L1x , L1 y , L1z ) and L2 ( L2 x , L2 y , L2 z ) be the unknown constraint forces of reaction at

respective points o1 and o2 , F ( Fx , F y , Fz ) be the resultant of given external forces acting on the body. The condition of (translation) equilibrium F + L1 + L2 = 0

is equivalent to the system of 3 scalar equations: Fx + L1x + L2 x = 0 , F y + L1 y + L2 y = 0 , Fz + L1z + L2 z = 0 which contains 6 unknowns. We are going to express the condition of rotation equilibrium, for instance at o1 . Since the (total) moment of forces about o1 is composed of: -

the (total) moment of given external forces M o(1e) ( M x , M y , M z ) ,

-

the moment of the constraint force of reaction L2 , namely: M o1 ( L2 ) = o1o 2 ∧ L2 = − d L2 y i + d L2 x j ,

then the conditions of rotation equilibrium are written: M x − d L2 y = 0 ,

M y + d L2 x = 0 ,

Mz = 0.

Be careful! This last equation imposes one condition, namely: the (total) moment of external forces about the fixed line o1o2 must be zero. The remaining conditions of equilibrium allow determining the components of constraint forces: L2 x = − M y d ,

L2 y = M x d .

Next, the first two equations of translation equilibrium allow obtaining L1x and L1 y . So, there is only one equation at our disposal: L1z + L2 z = − Fz , but we must calculate two unknowns L1z and L2 z ! The collection of forces is hyperstatic since there are more unknowns than equations. It is possible to make it isostatic by canceling, for instance, the component L2 z with the aid of a sliding guide supporting only forces normal to the guide. Example 2. The collection of forces which fix a rigid body at three points is hyperstatic in general.

Indeed, there are 3 constraint (reaction) forces, that is 9 unknowns for 6 equilibrium equations. 1

« Sens d’orientation » in French

Statics 1.5

47

STRESS AND CONTACT DYNAM

In reality, forces between bodies in contact are distributed over an area; that is, concentrated forces do not exist, but are distributed over a region: there is a force field! This remark also holds in the cases of lines and volumes. Let B1 and B2 be rigid bodies in contact, S be the contact surface, Π be the tangent plane to bodies at any point p ∈ S . The body B1 exerts a (contact) mechanical action on B2 at every point of S.

Fig. 9 Remark. We can also imagine any surface splitting two parts of a body.

1.5.1 Stress D

The stress of B1 on B2 , at p, is the force exerted by B1 on B2 per unit area.

It is denoted by l (p1→2)

or simply by l

(if no possible confusion).

The usual units for stress are: - the pascal or newton per square metre: Pa = N / m 2 , the megapascal: MPa = 10 6 Pa = N / mm 2 = 10 bars .

D

The vector component of stress perpendicular to the tangent plane Π is called the normal stress and the vector component of stress tangent to the plane Π is called the shear stress.1

If we denote the shear stress by t and the normal stress by n, at any p ∈ Π , then we express the corresponding stress by l = t + n. 1

Unlike the rigid bodies for which shear stresses exist, the fluids at rest show only normal stresses.

(1-4)

48

Chapter 1

Fig. 10 By considering a surface splitting two parts of a body, we also define: D

A compression (resp. a tension) is a stress perpendicular to the surface which pushes (resp. pulls) the material of each part towards the other.

The next figure shows various stresses between materials of part 1 and part 2 in the cases of compression ( C ), tension ( T ) and shear stress ( SS ).

Fig. 11 1.5.2 Contact Dynam

Let us define the dynam of the mechanical action of contact of B1 on B2 , simply called the contact dynam of B1 on B2 or the constraint dynam of B1 on B2 . Let o be a reference point. D

The contact dynam of B1 on B2 is defined by its elements of reduction, at o, namely: L(1→2) = ∫ l dS

(1-5)

M 0(1→2) = ∫ op ∧ l dS .

(1-6)

S

S

It is denoted by  L(1→2)   (1→2)  .  M  o

Statics

49

Remark 1. The contact dynam is obtained from the stress l of B1 on B2 known at every point of S. Remark 2. We denote by ∆L(1→2) the resultant of forces acting on an element of S (at p) and by ∆A the corresponding area. ∆L(1→2) The resultant per unit area is obviously ⋅ ∆A

The stress acting on the surface element is the vector defined by ∆L(1→2) ⋅ ∆A→0 ∆A

l (1→2) = lim

The situation is less simple than the one of remark 1. The stress l at p ∈ S depends upon the choice of particular surface element. Remark 3. We must distinguish body forces from surface forces.

Body forces are distributed forces applied over the matter space occupied by the body. An obvious example is given by the gravitational attraction. The acceleration due to gravity being g ( m / s 2 ) and the mass density of the body being ρ ( kg / m 3 ), then the corresponding

body force per unit volume is ρ g ( N / m 3 ). The terminology “body force density” is sometimes used for the preceding. Surface forces are forces acting on the elements of a surface region of a body. For example, stresses show such distributed forces over the contacting surfaces of two bodies or pressure exerted by fluids, etc. They are forces per unit area. The terminology “surface force density” is sometimes used for the preceding.

Now, we introduce the notions of (total) tangential force and (total) normal force for a contact surface S. D

The (total) tangential force, denoted by T (1→2) , is the tangential vector component of the resultant of the contact dynam. The (total) normal force, denoted by N (1→2) , is the normal vector component of the resultant of the contact dynam.

In other words, we denote: T (1→2) = ∫ t dS ,

(1-7)

N (1→2) = ∫ n dS .

(1-8)

S

S

D

The rolling moment, about p, denoted M tp , is the tangential part of the moment of contact dynam. The pivoting moment, about p, denoted M np , is the normal part of the moment of contact dynam.

50

Chapter 1

To sum up, we simply denote: L=T + N ,

(1-9)

M p = M tp + M np .

(1-10)

Remark. The contact between two bodies strictly at a point is an idealized situation; in reality, there is always a contact area. If such an ideal contact is thinkable, then we naturally assume that the moment of contact dynam vanishes at the contact point.

1.5.3 Dry Friction and Coulomb’s Laws

Perfectly smooth contact between two bodies is evidently an ideal situation with only normal forces of reaction. In most of cases, additional forces must be taken into account to describe the real process. D

Friction forces are forces tangent to the plane of contact between bodies and opposing the (impending) motion of one surface relative to the other.

Friction forces are classified according to the nature of contacting surfaces, there are dry, fluid, internal frictions for instance. In particular, a dry friction can be roughly considered as an effect resulting from microscopic irregularities, from molecular attraction, etc. present in (rough) surfaces of two solids in contact. This type of friction is sometimes called Coulomb friction because, in 1781, Charles de Coulomb developed elementary laws describing dry frictions.

1.5.3a Friction Force and a Classic Experiment

Before showing the Coulomb’s laws, we consider the following classic experiment. A solid block B2 of mass m is in rest on a horizontal plane B1 . The contacting surfaces show a certain extent of roughness.

Fig. 12

The experiment consists in exerting on B2 a given horizontal force F with an increasing magnitude in order to set the block in motion.

Statics

51

In this problem with friction, the reaction force L of B1 on B2 is the resultant of two forces exerted by the plane on the block: - a (tangential) friction force, denoted T, which is in a direction to oppose motion (or motion tendency), - a (normal) friction force, denoted N, which balances the weight mg. By enlarging details of the contact surfaces, we would see irregularities.

Fig. 13

At each hump there is a reaction force L(i ) = T(i ) + N (i ) . The (total) tangential friction force T is the sum of various T(i ) . First, we consider the block at rest. In this case of equilibrium, we have in particular: F =0

⇒

T =0.

The reaction force L is normal to B1 L=N

and it balances mg. As soon as F is different from zero, but the block being motionless, we have: T = −F .

In this case, T is called the static friction force and is denoted by Ts . There is a maximum value of Ts which corresponds to the maximum magnitude of F for which B2 remains at rest. It is denoted by Tsmax .

The situation of static friction is summed as follows: T = F ,

0 ≤ Ts ≤ Tsmax .

Second, for a sufficient force F the slipping motion of the block is possible. As soon as this motion starts then T is called the kinetic friction force and is denoted by Tk .

52

Chapter 1

The following graph shows both situations.

Fig. 14

Along the ordinate of this graph we plot the norm of the tangential force of friction and along the abscissa we plot the norm of the given force F. As soon as the sliding begins, the friction force magnitude falls for a very short time and next decreases slightly. So, there are two regions, the one of the static friction force and the other one of the kinetic friction force beginning with the drop of the friction force. A drop explanation is possible by considering the irregularities of contact forces [see e.g. Meriam (1975)]. If the surfaces are in relative motion, then the contacts occur near the tops of the humps where the action lines of various reaction forces L(i ) imply that the corresponding tangential friction forces T(i ) are smaller than when the surfaces are in relative rest. This can explain that the force F for maintaining the block motion is smaller than the one to set the block in motion. Remark. In the case of surfaces B1 and B2 sliding over each other, the sliding velocity of B2 relating to B1 at some point x belonging to the tangent plane is denoted by v; then the condition expressing that the friction force T must be in a direction opposite to the direction of relative motion is expressed as: T ∧v =0

and

T .v < 0 .

1.5.3b Coulomb’s Laws

We know that if the constraints are not smooth, besides the normal stresses supplementary unknowns must be taken into account in problems of frictional resistance; namely: the shear stresses. Therefore, it is necessary to introduce new relationships called the laws of friction. These laws are different according to the problem is concerned by equilibrium (statics) or not. Let us show the friction laws of Coulomb.

53

Statics

At rest, just before the impending motion, it is observed that the maximum value of the static friction force is proportional to the normal force; that is: Tsmax = f s N

(1-11)

where the (positive) coefficient of proportionality f s is called the coefficient of static friction. To conclude the case of no relative motion of surfaces, we write the following general condition: Ts ≤ f s N (1-12) where the equality sign corresponds to the impending motion. In the case of relative motion of contact surfaces, a simple law is also observed: The norms of the kinetic friction force and normal force are proportional Tk = f k N

(1-13)

and the coefficient of proportionality f k is called the coefficient of kinetic friction. Remark 1. Roughly speaking, coefficients of friction greatly depend on the type of materials in contact. For instance:

Material Steel on steel (dry) Steel on steel (greasy) Brake material on cast iron Rubber on wood Mild steel on mild steel Aluminum on mild steel Teflon on steel Cast iron on cast iron

fs

fk

0.6 0.1 0.4 0.4 0.74 0.61 0.04 1.1

0.4 0.05 0.3 0.3 0.57 0.47 0.04 0.15

All these values depend on the experiment conditions (lubrification, polishing, temperature,…) and the last example shows a coefficient higher than unity. Remark 2. In the majority of cases, the coefficient of kinetic friction is a lower value than the coefficient of static friction:

Tk ≤ Tsmax

⇒

fk ≤ fs .

To conclude this section, we consider very interesting angles. When two surfaces are in contact, experiments (and intuition) point up that the angle φ between L and N is larger for rough surfaces than for smoother surfaces. It is quite natural to consider the tangent of this angle: tan φ =

T N

⋅

(1-14)

54

Chapter 1

So, for the maximum value of the static friction force, there is a specific angle φ s such that tan φ s = f s ,

(1-15)

where the angle φ s is called the angle of static friction. In the same manner, in the case of relative motion, we consider the angle φ k defined by

tan φ k =

Tk N

= fk

(1-16)

which is called the angle of kinetic friction. Of course, we have:

φ ≤ φs . So, there is a limiting position of the reaction force L; a right circular cone of vertex angle 2φ s is defined and said to be the cone of static friction, the line of action of possible reaction force L, at impending motion, being a generating line of the cone boundary.

Fig. 15 When motion occurs, the reaction force line lies to the surface of a cone of vertex angle 2φ k called the cone of kinetic friction.

1.6

TYPES OF CONSTRAINTS

Let B1 and B2 be two rigid bodies in contact, S be the portion of contact surface, Π be the tangent plane to B1 and B2 at any point p ∈ S . In the following study of particular constraints, we will only consider constraints without friction.

55

Statics

We recall the following. D

A constraint between B1 and B2 is frictionless or smooth if the stress is such that ∀p ∈ S : l p ⊥ Π .

In theoretical mechanics, the constraints between rigid bodies (in relative motion) are modeled. Let { o; i , j , k } be an orthonormal frame of reference such that the origin belongs to each constraint, but we emphasize this frame is not fixed in connected bodies. If the constraint dynam of B1 on B2 is defined by L(1→2) = X i + Y j + Z k , M 0(1→2) = L i + M j + N k

(1-17)

then we denote this contact dynam by X Y Z  L M N  .  o

In order to make clear the various types of constraints, it is necessary to define the following. D

The number of degrees of freedom of a constraint is the minimum number of possible independent motions of translation and rotation, relating to x, y and z-axes, of B2 with respect to B1 .

Let us introduce different smooth constraints. 1.6.1 Punctual Constraint D

Rigid bodies B1 and B2 are said to be constrained at a point or to be connected by a punctual constraint if, during their relative motion, a point of one body remains in a plane “fixed” in the other one.

For instance, two cylinders such that their respective generating lines are not parallel show a punctual constraint. Let us search for the number of degrees of freedom. Let q be a point of B2 remaining in a plane Π of B1 . Let oxyz be a coordinate system such that the contact point q is at point o and z-axis is normal to the plane Π. The allowed motions of B2 relative to B1 are of 5 types, namely: x and y-translations, rotations about ox, oy and oz, since in these cases, the point q definitely remains in Π. So, we say:

56

PR7

Chapter 1

The number of degrees of freedom of a punctual constraint is 5.

Fig. 16 The hypothesis of a strictly punctual constraint implies that the moment of the contact dynam is zero. In addition, the hypothesis of a smooth connection means that the resultant L(1→2) of the contact dynam is normal to Π (since l o ⊥ Π ). So, we state: PR8

A punctual constraint is characterized by a contact dynam of the following type 0 0 Z  0 0 0   o

and is schematized as follows:

Fig. 17 1.6.2 Rectilinear Constraint D

Rigid bodies B1 and B2 are connected by a rectilinear constraint if, during their relative motion, a straight line D of one body remains in a plane “fixed” in the other one.

Two cylinders with parallel generating lines, a cone connected with a plane, etc. are examples of rectilinear constraints. Let oxyz be a coordinate system such that o belongs to the straight line D of B2 , x-axis is collinear to D and z-axis is perpendicular to Π.

Statics

57

Fig. 18 The allowed motions of B2 relative to B1 are of 4 types, namely: x and y-translations, rotations about ox and oz, since in these cases D definitely remains in Π. So, we say: PR9

The number of degrees of freedom of a rectilinear constraint is 4.

The hypothesis of a smooth connection materialized by the x-axis means that the resultant of the contact dynam L(1→2) = ∫ l p dx ox

is along oz. In addition, the moment of contact dynam about o

M o(1→2) = ∫ op ∧ l p dx ox

is collinear to the y-axis. So, we have: PR10 A rectilinear constraint shows a contact dynam of the following type 0 0 Z  0 M 0   o

and is schematized as follows:

Fig. 19

58

Chapter 1

1.6.3

Annular-Linear Constraint

D

Rigid bodies B1 and B2 are connected by an annular-linear constraint if, during their relative motion, a point q of one body remains on a straight line D “fixed” in the other body.

A spherical surface tangent to a cylindrical surface in contact along some great circle is an example of such a constraint. The straight line D is perpendicular to the plane of this circle.

Fig. 20 Let oxyz be a coordinate system such that the x-axis is collinear to the straight line D of B1 , the point q ∈ B2 being the origin (at the center of the sphere). The allowed motions of B2 relative to B1 are of 4 types, namely: x-translation, rotations about ox, oy and oz, since in these cases q definitely remains on D. So, we say: PR11 The number of degrees of freedom of an annular-linear constraint is 4. The hypothesis of a smooth constraint means that the line of action of the stress l p is through the origin o and perpendicular to the x-axis at each point p of the connection. Therefore, the resultant L(1→2) of the contact dynam belongs to the plane oyz and the moment of this dynam is zero. So we say: PR12 An annular-linear constraint shows a contact dynam of the following type: 0 Y 0 0 

Z 0  o

and is schematized as follows:

Statics

59

Fig. 21

1.6.4 Ball-and-Socket Joint D

Rigid bodies B1 and B2 are connected by a ball-and-socket joint if, during their relative motion, a point q connected to one body remains at a point r “fixed” in the other one.

Two concentric spherical surfaces of same radius exemplify such a constraint. Let oxyz be a coordinate system with origin at q ≡ r .

Fig. 22 The allowed motions of B2 relative to B1 are of 3 types: rotations about x, y and z-axes, since in these cases q and r coincide. So, we say: D

The number of degrees of freedom of a ball-and-socket joint is 3.

The hypothesis of a smooth constraint means that the line of action of the stress l p is trough o at each point p of the connection.

60

Chapter 1

So, we obtain: PR14 A ball-and-socket joint shows a contact dynam of the following type X 0 

Z 0 0  o

Y

and is schematized as follows:

Fig. 23 1.6.5 Plane Support D

Rigid bodies B1 and B2 are connected by a plane support if, during their relative motion, a plane connected to one body coincides with a plane connected to the other one.

Examples.

Rocker, roller, ball, … supports

Fig. 24 We consider two (coinciding) plane surfaces of connection. Let oxyz be a coordinate system such that o belongs to the common plane and z-axis perpendicular to this plane. The allowed motions of B2 relative to B1 are of 3 types, namely: x and y-translations, rotation about oz, since in these cases the coincidence of planes connected to B1 and B2 respectively is preserved. So, we get: PR15 The number of degrees of freedom of a plane support is 3.

Statics

61

Fig. 25 The hypothesis of a smooth constraint means that the line of action of the stress l p is collinear to the z-axis at each point p of the connection. Therefore the resultant L(1→2) of the contact dynam is normal to the supporting plane surface and the moment of this dynam is tangent to the plane oxy. So, we say: PR16 A plane support shows a contact dynam of the following type 0 0 Z  L M 0   o

and is schematized as follows:

Fig. 26 1.6.6 Sliding Pivot D

Rigid bodies B1 and B2 are connected by a sliding pivot or a sliding hinge if, during their relative motion, a straight line “fixed” in one body coincides (but slides!) with a straight line “fixed” in the other body.

The tangent surfaces of connection are revolution cylindrical surfaces. Example. A collar free to move smoothly along a rod.

62

Chapter 1

Let oxyz be coordinate system such that x-axis and cylinder axis coincide.

Fig. 27 The allowed motions of B2 relative to B1 are of 2 types: x-translation, rotation about ox, since in these cases the mentioned straight lines coincide. Thus we say: PR17 The number of degrees of freedom of a sliding pivot is 2. The hypothesis of a smooth guide means that the line of action of the stress l p (the resultant of action dynam as well) is parallel to the plane oyz and is through the x-axis, at each point p of the contacting surface. The moment of l p about the x-axis is zero and thus the moment of contact dynam has no xcomponent. Thus we have: PR18 A sliding pivot shows a contact dynam of the following type 0 Y 0 M 

Z N  o

and is schematized as follows:

Fig. 28

Statics

63

1.6.7

Sliding Guide

D

Rigid bodies B1 and B2 are connected by a sliding guide if, during their relative motion, a plane “fixed” in one body coincides with a plane “fixed” in the other one and if a straight line of the first plane coincides with a straight line of the other plane.

The surfaces of connection are ruled (but not of revolution) with generating lines parallel to a common straight line D. Example.

Fig. 29 Let oxyz be a coordinate system such that the x-axis is along the common straight line D.

Fig. 30 The allowed motion of B2 relative to B1 is only of type: x-translation, since in this case the mentioned straight lines and planes of B1 and B2 coincide respectively. Thus we can state: PR19 The number of degrees of freedom of a sliding guide is 1. The hypothesis of a smooth connection means that the line of action of the stress l p (the resultant of action dynam as well) is parallel to the plane oyz. But, unlike the sliding pivot, this line of action is not (necessarily) through the x-axis and the moment of contact dynam has a component in the x-direction.

64

Chapter 1

So, we get: PR20 A sliding guide shows a contact dynam of the following type 0 Y L M 

Z N  o

and is schematized as follows:

Fig. 31 1.6.8 Screw Joint D

Rigid bodies B1 and B2 are connected by a screw joint if, during their relative motion, a straight line of one body, e.g. B2 , coincides with a straight line of the other body B1 called the helicoidal axis D of a circular helix bound to B1 . In addition, a point connected to B2 follows the previous circular helix.

The surfaces of connection are helicoid surfaces of which the points, in relative motion, follow helixes. Examples of such a connection are well-known (jack,…). Let oxyz be a coordinate system such that x-axis is along the straight line D.

Fig. 32 The allowed motion of B2 relative to B1 is of type: x-translation combined with a rotation about x-axis. The rotation is not independent of the translation because there is an obvious equality:

Statics

x=

h 2π

65

θ

where h is the pitch of the helix and θ the rotation angle. Consequently, we have: PR21 The number of degrees of freedom of a screw joint is 1. The inclination angle α of the circular helix is defined by tan α =

h

(=

2π R

x ) θR

where R is the circle radius. The relationship between translation and rotation is expressed as x = θ R tan α .

By considering a smooth connection, from the expression of l p it can be proved the following M o(1→2) . i = −

h 2π

i . L(1→2) .

Thus we express: PR22 A screw joint shows a contact dynam of the following type

Y X − h  X M  2π

Z  N o

and is schematized as follows:

Fig. 33 1.6.9 Pivot D

Rigid bodies B1 and B2 are connected by a pivot if, during their relative motion, two different points of one body continuously coincide respectively with two points “fixed” in the other body.

It is the most frequent connection. For example, all the mechanisms which transform motions of rotation have pivots.

66

Chapter 1

The following pin connection is an example of pivot.

Fig. 34 The surfaces of pivot connection are non cylindrical surfaces of revolution. Let a2 and b2 be points of B2 that respectively coincide with a1 and b1 of B1 . Let oxyz be a coordinate system such that the z-axis is the rotation axis (a1b1 ) ≡ (a 2 b2 ) .

Fig. 35

The allowed motion of B2 relative to B1 are only rotations about oz. So, we say: PR23 The number of degrees of freedom of a pivot is 1. The hypothesis of a smooth connection means that the line of action of the stress l p (the resultant of action dynam as well) is through the z-axis and thus the moment of l p about this axis is zero. So, we express: PR24 A pivot shows a contact dynam of the following type X Y Z  L M 0   o

and is schematized as follows:

Fig. 36

Statics

67

1.6.10 Embedding or Welded Joint D

Two rigid bodies B1 and B2 are welded or embedded if, during their relative motion, three different points of one body continuously coincide respectively with three points “fixed” in the other body.

No relative motion is allowed. The mechanical actions on one body are transmitted to the other body.

Fig. 37

Thus we have: PR25 The number of degrees of freedom of an embedding is zero. PR26 An embedding shows a contact dynam of the following type X Y Z  L M N  .  o

A welded connection shows the greatest degree of fixity that a connection can impose to a body. It is generally not schematized. To conclude, we give the plane representations of the essential previous connections. 1. Annular linear constraint

2. Ball-and-socket joint

68

Chapter 1

3. Plane support

4. Sliding pivot

5. Sliding guide

6. Screw joint

7. Pivot

Fig. 38 1.7 FREE-BODY DIAGRAM

In classical methods of statics, it is essential to determine all the forces required to prevent a body or a system of bodies from moving. In equilibrium studies, any action cannot be forgotten. In order to take account of all given constraint forces, we must introduce the important method of isolation of body or system of bodies to consider. This so-called free body diagram method consists in choosing a body or bodies to be isolated from surroundings. The isolation choice is a very important stage of the method! Next, once a decision is taken about which bodies are to be considered then these connected bodies are viewed as a single body enclosed by a boundary materialized by the isolation. Finally, all the forces acting on the “isolated body” (that is, the forces applied by the removed contacting and attracting bodies) must be represented.

Statics

69

Example. Free-body diagram of a plane frame.

We consider the following structural elements that altogether form a rigid framework, that is considered as a single rigid body.

Fig. 39

The pivot at point a exerts, on the frame, a reaction force L with - a vertical component that must be directed up to be opposed to weights W1 and W 2 , - a horizontal component that must be directed to the right to be opposed to the tension T.

Fig. 40

It is advisable to show the chosen coordinate axes. Remark 1. In 3-dimensional problems, the free-body diagram can be shown in various coordinate planes. Remark 2. If the direction of a constraint force is unknown, then it is arbitrarily chosen, but the right direction is later deduced from calculus. Example. Draw the free-body diagram for a horizontal pole of mass m maintained in equilibrium under the action of two forces exerted by two cables fixed at points b and c of the pole; a ball-and-socket joint prevents the motion at point a.

Fig. 41

70

Chapter 1

In the free-body diagram, the x- and z-directions of La are arbitrarily chosen, but the ydirection is certainly correct. Remark 3. In the study of systems of interconnected rigid bodies, it is necessary to draw a free-body diagram for each body if we want to bring the internal forces of constraint into prominence. It is essential to make an appraisal of all the forces applied to each solid and it is highly advisable to be careful of various types of constraints present in the system.

The choice of the directional sense of unknown forces of constraint is not arbitrary since it must be consistent with the connections of rigid bodies. For example, we consider two rods pinned at both of their ends.

Fig. 42

2. METHOD OF VIRTUAL WORK One of the essential goals of the virtual work method is to establish the equilibrium positions of a system of particles (or of a particle) by introducing arbitrary vectors a priori; such vectors will next be called either virtual displacements 1 or virtual velocities according to the considered context. The method of virtual work lets determining the unknown forces that are necessary and sufficient to maintain the equilibrium state of systems, it permits to know a lot about stability of systems in equilibrium too. In many studies, the virtual work method leads to faster results than the classical method, because its main advantage consists in eliminating the part of unknown reaction forces. Both methods contribute to the understanding of the mechanics of systems, each having its more or less suitable application domain; however their simultaneous use can be profitable. An essential advantage of the virtual work method lies in the fact that the system as a whole is viewed, there is no dismemberment of the system and the analysis of the forces that must be considered is more reduced. So, in the case of interconnected systems of bodies for instance, the method of virtual work will be more advisable.

1

This concept was introduced by Jean Bernoulli in the 18th century (under another formulation obviously).

71

Statics

2.1

NUMBER OF DEGREES OF FREDOM – GENERALIZED COORDINATES

First, we recall the following definition. D

A system of particles (a particle in particular) is subject to constraints if it cannot freely move. In other words: if the motion of particles is constrained.

Constraints can be specified in different ways, either by a restriction on the coordinates, or on the velocities or both of them. Later, we will meet several types of constraints.

2.1.1 Number of Degrees of Freedom D

The number of degrees of freedom of a mechanical system is the minimum number of coordinates1 (or parameters) necessary to describe the motion of the particles of the system.

It is denoted by n. Example 1. We consider a particle moving on a surface whose equation (of constraint) is the following relation between Cartesian coordinates and time: f ( x, y , z ; t ) = 0 .

The particle has two degrees of freedom since two independent coordinates are required to specify the position of the particle, at any instant, on the (possibly moving) surface. In other words, two parameters are necessary and sufficient to determine the particle position at any instant. Example 2. constraint):

A particle moving along a curve defined by the following equations (of f ( x, y , z ; t ) = 0

g ( x , y , z; t ) = 0

has one degree of freedom. Only one parameter is necessary and sufficient to specify, at any instant, the position of the particle on the (possibly moving) curve. Example 3. Evidently, a particle free to move in space has three degrees of freedom, since three independent coordinates are necessary and sufficient to specify the position of the particle at any instant.

So, in a general way, the number of degrees of freedom of a particle in space is n = 3−m 1

That is: Independent coordinates

(0 ≤ m ≤ 2)

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Chapter 1

where m is the number of constraint equations of type: f ( x, y , z ; t ) = 0 .

Example 4. A rod supported by two springs has two degrees of freedom, since each spring which is assumed to oscillate in a vertical line shows one degree of freedom.

In the first two examples of a particle subject to one or two constraints, the Cartesian coordinates of the particle are not independent and thus are overabundant coordinates. We say: D

Overabundant coordinates are called primitive coordinates.

They are denoted by u j .

2.1.2 Generalized Coordinates D

The generalized coordinates 1 are the independent coordinates required to specify the position of each particle of a system.

The generalized coordinates let describe the motion of a mechanical system. They are as many as degrees of freedom. Generalized coordinates are denoted by qj

( j = 1,..., n ).

Remark 1. Generalized coordinates can just as well represent distance coordinates as angles; we will see examples of this. Remark 2. The choice of generalized coordinates can be multiple; however these coordinates will be chosen so as to simplify calculations. D

The space of generalized coordinates is called the configuration space.

Every configuration space of a mechanical system with n degrees of freedom has n dimensions. A point (q1 ,..., q n ) of the configuration space determines the “position” of the mechanical system at a given instant; in other words this (descriptive) point gives the configuration of the system. The evolution of the mechanical system in the course of time makes describe to (q1 ,..., q n ) a curve in the configuration space and we say:

1

Also called Lagrange coordinates.

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Statics

D

The generalized trajectory of a mechanical system is the curve followed by the point (q1 ,..., q n ) in the configuration space.

D

The generalized velocity associated with the generalized coordinate q k is the time derivative q
k .

For example, the generalized velocity associated with an angle φ is the angular velocity φ
. Example 5. Given a free particle moving in 3-dimensional space, the three chosen generalized coordinates are either the Cartesian coordinates, or the three spherical coordinates or the three cylindrical coordinates, the choice depending on the geometrical context (configuration!).

The configuration space is the 3-dimensional space R 3 . Example 6. A particle moving along an ellipse defined by x = a cos θ ,

y = b sin θ ,

z = 0,

( a, b ∈ R )

has one degree of freedom, the generalized coordinate being the polar angle θ. The configuration space is one-dimensional. Example 7. A particle moving on a spherical surface has two degrees of freedom. So is the pendulum bob constrained to move on a spherical surface of radius R.

The bob is free to swing through the entire solid angle about a point, it is located by two generalized coordinates which are the following spherical coordinates: the colatitude θ and the longitude φ. The overabundant Cartesian coordinates are related to generalized coordinates as follows: x = R sin θ cos φ ,

y = R sin θ sin φ ,

z = R cos θ .

The configuration space is a 2-dimensional space: the sphere S 2 . Example 8. A plane double pendulum is an example of mechanical system which has two degrees of freedom.

Generalized coordinates are the respective angles that measure the deviations of each of two weightless rigid rods from the vertical. The configuration space is a 2-dimensional space: the torus T 2 = S 1 × S 1 , Cartesian product of two circles. Example 9. A system of two particles linked together by a rigid weightless rod and moving in space has five degrees of freedom.

Generalized coordinates are the three coordinates of the center of mass of the system and two angles determining the inclination of the rod in space. The configuration space is a 5-dimensional space.

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Chapter 1

In another way, a system of two particles in space has 3 N = 6 degrees of freedom a priori, but there is a constraint expressing that the distance between particles remains constant. This reduces to 5 the number of degrees of freedom. This last example suggests that constraints reduce the number of degrees of freedom. In a general way, a set of N particles in R 3 subject to constraints defined by m equations f j ( x, y, z ) = 0 has n = 3N − m degrees of freedom. Exercise 10. A rigid body B is an obvious example of system of particles subject to constraints, since the mutual distances of all pairs of particles remain constant, that is:

∀a, b ∈ B : ab = ( xb − x a ) 2 + ( yb − y a ) 2 + ( z b − z a ) 2 = k

( ∈ R+ ).

It would be impossible to exactly describe the motion of any system of N ( ≥ 3 ) points if it was not a rigid body. In fact, it is sufficient to know the positions of three particles of a rigid body to know the ones of all particles of the solid. The rigid body would have nine (3 × 3 ) degrees of freedom a priori. But there are three constraints that are the three mutual distances ( 12 N ( N − 1) = 3 ). Therefore, every position of a free solid is completely determined by six parameters or generalized coordinates ( n = 9 − 3 = 6 ).

We can choose as generalized coordinates: the three coordinates of the center of mass and three parameters describing the body inclination for instance by the three Euler angles. Example 11. What is the configuration space of a set of two particles following a curve defined by { ( x, y, z ) ∈ R 3 : f ( x, y, z ) = 0, g ( x, y, z ) = 0 }?

A priori two particles of R 3 have 3 N = 6 degrees of freedom. But the constraint equations are m = 4 in number because the coordinates of particles ( x1 , y1 , z1 ) and ( x 2 , y 2 , z 2 ) must verify the following equations: f ( x1 , y1 , z1 ) = 0 ,

f ( x2 , y2 , z 2 ) = 0 ,

g ( x1 , y1 , z1 ) = 0 ,

g ( x2 , y 2 , z 2 ) = 0 .

The configuration space has thus 3 N − 4 = 2 dimensions, the system of two particles has two degrees of freedom. More quickly, the curvilinear coordinates of every particle allow seeing that the configuration space is R 2 such that q1 and q 2 are the curvilinear coordinates of the respective particles. Example 12. A bike moving along a road has seven degrees of freedom. A priori, the frame has six degrees of freedom, the back wheel has one, the handlebar has one and the rotation of the front wheel leads to one in addition. But the wheels must be in contact with the road and thus there are two constraints, namely: the distance between each wheel center and the corresponding contact point on the road remains constant.

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75

Example 13. Given two fixed points o and o′ , a thin rigid rod ab can move in space so that the distances oa and o′b remain constant. Find the number of degrees of freedom for the rod ab if oabo ′ is such that: (i) it remains in a fixed plane, (ii) it remains coplanar, (iii) it does not remain coplanar.

For (i), there is one degree of freedom since each of points a and b has two degrees of freedom; but there are three constraint equations expressing that the distances oa , ab and o′b are constant. For (ii), there are two degrees of freedom since the plane can rotate about the axis oo ′ (angle of rotation!) and thus there is a supplementary degree of freedom with respect to (i). For (iii), there are three degrees of freedom since each of points a and b has three degrees of freedom; but there are three constraint equations.

2.1.3

Types of Constraints

We can say: D

A system of N particles is subject to a constraint if the coordinates of positions of particles or their first derivatives or both of them are not independent.

Let r1 ,..., rN be position vectors of particles of a system. D

A constraint is bilateral if the positions or velocities of particles are related by an equation;1 it is unilateral if the relation is an inequation.

The previous examples show bilateral constraints. A particle that must stay inside a sphere of radius R is an example of unilateral constraint such that the inequation is of type r − R < 0,

where r is the radial distance of the particle. Another example of unilateral constraint is given by two particles a and b, of respective Cartesian coordinates ( x a , y a , z a ) and ( xb , yb , z b ) , which are connected by an unstretchable string of length l. The inequation of constraint is ( xb − x a ) 2 + ( y b − y a ) 2 + ( z b − z a ) 2 ≤ l 2 .

1

For instance :

f (r1 ,..., rN , r
1 ,..., r
N ; t ) = 0 .

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Chapter 1

D
A constraint is scleronomic if it is not explicitly dependent on time: ∂f = 0. ∂t

(1-18)

In the opposite case, the constraint is said to be rheonomic. All the above examples involve scleronomic constraints. If a pearl follows a ring rotating about a diameter with an angular velocity ω, then the constraint is rheonomic since this explicitly depends on the angle ω t of rotation of the ring. The simple pendulum with a mass suspended from a moving point gives another example of rheonomic constraint. In the same manner, if a pendulum is such that the mass suspended from a fixed point by a string of not constant length, then the constraint is rheonomic. A crane illustrates this case. D
A constraint is holonomic if it is expressed by equations of type

f (r1 ,..., rN ; t ) = 0

(1-19)

relating only the coordinates and possibly t. In the opposite case, it is said to be nonholonomic. Example 14. A cylinder of radius R is rolling down an inclined plane (without slipping).

If s represents the distance covered by the cylinder and θ the corresponding angle of rotation about the cylinder axis, then the constraint is expressed by a relation between velocities s
= Rθ
,

and the constraint is holonomic since this equation is immediately integrated: s − Rθ = k

( k ∈ R ).

Example 15. A sphere S of radius R is rolling without slipping on a fixed plane Π.

Let xc , y c , z c be (primitive) overabundant coordinates of the center c of the sphere, in a frame of reference = { o;1x ,1 y ,1z } such that the axes of coordinates x and y are located in Π.

R

Let { c;1X ,1Y ,1Z } be a frame “fixed” in the sphere. Let φ ,θ ,ψ be Euler angles that specify the position of the frame { c;1X ,1Y ,1Z } with respect to R .1 The constraint expressing the contact between the sphere and the plane is such that zc − R = 0 . This is a scleronomic and holonomic constraint. 1

The Euler angles are introduced in first courses of mechanics and are clearly shown in Fig. 43 of Sec. 2.2.3.

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Statics

Since the sphere is rolling without slipping on the plane (sufficient friction!), there are other equations of constraint. We express that the slipping velocity at the contact point i is zero: v i = M i (VSΠ ) = v c + ic ∧ ω 1x

1y

1z

= x
c 1x + y
c 1 y + z
c 1z + 0

0

R = 0.

ωx ω y ωz So, we obtain:

x
c − Rω y = 0 , y
c + Rω x = 0 , z
c = 0 . The last equation is the above mentioned holonomic constraint. We can express the two first equations from Euler angles. Let 1N be the unit vector along the line of nodes. We know that ω = φ
1z + θ
1N + ψ
1Z where the angular velocities φ
, θ
and ψ
are respectively called the precession velocity, the nutation velocity and the spin rotation (see also 2.2.3b). We have: ⇔ ⇔

v i = v c + R 1z ∧ (φ
1z + θ
1N + ψ
1Z ) = 0 v + Rθ
1 ∧ 1 + Rψ
1 ∧ 1 = 0 c

z

N

z

Z

v c + Rθ
1u + Rψ
sin θ 1N = 0

where 1u is obviously in the plane { i ; 1x ,1 y } and perpendicular to 1N . Projections onto the x-axis and y-axis lead to the respective equations x
c − R (θ
sin φ − ψ
sin θ cos φ ) = 0 , y
+ R (θ
cos φ + ψ
sin θ sin φ ) = 0 , c

which are not integrable. So they represent two nonholonomic constraints. The six overabundant coordinates verify three equations of constraint and the sphere has thus three degrees of freedom. In the preceding, we have assumed that the friction is sufficient to prevent any slipping. We note that for a frictionless contact the sphere has obviously five degrees of freedom. Remark. We only consider constraints without dissipation of mechanical energy, every motion being reversible. The study of constraints which imply a dissipation of mechanical energy (heat!) is important but the resultant irreversible motions are not here introduced.

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Chapter 1

2.2

VIRTUAL DISPLACEMENTS AND VIRTUAL VELOCITIES

We are going to introduce the notion of virtual displacement. Such an a priori arbitrary vector will be particularized so as not to take unknown forces of constraint into account. Let E3 be a Euclidean vector space, E 3 be the associated point space (or affine space), Ω be an open of E3 , = { o; e1 , e 2 , e3 } be an orthonormal frame of reference.

R

2.2.1

Generalized Coordinates

(i) For a free particle p in E 3 , it is often useful to consider coordinates different from Cartesian coordinates; so it is better to use three coordinates suited to the geometrical configuration of the problem (spherical, cylindrical,…). Let U be an open of R 3 , q = (q 1 , q 2 , q 3 ) be an element of U composed of generalized coordinates of p.

The position vector of p in R is defined by: x = r (q (t )) = r (q 1 (t ), q 2 (t ), q 3 (t )) .

(ii)

We consider a particle p belonging to a surface.

Let U be an open of R 2 , q = (q 1 , q 2 ) be an element of U, pair of generalized coordinates of p.

A vector function of class C 2 : r : U → Ω : q a r (q ) = r ( q 1 , q 2 )

is a parametric representation of the surface of constraint. This parametric representation is supposed regular, that is: ∀(q 1 , q 2 ) ∈ U :

where

∂r ∂q

1

and

∂r ∂q 2

∂r ∂q

1

∧

∂r ∂q 2

≠0

are linearly independent vectors and tangent to the surface; the first is

tangent to a curve such that q 2 is constant and the second is tangent to a curve such that q1 is constant. Thus there is a tangent plane at each point of the surface. The position vector of p in R is defined by: x = r (q (t )) = r (q 1 (t ), q 2 (t )) .

Statics

(iii)

79

We consider a particle p belonging to a curve.

Let U be an open of R, q be an element of U, generalized coordinate of p. A vector function of class C 2 r : U → Ω : q a r (q)

is a parametric representation of the curve of constraint. This parametric representation is supposed regular, that is there exists, at each point of the ∂r ⋅ curve, a tangent vector ∂q The position vector of p in R is defined by: x = r (q (t )) .

We consider a free rigid body B in E 3 .

(iv)

Its position is located by six generalized coordinates q j , for instance the three coordinates x , y , z of the center of mass and the three Euler angles φ ,θ ,ψ . The position of any point p of B in R is determined by the mapping (supposed of class C 2 ): r : U (⊂ R 6 ) → E3 : ( x , y , z , φ ,θ ,ψ ) a r ( x , y , z , φ ,θ ,ψ ) . (v) We consider a mechanical system S made up of k1 rigid bodies, of k 2 rectilinear rigid bodies and of k 3 particles. If all these elements are supposed free, then the system has n = 6k1 + 5k 2 + 3k 3 degrees of freedom. The position of any point of S in R is defined by the mapping r : U (⊂ R n ) → E3 : (q1 ,..., q n ) a r (q1 ,..., q n ) supposed of class C 2 . Remark. For every mechanical system in a moving frame of reference and for any system with rheonomic constraints too, it is necessary to introduce the variable t in addition to the generalized coordinates.

2.2.2

Definition and Expressions of Virtual Displacements

The following definition concerns general mechanical systems, free or constrained. Let q 1 ,..., q n be generalized coordinates of a mechanical system S, U be an open of R n , I be an open of R.

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Chapter 1

We denote q = (q 1 ,..., q n ) ∈ U .

Each particle is located by its position vector x defined by a function of class C 2 : r : I × U → Ω : (t , q ) a r (t , q ) .

The components of the position vector x of a mass point p in R are x1 = r 1 (t , q1 ,..., q n ) , x 2 = r 2 (t , q1 ,..., q n ) , x 3 = r 3 (t , q1 ,..., q n ) . Notation. According to usage, the vector function r will be identified with x.

The virtual work method consists in arbitrarily associating with p a neighboring point p ′. How? By introducing the notion of virtual displacement, that is an arbitrary and infinitesimal vector pp ′ resulting from arbitrary increments given to the only generalized coordinates. This is going to be made more explicit. We denote this vector by

δx

(or δ r ).

Arbitrary increments δ q j of generalized coordinates lead to arbitrary increments of the components of x, namely: n

δ xi = ∑

j =1

∂r i δqj j ∂q

i = 1,2,3.

So, the expression of a virtual displacement of p is 3

3

∂r i δ q j ) ei j q j =1 ∂ n

δ x = ∑ δ x i e i = ∑ (∑ i =1

i =1

n

3

j =1

i =1

= ∑ (∑

∂r e )δ q j j i ∂q i

and thus


n

δx=∑

j =1

∂r δqj. j ∂q

(1-20)

We set: D  A virtual displacement 1 of a point p is a vector which associates with p a neighboring point from the variation of the only generalized coordinates.

1

The concept of virtual displacement was introduced by Jean Bernoulli in 1717.

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81

A virtual displacement is a comparison between two neighboring positions at a same instant. It is a vector that not necessarily represents a displacement! A virtual displacement is performed at a fixed time. It is a fictitious displacement that would have been perhaps preferentially called “Bernoulli-Lagrange vector” although the vector notion was introduced later by Hamilton. A fixed time is imposed at any virtual displacement. This requirement will be ever-present in mind. We emphasize that a virtual displacement of p is denoted by δ x and not by dx, this last being a (real) infinitesimal displacement. Unlike virtual displacements, (real) displacements take place during a time interval for which the given applied forces and constraint forces change. We note that δ, which acts on generalized coordinates, behaves as a first order differential operator. In a general way, the notion of virtual displacement has nothing to do with the one of (real) displacement. Moreover, there is an infinite number of virtual displacements of p. However, it is not impossible that a virtual displacement is a real displacement. But, in statics, we notice that the only possible displacements are… virtual displacements! In the case of systems with constraints, the virtual displacement must be compatible with the constraint(s) (also called consistent with the constraint(s)). This notion will be specified in the following. For example, these virtual displacements are perpendicular to external reactive forces (e.g. tangent to smooth connections), preserve the rigidity of solids, etc. For a pivot, the virtual displacement of a point of one body is identical to the one of the contact point of the other body. Remark. In this chapter, most of the problems are solved by using generalized coordinates; the use of (overabundant) primitive coordinates would amount to define virtual displacements by giving arbitrary increments to primitive coordinates. This will later be made.

2.2.3

Virtual Velocity and Examples

According to usage, every mechanical system S (point, rigid body, system of rigid bodies) is identified with the part of space occupied by the system at a given time. 2.2.3a Free Particle In the case of a unconstrained motion of a particle p, the consideration of three primitive coordinates goes with the number of degrees of freedom. These coordinates are generalized coordinates. The particle position in a (continuously) moving frame of E 3 depends on time t and generalized coordinates q j . It is determined by the vector function

R 4 → E3 : (t , q1 , q 2 , q 3 ) a r (t , q1 , q 2 , q 3 ) .

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Chapter 1

Instead of the notion of virtual displacement, we can consider the concept of virtual velocity. We recall that the expression of the velocity of p is vp =

∂r ∂r 1 ∂r 2 ∂r 3 + q& + 2 q& + 3 q& . ∂t ∂q 1 ∂q ∂q

As for virtual displacements, we consider a fixed time. So, we can define the notion of virtual velocity of p denoted by v ∗p : D
A virtual velocity of p is an arbitrary vector 3

v ∗p = ∑



j =1

∂r ∗ j q& ∂q j

(1-21a)

where the various q& ∗ j are arbitrary reals. Notation. The arbitrary reals q& ∗ j will be denoted by q ∗ j and thus a virtual velocity of p is expressed as 3 ∂r ∗ j v ∗p = ∑  q . (1-21b) j q ∂ j =1 Remark.

We specify if the coordinates q j have the length dimensions, then the

corresponding q ∗ j have the velocity dimensions. In the same manner, if the coordinates q j are angles, then the corresponding q ∗ j have the angular velocity dimensions.

2.2.3b Unconstrained Rigid Body

The consideration of six primitive coordinates goes with the number of degrees of freedom. These coordinates are the generalized coordinates q 1 ,..., q 6 . The position of a rigid body B in an (inertial) frame of reference determined by a function of six generalized coordinates

R = { o;1 ,1 ,1 } is x

y

z

R 6 → E3 : (q1 ,..., q 6 ) a r (q1 ,..., q 6 ) .

In some problems, we know that the time t explicitly appears in addition. To introduce the notions of virtual displacement and virtual velocity, we recall that the position of the rigid body B is located by six generalized coordinates, which are the three coordinates x, y, z of a point O fixed in the moving rigid body B and the Euler angles φ, θ and ψ which specify the position of the frame {O ;1X ,1Y ,1Z }, “fixed” in the rigid body, with respect to the frame of reference R.

The reader will use the following representation of frames.

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Statics

Fig. 43

The position of any p ∈ B is defined by R 6 → E3 : ( x, y, z, φ ,θ ,ψ) a r ( x, y, z , φ ,θ ,ψ ) . Every position vector of p is expressed as: r = oO + Op = x 1x + y 1 y + z 1z + X 1X + Y 1Y + Z 1Z . We immediately have: ∂r ∂q 1

=

∂r = 1x , ∂x

∂r ∂q 2

=

∂r = 1y , ∂y

∂r ∂q 3

=

∂r = 1z . ∂z

We are going to calculate the other partial derivative vectors. We have:

1X = cosψ 1N + sin ψ 1 y′′ = cosψ (cos φ 1x + sin φ 1y ) + sin ψ (cosθ 1y′ + sin θ 1z )

but 1 y′ = − sin φ 1x + cos φ 1 y

thus X 1X = X (cosψ cos φ − sinψ cos θ sin φ ) 1x + X (cosψ sin φ + sinψ cos θ cos φ ) 1 y + X sinψ sin θ 1z .

In the same manner: 1Y = − sin ψ 1N + cosψ 1 y′′

⇒

Y 1Y = −Y (sinψ cos φ + cosψ cos θ sin φ ) 1x + Y (− sin ψ sin φ + cosψ cos θ cos φ ) 1 y + Y cosψ sin θ 1z

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Chapter 1

and also 1Z = cosθ 1z − sin θ 1 y Z 1Z = Z sin θ sin φ 1x − Z sin θ cos φ 1 y + Z cos θ 1z .

⇒

Then, we immediately obtain: ∂r ∂r = = X [(− cosψ sin φ − sin ψ cos θ cos φ )1x + (cosψ cos φ − sin ψ cos θ sin φ )1 y ] ∂q 4 ∂φ + Y [(sinψ sin φ − cosψ cos θ cos φ )1x + (− sin ψ cos φ − cosψ cos θ sin φ )1y ] + Z [(sin θ cos φ 1x + sin θ sin φ 1 y ]

= 1z ∧ Op and similarly: ∂r ∂r = = 1N ∧ Op , 5 ∂θ ∂q ∂r ∂r = = 1Z ∧ Op . 6 ∂ψ ∂q

In conclusion, in the case of a free rigid body, the six vectors

∂r = 1y , ∂y

∂r = 1x , ∂x

∂r ∂q j

are determined by:

∂r = 1z , ∂z

(1-22)

∂r = 1z ∧ Op , ∂φ

∂r = 1N ∧ Op , ∂θ

∂r = 1Z ∧ Op . ∂ψ

By taking account of the preceding, a virtual displacement is expressed as follows: 6

δx=∑

j =1

∂r δ q j. j ∂q

We can consider virtual velocities instead of virtual displacements. By considering a fixed time, we define: D

A virtual velocity of p ∈ B is an arbitrary vector 6

v ∗p = ∑

j =1

∂r ∗ j q ∂q j

where the various q ∗ j are arbitrary reals (also denoted q& ∗ j ) and r may explicitly depend on t.

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Statics

A field of virtual velocities on B is defined by the mapping

D

6

B → E3 : p a v ∗p = ∑

j =1

∂r ∗ j q . ∂q j

Example. We consider a rigid body B which rotates about a fixed point o. Let

R = { o;1 ,1 ,1 } be a fixed frame of reference, x

y

z

{ o;1X ,1Y ,1Z } be a frame “fixed” in the moving body.

The position of the frame “fixed” in the body is determined with respect to the frame of reference R by the Euler angles. The position vector of p ∈ B being constant in the frame “fixed” in the solid, the field of velocities of p is defined by: ∀p ∈ B : v p = ω ∧ op .

Since

ω = φ& 1z + θ& 1N + ψ& 1Z ,

we have:

(1-23)

v p = φ& 1z ∧ op + θ& 1N ∧ op + ψ& 1Z ∧ op .

But the followings ∂r = 1z ∧ op , ∂φ

∂r = 1N ∧ op , ∂θ

∂r = 1Z ∧ op ∂ψ

imply ∂r & ∂r ∂r v p = φ& +θ + ψ& ⋅ ∂φ ∂θ ∂ψ

From this expression, we introduce the notion of virtual velocity as follows. Given arbitrary reals φ ∗ ,θ ∗ and ψ ∗ , we consider v ∗p = φ ∗

∂r ∂r ∂r +θ ∗ +ψ ∗ ∂φ ∂θ ∂ψ

= (φ ∗ 1z + θ ∗ 1N + ψ ∗ 1Z ) ∧ op .

By defining D

The vector of virtual rotation of B is

ω∗ = φ ∗ 1z + θ ∗ 1N + ψ ∗ 1Z ,

(1-24)

the virtual velocity is expressed as v ∗p = ω∗ ∧ op .

(1-25)

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Chapter 1

2.2.3c Systems of Free Bodies

We consider a mechanical system S made up of free bodies. If there are k1 solids, k 2 rectilinear solids and k 3 particles, we know that the system has n = 6k1 + 5k 2 + 3k 3 degrees of freedom. Let p be a mass point of S. By considering a fixed time, we define: D

A virtual velocity of p is an arbitrary vector n

v ∗p = ∑

j =1

∂r ∗ j q ∂q j

where the various q ∗ j are arbitrary reals. A field of virtual velocities on S is defined by the mapping: n

S → E3 : p a v ∗p = ∑

j =1

∂r ∗ j q . ∂q j

2.2.3d Particle Subject to a Constraint

For instance, we deal with a particle moving along a circle which rotates about a vertical diameter. In more concrete terms, we consider a small ring (particle p) frictionless running along a circular rail of radius R and center o. The rails rotates about a vertical diameter with a constant angular velocity ω.

Fig. 44 Let { o;1x ,1 y ,1z } be a fixed frame of reference such that 1z is collinear to the vertical diameter.

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87

Let θ be the (counterclockwise) angle (1z , op) , 1θ be the unit vector tangent to the circle, at p, of positive sense (counterclockwise), 1φ be the unit vector perpendicular to the plane of the circle and such that (1z ,1θ ,1φ ) is right ordered, 1r be the unit vector in the direction of r = op, φ be the positive angle (1x ,1φ ) . We assume that 1x and 1φ coincide at t = 0 , that is

φ = ω t. The small ring has one degree of freedom and we can choose the angle θ giving the position of p as generalized coordinate. By definition, we know that the velocity of transport of p is the absolute velocity of p considered as fixed in the moving frame. The expression of this vector is obviously:

Rφ& sin θ 1φ . Hence, the absolute velocity of p is (with φ& = ω ): v p = Rθ& 1θ + Rω sin θ 1φ . This vector is obviously not tangent to the circular rail. A (real) infinitesimal displacement of p is dop = R dθ 1θ + R dφ sin θ 1φ .

We are going to search for virtual displacements and virtual velocities. First of all let us introduce primitive coordinates, namely the radial distance r, the colatitudeθ (or nutation angle) and the longitude φ (or precession angle). These coordinates are evidently not independent since there are equations of constraint:

φ =ωt .

r=R

In this example, the Eulerian spin angle (or angle of proper motion previously denoted ψ ) is zero and 1φ is directed along the line of nodes. So, in a general way, if we give arbitrary increments to primitive coordinates, a virtual velocity of p is written a priori as follows: v ∗p =

∂r ∗ ∂r ∗ ∂r ∗ r + θ + φ ∂r ∂θ ∂φ

and a virtual displacement as follows:

δx=

∂r ∂r ∂r δ r + δθ + δφ ∂r ∂θ ∂φ

where r ∗ ,θ ∗ , φ ∗ , δ r , δθ and δφ are arbitrary reals.

88

Chapter 1

Since the position vector op of p is indiscriminately written:

x = r = r1r , from (1-22) we deduce the followings: ∂r = 1r , ∂r ∂r = 1φ ∧ r 1r = r 1θ , ∂θ ∂r = 1z ∧ r 1r = r sin θ 1φ . ∂φ

A priori and in an general manner we can thus write:

v ∗p = r ∗ 1r + rθ ∗ 1θ + r sin θ φ ∗ 1φ and

δ x = δ r 1r + rδ θ 1θ + r sin θ δ φ 1φ where the respective terms of this sum are called the radial, meridian and longitudinal terms. In this problem with one degree of freedom for which θ is the generalized coordinate, any virtual displacement or virtual velocity consistent with the constraint has the direction of 1θ . These vectors are respectively written:

δ x = R δ θ 1θ and

v ∗p = Rθ ∗ 1θ . We also say that these virtual displacement and virtual velocity are compatible with the constraint equations r = R and φ = ω t . Remark 1. The virtual displacements (resp. virtual velocities) consistent with the constraint are obtained from virtual displacements (resp. virtual velocities) written with the help of primitive coordinates; that is:

δ x = δ r 1r + rδθ 1θ + r sin θ δφ 1φ , v ∗p = r ∗ 1r + rθ ∗ 1θ + r sin θ φ ∗ 1φ . How? Immediately if we derive the two constraint equations r = R and φ = ω t , the time being fixed! This leads to

δ r = 0,

δφ = 0

r∗ = 0 ,

φ∗ = 0.

or

So, virtual displacements and virtual velocities consistent with the constraint are really expressed as δ x = R δθ 1θ , v ∗p = R θ ∗ 1θ .

Statics

89

Remark 2. From the previous classical example, we really notice that an infinitesimal displacement dx (resp. velocity v p ) of a particle p is not necessarily a virtual displacement (resp. virtual velocity) consistent with the constraint. This example confirms the fact that the set of virtual displacements (resp. virtual velocities) does not necessarily contain the real displacements (resp. velocities). Indeed, the (real) displacement Rθ
1θ + Rω sin θ 1φ

is not of the following type: Rθ ∗ 1θ .

2.2.3e Systems of Constrained Bodies

We consider a system S of rigid bodies. Primitive coordinates u 1 ,..., u N give the position of S. So, the position of some p ∈ S is determined by a vector function R N +1 → E3 : (t ; u 1 ,..., u N )  r (t ; u 1 ,..., u N ) of class C 2 .

If the system S has n degrees of freedom, we recall that the expressions of any virtual displacement or any virtual velocity are respectively: n

δx=∑

j =1 n

v ∗p = ∑

j =1

∂r δqj, j ∂q ∂r ∗ j q ∂q j

where the various q j are the n generalized coordinates. The existence of constraints between solids implies that the N primitive coordinates u i are not independent, they are not generalized coordinates. (i)

First, we consider a system S subject to a holonomic constraint of equation: f (t , u 1 ,..., u N ) = 0 ,

the function f being supposed of class C 2 . So we have N ∂f ∂f + ∑ i u
i = 0 . ∂t i =1 ∂u

Before introducing virtual displacements or virtual velocities consistent with constraint equations, we consider again the example of a small ring following a circle which rotates about a vertical diameter (see Sect. 2.2.3d).

90

Chapter 1

The constraint equations being g ≡) φ − ω t = 0 ,

f ≡) r − R = 0 ,

we have immediately the following implications: ∂f ∗ r =0 ∂r

⇒

r∗ = 0 ,

∂g ∗ φ =0 ∂φ

⇒

φ ∗ = 0.

This example shows that the expressions of general virtual displacements (and virtual velocities) are made simpler since constraint equations exist, these vectors being written:

δ x = R δ θ 1θ and v ∗p = R θ ∗ 1θ .

In the case of a holonomic constraint defined by f (t , u 1 ,..., u N ) = 0 ,

we say: D

Virtual displacements (resp. virtual velocities) on S are consistent or compatible with a constraint of Eq. f = 0 if reals δ u i (resp. u ∗i ) verify the following: N

∂f

N

∑ ∂u i δ u i = 0

[ resp.

i =1

∂f

∑ ∂u i u ∗i = 0 ]. i =1

If one primitive coordinate, for instance u N , is expressed in the following explicit form: u N = ϕ (t , u 1 ,..., u N −1 ) ,

(1-26)

where ϕ is of class C 2 , then we express: D

Virtual displacements (resp. virtual velocities) on S are consistent (or compatible) with the constraint of Eq. (1-26) if:

δuN =

N −1

∑ i =1

∂ϕ δ ui i ∂u

( u ∗N =

N −1

∑ i =1

∂ϕ ∗i u ). ∂u i

(1-27)

So, for S, given holonomic constraints, the field of virtual velocities is N

S → E3 : p
v ∗p = ∑ i =1

(ii)

∂r ∗i u . ∂u i

Second, we consider a system S subject to nonholonomic constraints.

We assume that S is subject to the nonholonomic constraint of equation N

∑ ai (t , u 1 ,..., u N ) u
i + b(t , u1 ,..., u N ) = 0 i =1

ai , b ∈ R .

(1-28)

91

Statics D

Virtual displacements (resp. virtual velocities) on S are consistent with a constraint of Eq. (1-28) if reals δ u i (resp. u ∗i ) verify the equation N

∑

N

ai δ u i = 0

(

i =1

∑ ai u ∗i = 0 ) .

(1-29)

i =1

Now, we assume that S is subject to s nonholonomic constraint equations: N

∑ a ji (t, u1 ,..., u N ) du i + b j (t , u1 ,..., u N ) dt = 0

(1-30)

i =1

with j = 1,..., s ; ∀a ji , b j ∈ R . We say: D

Virtual displacements (resp. virtual velocities) are consistent with a nonholonomic constraint of Eqs. (1-30) if reals δ u i (resp. u ∗i ) verify the s equations: N

N

∑ a ji (t , u1 ,..., u N ) δ u i = 0

∑ a ji u ∗i = 0 )

(

i =1

(1-31)

i =1

with j = 1,..., s. In conclusion, by combining the definitions of virtual displacements (resp. velocities) subject to both holonomic and nonholonomic constraints, the reader will obviously define fields of virtual displacements (resp. velocities) in the case of systems subject to any constraints. Remark. We note that the constraints are holonomic if the differential system (1-30) is completely integrable; that is, if there are s functions ψ j such that:

ψ j (t , u 1 ,..., u N ) = c j or

dψ j =

∂ψ j ∂t

j = 1,..., s

N

∂ψ j

i =1

∂u

dt + ∑

i

du i = 0

(so written in the form of nonholonomic constraint equations).

2.2.4

Virtual Fields and Dynams We consider a (free) rigid body B located by six generalized coordinates q j .

2.2.4a Field of Moments of Dynam PR27 The vector field of virtual displacements (or virtual velocities) of any p ∈ B : n

δx=∑

j =1

∂r δqj j ∂q

is the field of moments of a dynam.

(

v ∗p

n

=∑

j =1

∂r ∗ j q ) ∂q j

92

Chapter 1

Proof. First, we prove that the vector field B → E3 : p


∂r ∂q j

is equiprojective. Indeed, the distance between any two particles of the moving solid B remains constant, that is:

k ∈ R+

∀a, b ∈ B : ab = k ⇒

∀q j ∈ R :

⇒

ab . (

∂rb ∂q

−

j

∂ ∂q j

∂ra ∂q j

ab

2

=0=

∂ ab ∂q j

⋅ ab

)= 0.

This equation really expresses that the vector field p


∂r ∂q j

is equiprojective and thus it is a

field of moments of a dynam. Therefore, each following vector field ( j = 1,..., n) : p


∂r ∂q

j

δqj

(resp. p


∂r ∂q

j

q∗ j )

and thus their linear combinations, as δ x (resp. v ∗p ), are also fields of moments of a dynam. A virtual displacement of a rigid body consists in “virtually” moving each of points of this solid; that is, there is a virtual displacement for each point. We must particularize these virtual displacements so as to “conserve” the rigid body. Later we will say that such virtual displacements characterize a rigid body motion. In order to express such particular virtual displacements, we first introduce the notion of dynam of virtual velocities.

2.2.4b Dynam of Virtual Velocities and Rigid Body Motion

What is the dynam of which the field of moments was just mentioned? Let O be an arbitrary reference point fixed in a rigid body B, p be some material point of B. The fields of velocities of points of a rigid body are continuous1 since we have: ∀p, p ′ ∈ B : v p′ − v p = p ′p ∧ ω ⇒

1

v p′ − v p ≤ ω p ′p .

A vector field w, defined at the instant t, is continuous on B if, for any neighboring points p and have :

∀ε ∈ R+ : w ( p ′) − w ( p ) < ε .

p ′ of B, we

Statics

93

The well-known formula of field velocities: ∀p ∈ B : v p = v O + pO ∧ ω

leads to the following expression of the virtual velocity: ∀p ∈ B : v ∗p = v O∗ + pO ∧ ω∗ . It is the moment of a dynam whose resultant is the vector ω∗ which is the velocity of virtual rotation of B also called the virtual angular velocity vector. D

The dynam of virtual velocities or virtual kinematic dynam is defined by its following elements of reduction

ω∗   ∗ . v  p From the viewpoint of virtual displacements, we say: PR28 Every virtual displacement of p ∈ B is composed of - a vector of virtual translation denoted by δ O, - a vector of virtual rotation corresponding to a virtual angular displacement δ θ expressed, with respect the basis of the frame of reference, as

δ θ = δ θ 1 e1 + δ θ 2 e 2 + δ θ 3 e 3 . We denote such a virtual displacement by

δ x = δ O + pO ∧δ θ . We note the consisting of dimensions. D

The dynam of virtual displacements of every p ∈ B is defined by its following elements of reduction (at p): δ θ  δ x  .  

We say: PR29 Every field of virtual displacements (resp. velocities) restricted to B and associating to p the vector n n ∂r ∂r ∗ j j ∗ v δx=∑ δ q (resp. q ) = ∑ p j j j =1 ∂q j =1 ∂q defines a dynam. Finally, we say: D

A field of virtual displacements (resp. velocities) on a solid B, at t, is a rigid body field if it is the moment of a dynam.

94

Chapter 1

So, the following proposition is obvious: PR30 Every field of virtual displacements (resp. velocities) on B: n

p
δ x =∑

j =1

∂r δqj j ∂q

n

( p
v ∗p = ∑

j =1

∂r ∗ j q ) ∂q j

is a rigid body field on B (at t). This notion can be extended to systems of points and solids by considering the restriction of field of virtual displacements (or virtual velocities) to each of subsystems.

2.2

VIRTUAL WORK

2.3.1

Definitions, Rigid Body Motion and Ideal Constraint

First we are going to define the virtual work and the virtual power of concentrated and distributed forces. We consider a mechanical system S in E 3 . 2.3.1a Definitions D
The virtual work (resp. virtual power) done by a force f acting on a particle p for the field of virtual displacements δ x (resp. field of virtual velocities v ∗ ) is the real  

(resp.

δ W = f .δ x ,

(1-32)

P ∗ = f . v ).

(1-33)

If the virtual work is not done by any concentrated force f during the virtual displacement but on the contrary by any distributed force, then we consider the following. Given a distributed force of continuous density Ψ, we consider at any mass point p of an elementary open of S (of volume dµ ), the force Ψ ( p) dµ as concentrated at p. So, we say: D

The virtual work (resp. virtual power) done by a distributed force of density Ψ on S for the field of virtual displacements δ x (resp. field of virtual velocities v ∗ ) is

δ W = ∫ Ψ.δ x dµ ,

(1-34)

P ∗ = ∫ Ψ . v ∗ dµ ).

(1-35)

Σ

(resp.

Σ

In the case of concentrated forces f h of various p h ∈ S , the virtual work is written:

δ W = ∑ f h .δ x h . h

95

Statics

In the case of forces distributed over any subsystem of S, with a force density Ψ, the virtual work is written:

δ W = ∫ Ψ dµ .δ x . Σ

By denoting df = Ψ dµ , we gather the two previous notions together in only one writing:


δ W = ∫ df .δ x .

(1-36)

Σ

In the same manner, the virtual power of concentrated forces, that is:

P ∗ = ∑ f h . v ∗h h

and the virtual power of distributed forces are notions that we can put together in the following writing:


∗

P ∗ = ∫ df . v .

(1-37)

Σ

2.3.1b Rigid Body Fields

We consider a rigid body field of virtual displacements (resp. of virtual velocities). We know that a dynam, the resultant of which is δ θ (resp. ω∗ ), is automatically associated Let q be an arbitrary point of reference. The virtual displacement of p ∈ S is such that:

δ p = δ q + pq ∧ δ θ and v ∗p = v q∗ + pq ∧ ω∗ .

The corresponding virtual power is written: P ∗ = ∫ v ∗p . df = ∫ v q∗ . df + ∫ ω∗ ∧ qp . df S

=

S

v q∗ .

S

∗

∫S df + ω . ∫S qp ∧ df .

But, the dynam of forces is defined at point q as follows:

   qp ∧ df  q  ∫S 

[F ] = 

∫S df

and the dynam of virtual velocities is defined, at point q, by ω  . ∗ v  q

[V ] = 

96

Chapter 1

Therefore, the expression of P ∗ is a product of two dynams; more precisely, given an arbitrary point q, we have: PR31 The virtual power of forces acting on the mechanical system S is the product of the dynam of forces and the dynam of virtual velocities:

[ ]

P ∗ = [F ]. V ∗ .

(1-38)

The reader will immediately deduce a similar statement about the virtual work since we can write, given an arbitrary point q:

δ W = δ q . ∫ df + δ θ . ∫ qp ∧ df S

⇔

S

δ W = [F ] . [δ x ] .

(1-39)

2.3.1c Ideal Constraints D
A constraint between rigid bodies is ideal if the virtual work (resp. virtual power) done by constraint forces for every field of virtual displacements (or virtual velocities) consistent with the constraint is zero.

So, in the case of rigid bodies and interconnected systems of rigid bodies, the virtual displacements and virtual velocities are particularized so as to be compatible with the (ideal) constraints. For a rigid body B rotating about a fixed point, any virtual displacement compatible with the constraint is necessarily the null vector at this point. Any other material point p h of the solid will have a consistent virtual displacement of rotation about the fixed point. This is mathematically expressed as: ∀O ∈ B : δ O = 0

⇒

δ x h = ph O ∧ δ θ .

We also recall that if a connection constraints two rigid bodies to have a contact point continuously, then the reactive force of constraint (internal to the system) does not contribute to the virtual work if a same virtual displacement is chosen at the common point. PR32 A constraint corresponding to frictionless contact is ideal. Proof. Let q be a point of contact between two rigid bodies B1 and B2 . Let Π be the common tangent plane at q. Since there is no friction, the contact dynam at q has a resultant perpendicular to Π. But every virtual velocity v q∗ compatible with the constraint has, at q, a direction of the

tangent plane Π and thus the scalar product of the resultant of the dynam and v q∗ is zero. So, the corresponding virtual power (or work) is zero; the constraint is ideal. The converse of this proposition is false; that is:

Statics

97

PR33 An ideal constraint can be with friction. Proof. A constraint without sliding, at q, is ideal since the field of virtual velocities compatible with the constraint is zero: v q∗ = 0 . Therefore the corresponding virtual power (or work) is zero and nevertheless there is friction because there is no sliding. Example 1. Characterize the dynam of virtual velocities and the dynam of constraint forces of a pivot from the definition of an ideal constraint between two rigid bodies B1 and B2 .

Let 1z be the unit vector in the direction of the existent axis of rotation. At any point a of the axis, the dynam of virtual velocities is the following: θ ∗ 1z  ∀θ ∗ ∈ R :   . 0  a The dynam of forces exerted by B1 on B2 , at a:  f (1→2)   (1→2)   M  a

is opposite to the one exerted by B2 on B1 . The condition of ideal constraint existence:

∀θ ∗ ∈ R : P ∗ = θ ∗ 1z . M a(1→2) = 0 implies: M a(1→2) . 1z = 0 . In conclusion, the moment of the dynam of constraint forces is perpendicular to the axis of the pivot. This is a well-known result. Example 2. Characterize the dynam of virtual velocities and the dynam of constraint forces of a ball-and-socket joint from the definition of an ideal constraint between two rigid bodies B1 and B2 .

Let o be the center of the joint. The dynam of virtual velocities is ω∗     0 o

where ω∗ is an arbitrary vector. The condition of ideal vector existence: P ∗ = ω∗ . M o(1→2) = 0 implies that the moment, at o, of the dynam of constraint forces is zero. This is a well-known result.

98

2.3.2

Chapter 1 Principle of Virtual Work (First Expression)

The goal of the present developments is to formulate the conditions of equilibrium in terms of generalized coordinates; that is, without reference to vector coordinates of any material point of a mechanical system. Let t be an instant of a given interval I. 2.3.2a Particle

Let us prove the following principle of virtual work. PR34 A particle at rest, at time t, and subject to ideal constraints is in equilibrium with respect to a frame of reference iff, for every virtual displacement compatible with the constraints, the virtual work of the resultant of only given forces acting on the particle vanishes (at any instant t ∈ I ). In brief: Let x be the position vector of a particle p of mass m, F be the resultant of given forces applied to p. The principle is written for every δ x compatible with the ideal constraints as follows:


p in equilibrium iff δ W = F .δ x = 0 .

(1-40)

Proof. The necessary condition is direct. Indeed, let f be the resultant of all the forces acting on p: f = F + L.

The hypothesis of equilibrium of p, namely f = 0 , implies: f .δ x = ( F + L).δ x = 0 .

But, the virtual displacements being compatible with the ideal constraints (that is L .δ x = 0), we have necessarily: F .δ x = 0 .

Let us prove the sufficient condition. By assumption, we have: F .δ x = 0

and for every virtual displacement δ x compatible with the ideal constraints we know that L .δ x = 0 .

Consequently we have: F .δ x + L . δ x = f . δ x = 0 .

This last result implies that p is in equilibrium. Indeed, if the particle p was not in equilibrium then it should necessarily be accelerated since it would automatically leave a position of rest (occupied at a given time). Thus, we would have for a consistent virtual displacement δ ′ x :

99

Statics a . δ ′x ≠ 0

where a is the acceleration of p. The obtained result, namely: ma . δ ′ x ≠ 0

is absurd since it is opposite to the hypothesis. Consequently, the particle p is necessarily in equilibrium. Remark. The principle of virtual work can be widespread to systems of particles as solids or interconnected rigid bodies. For that, we recall that the resultant f h of forces on each particle p h of a system in equilibrium is zero. So, for any virtual displacement (which is not even compatible with the constraint!) the virtual work of all forces vanishes in the case of equilibrium:

δ W = ∑ f h .δ x h = 0 . h

However, in the principle of virtual work, the virtual displacements are particularized in order to cancel the virtual work of unknown forces! This last particularity confers great power on the method! First, we consider the case of rigid body.

2.3.2b Rigid Body Preliminary proposition PR35 The virtual work of internal forces done on a rigid body B is zero. Proof. The virtual work of internal forces f h(i ) done on a rigid body is

δ W (i ) = ∑ f h(i ) .δ x h = ∑ f h(i ) .δ O + ∑ f h(i ) . ph O ∧ δ θ h

=∑ h

h

f h(i ) .δ O

+∑

h

f h(i )

∧ p h O .δ θ

h

where O is an arbitrary point fixed in the rigid body. We have thus obtained:

δ W (i ) = f (i ) .δ O + M O(i ) .δ θ . This means that the virtual work of internal forces is the product of dynams of internal forces and virtual displacements:

δW

(i )

 f (i )  δ θ  =  .  . (i )  M  O δ O 

But the dynam of internal forces of every solid is zero, which implies:

δ W (i ) = 0 .

100

Chapter 1

In the same manner, we have:

P ∗( i ) = 0 . In conclusion, the internal forces do not contribute to the virtual work done on a rigid body for virtual displacements consistent with the rigidity (that is, in accordance with the existence of forces which hold the particles of the solid at fixed distance from one another). Now, we can state the following Principle of virtual work

PR36 A rigid body B at rest, at time t, and subject to ideal constraints is in equilibrium with respect to a frame of reference iff, for every virtual displacement compatible with the constraints, the virtual work of the resultant of given forces done on the rigid body vanishes (at any instant t ∈ I ). In brief: Let p h be a particle of the solid, Fh(e) be the resultant of given forces applied to p h .

The principle is written for every δ x h compatible with the ideal constraints as follows:


B in equilibrium iff δ W ( e) = ∑ Fh(e ) .δ x h = 0 .

(1-41)

h

The reader will state this principle in the power context. Example. Find again the equilibrium conditions for a rigid body B rotating about a fixed point O (ball-and-socket joint). Answer. Any virtual displacement of some p h ∈ B :

δ x h = δ O + ph O ∧ δ θ is chosen compatible with the constraint, that is zero at the fixed point O. Since there is no possible virtual translation, we have necessarily:

δ x h = ph O ∧ δ θ . The principle of virtual work means that

δ W = ∑ Fh(e ) .δ x h = ∑ Fh( e) . ( ph O ∧ δ θ ) h

=∑

h

Fh(e )

∧ p h O .δ θ = 0

h

and thus

∑ Oph ∧ Fh(e) = M O(e) = 0 . h

101

Statics 2.3.2c System of Rigid Bodies

In the case of systems of rigid bodies, it is necessary to take given internal forces Fh(i ) into account (if existence). For example, such a force is due to a spring of given torsional stiffness at the hinge. We know that the virtual displacements are chosen compatible with the ideal constraints. For example, a constraint which forces two rigid bodies to keep a common contact point imposes, for each solid, identical virtual displacements at the common point. We can state: Principle of virtual work

PR37 A system S of rigid bodies at rest, at time t, and subject to ideal constraints, is in equilibrium with respect to a frame of reference iff, for every virtual displacement compatible with the constraints, the virtual work of all the given external and given internal forces done on the system vanishes (at any instant t ∈ I ). In brief: Let p h be a particle of the system S, Fh( e) be the resultant of given external forces acting on p h , Fh(i ) be the resultant of given internal forces acting on p h . The principle is written for every δ x h compatible with the ideal constraints as follows: S in equilibrium iff δ W = ∑ ( Fh( e) + Fh(i ) ) .δ x h = 0 .

(1-42)

h

The reader will express this principle in the power context. Remark 1. In the case of a particle or a rigid body, the method of virtual work shows no great advantage with respect to the classical method. But the method of virtual work is tremendously profitable in the study of interconnected systems of rigid bodies. Remark 2. If the system of rigid bodies contains also elastic bodies, then we take these ones into account when we write the internal forces in the expression of the virtual work principle.

Before extending this principle, until now viewed for systems without elastic members, we recall that if k designates the stiffness of a spring then the spring exerts a restoring force: F = − kx i

where x measures the distance from the relaxed position of the spring (along the direction of a unit vector i). The virtual work done by F for a virtual displacement δ x is expressed as

− kx δ x = −δ (k x 2 2) = −δ Vel that is the virtual change in potential energy of the spring. The principle of virtual work done by given external forces and by elastic internal forces acting on a mechanical system means that, in equilibrium, we have the following relation:

102

Chapter 1

δ W ( e) + δ Wel = 0 where δ Wel represents the virtual work done by elastic members. This is obviously written:

δ W ( e) = δ Vel .

(1-43)

So, we say: PR38 A system of interconnected rigid bodies constrained by elastic internal forces is in equilibrium iff the virtual work done by all given external forces during any virtual displacement compatible with the constraints equals the virtual change in the elastic energy of the system. Remark 3. In any study of equilibrium in a reference system moving with respect to an inertial frame of reference, it is necessary to take account of forces of transport in the virtual work expression, but not of Coriolis forces [see Remark in Sect. 1.3.1a].

2.3.2d Torricelli Theorem

PR39 In gravitational field, any virtual displacement1 of the center of mass G of a rigid body in equilibrium is horizontal. Proof. Given any particle p h of mass mh and a corresponding (compatible) virtual displacement δ x h , in equilibrium, the virtual work is

δ W = ∑ mh g .δ x h = 0 . h

But the definition of the center of mass which occupies two neighboring positions G and G ′ leads to M GG ′ = ∑ mh p h p ′h . h

Denoting a virtual displacement of G by δ G , the expression of virtual work becomes: M δG.g = 0

which proves the theorem. PR40 In gravitational field, the center of mass G of a rigid body, at rest, fixed at a point belongs to the vertical line through this point necessarily. Proof. This proposition follows from the previous theorem. Indeed, if G did not belong to the vertical line then any virtual displacement consistent with the fixed point would not be horizontal. Such a conclusion would be absurd seeing the previous theorem. To conclude this section, we emphasize the following remarks. 1

Compatible with the constraints.

Statics

103

Remark 1. We recall that the generalized coordinates q j do not necessarily have the dimensions of length (L), they may be angles for instance. However, virtual work (scalar product) has the dimensions of work, that is ML2T −2 and has the units of force times displacement, namely: newton times meter (called the joule). The virtual work can be the scalar product of any force and vector of dimension L. However, we know that a virtual displacement can be a virtual rotation of a body. In this case, the virtual work done by the moment M of a couple (M being a free vector of type r ∧ F ) during a virtual angular displacement δ θ is

δ W = M .δ θ

(1-44)

The virtual work has the dimensions of work; the units of M are the ones of the work (Nm) and the angular displacement is expressed in radians. Remark 2. The number of independent virtual displacements equals the number of degrees of freedom of any mechanical system. But the reader will be careful of the relations of dependence between virtual displacements if necessary. Remark 3. The essential advantage of the method of virtual work is that no reference is made to unknown reactive forces of constraint. But, if we want to know these forces as well as the equilibrium positions of a particle or a mechanical system, it is necessary to make the corresponding constraint forces virtually work. In this case, the virtual displacements are not chosen consistent with the constraints! This is illustrated with the following example which also compares the classical and virtual work methods. Example. A homogeneous ladder ab of mass m and length l rests against an inclined wall (angle α as shown in Fig. 45 ) and on a horizontal floor, the surfaces being smooth. (i) Let us calculate the horizontal force F, along the x-axis and acting on a, which is able to hold the ladder in equilibrium; first by the classical method and second by the method of virtual work. (ii) Let us find again the reactive forces of constraint from the method of virtual work.

Fig. 45 This problem has one degree of freedom. We choose as generalized coordinate the angle θ between the ladder and the floor.

104

Chapter 1 l

l

2

2

(i) The weight − mg j of the ladder is applied at point G ( cos θ − l sin θ cot α , sin θ ) . The force F = − F i is applied at point a( l cos θ − l sin θ cot α , 0 ). The reactive force of constraint at point a is La = La j . The reactive force of constraint a point b( − l sin θ cot α , l sin θ ) is Lb = Lb sin α i + Lb cos α j . The first condition of equilibrium, namely the resultant of forces is zero, leads to the respective equations of projection: Lb sin α − F = 0 , Lb cos α − mg + La = 0 . In the same manner, the total moment, for instance at a: M a = aG ∧ m g + ab ∧ Lb introduces a third equation, namely: l

mg cos θ − Lb l cos(α − θ ) = 0 . 2

In conclusion, we have: F=

mg cosθ sin α , 2 cos(α − θ )

La = mg ( 1 − Lb =

cosθ cos α ), 2 cos(α − θ )

mg cos θ ⋅ 2 cos(α − θ )

Second, any virtual displacement of the ladder, compatible with the constraints, implies that δ a is directed along ox and δ b along ob since the constraints are smooth. This implies: La .δ ra = 0 , Lb .δ rb = 0 . The conditions of equilibrium of the ladder follow from the principle of virtual work for which only the given active forces must be considered:

δ W = m g .δ rG + F .δ ra = 0 . We have

δ ra = δ xa i and

δ rG = δ xG i + δ yG j where the first term of δ rG does not contribute to δ W . Since l

yG = sin θ 2

105

Statics

and since the component of δ rG along j is positive whenever θ increases by δ θ , then we deduce: l

δ y G = cosθ δ θ . 2

Since x a = l cos θ − l sin θ cot α and since the component of δ ra is negative whenever θ increases by δ θ , then we deduce:

δ x a = −(l sin θ + l cosθ cot α ) δ θ . The condition of equilibrium is thus l (−mg j ). ( cosθ δ θ j ) + (− F i ) .(−l sin θ − l cos θ cot α )δ θ i = 0 2

⇔

sin θ sin α + cos θ cos α l − mg cos θ ) δ θ = 0 . sin α 2

( Fl

The equation holds for arbitrary δ θ which implies: F=

mg cos θ sin α ⋅ 2cos(θ − α )

(ii) In order to calculate the reactive forces of constraint from the method of virtual work, we are going to make the forces of constraint virtually work. To obtain La , we choose a virtual displacement compatible with the constraint at point b and which makes the reactive force La virtually work. For example, we choose δ γ 1τ as virtual displacement tangent to the circle of center b, this vector corresponding to a virtual rotation of (positive) angle δ γ about b. For this virtual displacement, the condition of equilibrium is written: l

δ W = m g . δ γ 1τ + La . l δ γ 1τ + F . lδ γ 1τ 2

l

= (− mg cos θ + La l cosθ − F l sin θ ) δ γ = 0 2

and thus 1

1

2

2

La = mg + F tan θ = mg (1 + = mg ( 1 −

sin θ sin α ) cos(α − θ )

cos θ cos α ). 2 cos(α − θ )

Similarly, to obtain Lb , we choose δ θ 1τ ′ as virtual displacement tangent to the circle of center a, this vector corresponding to a virtual rotation of (positive) angle δ θ about a. For this virtual displacement, the condition of equilibrium is written:

106

Chapter 1 l

δ W = m g ⋅ δ θ 1τ ′ + Lb . l δ θ 1τ ′ 2 l

= (mg cos θ − Lb l cos(α − θ )) δ θ = 0 2

and thus Lb =

mg cos θ ⋅ 2cos(α − θ )

This example shows that the method of virtual work used in the case of a unique rigid body unlike the one of interconnected systems of rigid bodies is not very profitable with respect to the classical method.

2.3.3

Principle of Virtual Work (Second Expression)

2.3.3a Generalized Force and Principle of Virtual Work (i) Particle Given n generalized coordinates, we know that any virtual displacement of a particle p is defined by n

δx=∑

j =1

∂r δqj j ∂q

and the virtual work of a force F is n

δ W = F .δ x = ∑ F ⋅ j =1

∂r δ q j. j ∂q

D
The generalized force associated with the generalized coordinate q j and relating to F is the function

R 2 n +1 → R : (t , q, q& ) a Q j (t , q, q& ) such that 

Qj = F ⋅

∂r ∂q j

⋅

(1-45)

We assume that the virtual displacements are compatible with the constraints. Therefore, the principle of virtual work is written: n

δW = ∑ Qj δ q j = 0 j =1



⇔

Qj = 0

∀j = 1,..., n.

(1-46)

The previous equivalence follows from the independence of the n generalized coordinates. The n equations (1-46) determine the n values of generalized coordinates in equilibrium.

107

Statics

So, we can state the second formulation of the principle of virtual work: PR41 A particle at rest at a given instant and subject to ideal constraints is kept in equilibrium with respect to an inertial frame of reference iff all the generalized forces are zero.

(ii) System of particles Let rh be the position vector of any particle p h of a system of n degrees of freedom. Any virtual displacement of p h is defined by n

∂rh

j =1

∂q

δ xh = ∑

j

δqj

and the virtual work of the forces acting on the system is

δ W = ∑ ( Fh( e) + Fh(i ) ) .δ x h h

n

∂rh

j =1

∂q

= ∑ ( ∑ ( Fh(e ) + Fh(i ) ) ⋅

D

h

j

)δ q j .

The generalized force associated with the generalized coordinate q j and relating to the given forces acting on the system is the function R 2 n +1 → R : (t , q, q& ) a Q j (t , q, q& ) defined by the scalar product: Q j = ∑ ( Fh(e ) + Fh(i ) ) ⋅




h

∂r ∂q j

⋅

(1-47)

We assume that the virtual displacements are compatible with the constraints. Therefore, the principle of virtual work is written: n

δW = ∑ Qj δ q j = 0 j =1

⇔


Qj = 0

∀j = 1,..., n.

(1-48)

The previous equivalence follows from the independence of the n generalized coordinates. The n equations (1-48) determine the n values of generalized coordinates in equilibrium. So, we can state the second formulation of the principle of virtual work: PR42 A system of particles at rest at a given instant and subject to ideal constraints is kept in equilibrium with respect to an inertial frame of reference iff all the generalized forces Q j are zero.

108

Chapter 1

Remark 1. We point out that the forces of transport must be taken into account if the frame of reference is not inertial. Remark 2. Since the various generalized coordinates q j have not necessarily the dimensions of length, then the generalized forces Q j have not necessarily the dimensions of force.

More precisely, if q k has the dimensions of length, then the dimensions of Qk are MLT −2 .

If q k is an angle, then the dimensions of Qk are ML2T −2 . So, the virtual work has always the dimensions of work.

2.3.3b Conservative Force Fields and Principle of Virtual Work

(i) Particle Let (e1 , e 2 , e 3 ) be an orthonormal basis of a Euclidean space E3 . We recall that a given force field F is conservative iff there exists a continuously differentiable function named the potential, namely1: V : E 3 → R : p a V ( p) such that F = −∇ V ,

or similarly: iff there is a force function U = −V such that F = ∇U .

The expression F=

∂U ∂U ∂U e + 2 e 2 + 3 e3 1 1 ∂x ∂x ∂x

implies: Qj = F ⋅

∂r ∂q j

∂U ∂U ∂U ∂x1 ∂x 2 ∂x 3 = ( 1 e1 + 2 e 2 + 3 e 3 ) ⋅ ( j e1 + j e 2 + j e 3 ) ∂x ∂x ∂x ∂q ∂q ∂q =

∂U ∂x1 ∂x1 ∂q j

1

∂U ∂x 2 ∂x 2 ∂q j

3

∂U ∂x i

i =1

∂x i ∂q j

=∑ =

+

∂U ∂q j

+

∂U ∂x 3 ∂x 3 ∂q j

⋅

In rheonomic problems, potentials depend explicitly on time.

109

Statics

In conclusion, in the conservative case, the generalized forces are expressed as


Qj =

∂U ∂q

=−

j

∂V

⋅

∂q j

(1-49)

This result is valid in rheonomic problems [where explicitly: Q j (t , q )] . So, we state the principle of virtual work: PR43 In the case of conservative given forces, a particle is kept in equilibrium with respect to an inertial frame of reference iff the potential is a function V such that


∂V ∂q j

j = 1,..., n .

=0

(1-50)

(ii) System of particles The resultant of given forces acting on any particle p h is written1: Fh = −∇ hV

and thus Q j = ∑ Fh ⋅ h

=−

∂V ∂q j

∂rh ∂q

j

= −∑ ∇ hV ⋅ h

∂rh ∂q j

⋅

Therefore, we state the principle of virtual work: PR44 In the case of conservative given forces, a system of particles is kept in equilibrium with respect to an inertial frame of reference iff the potential is a function V such that:


∂V ∂q j

=0

∀j = 1,..., n .

(1-51)

Sometimes these equations are said to define the conditions for a stationary value of V.

2.3.3c Types of Equilibrium

(i) Particle

Let f be the resultant of forces acting on a particle, dr be any slight displacement (or differential displacement) away from the equilibrium position toward a neighboring point and taking the constraint into account.

1

Gradient at point p h : ∇ hV = (

∂V ∂x

1

,

∂V ∂x

2

,

∂V ∂x

3

).

110

D

Chapter 1

An equilibrium position of a particle is said to be stable if the work done by f acting on the particle along any differential displacement dr is negative: dW = f . d r < 0 ;

it is said to be unstable if: dW = f . d r > 0 ;

it is said to be neutral if: dW = f . d r = 0 .

In other words, an equilibrium position is stable iff the vector of projection of f along any differential vector dr is directed in the sense opposite to dr as shown as follows:

Fig. 46

In the same manner, the reader will consider the unstable and neutral cases.

(ii) Conservative mechanical systems

First, we consider the case of systems having one degree of freedom. Let q be the generalized coordinate. The condition of equilibrium is dV = 0. dq

This requirement defines the condition for a stationary value of V; that is, a minimum, a maximum, a stationary point of inflection or a constant value, in the relation of V versus q. This type of problem being studied in courses of differential calculus, we only consider the example of a cylinder on a surface which clearly illustrates the different types of equilibrium positions.

Statics

111

Fig. 47

In the first case, any slight displacement away from the equilibrium position implies an increase of potential energy ( ∆V > 0 ), so the cylinder will go back to its equilibrium position where V is minimum. It is a position of stable equilibrium. In the second case, any slight displacement away from the equilibrium position implies a decrease in potential ( ∆V < 0 ), so the cylinder will irremediably move away from the equilibrium position where V is maximum. It is a position of unstable equilibrium. In the third case, any slight displacement away from the equilibrium position (inflection point) means the cylinder will definitely leave the equilibrium position. It is a position of unstable equilibrium. In the last case, any slight displacement away from the equilibrium position does not lead the cylinder to move away from the previous equilibrium position or to go back to this. Any new position is an equilibrium position. This case, where ∆V = 0 , represents a neutral equilibrium. The sign of ∆V is obtained from a Taylor expansion. Let q0 = 0 be the value of the general coordinate at the equilibrium position. We mention that if this value is different from zero, then a new variable will be chosen, namely: q = q − q 0 . At a neighboring point of the equilibrium position, the potential is expressed as: V (q) = V0 + (

d 2V q 2 d 3V q 3 dV )0 q + ( 2 )0 + ( 3 )0 + ⋅⋅⋅ 2! 3! dq dq dq

where the subscript zero refers to the value of each function at q0 = 0 . In equilibrium, the condition (

dV )0 = 0 dq

implies ∆V = (

d 2V dq 2

)0

q2

d 3V

q3 + ( 3 )0 + ⋅⋅⋅ 2 3! dq

112

Chapter 1

The smallness of q implies that the sign of the previous expansion is the one of the lowest order nonzero term which remains. If it is the second term, then: d 2V

(i)

The equilibrium is stable ( ∆V > 0 ) iff (

(ii)

The equilibrium is unstable ( ∆V < 0 ) iff (

(iii)

dq 2

)0 > 0 .

d 2V dq 2

)0 < 0 .

d 2V

d nV

) 0 = 0 , then the sign of the nonzero derivative ( n ) 0 of the lowest order dq dq 2 must be examined.

If (

- If this order n is even, then we conclude: d nV Stable equilibrium iff ( n ) 0 > 0 dq d nV

)0 < 0 . dq n - If this order n is odd, then we conclude:

Unstable equilibrium iff (

Unstable equilibrium (inflection point). Second, we consider a system possessing two degrees of freedom. The sign of the increment ∆V is obtained from the Taylor expansion for the function V of two generalized coordinates q1 and q 2 . Indeed, let (0,0) be the equilibrium position, after changes of q1 and q 2 if necessary. In the neighborhood of the equilibrium position, the expansion for V is V ( q 1 , q 2 ) = V0 + (

∂V

∂V

1

∂q

∂q

2

) q1 + ( 1 0

) q2 + 2 0

( A(q1 ) 2 + 2 Bq1q 2 + C (q 2 ) 2 ) + ⋅ ⋅ ⋅

where A=(

∂ 2V ∂ (q1 )

B=(

) , 2 0

∂ 2V ∂q 1∂q

) , 2 0

C=(

∂ 2V ∂ (q 2 ) 2

)0 .

In equilibrium, the conditions (

∂V ∂q

) 1 0

=(

∂V ∂q 2

)0 = 0

imply: 1

∆V = ( A(q1 ) 2 + 2 Bq1q 2 + C (q 2 ) 2 ) + ⋅ ⋅ ⋅ 2

By neglecting the derivatives of order higher than 2, then the previous equation becomes the one of a surface, by expressing ∆V as a function of two independent variables q1 and q 2 .

Statics

113

The differential calculus leads to the following conclusions: - Stable equilibrium ( ∆V > 0 ) iff B 2 − AC < 0 and A + C > 0 . - Unstable equilibrium ( ∆V < 0 ) iff B 2 − AC < 0 and A + C < 0 . - Unstable equilibrium (saddle point) iff B 2 − AC > 0 . - Undetermined equilibrium iff B 2 − AC = 0 .

Fig. 48 The equilibrium is indifferent if V is constant ( ∆V = 0 ). Example. A mass m slides in a vertical cylinder and is connected by a negligible mass rod of length l to an end of a spring of stiffness k which slides in a horizontal guide. The other end of the rod is aligned with the cylinder axis when the spring is unstretched.

Fig. 49 We assume there is no friction and we are going to characterize the equilibrium positions. The system has one degree of freedom, the angle θ between the rod and the cylinder axis is the chosen generalized coordinate. The spring is unstretched when θ = 0 o .

114

Chapter 1

From a suitable choice of constants, the potential of the system is k V = l 2 sin 2 θ + mg l cos θ . 2

The condition of equilibrium: dV = k l 2 sin θ cos θ − mg l sin θ = 0 dθ

is fulfilled for the values:

θ = 0o

and

θ = cos −1 (

mg kl

).

The type of stability depends on the sign of d 2V = k l 2 (cos 2 θ − sin 2 θ ) − mg l cos θ . 2 dθ For θ = 0 o , the equilibrium is stable if k > For θ = cos −1 (

mg kl

mg mg and is unstable if k < ⋅ l l

) , the equilibrium is stable if k
⋅ l l

3. EXERCISES Exercise 1.

Is a vertical wheel of radius R, rolling (without slipping) on a horizontal plane oxy, subject to holonomic constraints? What is the number of degrees of freedom of this wheel which cannot pivot?

Fig. 50

115

Statics

Answer. We choose four overabundant coordinates, namely: - the coordinates xc , y c of the wheel center, - the angle φ between the (horizontal) axis of the wheel and x-axis, - the angle θ of rotation of the wheel about its axis. Since the wheel never slips, we have the well-known speed of the contact point: v = s& = Rθ& .

The components of the velocity of the wheel center are: x& c = Rθ& sin φ ,

y& c = Rθ& cos φ

or in an equivalent manner: dxc − R sin φ dθ = 0 ,

dy c − R cos φ dθ = 0 .

The angle φ varies and thus the previous equations cannot be integrated; in consequence, the constraints are nonholonomic. There are two ( 4 − 2 ) degrees of freedom, the angles φ and θ are the generalized coordinates for instance. In this case, the two previous equations determine the overabundant coordinates xc and y c (if the wheel does not slip!) Exercise 2.

Find the number of degrees of freedom of a rear axle for which the two wheels, of radius R, can neither pivot nor slip on a plane. Characterize the constraints by choosing five overabundant primitive coordinates that are the coordinates x m , y m , z m of the middle m of the axle, the angle θ of rotation of wheels and the angle φ indicating the direction of the axle. Answer. The contact points of wheels are i and j, their distance is equal to 2l .

Fig. 51

116

Chapter 1

Since the wheels are rolling without slipping, the following velocities vanish: v i = v m + im ∧ ω = 0 , v j = v m + jm ∧ ω = 0

(1) (2)

where the angular velocity vector ω is a priori:

ω = φ& 1z + θ& 1ij and im = R 1z + l 1ij

jm = R 1z − l 1ij

and thus:

im ∧ ω = ( Rθ& − lφ&)(−1vm ) , jm ∧ ω = ( Rθ& + lφ&)(−1 ) . vm

The x,y and z-components of (1) and (2) are immediately: x& m − ( Rθ& − lφ&) sin φ = 0 , y& − ( Rθ& − lφ&) cos φ = 0 , m

x& m − ( Rθ& + lφ&) sin φ = 0 , y& − ( Rθ& + lφ&) cos φ = 0 , m

z& m = 0 . The integration of the last equation leads to the (constraint) equation: zm = R which means that the wheels stay in contact with the plane oxy. The other equations: x& m = ( Rθ& − lφ&) sin φ = ( Rθ& + lφ&) sin φ ,

y& m = ( Rθ& − lφ&) cos φ = ( Rθ& + lφ&) cos φ

imply obviously:

φ& = 0

that is the equation of constraint:

φ = c1

( c1 ∈ R ).

Therefore, we have the equations: x& m = Rθ& sin φ ,

y& m = Rθ& cos φ

which could have been directly obtained. Since the angle φ is constant, we deduce: x m = Rθ sin φ + c 2 , where c2 , c3 ∈ R .

y m = Rθ cos φ + c3

117

Statics

In conclusion, the five primitive coordinates are linked together by four scleronomic and holonomic constraint equations which are: zm = R ,

φ = c1 ,

x m = Rθ sin φ + c 2 ,

y m = Rθ cos φ + c3 .

The system has one degree of freedom. Exercise 3. A disk is rolling and pivoting, but without slipping, on a plane in (known) translation. Characterize the virtual velocities and virtual displacements which are compatible with the constraints. Answer. Let Π be a plane in given translation, D be a disk of radius R, i be the point of contact of D with Π, = { o;1x ,1 y ,1z } be a frame of reference,

R

R D = { c;1X ,1Y ,1Z } be a frame “fixed” in the disk where c is the center of D and 1Z is perpendicular to the plane of D. The plane Π being in translation, we choose the vector 1z perpendicular to Π. Let h(t ) be the height of Π in R. We recall that the field of velocities in the motion of translation is constant at any time. The vector of this field, at instant t, is denoted by w (t ) = w x (t ) 1x + w y (t ) 1y + h&(t ) 1z .

We choose the following primitive coordinates: - the coordinates x, y and z of the center c, - the Euler angles φ , θ and ψ specifying the position of basis vectors of R D with respect to the ones of R. The nutation angle (or angle of inclination) is such that θ ∈] 0, π [ .

Fig. 52

118

Chapter 1

A first equation of holonomic and rheonomic constraint is obviously: z = h(t ) + R sin θ

and it expresses the permanence of contact. In addition and in general, we recall that the velocity of some point p of D is v p = v i + pi ∧ ω + w

where w represents the velocity of transport corresponding to the translation of Π. But in this example, the disk is rolling without slipping, thus there is no slip velocity; that is, the velocity v i (relative to Π ) of the point of D in contact with Π at instant t, is zero. So, by making reference to point c, the slip velocity is v i ≡) v c + ω ∧ ci − w = 0

where v c is the absolute velocity of c (that is relative to R). The (scalar) projection of terms of the vector equation onto the unit vector 1τ along the tangent to D at point i leads to the following v c . 1τ + (φ& 1z + θ& 1τ + ψ& 1Z ) ∧ ci . 1τ − w . 1τ = 0

⇔

x& cos φ + y& sin φ + Rφ& cosθ + Rψ& − w x cos φ − w y sin φ = 0 ,

that is the equation of a nonholonomic constraint. In the same manner, the (scalar) projection of terms of the already used vector equation onto the unit vector 1n = 1z ∧ 1τ leads to the following v c . 1n + (φ& 1z + θ& 1τ + ψ& 1Z ) ∧ ci . 1n − w . 1n = 0

⇔

− x& sin φ + y& cos φ + Rθ& sin θ + w x sin φ − w y cos φ = 0 ,

that is the equation of a nonholonomic constraint. So, the disk D has three ( 6 − 3 ) degrees of freedom. The most general fields of virtual velocities and virtual displacements are respectively: p a v ∗p =

∂r ∗ ∂r ∗ ∂r ∗ ∂r ∗ ∂r ∗ ∂r ∗ x + y + z + φ + θ + ψ , ∂x ∂y ∂z ∂φ ∂θ ∂ψ

paδx =

∂r ∂r ∂r ∂r ∂r ∂r δ x + δ y + δ z + δφ + δθ + δψ . ∂x ∂y ∂z ∂φ ∂θ ∂ψ

They are compatible with the constraints for x ∗ , y ∗ , z ∗ , φ ∗ ,θ ∗ and ψ ∗ verifying the following relations: z ∗ = R θ ∗ cos θ [because (1-27)], x ∗ cos φ + y ∗ sin φ + Rφ ∗ cos θ + Rψ ∗ = 0

[because (1-29)],

− x ∗ sin φ + y ∗ cos φ + Rθ ∗ sin θ = 0

[because (1-29)].

Statics

119

Exercise 4.

A hoist is a system of two pulleys having several grooves, the upper pulley being fixed and the lower being moving. A weight W is acting on the lower pulley. An inextensible rope passes n times over pulleys, one of its ends is fastened to the lower pulley, while at its other end a vertical force F is exerted down. Determine the force F required to hold W in equilibrium. Answer. This system has one degree of freedom. By assuming there is no friction, the only forces are the reactive force L exerted on the fixed pulley, the given force W and the force F.

Fig. 53 Any virtual displacement compatible with the constraints must maintain fixed the upper pulley in order to eliminate the contribution of L to the virtual work. To a vertical virtual displacement of length δ y of the lower pulley corresponds a virtual displacement of length n δ y along the line of action of F. Indeed, the length of the rope passing over the pulleys being constant, then the virtual displacement associated with F is divided between n parallel segments of the rope. The chosen virtual displacement of the point of action of F has the same directional sense as F and the directional sense of the corresponding virtual displacement of W is thus opposite to the directional sense of W. Thus, by introducing the magnitudes F and W, the equilibrium condition is written: −W δ y + n F δ y = 0 ,

that is: F = W⋅ n Exercise 5.

A rope, passing over a vertical pulley on which a mass m is hung, is fixed to a ceiling at one of its ends. The other end p is fastened to a spring which is fixed to the ceiling. The radius of the pulley is R and the spring stiffness is k. If the pulley is released initially from the position where the force in the spring is zero, calculate the angle θ of rotation of the pulley when the equilibrium is reached.

120

Chapter 1

Answer. In this problem of one degree of freedom, let δ y be an arbitrary increase of the ordinate of the center c of the pulley. The virtual displacement of c being vertical and down, the virtual work done by the force acting on the pulley is m g .δ rc = mg δ y .

Fig. 54

Consequently, the fastening point p undergoes a virtual displacement of increase δ ( 2 y ) since the point i represents the instantaneous center of zero velocity (Fig. 54). Then, the virtual work done by the (elastic) restoring force is written:

δ Wel = −k 2 y δ (2 y ) = −4ky δ y but

y = Rθ

and thus the following condition of equilibrium

δ W = mgR δ θ − 4kR 2θ δθ = 0 implies:

θ=

mg ⋅ 4kR

Exercise 6.

Given a weight W exerted on a truss at a point p as shown in Fig. 55, express the magnitude F of the force Fq or Fs in the top member of the truss composed of equilateral triangles, the length of each structural element being l. Answer. First, we recall that in the classical method we must be careful to the constraints, namely here a pin connection, at o, which is capable of supporting a force in any direction in the plane normal to the pin axis, while the rocker at point a can support a vertical force only. But the method of virtual work allows to find F without such considerations.

The friction forces are neglected.

Statics

121

Fig. 55

The angle θ is the chosen generalized coordinate for the framework, this problem having one degree of freedom. We choose the virtual displacements compatible with the constraints; namely: there are no virtual displacement of the fixed point o, no deformation of structural elements and no friction; it is so that we consider any increase δθ . The condition of equilibrium is written:

δ W = Fq .δ rq + Fs .δ rs + W .δ r p = 0 . The only horizontal components of δ rq and δ rs must be taken into account. Since x q = l cos θ , the (negative) component of δ rq in the i-direction is δ x q = −l sin θ δθ and the (positive) component of δ rs in the i-direction is δ x s = l sin θ δθ . The one and only component of δ r p is the (positive) j-component:

δ y p = l δθ. In conclusion, the principle of virtual work is written: (−W j ) . (l δ θ j ) + (− F i ) . (−l sin θ δθ i ) + ( F i ) . (l sin θ δθ i ) = 0

⇔ − W l δθ + 2 Fl sin θ δθ = 0

that is F=

W ⋅ 2 sin θ

122

Chapter 1

Exercise 7. A slider-crank mechanism is composed of a rod op of length R, rotating about a fixed point o, and a rod pq of length l where p represents the crank pin.1 The particle q slides along a horizontal rail. All the frictions are neglected. What force F along the rail is sufficient: (i) to hold a weight W in equilibrium at p; (ii) to hold in equilibrium the moment M of a couple about o; more precisely deal with this question when a piston P is pinned to the connected rod and slides in a cylinder. (iii) Find again the result of this last case by introducing virtual velocities.

Fig. 56 Answer. (i) The chosen generalized coordinate of the mechanical system of one degree of freedom is the angleθ between the rail and the rod op. Any virtual displacement must take the fixity of o into account; that is, δ o = 0 . We have thus for the reactive force L at o: L .δ o = 0 .

Since o is fixed and the rod op is rigid, any virtual displacement δ p compatible with the constraints is necessarily tangent to the circle of radius op = R . In addition, the reactive force of constraint acting on the particle q is perpendicular to the rail, so that any consistent virtual displacement δ q will be horizontal. Thus the principle of virtual work leads to

δ W = W .δ p + F .δ q = 0 . Now, we calculate the only component of δ q as follows. To the abscissa of q:

x q = R cos θ + l cos φ = R cos θ + l 1 − ( R l ) 2 sin 2 θ (where φ is the angle between the connecting rod and the rail) corresponds the following virtual change: 1

In French, the crank op is called the “manivelle” and pq is called the “bielle.”

123

Statics

sin θ cos θ

R2 δ x q = (− R sin θ − l

1 − ( R l ) 2 sin 2 θ

) δθ .

This expression is negative for any increase δθ and thus we have

δ q = (− R sin θ −

R2 l

sin θ cosθ 1 − ( R l ) 2 sin 2 θ

) δθ i .

The virtual work done by W is only due to the vertical component of any virtual displacement δ p. Since the ordinate of p is y p = R sin θ , the j-component of δ p is

δ y p = R cosθ δθ . The vertical component δ y p is positive for any increase δθ , so that the principle of virtual work is immediately written: R2 (−W j ) . ( R cosθ δθ j ) + (− F i ) . (− R sin θ − l

sin θ cos θ 1 − ( R l ) 2 sin 2 θ

) δθ i = 0

⇔

− W cos θ δθ + F sin θ (1 +

R l

cosθ 1 − ( R l ) 2 sin 2 θ

) δθ = 0 .

The increase δθ being arbitrary, we deduce: F=

W cos θ ⋅ R cos θ sin θ (1 + ) l 1 − ( R l ) 2 sin 2 θ

In particular, if R = l , we have: F=

W cot θ . 2

(ii) A piston P slides in a horizontal cylinder as shown in Fig. 57. If we consider the force f tangent to the circle (o, R) , then the moment M of the couple is M = op ∧ f = − R f k .

From previous comments about reactive forces of constraint, we can state here the principle of virtual work as follows: − Mk .δθ k + (− F i ) .δ x q i = 0

⇔ − M δθ + FR sin θ (1 +

R cos θ ) δθ = 0 . l 1 − ( R l ) 2 sin 2 θ

124

Chapter 1

Fig. 57 Thus the equilibrium exists for F=

M

=

R cosθ R sin θ (1 + ) l 1 − ( R l ) 2 sin 2 θ

f R cos θ sin θ (1 + ) l 1 − ( R l ) 2 sin 2 θ

⋅

We point out that: The previous result is immediately obtained by expressing the virtual work done by f, that is: f .δ s = − fR δθ

where δ s is any consistent virtual displacement of p. The dimensions of the virtual work done by M are really the ones of a work since the chosen virtual displacement is a virtual angular displacement δθ k . By introducing virtual velocities, the principle of virtual work takes the following form:

f . v∗p + F. vq∗ = f v∗p − F vq∗ = 0 and thus we have: F= f

v ∗p v q∗

= f

ip ω ∗ iq ω ∗

⋅

From ip sin(90 − φ ) o

=

iq sin(θ + φ )

we deduce: F=

f cos φ = sin(θ + φ )

f cos φ f = ⋅ R cos θ sin φ sin θ (cos φ + cos θ ) sin θ (1 + ) l cos φ sin θ

Statics

125

Exercise 8.

The platform of mass m of a vehicle goes up by the application of the moment M of a couple acting about the lower end a of a jointed rod of length l = 8 (m). There are horizontal slots which allow the linkage to unfold as the platform is elevated. In equilibrium, express the norm of M as a function of the height h of the platform. Calculate this norm M if the platform 4 meters high has a mass of 8 tons. Any friction is neglected.

Fig. 58 Answer. The system has one degree of freedom and the chosen generalized coordinate is the angle θ of inclination of the rod at a. The frame of reference { a; i , j , k }, the weight mg and the moment M are shown in Fig. 58. The condition of equilibrium is

δ W = mg .δ rG + M k .δθ k = 0 where δ rG and δθ k are virtual displacements compatible with the constraints. Since the weight m g is vertical, only the j-component of rG matters, namely: l sin θ + k

where k is the height of G with respect to the platform. For any (positive) increase δ θ , the virtual work done by m g is negative and the one of M is positive. We have thus:

δ W = −mg l cos θ δθ + M δθ = 0 which implies

M = mg l cosθ

= mg l 1 − (h l ) 2 = 543171 (Nm).

126

Chapter 1

Exercise 9.

An industrial tool fastened to a plate goes up and down when a handle rotates. A rotation of constant angular velocity (uniform rotation) generates a uniform translation of the plate through mechanisms not to be described. When the handle, the arm of which is 0.18 (m) in length, turns once then the plate goes up of 0.1 (m). What force F perpendicular to the handle is necessary to hold the plate and tool of 250 (N) weight in equilibrium? Answer. This system has one degree of freedom. We choose a virtual displacement compatible with the constraints such that to any increase δ y of the height of the plate corresponds an increase of angle δθ . These increases are linked by the relation:

δθ δ y = ⋅ 2π 0 .1 The principle of virtual work, which is written as follows:

δ W = 0.18 F δθ − 250 δ y = 0 ⇔ (0.18 F

2π 0 .1

− 250) δ y = 0 ,

leads to the result: F = 22.1 (N).

This exercise proves again the efficiency and the simplicity of the method of virtual work since it is not necessary to describe the (unknown) internal mechanisms.

Exercise 10.

Two bars oa and ob of length l, in a vertical plane, are pivoted about a pin at fixed point o. An interconnected system of four vertical bars ad, db, bc, ca of same length L is jointed to the first two bars at a and b. All the massless bars, in the same plane oxy, are connected by frictionless pins. Given a force Fd applied at d, find the force Fc to apply at c in order to hold the system in equilibrium. Answer. A set of four moving particles in a plane has eight degrees of freedom a priori; but here the number of degrees of freedom is 2 because there are six constraint equations. Let { o;1X ,1Y } be the moving frame such that the X-axis is through c and d. We choose the following generalized coordinates: - the angle θ which determines the direction of oX relative to the x-axis, - the angle φ between oa (or ob) and the X-axis. We are going to introduce virtual displacements compatible with the constraints. On the one hand, we choose an arbitrary increase δθ , while φ remains constant.

127

Statics

Fig. 59 In this case, where the distance between o and c is necessarily constant (and also between o and d), from rc = oc 1X , rd = od 1X we deduce respective virtual displacements:

δ rc =

∂rc δθ = oc δθ 1Y , ∂θ

δ rd =

∂rd δθ = od δθ 1Y . ∂θ

Thus the corresponding generalized force is Qθ = Fc ⋅

∂rc ∂r + Fd ⋅ d ∂θ ∂θ

that is: Qθ = ( Fc ) Y oc + ( Fd ) Y od where the notation ( ) Y represents the projection onto the Y-axis. In equilibrium, the condition Qθ = 0 implies: ( Fc ) Y = −

od ( Fd ) Y . oc

(1)

On the other hand, we choose an arbitrary increase δφ , while θ remains constant. In this case, where c and d move closer (along X-axis), we have:

δ rc =

∂rc δφ . ∂φ

128

Chapter 1

From rc = ( l cos φ − L2 − l 2 sin 2 φ ) 1 X

we deduce the virtual displacement:

δ rc = (−l sin φ +

l 2 sin φ cos φ L − l sin φ 2

l sin φ δφ

=

L2 − l 2 sin 2 φ

2

2

) δφ 1X

( l cos φ − L2 − l 2 sin 2 φ ) 1X

and so: l oc sin φ

δ rc =

L2 − l 2 sin 2 φ

δφ 1X .

In the same manner, since od = l cos φ + L2 − l 2 sin 2 φ ,

we have: l od sin φ

δ rd = −

L2 − l 2 sin 2 φ

δφ 1X

and the generalized force is expressed as Qφ = Fc ⋅

that is Qφ =

∂rc ∂r + Fd ⋅ d ∂φ ∂φ

l sin φ L2 − l 2 sin 2 φ

( oc ( Fc ) X − od ( Fd ) X ) .

In equilibrium, the condition Qφ = 0 implies: ( Fc ) X =

od ( Fd ) X . oc

(2)

So we have: Fd = [( Fd ) X cos θ + ( Fd ) Y sin θ ] 1x + [(− Fd ) X sin θ + ( Fd ) Y cos θ ] 1 y ,

Fc =

od od [( Fd ) X cosθ − ( Fd ) Y sin θ ] 1x − [( Fd ) X sin θ + ( Fd ) Y cosθ ] 1 y oc oc

and, for instance, Fc is obtained from Fd in equilibrium.

Statics

129

Exercise 11.

A block of mass M supported by two rollers of same mass m is rolling (without slipping or pivoting) down an inclined plane, α being the angle of inclination. We assume the rollers neither slip nor pivot. Calculate the force F, applied to the block parallel to the inclined plane, sufficient to hold the system in equilibrium.

Fig. 60 Answer. The problem would be complex by using the classical method since there are four reactive forces of friction which prevent slipping. In this problem of one degree of freedom, the chosen generalized coordinate is the coordinate x of the center of mass G of the block such that δ x is positive upwards. Let us consider virtual displacements compatible with the constraints. At time t, the point i1 of a roller in contact with the incline at fixed point i2 has no velocity since the roller is not slipping. It is the instantaneous center of zero velocity. So, the virtual velocities and displacements compatible with the frictional constraints are zero:

δ ri1 = δ ri2 = 0 . It is an example of ideal constraint with friction. In the same manner, the point j 2 of the block in contact with the point j1 of the roller is such that: δ r j1 = δ r j2 . The reactive forces of constraint do no virtual work, and the principle of virtual work is expressed as follows:

δ W = Mg .δ rG + F .δ rG + m g .δ rc1 + m g .δ rc2 = 0 where c1 and c2 belong to respective roller axes.

130

Chapter 1

But for this pure rolling motion, we have at time t: 1

1

2

2

δ rc1 = δ r j2 = δ rG which is δ rc2 too. The virtual displacements are not independent since there is only one degree of freedom. The principle of virtual work is written:

δ W = − Mg sin α δ xG + F δ xG − mg sin α δ xG = 0 which implies:

F = ( M + m) g sin α .

Exercise 12.

Two particles of respective masses m1 and m2 are located on a frictionless double incline. The constant angle between the inclined planes is denoted by α. Both planes rotate about a smooth hinge, the axis of this pivot being horizontal. The particles are connected by an inextensible string (of length l and negligible mass) passing over the smooth hinge. Find the conditions of equilibrium of the particles by using the principle of virtual work.

Fig. 61 Answer. We choose the system of coordinates oxz in the vertical plane of motion. The system made up of particles has two degrees of freedom. Let us introduce the following generalized coordinates: -

the angle θ between a plane and the vertical as shown in Fig. 61, the length l1 = op1 .

We have obviously op 2 = l − l1 . The principle of virtual work is written:

δ W = m1 g .δ r1 + m2 g .δ r2 = −m1 g δ z1 − m2 g δ z 2 = 0

131

Statics

with z1 = −l1 cosθ ,

z 2 = −(l − l1 ) cos(α − θ ) .

The constraints imply the following relations:

δ (l − l1 ) = −δ l1 , δ (α − θ ) = −δθ and

δ z1 = −δ l1 cosθ + l1 sin θ δθ , δ z 2 = δ l1 cos(α − θ ) + (l − l1 ) sin(α − θ )(−δθ ) . The principle of virtual work is written:

δ W = [m1 g cosθ − m2 g cos(α − θ )]δ l1 + [−m1 g l1 sin θ + m2 g (l − l1 ) sin(α − θ )]δθ = 0. Since the general coordinates l1 and θ are independent, we deduce the two conditions of equilibrium: m1 g cos θ = m2 g cos(α − θ ) , m1 g l1 sin θ = m2 g (l − l1 ) sin(α − θ ) which express that the forces along the string have the same norm and the moments of forces about the origin o have the same norm.

Exercise 13.

A particle p of mass m, which is travelling along a frictionless vertical circle of radius R, is the end of a spring; the other end is fastened to the fixed point a( R,0) in the frame of reference { o; i , j } in the plane of the circle. Knowing that the spring of stiffness k is 2 R in length when relaxed, do equilibrium positions of p exist?

Fig. 62 Answer. The chosen generalized coordinate is the angle θ = (i , op) . The spring is always compressed and thus the elongation x is always negative except if ap = 2 R , that is x = 0 .

132

Chapter 1

The restoring force F is directed from point a toward p and is expressed as F = −k ( ap − 2 R ) 1ap . An arbitrary virtual displacement of p will be chosen so that the reactive force of constraint, which is normal to the circle, does not virtually work. Therefore, such a displacement must be tangent to the circle. From

ρ = ap = ( R cos θ − R ) i + R sin θ j

we deduce: ∂ρ = − R sin θ i + R cos θ j . ∂θ

This last vector being perpendicular to the position vector op, it is really tangent to the circle ∂ρ and so is δ p = δθ . ∂θ Let us use the second formulation of the principle of virtual work. From ∂ρ ⋅ (− mg j + 2kR 1ap − k ap) ∂θ kR 2 sin θ = − mgR cosθ + 2 − kR 2 sin θ = 0 1 − cos θ

Qθ =

and since sin θ = 2 cos

θ 2

2 sin

θ 2

= 2 cos

θ 2

1 − cos θ

we deduce: Qθ = −

mg θ cos θ − sin θ + 2 cos = 0 . kR 2

θ By letting x = cos , the previous equation is written: 2

mg (2 x 2 − 1) + 2 x 1 − x 2 − 2 x = 0. kR

By letting C =

mg , the following equation kR (1 + C 2 ) x 4 − 2C x 3 − C 2 x 2 + C x +

C =0 4

allows to obtain the equilibrium positions of the particle p. If we let f (θ ) = C cos θ + sin θ − 2 cos

we have:

f (π ) = −C < 0

θ 2

,

Statics

and f (3π 2) = −1 + 2 > 0 ,

which means there is really an equilibrium position for θ ∈ (π , 3π 2) . In the same manner, since f ( 0) = C − 2

and f (π 2) = 1 − 2 < 0 ,

if C > 2 , that is if k
0 ).

The coordinates ( x, y, z ) of this point p [with respect to the orthonormal basis (e1 , e 2 , e 3 ) ] are written: cos α cos β cos γ x = op . e1 = , y= , z= . I∆ I∆ I∆ So the expression (3-31) of I ∆ leads to the equation of a quadric of center o: I x x 2 + I y y 2 + I z z 2 − 2 Pyz yz − 2 Pzx zx − 2 Pxy xy = 1

(3-32)

that is, by letting x1 = x , x 2 = y , x 3 = z :


I ij x i x j = 1 .

Remark 1. The reader will immediately bring together

I ∆ = e .I ⋅ e and I ∆ = I ij ( I ∆ x i )( I ∆ x j ) where the expressions between brackets are the components of e.

(3-32’)

Mass Geometry, Inertia Tensor

279

Remark 2. We note that there was no dimensional homogeneity when we defined op leading to the equation of the quadric. In fact, we must introduce a scale factor a for dimensional reasons: op =

ae I∆

.

But it is usual to set this factor equal to 1. So, the construction of the ellipsoid, but not its shape is subject to the choice of units. Remark 3. We emphasize that in every change of coordinates the coefficients of a quadric ( I ij x i x j = 1 here) are the covariant components of a symmetric tensor of second order (the inertia tensor here).

Indeed, first given any change of coordinates x i = α ki x ′ k ,

the equation of the quadric becomes I ijα ki α rj x ′ k x ′ r = 1 .

Since ′ x′ k x′ r = 1 , I kr we deduce that ′ = α ki α rj I ij . I kr Inversely, the various components of a symmetric tensor ( I ij of the inertia tensor here) define the coefficients of a quadric (of equation I ij x i x j = 1 here).

4.3

NATURE OF THE QUADRIC

We are going to specify the nature of the quadric of center o which is associated with the material system S whatever the system of coordinates. In a general manner, the quadric is an ellipsoid. Indeed, if all the particles do not belong to a straight line through o, then op = 1 I ∆ is always finite and we conclude that the quadric is an ellipsoid since it is the only quadric having no point to infinity and this quadric is called: the inertia ellipsoid. In the particular case where all the mass of a material system is distributed along a straight line through o, for example the z-axis, then the only nonzero components of the inertia tensor are I x = I y = ∫ z 2 dm . S

280

Chapter 3

Except for z-axis, the previous construction of the quadric is possible and, since I ∆ = I x (cos 2 α + cos 2 β ) ,

we obtain the following equation of the quadric: I x (x2 + y 2 ) = 1 .

It is the equation of a cylinder of revolution about the z-axis. In the particular case where all the mass of a material system is distributed in a plane, for instance the (y,z)-plane, then we have: Pyz = Pzx = 0

and Iz = Ix + Iy

since I x = ∫ y 2 dm ,

I y = ∫ x 2 dm .

S

S

Thus, the equation of the inertia ellipsoid is I x x 2 + I y y 2 + ( I x + I y ) z 2 − 2 Pxy xy = 1 .

So, the intersection of this ellipsoid with the (y,z)-plane is the inertia ellipse of equations:  I x x 2 + I y y 2 + ( I x + I y ) z 2 − 2 Pxy xy = 1   z = 0 .

4.4

RADIUS OF GYRATION

D

The radius of gyration of a system S with respect to a straight line ∆ is the distance K from ∆ such that I∆ = M K 2

⇔

K = I∆ M .

(3-33)

In other words, K is the distance such that, if all the mass of S were situated at a distance K from ∆ , then the moment of inertia would be I ∆ . Example. Express the radius of gyration of a homogeneous cylinder of mass M, height h and radius R, with respect to the axis of the cylinder (z-axis).

We have immediately: I z = ∫ ( x 2 + y 2 ) dm = ρ S

and thus K=R

2.

R

∫0

2π r 3 h dr =

π 2

ρ hR 4 = M R 2 2

281

Mass Geometry, Inertia Tensor

5. PRINCIPAL AXES We know that a tensor which is in diagonal form in a particular coordinate system will generally no longer be so after a change of coordinates; but the following fundamental proposition can be proved. We are going to consider any symmetric second-order tensor T that will be the inertia tensor in particular. 5.1

FUNDAMENTAL THEOREM ABOUT SYMMETRIC TENSORS

PR5

Every real and symmetric tensor can be brought into diagonal form by an orthogonal transformation. The resulting diagonal tensor is unique except for the presentation order of its diagonal elements.

In other words, this proposition states that by a suitable choice of coordinate axes a symmetric tensor can be made diagonal. Diagonalization of a tensor is an eigenvalue problem of linear algebra and we recall: D

An eigenvector of a tensor T is any vector v parallel to the vector T ⋅ v , that is: T ⋅v = λv

(3-34)

where the real λ is the corresponding eigenvalue. Before demonstrating the fundamental proposition, we are going to prove the following property. P1

The eigenvalues of a symmetric real tensor are real.

Proof. By considering T ⋅ v = λ v , first we let the various eigenvalues λ and components of corresponding v be complex.

[The complex vector norm is v =

∗

v .v where ∗ v denotes the complex conjugate of v.]

The complex conjugate of the following equation ∗

v .T ⋅ v = λ ∗ v . v

(1)

v .T ⋅ ∗v = ∗λ ∗ v . v

(2)

is (because T is real and the scalar product is commutative). Since, from the definition of the transpose t T of T , we deduce: ( ∗ v ⋅ T ) . v =( t T ⋅ ∗ v ) . v = v . t T ⋅ ∗ v ,

and since T is symmetric, we have obtained:

282

Chapter 3 ∗

v ⋅ T . v = v .T ⋅ ∗ v ;

that is (1) = (2) and thus any eigenvalue is such that

λ = ∗λ . Now, we prove the proposition by following the general method of obtention of principal axes; that is, by making explicit (T − λ I ) ⋅ v = 0 ,

(3-34’)

 (T11 − λ ) v1 + T12 v 2 + T13 v3 = 0 ,  T21 v1 + (T22 − λ ) v 2 + T23 v3 = 0 ,  T v + T v + (T − λ ) v = 0 . 33 3  31 1 32 2

(3-35)

namely:

This linear system of equations has a nontrivial solution provided: det(T − λ I ) = 0 .

This cubic equation called the characteristic equation of the tensor has in general three solutions: the three eigenvalues λ(1) , λ( 2) and λ (3) . Any eigenvalue is substituted in the system of equations (3-35), which leads to the direction of the corresponding eigenvector v (only the direction since we obtain the ratios of values v1 , v 2 and v 3 ). For each eigenvalue λ(1) , λ ( 2) , λ (3) we consider a unit vector E1 , E 2 , E 3 respectively along the direction of the corresponding eigenvector, and we say: D


The axes defined by the unit vectors E i along the directions of eigenvalues of T are called the principal axes of T.

So, we emphasize that:


The tensor T is in diagonal form in the frame of its principal axes.

We must prove again that the previous vectors E j are perpendicular; that is: P2

The eigenvectors of a symmetric tensor corresponding to different eigenvalues are orthogonal. In other words: The principal axes of T are associated with the vectors E i of an orthonormal basis.

Proof. Given two eigenvalues λ(i ) ≠ λ( j ) of T ; that is: T ⋅ v (i ) = λ ( i ) v (i ) ,

we have:

T ⋅ v ( j ) = λ( j ) v ( j ) ,

283

Mass Geometry, Inertia Tensor v ( j ) .T ⋅ v ( i ) = λ ( i ) v ( j ) . v ( i ) , v ( i ) .T ⋅ v ( j ) = λ ( j ) v (i ) . v ( j ) .

The left-hand members of previous equations are equal since

v ( j ) .T ⋅ v ( i ) = ( t T ⋅ v ( j ) ) . v ( i ) = v ( i ) .T ⋅ v ( j ) . Therefore, from (λ ( i ) − λ ( j ) ) v ( i ) . v ( j ) = 0

we deduce: v (i ) . v ( j ) = 0

since λ(i ) ≠ λ( j ) . To conclude, we have obtained vectors E j which are expressed with respect to the vectors of the frame of reference as follows:

E j = α ij e i where the various α ij = E j . ei are the coefficients of the orthogonal transformation from { o; e i } to { o; E j }. The so-proved fundamental proposition can be particularized in the inertia tensor study; the principal axes are called the principal axes of inertia of S (at point o). PR6

The eigenvalues of the inertia tensor are the moments of inertia about the principal axes of inertia.

Proof. It is obvious since if ∆ is the axis defined by a unit eigenvector E corresponding to an eigenvalue λ , then we have: E .I ⋅ E = E . λ E = λ .

The expression of the inertia tensor, at o, with respect to the principal X,Y Z-axes, is I X 0 0  0 I 0  . Y   0 0 I Z 

(3-36)

D

A moment of inertia about a principal axis is called a principal moment of inertia.

PR7

Any principal moment of inertia cannot be higher than the sum of the two other ones.

Proof. It is obvious from the definition of the principal moment of inertia.

284

5.2

Chapter 3 EQUAL EIGENVALUES

Until now, we have dealt with the case of different eigenvalues. The so-obtained respective principal axes are unique except for degeneracy. So, the cases of double and triple roots of the characteristic equation must be to consider. The following developments are valid (as before) for any real symmetric second-order tensor; but we will particularize this tensor by considering the inertia tensor I. - First, we consider a double root: λ(i ) = λ ( j ) ≠ λ ( k ) . For instance, the principal axis corresponding to the eigenvalue λ (3) is chosen as Z-axis without restriction. PR8

If two eigenvalues are equal ( λ(1) = λ ( 2) ), then the inertia tensor is of type

I X 0 0  0 I  X 0 .   0 0 I Z 

(3-37)

Proof. In this case where v = E 3 , in the system of equations (3-35) we successively have v1 = cos α = 0, v 2 = cos β = 0, v3 = cos γ = 1 , and this system becomes:

 ( I X − λ )cos α − PXY cos β − PXZ cos γ = − PXZ = 0 ,  − PXY cos α + ( I Y − λ ) cos β − PYZ cos γ = − PYZ = 0 , − P cos α − P cos β + ( I − λ ) cos γ = I − λ = 0. YZ Z Z  XZ So, we have obtained:

λ ( 3) = I Z ,

PXZ = PYZ = 0 .

Conversely, if the products of inertia PXZ and PYZ are zero, then the Z-axis is a principal axis of inertia. So, the inertia tensor is a priori:  IX − P  XY  0

−

PXY IY 0

0 0  I Z 

with the corresponding characteristic equation: 2 ( I Z − λ ) [( I X − λ )( I Y − λ ) − PXY ] = 0.

The eigenvalues λ(1) and λ( 2) are the roots of 2 λ2 − ( I X + I Y )λ + I X I Y − PXY = 0.

Mass Geometry, Inertia Tensor

285

Since these two roots are equal, the discriminant vanishes: 2 ( I X − I Y ) 2 + 4 PXY =0

which implies: I X = IY ,

PXY = 0

and the predicted inertia tensor is found. PR9

Given a double eigenvalue, every unit vector E perpendicular to the principal vector corresponding to the other eigenvalue is an eigenvector associated with the double eigenvalue.

Proof. If λ(1) = λ ( 2) , we choose the eigenvector corresponding to λ (3) along the Z-axis. We have:  I X 0 0   cos α     I ⋅ E =  0 I X 0   cos β  = I X E  0 0 I  0  Z   

where the expression of E with respect to an orthonormal basis shows no component along the E3-axis. We have proved that E is the eigenvector associated with I X . The existence of principal axes of I follows from the preceding proposition. -

Second, we consider a triple root: λ(i ) = λ ( j ) = λ ( k ) .

We have immediately: PR10 In the case of a triple eigenvalue, every unit vector E is a unit eigenvector associated with the triple eigenvalue. Proof. This is obvious from the previous case where I X = I Z and the inertia tensor is “spherical”, namely: IX 0   0 IX  0 0 

0   0  I X 

where I X is the common eigenvalue. The existence of principal axes of I follows from this proposition.

(3-38)

286

5.3

Chapter 3 INERTIA ELLIPSOID AND PRINCIPAL AXES

The reduction of the equation of a quadric to its canonical form is a well-known problem in analytical geometry (diagonalization). We recall that the inertia ellipsoid (at o) is the locus of the extremity of the vector along every axis ∆ through o and of norm 1 I ∆ .

At a given point o, the inertia ellipsoid shows (at least) three orthogonal axes of symmetry: the principal axes of inertia of S (at point o). The canonical equation of the inertia ellipsoid with respect to the system of its principal axes oXYZ is I X X 2 + IY Y 2 + I Z Z 2 = 1 .

(3-39)

We specify that the distance from o to some point p of the ellipsoid, namely op = 1 1 I X , 1 IY , 1 of the ellipsoid.

I ∆ is

I Z for the respective principal X,Y,Z-axes. These values are the semi-axes

(i) We consider the general case where I X ≠ I Y ≠ I Z .

If for instance I X > I Y > I Z ( 1 axis), we have:

I X is thus the semi-minor axis and 1

IZ

the semi-major

I X (cos 2 α + cos 2 β + cos 2 γ ) ≥ I X cos 2 α + I Y cos 2 β + I Z cos 2 γ

that is

I X ≥ I∆ . The equality takes place if I X cos 2 β = I Y cos 2 β ,

I X cos 2 γ = I Z cos 2 γ ,

that is if cos β = cos γ = 0 , that is the X-axis. Therefore, I ∆ shows a strict maximum I X for the axis corresponding to the semi-minor axis. In the same manner, I Z (cos 2 α + cos 2 β + cos 2 γ ) ≤ I X cos 2 α + I Y cos 2 β + I Z cos 2 γ

that is

IZ ≤ I∆ and I ∆ shows a strict minimum I Z for the axis corresponding to the semi-major axis.

(ii) If two principal moments of inertia are equal, for example I X = I Y , then the equation of the quadric (at o):

Mass Geometry, Inertia Tensor

287

I X (X 2 + Y 2) + IZ Z 2 = 1

is the one of an ellipsoid of revolution about the Z-axis if I Z ≠ 0 , and the one of a cylinder if IZ = 0.

Every axis through o orthogonal to the Z-axis is a principal axis of inertia. Finally, if the three principal moments of inertia are equal I X = I Y = I Z , then the inertia ellipsoid (at o) is the sphere of equation X 2 +Y 2 + Z2 =

1 IX

⋅

Every axis through o is a principal axis of inertia. To conclude, we mention that the inertia ellipsoid (at o) is determined from the knowledge of the principal axes and the calculation of the principal axes of inertia. Next, every moment of inertia about any axis ∆ through o is known from op = 1 I ∆ . On the other hand, if the principal axes are unknown, it is necessary to calculate the six moments and products of inertia which are the coefficients of the equation of the quadric with respect to a coordinate system. Next, the equation of the quadric is reduced to canonical form (with respect to the system of principal axes).

5.4

MATERIAL SYMMETRIES

We recall that a material symmetry means a geometric symmetry as well as a mass distribution one. This well-known notion signifies, for instance, that a density is the same at two symmetric points or that two symmetric points have the same mass. In this case, the center of mass coincides with the center of symmetry. Indeed, given two particles ( x1 , m) and ( x 2 , m) , we have immediately: x1G =

1 2m

(m x1 x1 + m x1 x 2 ) =

1 2

x1 x 2 .

In the same manner, if a system has an axis of material symmetry or an axial symmetry (that is, the material system is symmetric under rotations about an axis), then the center of mass belongs to the axis. Likewise, if a system has a plane of material symmetry, then the center of mass belongs to it. For instance, for a material domain D of a (x,y)-plane of material symmetry, we have:

∫D ρ z dµ = 0 since ρ ( x, y, z ) = ρ ( x, y,− z ).

288

Chapter 3

We emphasize the following. PR11 Every plane of symmetry of a material system is orthogonal to a principal axis. Every axis of symmetry of a material system is a principal axis and the plane of other principal axes is orthogonal to this axis. Proof. First, we choose the (x,y)-plane as the plane of symmetry and we have thus ρ ( x, y, z ) = ρ ( x, y,− z ). We deduce immediately that: Pxz = Pyz = 0

and the corresponding inertia tensor is  I x − Pxy 0    I = − Pxy I y 0  .  0 0 I  z 

But a rotation about the z-axis makes this tensor diagonal and thus the principal axis is really orthogonal to the plane of symmetry corresponding to the other principal axes. In another manner, the characteristic equation is written (I z − λ)

Ix − λ −

Pxy

−

Pxy

Iy −λ

= 0,

which shows that I z is a root of this equation and thus an eigenvalue of the inertia tensor. Therefore, the z-axis associated with this value I z is really a principal axis of inertia. Second, in an analogous manner, the z-axis is principal and the two other principal axes belong to the (x,y)-plane perpendicular to the axis of symmetry.

6. STEINER’S THEOREM First, we say: D

Principal axes of inertia of a material system S with respect to the center of mass G are called central axes of inertia. The inertia ellipsoid of S at G is called the central ellipsoid of inertia.

We are going to prove that the knowledge of central moments of inertia (that is at G) determines the inertia tensor at every point. It is obvious that the moments and products of inertia vary from one to another point of the system and Steiner’s theorem is relevant to the variation of moments of inertia when parallel axes are considered.

289

Mass Geometry, Inertia Tensor

Given an orthonormal basis, we denote by { o; xi } the corresponding frame of reference. We consider another system of axes { G; X i } parallel to the previous and we set out to calculate the second-order moments at point o in function of the ones at G.

Fig. 71 Let xi(G ) denote the (fixed) coordinates of G with respect to { o; xi }. The coordinates of any point with respect to the two frames are such that

xi = xi(G ) + X i . Before Steiner’s theorem, we prove the following general formula:


I (pqo ) = I (pqG ) + M (δ pq d 2 − x (pG ) x q(G ) )

(3-40)

where I (pqo ) and I (pqG ) denote the components of the inertia tensor of S with respect to the respective coordinate systems { o; xi } and { G; X i }, M the mass of S and d = oG . Indeed, we have successively: 3

I (pqo ) = ∫ [δ pq ∑ ( xi ) 2 − x p xq ] dm S

i =1

= ∫ [δ pq ∑ ( xi(G ) + X i ) 2 − ( x (pG ) + X p )( x q(G ) + X q )] dm S

i

= ∫ [δ pq ∑ ( X i ) 2 − X p X q ] dm + Mδ pq ∑ ( xi(G ) ) 2 − Mx (pG ) xq(G ) S

i

+ 2δ pq ∑ i

xi(G )

i

∫S X i dm −

x (pG )

∫S X q dm − xq ∫S X p dm . (G )

The three last integrals vanish since they are just the components of M X (G ) with respect to { G; X i } which obviously vanish.

290

Chapter 3

Thus we have : I (pqo ) = ∫ [δ pq ∑ ( X i ) 2 − X p X q ] dm + Mδ pq d 2 − M x (pG ) x q(G ) . S

i

The previous integral represents every moment of second order relative to the frame { G; X i } of which the axes are parallel to the corresponding ones of { o; xi } and we have obtained consequently: I (pqo ) = I (pqG ) + M (δ pq d 2 − x (pG ) x q(G ) ) . Now we can state the Steiner’s theorem. PR12 The moment of inertia about any given axis ∆ is the moment of inertia about the parallel central axis C plus the moment of inertia about the given axis if all the mass were located at the center of mass. It is expressed as


I ∆ = IC + M d 2

(3-41)

where d is the distance between o and G. Proof. The moments and products of inertia about any axis parallel to the corresponding central axis are easily obtained from the ones about the central axis. For instance, if p = q = 1 : (o) (G ) (G ) I11 = I11 + M (d 2 − x1(G ) x1(G ) ) = I11 + M ( yG2 + xG2 )

where yG2 + z G2 is the square of the distance between the parallel axes ox and GX. Since there is such an equality for every coordinate axis, we have proved (3-41). The following is obvious: PR13 Among the moments of inertia relative to a set of parallel axes, the smallest is the one about the central axis. Example. The moment of a thin homogeneous rod S of length 2l about a central axis C perpendicular to the rod is l

l2

−l

3

I C = ∫ x 2 dm = ∫ ρ x 2 dx = M S

.

Steiner’s theorem confirms this result since, given the axis ∆ through the origin of the rod and parallel to C, we have: 2l

4

0

3

I ∆ = ∫ ρ x 2 dx =

Ml2

and thus IC = I∆ − M l 2 = M

l2 3

.

Mass Geometry, Inertia Tensor

291

7. EXERCISES We consider only a few exercises because the inertia tensor notion is developed in intermediate courses in mechanics at the undergraduate level (in the matrix context). Exercise 1. Given the symmetric tensor

 7 −3 − 3 − 1   0 0

0 0 8 

with respect to an orthonormal frame {o;x,y,z}, find the principal axes and the orthogonal transformation from the initial frame to the one of principal axes. Answer. The characteristic equation is immediately (λ − 8)(λ2 − 6λ − 16) = 0

and the eigenvalues are the double root λ1 = λ 2 = 8 and λ3 = −2 .

For the double eigenvalue, every eigenvector v = vi e i is obtained by solving the system − v1 − 3v2 = 0 ,  − 3v1 − 9v 2 = 0 , 0 = 0 

and from this system, we deduce v1 = −3v 2 while v3 is arbitrary. For this double eigenvalue there is obviously a two-parameter family of possible eigenvectors: v = −3v 2 e1 + v 2 e 2 + v3 e3 .

Within this family of eigenvectors, first we choose the unit eigenvector e 3 denoted E 3 and second we choose the unit eigenvector

−3 10

e1 +

1 10

e 2 denoted E 2 .

We point out that these orthogonal vectors are really eigenvectors associated with λ = 8 since we have directly: I ⋅ E2 = 8 E2 .

I ⋅ E3 = 8 E3

For the eigenvalue λ3 = −2 , every eigenvector v = vi e i is obtained by solving the system  9v1 − 3v 2 = 0 ,  − 3v1 + v2 = 0 ,  10v = 0  3

⇔

v 2 = 3v1 ,  v 3 = 0 .

292

Chapter 3

Within the one-parameter family of possible eigenvectors, we choose the unit eigenvector E1 =

1 10

e1 +

3 10

e2

so as to obtain a direct basis ( E1 , E 2 , E 3 ) or right-handed coordinate system oXYZ. We point out this vector E1 is really an eigenvector corresponding to the eigenvalue λ3 = −2 since we have immediately

I ⋅ E 1 = −2 E 1 . In conclusion, the three unit vectors E1 , E 2 , E 3 define principal axes of I. In addition, let us prove that PR9 is verified; namely: Every unit vector perpendicular to the principal vector E1 ( λ3 = −2 ) is an eigenvector corresponding to λ1 = λ 2 = 8. Indeed, E 3 = e 3 as well as E 2 =

−3 10

e1 +

1 10

e 2 are perpendicular to E1 and thus any vector

of the plane determined by E 2 and E 3 is so. Such a unit vector is written ∀a, b ∈ R : − 3a 10a 2 + b 2

e1 +

a 10a 2 + b 2

e 2 + b e3 .

It is really an eigenvector corresponding to the double eigenvalue 8 since 2 2  2 2     7 − 3 0   − 3a 10a + b   − 3a 10a + b       2 2 2 2  − 3 − 1 0   a 10a + b  = 8 a 10a + b .  0 0 8    b b         

Finally, the orthonormal basis (e1 , e 2 , e 3 ) of the initial frame oxyz is transformed into the orthonormal basis ( E1 , E 2 , E 3 ) of the frame oXYZ of principal axes. The corresponding orthogonal transformation is defined by  E1  1 10 3 10 0   e1       E 2  =  − 3 10 1 10 0   e 2  .   E   0 0 1   e 3   3  

Exercise 2.

Given a material system S determined by the illustrated homogeneous tetrahedron (of density ρ ), find the inertia tensor of S at o, (i) with respect to the frame oxyz,

293

Mass Geometry, Inertia Tensor

(ii)

with respect to the frame oXy ′z , the y ′ -axis being along the bisector of the angle (xoy ) and X –axis perpendicular to the previous.

Answer.

Fig. 72 (i) Since the element of mass of the tetrahedron is x2 dm = ρ dµ = ρ dz 2

and since x z + =1 a 2a

x = a − z 2,

⇒

we deduce that the mass of the tetrahedron is M =

ρ

2a

2 ∫0

z ρ (a − ) 2 dz = a 3 . 2

3

We note that the axes of the frame oxyz are not principal axes and thus we will have to calculate the products of inertia as follows. We note that there is a simplification since Ix = Iy,

Pxz = Pyz .

First, let us calculate:

I x = ∫ ( y 2 + z 2 ) dm = I xoy + I zox . S

From I zox = ∫ y 2 dm = ∫ x 2 dm = S

and I xoy = ∫ z 2 dm = S

we deduce:

S

ρ

2a

2 ∫0

ρ

2a

2 ∫0

z ρ 3 (a − ) 4 dz = a 5 = Ma 2 2

z 2 z 2 (a − ) 2 dz = Ma 2 2

5

5

5

294

Chapter 3

I x = Ma 2 and also I z = ∫ ( x 2 + y 2 ) dm = 2 ∫ x 2 dm = S

S

6 5

Ma 2 .

Now, we calculate the products of inertia: 2a

a− z 2

0

0

Pxy = ∫ xy dm = ρ ∫ dz ∫ S

a5

=ρ

40

3

=

40

S

19

a− x

xy dm =

0

ρ

2a

2 ∫0

dz ∫

2a

z dz ∫

a− z 2

0

x(a − x) 2 dx

Ma 2 2a

a− z 2

0

0

Pyz = ∫ yz dm = ρ ∫ dz ∫ =

dx ∫

dx ∫

a− x

0

yz dy =

ρ

2 ∫0

a− z 2

0

(a − x) 2 dx

Ma 2

80

The inertia tensor with respect to oxyz is thus written:

1

3 40

2

19 80 

 I = Ma 3 40 1 19 80 .  19 80 19 80 6 5  It is better to take the symmetry into account and to use the frame with the y ′ -axis.

(ii) The X-axis is a principal axis of inertia since it is orthogonal to the ( y ' , z )-plane which is a plane of symmetry. Let us determine the inertia tensor at o with respect to oXy ′z . The moment of inertia about the X-axis is obtained from (3-31), that is (since cos α = 2 2 , cos β = − 2 2 , cos γ = 0 ): IX =

1 2

1

43

2

40

I x + I y + Pxy =

Ma 2 .

The moment of inertia about the y ′ -axis is (since cos α = 2 2 , cos β = 2 2 , cos γ = 0 ): I y' =

1 2

1

37

2

40

I x + I y − Pxy =

Ma 2 .

Now, to find the product of inertia Py ' z , we calculate the moment of inertia about the b-axis along the bisector of the angle ( y ' oz ) in two ways. First, with respect to oxyz, since cos α = 1 2 , cos β = 1 2 , cos γ = 2 2 , we have:

295

Mass Geometry, Inertia Tensor

Ib =

1 4

=(

1

1

1

4

2

2

19 2

) Ma 2 .

I x + I y + I z − Pxy −

85 80

−

80

2 2

Pyz −

2 2

Pzx

Second, with respect to oXy' z , since cos α = 0 , cos β = 2 2 , cos γ = 2 2 , we have: Ib = =

1 2

1

I y ' + I z − Py ' z 2

85 80

Ma 2 − Py ' z .

The comparison between the two results implies: Py ' z =

19 2 80

Ma 2 .

Since the ( y ' , z ) -plane, orthogonal to the principal X-axis, is a plane of symmetry, we have: PX y ' = PX z = 0

and the tensor of inertia with respect to oXy ' z is written:  43  40   0   0 

   37 19 2  Ma 2 . −  40 80  19 2 6  −  80 5  0

0

From the corresponding characteristic equation, the reader will calculate the two last principal axes of inertia.

Exercise 3.

As shown in Fig. 73, a material system S is made up of particles of -

a conic system S1 of height h, radius R and constant density ρ1 , a cylindrical system S 2 of height H, radius R and constant density ρ 2 .

(i) Calculate the distance between the origin o and the center of mass G of S, (ii) Find the central tensor.

296

Chapter 3

Answer.

Fig. 73 (i) The height of the center of mass of the cone is easily obtained as follows. The mass of the conic system S1 being M 1 , we have:

M 1 z G1 = ∫ z dm = ρ1 ∫ S1

2π

0

R

= ρ1 2π ∫ r 0

h2 2

R

h

0

h

dθ ∫ r dr ∫

(1 −

r R

z dz

2 r2 2 R ) dr = h ρ π 1 4 R2

3

= M1 h. 4

So, with respect to (e1 , e 2 , e 3 ) , the center of mass G1 is such that oG1 =

3 4

h e3

and the center of mass G2 of S 2 is such that oG 2 = (h +

H ) e3 . 2

From

( M 1 + M 2 ) oG = M 1 oG1 + M 2 oG 2 we deduce that oG =

1 H 3 ( M 1 h + M 2 (h + )) . 4 2 M1 + M 2

(ii) The axisymmetric system about the z-axis is a principal axis and we know that I x = I y .

First, we calculate the inertia moment of S1 about the z-axis:

297

Mass Geometry, Inertia Tensor

I z(1)

= ∫ ( x + y ) dm = ρ1 ∫ 2

2

S1

=

3 10

2π

R

h hr R

dθ ∫ r dr ∫

0

3

0

dz = ρ1π

R 4h 10

M1R 2 .

Likewise, the inertia moment of S 2 about the z-axis is I z( 2) = ∫ ( x 2 + y 2 )dm = ρ 2 ∫ S2

2π

R 3

∫0 r

dθ

0

H

1

0

2

dr ∫ dz =

ρ 2π R 4 H

1

= M 2R2. 2

In conclusion, the moment of inertia of S about the x-axis is R2

I z = I z(1) + I z( 2) = (3M 1 + 5M 2 )

10

.

Next, we calculate the moment of inertia of S1 about the x-axis: I x(1) = ∫ ( y 2 + z 2 ) dm = S1

∫x

(since I x(1) = I y(1) ⇒

2

1 (1) Iz 2

+ ∫ z 2 dm S1

dm = ∫ y 2 dm ).

From

∫S

1

z 2 dm = ρ1 ∫

2π

0

R

h

1

R

5

dθ ∫ r dr ∫h r z 2 dz = ρ1 π R 2 h 3 0

3

= M 1h 2 , 5

we deduce: I x(1) = I y(1) =

3 20

M 1 ( R 2 + 4h 2 ) .

In the same manner, the moment of inertia of S 2 about the central X-axis (parallel to x-axis) is such that I X( 2) =

1 ( 2) Iz + z 2 dm . S 2 2

∫

Since 2π

I z( 2) = ∫ ( x 2 + y 2 ) dm = ∫ dθ S2

0

R

∫0

ρ r 3 dr ∫

H 2

−H 2

dz = M 2

and

∫S we deduce:

z dm = ρ 2 ∫ 2

2

2π

0

dθ

R

∫0

r dr

H 2

∫−H 2 z

2

dz = M 2

H2 12

,

R2 2

298

Chapter 3

I X( 2)

= M2(

R2 4

+

H2 12

).

Steiner’s theorem leads to the moment of inertia of S 2 about the x-axis: I x( 2) = M 2 [

R2 4

+

H2 12

+ (h +

H 2

)2 ] .

Therefore, the moment of inertia of the system S about the x-axis is I x = I x(1) + I x( 2) =

3 20

M 1 ( R 2 + 4h 2 ) + M 2 [

In conclusion, the central tensor (that is at G) is

IX 0 0     0 IX 0  0 0 I  Z   where

I X = I x − ( M 1 + M 2 ) oG and I Z = I z obviously.

2

R2 4

+

H2 12

+ (h +

H 2

)2 ] = I y .

CHAPTER 4

KINETICS AND DYNAMICS OF SYSTEMS

Kinetics, which is elaborated from kinematics and mass geometry, is dealt with notions of momentum, angular momentum and kinetic energy. Dynamics analyzes the relationships between motions and mechanical actions that are the forces. From works of Stevinus, Galileo, Kepler and Huygens, Newton stated the fundamental laws of classical mechanics in his “Philosophiae Naturalis Principia Mathematica” (1687). Later, d’Alembert, Euler, Lagrange, Laplace, Poinsot, Poisson, Jacobi, Hamilton, Poincaré (and others) made mechanics progress in connection with astronomy notably. However, atomic or galactic phenomena with velocities approaching the speed of light, enormous masses, etc. cannot be described in classical mechanics and thus require new mechanics. But innumerable applications of engineering and astronautics for instance reveal the importance of classical dynamics nowadays. The following postulates form the axiomatics of this branch.

1. NEWTON’S POSTULATES We recall that the frame of classical mechanics is an affine Euclidean space, which is homogeneous (no privileged point a priori) and isotropic (no privileged direction a priori). The position of every particle of a material system S is a point of this space at a given time. A corresponding vector space is defined by relating any position to a reference point (observer at the same time); that is, by choosing an origin of the space. Let E be a Euclidean point space, E be the vector space associated with E, R be a frame of reference, Ω be an open of E, I be an interval of R.

299

300

Chapter 4

Given a system S of N particles p h defined by N points x1 ,..., x N of respective position vectors ox h (t ) = rh (t ) and respective masses m1 ,..., m N , we say: D

A motion of S is a set of N vector functions of class C 2 rh : J (⊆ I ) → Ω : t
rh (t ) = x h such that: ∀t ∈ J , ∀h ≠ k : x h ≠ x k .

D

A force acting on p h (with respect to R ) is a function of class C 1

f h : I × Ω N × E N → E : (t , x1 ,..., x N , x
1 ,..., x
N )  f h (t , x1 ,..., x
N ) . A material domain of N particles is so defined. Notation. The velocity of any xh with respect to R is denoted by v h .

1.1

EXPERIMENTAL LAWS

Let r( q ) (t ) denote the position vector of a particle (or body), at the time t, of a qth experiment. We consider the three following experimental phenomena. (i) A body is fastened to the end of a vertical spring of which the other extremity is fixed. If the body has various vertical motions, we observe that the acceleration is always proportional to the elongation. With another body, we observe the same experimental law but the acceleration is multiplied by another coefficient. For every experiment, this law is expressed as follows: m
r
( q ) (t ) = − k r( q ) (t )

where the positive coefficient m, peculiar to each body, is called the mass of the body and the constant k concerning the spring is called the stiffness. An inductive reasoning, which postulates that such experimental results will be always obtained, leads to the following conclusion: A motion r : I (⊆ R) → E : t
r (t ) ,

with initial conditions r (t 0 ) = r0 and r
(t 0 ) = v 0 (given r0 and v 0 ), mathematically verifies the differential equation: m r

= − k r .

Kinetics and Dynamics of Systems

301

(ii) At the beginning of the 17th, Galileo which made observations concerning bodies in free fall and motions on inclined planes concluded that heavy bodies fall with the same acceleration. For every experiment, this law is expressed with the gravity acceleration g as follows:
r
( q ) (t ) = g

An inductive reasoning leads to a theoretical law given by the differential equation: m
r
= m g .

(iii) From Kepler’s laws (deduced from Tycho Brahe’s observations of planetary motions), Newton found an experimental law which holds for all the planets of the solar system: r( q ) (t )


r
( q ) (t ) = − k

r( q ) (t )

3

where k is a constant of proportionality. This is valid in the frame of reference with origin at the center of mass of the solar system and axes in the directions of three “fixed” stars. An inductive reasoning leads to the following differential equation m r

= −k m

1r r

2

⋅

To sum up, the three previous experiments lead to a common experimental law, namely: Given the observed motions, there is a vector function f such that m
r
( q ) (t ) = f (m, r( q ) (t ))

where the vectors are referred to a well-determined frame of reference (to be specified). An inductive reasoning introduces in all the cases a differential equation of type: m
r
(t ) = f (m, r (t ))

with the initial conditions r (t 0 ) = r0 and r
(t 0 ) = v 0 . In other experiments, the velocity-vector r
(t ) can occur and time can explicitly appear too. Therefore, the differential equation ensuing from experiments is written (with respect to R ): m
r
(t ) = f (t , r (t ), r
(t ))

(4-1)

where the coefficient m is removed from f since it is constant during the motion.

1.2

POSTULATES

(i)

Postulate of Initial Conditions

The conclusions of works of Galileo (1638) can be stated as follows: PO

The position vector of any particle is determined at every time if the position and the velocity of the particle are known at initial time.

302

Chapter 4

So, the acceleration vector of the particle is determined at every time too.

(ii)

Newton’s Second Law (Motion Law)

In a way, Newton introduced the notion of momentum. Newton’s second law also called the fundamental principle of dynamics can be stated as follows: PO

The time derivative of the linear momentum of every particle is collinear to the resultant force acting on the particle in an absolute frame of reference.1

We denote: f =

d (m r
) . dt

(4-2)

If the mass is constant, this postulate expresses that the differential equation f (t, r(t), r
(t)) = m
r
(t)

must be verified for every motion defined by r : I → Ω : t
r (t ) .

The coefficient of proportionality between the force f and the acceleration r

defines the mass. This law is valid in relativistic mechanics. (iii)

Newton’s First Law (Inertia Law)

The previous motion law of Newton has the following corollary also called the law of inertia, namely: PO

In the absence of forces acting on a particle, its linear momentum is constant.

We denote: f =0

⇒

m r
= c

(constant vector c).

(4-3)

This law is valid in relativistic mechanics. If the mass is invariable, this law is written: f =0

⇒

r
= k

(constant vector k)

and it is the famous Galilean principle of inertia: PO

A particle remains at rest or continues to move in a straight line with a uniform velocity if no force acts on it.

With this principle, Galileo puts an end to the false Aristotelian principle which claimed that every motion (even uniform rectilinear motions) must be caused. 1

This well-known notion is recalled later.

Kinetics and Dynamics of Systems

303

(iv)

Newton’s Third Law (Principle of Action and Reaction)

PO

The forces of action and reaction between interacting particles are collinear, opposite in direction and equal in norm.

Explicitly, if f hk denotes the force exerted by the particle p k = ( x k , mk ) on the particle p h = ( x h , mh ) , we have: ∀x k ≠ x h , ∃ c hk ∈ R+ :

f hk = c hk x h x k , f hk = − f kh .

(4-4)

This principle is not valid in relativistic mechanics. (v)

Law of “Parallelogram of Forces”

Stevinus introduced the well-known “law of parallelogram of forces” in statics (1586) and Newton expressed it in dynamics later. It is a corollary for Newton: CO

If several forces act on a particle they are combined according to the “parallelogram law of addition of forces.”

1.3

GALILEAN RELATIVITY AND INERTIAL FRAMES

The postulates of dynamics are implicitly referred to a frame given the presence of velocities and accelerations. More precisely, the fundamental law of dynamics of N particles: mh
r
h = f h (t , r1 ,..., rN , r
1 ,..., r
N )

h = 1,..., N

defines a motion of the material system corresponding to the 2N initial conditions:

rh (t 0 ) = rh0 ,

r
h (t 0 ) = v h0 ,

given rh0 and v h0 at t 0 . If this law holds in a given frame, then it will not be verified in another frame accelerated with respect to the previous one. An essential problem is to find a class of frames of reference for which the three Newton’s laws hold. D

Every frame of reference such that Newton’s laws hold is called an absolute frame of reference.

This cannot be checked directly, this can be made by calculating the motions from the postulates and next by comparing with experimental results.

304

Chapter 4

So a local reference frame attached to the surface of the earth; that is, a trihedron with axes in the apparent vertical, south and east directions, is suitable for most problems of classical mechanics. Time is measured from the rotation of earth (sideral time). But there are discrepancies between calculations from postulates and observations. For example, there exists deviations of landing places of falling bodies, of rockets and so on. Discrepancies between observations and theoretical predictions vanish if a heliocentric trihedron is chosen; that is, such that its origin is the center of mass of the solar system and the directions of axes are fixed with respect to the stars. This heliocentric trihedron, also called the trihedron of Copernicus, and the sideral time explain almost all the phenomena. However, for instance, we observe that the moon is early with respect to the calculated time. This anomaly follows from the variation of the rotation of earth (tides, etc.) and vanishes if the atomic clock is chosen. The “atomic time” is more accurate than astronomical time (earth’s period); it corresponds to the period of an atomic vibration (see physics and astronomy). We conclude: PR1

The truth of the postulates of classical mechanics was always held, but successive changes of frames of reference were necessary. The trihedron of Copernicus and atomic time define an absolute frame of reference.

Now let us recall the essential notion of inertial frame. Let

R = { o; e , e , e } denote an absolute frame of reference. 1

2

3

Question. How is the expression of the fundamental principle of classical dynamics in an orthonormal frame R E = { O(t ); E1 (t ), E 2 (t ), E 3 (t ) } moving with respect to R ? We assume that the functions O(t ) and E i (t ) are of class C 2 .

We know that the acceleration of any point x in R (called the absolute acceleration) is written: a A = a R + aT + aC where the relative acceleration a R is the acceleration of x with respect to the moving frame R E , a T denotes the transport acceleration of x and a C is the well-known Coriolis acceleration a C = 2ω ∧ v R where v R is the relative velocity of x. If f A denotes the resultant of forces acting on x for an observer of R, then the fundamental principle of dynamics, namely: f

is written:


A

= ma A

maR = f

A

− m aT − m aC .

(4-5)

D  A frame of reference is said to be inertial if the fundamental principle of dynamics has the same form as in the frame of Copernicus (with the same clock):


R E inertial iff

maR = f A .

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305

There is a necessary and sufficient condition for inertial frame existence which is called the principle of Galilean relativity. More precisely, let us prove the following: PR2

A frame is inertial iff it moves with constant velocity vector with respect to the Copernicus frame R (uniform rectilinear translation).

Proof. First, the hypothesis maR = f

A

implies ⇔

aT + aC = 0

d 2 oO dt 2

R

(since m ≠ 0 )

dω dt

+

R

∧ Ox + ω ∧ (ω ∧ Ox ) + 2ω ∧ v R = 0 .

This equation being verified for any point x, in particular for O, we have: Ox = v = 0

⇒

R

d 2 oO dt 2

R

= 0.

Thus, we have for any x: dω dt

R

∧ Ox + ω ∧ (ω ∧ Ox ) + 2ω ∧ v R = 0 .

This equation being verified for any point x, in particular for x such that Ox = E1 , we have dω v R = 0 in this case; thus, by letting ω
= R , this equation becomes: dt

ω
∧ E1 + ω ∧ (ω ∧ E1 ) = 0 that is: E1

E2

E3

ω


ω


ω
3 + (ω. E1 )ω − ω E1 = 0 .

1

1

2

0

2

0

By considering the components of the vectors of this equations with respect to the moving basis of R E , the resulting system: 0 + ω 1ω 1 − ω

2

= 0,

ω
3 + ω 1ω 2 = 0 , − ω
2 + ω 1ω 3 = 0

has the following solution:

(ω 1 ) 2 = ω

2

,

ω2 =ω3 = 0.

306

Chapter 4

In the same manner, if x is such that Ox = E 2 in particular, we obtain ω 1 = 0 . Thus the various results lead to:

ω= 0. In conclusion, a moving inertial frame R E = { O; E1 , E 2 , E 3 } is such that: d 2 oO dt 2

R

=0

ω=0,

which means that R E moves in translation with constant velocity vector with respect to R, it is said to be in uniform rectilinear translation. Conversely, if the motion of R E is a uniform rectilinear translation with respect to a frame of Copernicus, that is: d 2 oO dt 2

R

=0

ω= 0,

then we deduce: aT = a C = 0 .

Therefore, the frame R E is inertial since the fundamental principle of dynamics is written as in the frame of Copernicus, namely: maR = f a .

From the previous proposition, we conclude: PR3

All the inertial frames (which are in uniform rectilinear translation) are equivalent.

Inertia law (or Newton’s first law) being held, no unique frame can be “privileged” and a class of inertial frames is so defined. In other words, a frame moving in uniform rectilinear translation with respect to an absolute frame of reference is absolute. The conclusions of classical mechanics are the same in both frames. So, for various observers of different absolute frames, the force exerted on any particle is invariable and we recall: PR4

The laws of classical mechanics are invariable through any change of inertial frame.

Remark. Concerning noninertial frames of reference, the principle of action and reaction is not valid!

Indeed, given any pair (a,b) of isolated mass points, this principle f ab + f ba = 0

does not hold with respect to a noninertial frame since, in such a frame, the forces are expressed as Fab = f ab − f T − f C , Fba = f ba − f T − f C

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307

and thus we have:

Fab + Fba ≠ 0 . In practice, noninertial frames of reference are often used provided the forces of transport and of Coriolis can be neglected. This is not always possible as the pendulum and gyroscope of Foucault prove it.

2. KINETICS Let us introduce the well-known notions of linear momentum and angular momentum, elements of a dynam, as well as the one of kinetic energy.

R

Given an inertial frame of reference = { o; e1 , e 2 , e3 }, we denote by v h the velocity of any particle ( x h , mh ) of a system S of N particles.

2.1

KINETIC DYNAM

D

The linear momentum of S with respect to R is the vector sum of momenta of all particles: N

P = ∑ mh v h

[or

h =1

∫S v ( x) dm ] .

(4-6)

By denoting any position vector rh = ox h , we say: D

The angular momentum of S about o with respect to R is the vector sum of momenta of all particles: N

Lo = ∑ rh ∧ mh v h

[or

h =1

D

∫S r ∧ v ( x) dm ] .

The kinetic dynam of S with respect to R is defined by its following elements of reduction at point o: P L .  o

Remark. We really have:

La = Lb + ab ∧ P since

(4-7)

∑ ax h ∧ mh v h = ab ∧ ∑ mh v h + ∑ bx h ∧ mh v h . h

h

h

308

Chapter 4

Angular momentum expression

From Lo = ∑ rh ∧ mh h

drh d = ∑ (oG + Gx h ) ∧ mh (oG + Gx h ) dt dt h d d Gx h + ∑ Gx h ∧ mh v G + ∑ Gx h ∧ mh Gx h dt dt h h

= ∑ oG ∧ mh v G + ∑ oG ∧ mh h

h

and because (invariable masses): oG ∧

d (∑ mh Gx h ) = 0 , dt h

∑ mh Gx h ∧ v G = 0 , h

we deduce: Lo = oG ∧ M v G + ∑ Gx h ∧ mh (v h − v G ) .

(4-8)

h

Koenig expression of angular momentum

Let R e′ = { G; e1′ (t ), e ′2 (t ), e 3′ (t ) } be a frame in translation with respect to R and the origin of which is the moving center of mass G of S. PR5

The angular momentum of S about o is the moment about o of the linear momentum of the total mass concentrated at G plus the angular momentum, about G, of S in relative motion with respect to R e′ : N

Lo = oG ∧ M v G + ∑ Gx h ∧ mh v hR .

(4-9)

h =1

The expression “in relative motion with respect to R e′ ” means that the angular momentum is expressed in function of velocities v hR relative to R e′ . Proof. The velocity v h of any point xh with respect to R is

v h = v G + v hR since the transport velocity is the velocity of the origin of R e′ in this particular case. From (4-8) we deduce the announced expression (4-9)

2.2

KINETIC ENERGY

D

The kinetic energy of a system S with respect to R is the sum of kinetic energies of all particles: T=

1 N

∑ 2

h =1

mh v h2

(or T =

1

∫ v 2 S

2

( x) dm ).

(4-10)

309

Kinetics and Dynamics of Systems

Koenig expression of kinetic energy We consider again the translating frame R 'e . PR6

The kinetic energy of S equals the energy of mass-center translation of S as a whole plus the energy due to motion of all particles relative to the center of mass: T=

1 2

1

M v G2 +

2

N

∑ mh (v hR ) 2 .

(4-11)

h =1

Proof. Kinetic energy is written: T=

d

1

mh [ (oG + Gx h )]2 ∑ dt 2 h

=

1

∑ mh [( 2 h

dGx h 2 doG 2 doG dGx h ) +2 ) ]. ⋅ +( dt dt dt dt

The second term of the sum vanishes since:

∑ mh h

doG dGx h doG d ⋅ = ⋅ ∑ mh Gx h = 0 dt dt dt dt h

and the proposition is thus proved if we recall that the velocity of any xh relative to the translated frame R e′ is denoted v hR .

3. THEOREMS OF MECHANICS OF SYSTEMS Theorems of linear momentum, angular momentum and kinetic energy follow from the postulates of classical mechanics. In an inertial frame of reference

3.1

R = { o; e , e , e } we consider a system S of N particles. 1

2

3

FIRST INTEGRALS OF A SYSTEM OF PARTICLES

The exact description of individual motions of N particles in gravitational interaction is generally impossible. The three-body problem cannot be already solved in the general case. The N bodies obey to the Newton’s law of gravitation. We recall that the long-range forces of attraction of gravity cannot be removed (for example with the help of a screen as for electromagnetic interactions). The N-body problem consists in determining the position and velocity of every particle, at each instant, from initial conditions that are the position and velocity of every particle at a given time t 0 .

310

Chapter 4

This problem is fundamental in planetary astronomy, celestial mechanics, stellar dynamics, plasma physics for instance. Let us show its equations. In the inertial frame of reference R, we consider N points xh of respective masses mh and coordinates ( x h , y h , z h ) . Each particle p h is attracted by the N − 1 other particles with a force: fh =

N

∑

G

mh mk rhk2

 k =1 k ≠h 

(4-12)

1hk

where

1hk =

rhk rhk

( rhk = x h x k ).

We have thus N differential equations: mh

d 2 rh dt

2

=

N

∑G

 k =1 k ≠ h 

mh mk rhk3

rhk .

Projections onto the coordinate axes lead to 3N differential equations, namely: mh

d 2 xh dt

2

=

G mh mk ( x k − x h )

N

∑

 k =1 k ≠ h 

(( x k − x h ) + (( y k − y h ) 2 + ( z k − z h ) 2 ) 3 2 2

and similar others in y and z. So, we have obtained a system of 3N second-order differential equations or 6N first-order equations to be solved with 6N unknowns which are the positions and velocities of particles. Therefore, 6N integrations are necessary, unless we find first integrals. We recall the following D

A first integral of motion equations is a function of time, positions and velocities which remains constant during the motion.

3.2

LINEAR MOMENTUM THEOREMS

3.2.1 Linear Momentum Theorem PR7

Given an inertial frame, the time rate of change of linear momentum of S equals the resultant of external forces on S: dP = f (e ) . dt

Proof. With respect to an inertial frame, we have:

(4-13)

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311

dP d = ∑ (mh v h ) dt h dt

and from the fundamental principle of dynamics (Newton’s second law) we deduce: dP = ∑ f h(e ) dt h

where only the external forces must be taken into account since we assume that the internal forces act along the line joining every pair of particles and are equal in norm and opposite in direction (Newton’s third law). The theorem is thus proved. Remark. We note that the impulse delivered by the resultant of external forces is the following increase: P (t ) − P (t 0 ) = ∫

t

t0

f ( e ) dt .

(4-14)

3.2.2 Theorem of Conservation of Mass PR8

In the absence of external forces, the linear momentum of S remains unchanged in an inertial frame.

Proof. The implication f (e ) = 0

⇒

d dt

∑ mh v h = 0 h

leads to the following first-order differential equation: P = ∑ mh v h = c

(constant vector).

(4-15)

h

Example. A gun recoils as soon as the explosion occurs since only internal forces play a role and the system weapon-projectile is initially at rest ( c = 0 ).

3.2.3 Theorem of Motion of Mass Center First, the linear momentum of a system S of constant masses is the product of the mass of S and the velocity of mass center:1


P = M vG

(4-16)

since P = ∑ mh v h = h

1

We denote

v (G ) by v G .

d d (∑ mh rh ) = ( M oG ) = M v G . dt h dt

312

Chapter 4

Second, we say: PR9

In an inertial frame, the center of mass of S moves as a particle whose mass is the one of S and on which acts a force equal to the resultant of external forces exerted on S:




M

dv G = f (e ) . dt

Proof. This is obvious from the equation M

(4-17) dv G dP and PR7. = dt dt

Remark. Since the internal forces do not act, this theorem is interesting to study complicated systems as the human body for instance. Examples. The center of mass of an exploded space capsule continues to follow its orbit since the explosion gives rise to only internal forces. In the same manner, internal forces lead the driving wheels of a vehicle (at rest) to spin on the ice, since the center of mass must remain at rest. The moving off is possible if there are external friction forces (opposed to the rotation forces).

3.2.4

Special Case of Rigid Bodies

We consider a rigid body B which is free to translate or rotate with respect to an inertial frame of reference R. Let R E = { O(t ); E1 (t ), E 2 (t ), E 3 (t ) } be a frame “fixed” in the body where O ∈ B and the various E i depend on time (with respect to R ). Notation. Unless otherwise specified, the times derivatives (and thus velocities) are related to the inertial frame of reference R. Angular velocity tensor We know that the various vectors

dE i can be expressed with respect to the moving basis dt

( E i ) as dE i = ω ij E j dt

i, j = 1,...,3.

PR10 The components ω ij are the components of a tensor, denoted by ω . Proof.

(4-18)

( )-tensor called the angular velocity 1 1

Given any change of bases of the moving frame (said to be a change of basis connected to B): E ′j = α ij E i ,

E k = β ks E s′

Kinetics and Dynamics of Systems

where the various α ij are constants, then the vectors respect to the basis ( E ′j ) as follows: dE ′j dt

dE ′j dt

313

are obviously expressed with

= ω ′jk E k′ .

From

ω ′jk E k′ =

dE ′j dt

= α ij

dE i = α ij ω ir E r = α ij ω ir β rk E k′ dt

we deduce the expected result:

ω ′jk = α ij β rk ω ir . This proposition being proved, let us make explicit the components of angular velocity tensor.

Fig. 74 Let ox be the position vector of any x ∈ B : ox = oO + Ox

where O is a “fixed” point in B. We have:

ox = oO + X i E i where the components X i of X = Ox are fixed with respect to ( E i ) . At time t, the velocity of point x (with respect to R) is v ( x) = v (O ) + X i

dE i dt

also denoted: v x = vO + X i

dE i , dt

i = 1,2,3

314

Chapter 4

and thus, the components of this velocity vector relative to the basis ( E i ) of R E “fixed” in B are expressed as

v j ( x) = v j (O) + ω ij X i where the various ω ij are the components of the

(4-19)

( )-tensor of angular velocity. 1 1

The covariant components of ω are

ω ij = g hj ω ih = ω ih E h . E j which implies


ω ij =

dE i ⋅Ej dt

(4-20)

Remark. It is obvious that no confusion is possible between the angular velocity tensor ω and the symbol of components ω i j = Γ ikj du k encountered in Section 5.3 of Chapter 2; in this

last case the corresponding elements α ij are not constant. PR11 Angular velocity tensor ω is antisymmetric. Proof. Since E i . E j is constant (for example E i . E j = δ ij for an orthonormal basis), we deduce: dE j dE i d ( Ei . E j ) = ⋅ E j + Ei ⋅ =0 dt dt dt that is:

ω ij = −ω ji .

(4-21)

Angular velocity vector

Let us consider a basis ( E1 , E 2 , E 3 ) of a Euclidean space connected to B.

()

We know that the adjoint of the 02 -tensor of components ω ij is a vector, denoted Ω, such that 1 Ω i= ε ( jk )i ω ( jk ) det g and called the angular velocity vector. We note:

ω jk = det g ε ijk Ω i . In Section 5.5.2 of Chapter 2 we defined the adjoint of exterior product of two vectors. By considering formula (2-114), given vectors of components Ω i and X j , the following expression:

Kinetics and Dynamics of Systems det g ε ijk Ω i X

315

j

represents the covariant components of the vector product Ω ∧ Ox

(also denoted Ω ∧ X ).

Notation. According to usage, we denote the angular velocity vector Ω by ω, no confusion with the angular velocity tensor being possible. In conclusion, we have shown the general coordinate presentation as well as the vector presentation of velocity field of points of a rigid body, which is really:


v ( x) = v (O ) + ω ∧ X

(4-22)

that is: v ( x) = v (O) + X i (ω ∧ E i ) = v (O) + X i

dE i dt

that is also written:


v ( x) = v (O) + ω ij X i E j

(4-23)

[cf. (4-19)]. We note that vector ω is an axial vector; that is, a vector whose directional sense depends on the handedness of the frame, just as the vector product! But, the use of the angular velocity tensor does not require any convention as regards to the frame orientation and lets use any coordinate system.. Example. Make explicit the components of the linear momentum P in function of v G for any rigid body. The linear momentum is written: P = ∫ v ( x) dm S

= M v (O) + ∫ ω ∧ X dm . S

Since OG =

we have:

1

∫

M S

X dm ,

P = M v (O) + M ω ∧ OG .

From the following derivative (with respect to R): doG = v G = v (O) + ω ∧ OG , dt

we find again: P = M vG .

316

Chapter 4

To conclude, we emphasize that in any coordinate system, the covariant components of ω ∧ OG are: det g ε ijk ω i X Gj . But, in an orthonormal basis of Euclidean space, contravariance and covariance are indistinguishable (variance is indifferent) and det g = 1 ; the components of P are

P k = M (v k + ε ijk ω i X Gj ) .

(4-24a)

O

For instance:

P1 = M [v1o + (ε 231 ω 2 X G3 + ε 321 ω 3 X G2 )] = M [v O1 + (ω ∧ OG )1 ] . In an equivalent manner, we have:


i P i = M (vO − ε ijk X Gj ω k )

(4-24b)

since i i P i = M (vO + ε jkiω j X Gk ) = M (vO + ε kji X Gj ω k ) .

3.3

ANGULAR MOMENTUM THEOREMS

Let us consider any system S of N mass points ( x h , mh ) . 3.3.1 Angular Momentum Theorem

PR12 In an inertial frame, the time derivative of angular momentum of S about any point a is equal to the moment about a of the external forces plus the vector product of the linear momentum of S and velocity of a: N d La = ∑ (ax h ∧ f h( e ) ) + P ∧ v a . dt h =1

Proof. Given an inertial frame of reference principle of dynamics is written:

(4-25)

R = { o; e , e , e }, we know the fundamental 1

2

3

d (mh v h ) = f h(e ) + f h(i ) . dt

Given any a in R, we have: d d La = ∑ (ax h ∧ mh v h ) dt h dt

= ∑ [( h

d ox h d oa d − ) ∧ mh v h ] + ∑ [ax h ∧ (mh v h )] dt dt dt h

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317

= −v a ∧ ∑ mh v h + ∑ [ax h ∧ ( f h( e) + f h(i ) )] h

h

= −v a ∧ P + ∑ (ax h ∧ f h(e ) ) , h

since we assume that the internal forces act along the line joining every pair of particles and are equal in norm and opposite in direction (Newton’s third law).

3.3.2 Relation between Kinetic Dynam and Dynam of Forces

The applications for which the term P ∧ v a vanishes are essential. It is the case where the reference point a is fixed (and thus chosen as origin o of R); then, expression (4-25) becomes the following d Lo = ∑ rh ∧ f h(e ) dt h

also written: d Lo = M o(e ) . dt

(4-26a)

It is also the case where the mass center G is chosen as reference point a. Indeed, if the masses are invariable, theorem of mass center ( P = M v G ) leads to the similar result: d LG = M G(e ) . dt

(4-26b)

Therefore, we say: PR13 In an inertial frame, the time rate of change of angular momentum of S about a fixed point or about the mass center equals the moment about the concerned point of the resultant of external forces acting on S. We note that the term P ∧ v a also vanishes if v a is collinear to v G , but this case is less interesting. In conclusion, there is the following relationship between kinetic dynam and dynam of forces:


d P 

 

dt  L  o

G

 f e)  =   . (e)  M  o G

Remark. Later, we will consider the notion of dynamic dynam defined as following.

(4-27)

318

D

Chapter 4 The dynamic dynam is the dynam defined by its elements of reduction at o fixed in the inertial frame: n   ∑ mh a h   h=1   n   ∑ (oa h ∧ mh a h )  o  h =1

where a h is the acceleration of the point xh of mass mh , or more generally:  a dm   ∫S   r ∧ a dm  ∫S o

where r is the position vector of any point x with acceleration a. We have:

 a dm  d P  .  ∫S = dt  L  o  r ∧ a dm G  Go  ∫S We find again Eq. (4-27) since the dynam of internal forces vanishes.

3.3.3 Conservation of Angular Momentum

PR14 In an inertial frame, the angular momentum of S about a fixed point or the mass center is constant if the sum of moments of external forces is zero:

M o(e) = 0

⇒

Lo = c

G

(constant).

G

Proof. It is obvious from the previous result. This theorem of conservation means that the internal forces of S cannot change the angular momentum of S, but we recall this is valid if the principle of action and reaction holds. Remark. The external moment M o(e) must not necessarily be zero so that a first integral exists; it is sufficient that one of its components vanishes. Example. Let us consider a system which rotates about the axis oz with an angular velocity ω.

We have:

Lo = ∑ (mh ox h ∧ v h ) = ∑ (mh ox h ∧ (ω ∧ ox h )) h

h

= ∑ (mh ox h ∧ (ω ∧ o h x h )) h

where oh is the projection of xh on the z-axis.

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Kinetics and Dynamics of Systems

From

Lo = ∑ mh [(ox h . o h x h )ω − (ox h .ω) o h x h ] h

and since o h x h is perpendicular to the z-axis, we deduce that the projection of Lo onto this axis is ( Lo ) z = ∑ mh (ox h . o h x h ) ω = ∑ mh o h x h ω 2

h

h

= I zω . We find again a well-known result where I z is the moment of inertia about the z-axis. If the component of the moment of all external forces along the z-axis (also called the total external torque about the z-axis) is zero, then the previous remark implies the following constant: Izω = c and two conclusions are possible: (i) If the system is a rigid body, then the moment of inertia about the axis is constant and the rigid body necessarily rotates with a constant angular velocity. (ii) If masses of the system move away from the axis, then I z increases and thus ω decreases in the same proportion; on the other hand, if I z decreases then ω increases in the same proportion. Examples illustrate these conclusions, as rotation of skaters, rotation of rotor blades of helicopters, etc. In first approximation, if we consider that the solar system is isolated from the rest of the galaxy, then the (total) angular momentum of planets about the mass center of the solar system is conserved.

3.3.4

Special Case of Rigid Bodies

We consider a rigid body B which is free to translate or rotate with respect to an inertial frame of reference = { o; e1 , e 2 , e3 }.

R

The expression of the angular momentum of B about O is

LO = ∫ Ox ∧ v x dm S

where v x denotes the velocity of x with respect to R. The components of LO relative to the basis (e i ) are

(LO ) i = ε ipq ∫ X p v q dm , S

where the components of the velocity v x = v O + ω ∧ Ox relative to (e i ) are v q = vOq + ε qrs ω r X s . Thus we have

320

Chapter 4

( LO ) i = ε ipq ∫ X p (vOq + ε qrsω r X s ) dm S

= ε ipq vOq ∫ X p dm + ε ipq ε qrs ∫ X pω r X s dm . S

S

By choosing an orthonormal frame of reference R, the first term is obviously:

ε ipq M X Gp vOq = M (OG ∧ v O ) i . Since

ε ipq ε qrs = δ ir δ ps − δ isδ pr , the second term is written:

δ ir δ ps ∫ X pω r X s dm − δ is δ pr ∫ X pω r X s dm S

=ω

r

∫S (δ ir X s X

S

s

− X r X i ) dm

and the inertia tensor I ir appears. In conclusion, we have obtained: (LO ) i = ε ipq M X Gp vOq + I ir ω r

(4-28)

( LO ) i = M (OG ∧ v O ) i + I ir ω r .

(4-29)

that is:

These general equations relate components of L and ω with respect to R. If the rigid body has a fixed point, this is chosen as reference point and thus v O = 0 . Otherwise, the center of mass G is taken as reference point O and thus OG = 0 . In both situations, by denoting the reference point by O, we say: PR15 In a frame of reference R, each component of the angular momentum about a fixed point of a rigid body or, failing this, about the center of mass is a linear function of all components of the angular velocity vector: ( LO ) i = I i1ω 1 + I i 2ω 2 + I i 3ω 3 .

(4-30)

Proof. The hypothesis reduces Eq. (4-29) to the following (LO ) i = I ir ω r and thus ( LO ) i e i = I ir ω r e i

(

∑ ), i

that is explicitly: r 1  I 11 I12 I13   ω   I1r ω   L1    2   r  L2  =  I 21 I 22 I 23   ω  =  I 2 r ω  .     I  3   L  r  I I ω I ω 3 33    O  31 32    3r 

Kinetics and Dynamics of Systems

321

We have obtained the essential result:


LO = I ⋅ ω .

(4-31)

This relation shows the linear transformation defined by the inertia tensor I and expresses the angular momentum LO in function of ω. We know that vectors LO and ω have not the same dimensions, more precisely we recall that the dimensions of the second-order tensor I are ML2 . The three equations (4-30) refer to axes considered as fixed in space. The inertia tensor is obviously not constant with reference to these axes, it changes when the body rotates. This last difficulty may be avoided by choosing a set of axes “fixed” in the rigid body; then the corresponding components of I remain evidently constant during the motion of B. This is all the more helpful since angular momentum theorem makes use of the time derivative of angular momentum. The problem is more simplified if these axes are chosen as principal axes since the inertia tensor is diagonal. The inertia tensor is written with respect to the frame R E = { O; E1 , E 2 , E 3 } “fixed” in B and made up of principal X,Y,Z-axes as following:

I X 0 0  0 I 0  Y   0 0 I Z  where I X , I Y and I Z are constants.

In general use, the angular velocity vector is written with respect to the frame of principal axes as: ω = p(t ) E1 (t ) + q(t ) E 2 (t ) + r (t ) E 3 (t ) and the angular moment about O is expressed as


LO = I X p(t ) E1 (t ) + I Y q(t ) E 2 (t ) + I Z r (t ) E 3 (t ) .

(4-32)

The last two expressions clearly show that ω and LO are not in general parallel. However, there are two important special cases where LO and ω are collinear: (i) If the three principal moments of inertia are equal (the corresponding ellipsoid of inertia is a sphere), then: LO = I ω

(4-33)

for any direction of ω . (ii) If the rigid body rotates about one of the principal axes of inertia (for instance the Z-axis), then two components of ω vanish (for instance p and q) and we have necessarily: LO = I Z ω

where I Z is the principal moment of inertia about the axis of rotation.

(4-34)

322

Chapter 4

Remark and Example. A principal axis of inertia is sometimes defined as any axis of rotation parallel to the angular momentum. Let us illustrate this seeing by considering a system composed of two spheres of mass m which turn about the axis OZ perpendicular to the straight line joining the centers of spheres, with angular velocity vector ω as shown in the first following figure.

Fig. 75 The distance between the mass center c of any sphere and the axis being denoted by R, the angular momentum due to the rotation of one sphere is Oc ∧ m v ,

that is a vector along the Z-axis (thus collinear to ω) and of norm equal to R mω R . The angular momentum of the system about O is thus

LO = 2mR 2ω . This illustrates the previous case (ii) where the axis of rotation is a principal axis of inertia given the symmetry of the system. On the other hand, if the angle between the straight line joining the centers of spheres and axis of rotation is φ ≠ 90 0 , then the principal axis of inertia is not parallel to the axis of rotation. The angular momentum of a sphere is a vector perpendicular to the straight line joining the spheres and the norm of which is equal to R mω ( R sin φ ) . The angular momentum of the system about O is 2 mR 2ω sin φ 1Z .

The system does not rotate about a principal axis of inertia, but in this case LO and ω are not collinear. This concludes the special case of rigid bodies a detailed study of which will be developed in a next volume.

Kinetics and Dynamics of Systems 3.4

KINETIC ENERGY THEOREMS

3.4.1

Kinetic Energy Theorem

323

3.4.1a Noninertial Frame

Let us consider the motion of a system S of N particles with respect to a noninertial frame. PR16 In a noninertial frame R E , the change of kinetic energy due to the motion of S during a time interval equals the corresponding work done by external, internal and transport forces exerted on S. Proof. Given a noninertial frame of reference, we know that the force acting on a particle p h of S is d (mh v h ) = f h(e ) + f h(i ) + f hT + f hC dt where we recall that f h( e ) , f h(i ) , f hT and f hC are successively the resultants of external, internal, transport and Coriolis forces exerted on p h . From d 1 ( 2 mh v h2 ) = ( f h( e) + f h(i ) + f hT + f hC ) . v h dt

we deduce the following expression of kinetic energy theorem:


dT = P (e) + P (i ) + P T dt

(4-35)

where T is the kinetic energy of S; P ( e) , P (i ) and P T are respectively the external, internal and transport powers. No Coriolis power appears since, for any p h , Coriolis acceleration a hC is perpendicular to velocity v h in any relative frame. From N

dT = ∑ ( f h( e) + f h(i ) + f hT ) . drh , h =1

where drh = v h dt , we deduce: T − T0 = ∫

t

t0

∑ ( f h(e) + f h(i ) + f hT ) . v h dt .

(4-36)

h

3.4.1b Inertial Frame

Let us consider the motion of a system S of N particles with respect to an inertial frame R.

324

Chapter 4

PR17 In an inertial frame, the change of kinetic energy due to the motion of S during a time interval equals the corresponding work done by external and internal forces exerted on S: T − T0 = ∫




t

t0

∑ ( f h(e) + f h(i ) ) . drh

(4-37)

h

Proof. It is obvious since, in an inertial frame, the kinetic energy theorem has the following expression: dT = P (e) + P (i ) . dt




(4-38)

Remark. Even if the system of internal forces is equivalent to zero, the power developed by these forces does not vanish obviously. Therefore, the condition of zero external power is not sufficient to obtain a theorem of conservation of kinetic energy.

We are going to consider the Koenig expression of kinetic energy encountered in Section 2.2. PR18 The kinetic energy due to the motion of S relative to a frame R 'e = { G; E1′ , E 2′ , E 3′ } translating with respect to an inertial frame of reference R has a form analogous to the one in R : d 1 [ 2 ∑ mh (v hR ) 2 ] = ∑ ( f h( e ) + f h(i ) ) . v hR . dt h h

(4-39)

Proof. The expression (4-11) of kinetic energy in R leads to the following derivative (with respect to R): dT d 1 d = ( 2 M v G2 ) + 12 [ ∑ mh (v hR ) 2 ] dt dt dt h

but it is also

= ∑ ( f h( e ) + f h(i ) ) . (v G + v hR ) h

and thus

= ∑ f h( e) . v G + h

∑ f h(i) . v G + ∑ ( f h(e) + f h(i ) ) . v hR h

.

h

From this last equality, we really deduce Eq. (4-39) since dv d 1 ( 2 M v G2 ) = M v G ⋅ G = ∑ f h( e ) . v G dt dt h and since

∑ f h(i) = 0 . h

[because Eq. (4-17)]

Kinetics and Dynamics of Systems

325

3.4.1c Internal and External Forces Derivable from Potentials

Let

R = { o : e , e , e } be an inertial frame of reference. 1

2

3

First, we consider the case where the internal forces are derivable from a potential function V (i ) , but the external forces are assumed not to be derivable from a potential. The internal force exerted by the particle of position vector x k on the particle of position vector xh is denoted f hk = c hk rhk where chk ∈ R , rhk = x k − x h . The power developed by the internal forces acting on any particle p h is N

∑ chk rhk . v h .

k =1 k ≠h

Thus, the power developed by all the internal forces is N

N

h =1

k =1 k ≠h

P (i ) = ∑ ( ∑ c hk rhk . v h ) .

(4-40)

By forming pairs {h,k} such that {1,2}, {1,3}, {2.3} and so on, the previous power is written: N

N

h =1

k =1 h< k

P ( i ) = ∑ ( ∑ c hk rhk . (v h − v k )) N

= −∑

h =1

(4-41a)

N

d ( ∑ c hk rhk ⋅ rhk ) dt k =1 h< k

also written:

P (i ) = − The factor

1 2

1

∑ chk rhk . r&hk

2 h ,k

.

(4-41b)

is necessary since by summing over k and h, each point of pair is twice

considered. PR19 In an inertial frame, if the internal forces of a system of N particles derive from an (internal) potential V (i ) , then the time derivative of the sum of kinetic energy and internal potential equals the power of external forces: N d (i ) (T + V ) = ∑ f h( e) . v h . dt h =1

Proof. If we introduce the following function internal potential defined by

(4-42)

326

Chapter 4 N

V (i ) ( x1 ,..., x N ) = ∑

N

∑ ∫ chk rhk . drhk

h =1 k =1 h c .

Given any function ν of class C 1 , positive on ] x ∗ − a, x ∗ + a [ and vanishing elsewhere, but more particularly equal to unity on ] x ∗ − a 2 , x ∗ + a 2 [ , we have: x2

∫x

1

x∗ + a 2

ν ( x) g ( x) dx = ∫ ∗

x −a 2

g ( x) dx ≥ c a > 0

which is at variance with the assumption. Thus it is absurd to assume that g is positive for every x ∗ ∈]x1 , x 2 [ . A similar conclusion would be obtained by choosing a strictly negative function. So g (x) vanishes on ]x1 , x 2 [ and the lemma is proved.

By applying this lemma to the necessary condition (5-34) of extremum for I, for which the lemma hypotheses are fulfilled, we obtain the famous Euler equation of variational calculus. By denoting

δ f ∂f d ∂f = − δ y ∂y dx ∂y ′

the so-called variational derivative of f, then Euler’s equation is written:

δf = 0. δy We emphasize that every function y of class C 1 , introduced as before, making I extremum, must necessarily verify the Euler equation. We say: D

A curve x
y (x) solution of Euler’s equation is called an extremal.

An extremal does not necessarily produce an extremum since it fulfills only a necessary (and not sufficient) condition. For instance, saddle points are possible.

366

Chapter 5

First integrals

Euler’s equation is a second-order differential equation, namely: ∂f ∂2 f ∂2 f ∂2 f ′ ′ ′ = 0. − + + y y 2 ∂x ∂y ′ ∂y ∂y ∂y ′ ∂y ′

(5-36)

It defines functions depending on two arbitrary constants: y = y ( x; c1 , c 2 ). Any problem of the variational calculus is obviously different from a (local) Cauchy’s problem because it consists in obtaining an extremal through two points (two arbitrary constants being determined by limits). Even for simple functions, the problem of existence and unicity of solutions is not always obvious! It is possible that there are one or several extremals through two limit points or none! But the integration of the Euler equations is simplified if there exists first integrals. So, (i)

If f does not explicitly depend on y, then

d ∂f ∂f = 0 and thus is a first integral of dx ∂y ′ ∂y ′

Euler’s equation. (ii)

∂f is a first integral since ∂y ′ d d ∂f ∂f ∂f ∂f ∂f y′ + y ′′ − y ′′ ( f − y′ ) = − y′ = 0. dx ∂y ′ ∂y ∂y ′ ∂y ′ dx ∂y ′

If f does not explicitly depend on x, then f − y ′

System of Euler’s equations

The variational method is immediately generalized to an integral of type x2

∫x

f ( x, y i ( x), y ′ i ( x)) dx

1

where f is a C 2 function of 2n + 1 variables x, y i , y ′ i . The various y i belong to the space of C 2 real functions (or piecewise C 1 ) defined on [ x1 , x 2 ] ⊂ R . By analogy with previous developments, we let: yαi ( x) = y i ( x) + α η i ( x)

i = 1,..., n ,

δy i = yαi ( x) − y i ( x) where y i ( x) = y 0i ( x) . It is easily proved that the (first) variation of I (α ) is (sum over i): x2

∂f

x1

∂y

δ I = ∫ δy i (

i

−

d ∂f ) dx . dx ∂y ′ i

(5-37)

Lagrangian Dynamics, Variational Principles

367

The following is immediately proved. A necessary and sufficient condition for a curve of equations y i = y i (x) fulfills the necessary condition of extremum δ I = 0 is provided by the following system of Euler’s equations:

δf ∂f d ∂f ≡) i − = 0. i dx ∂y ′ i ∂y δy

i = 1,..., n .

(5-38)

We note this is a system of n second-order differential equations with 2n conditions y i ( x1 ) = a i

y i ( x2 ) = b i

where a i and b i are 2n given reals. Example of the brachistochrone problem (Jacques Bernoulli).

A wire in a vertical plane links up two points a and b. A particle of mass m, at rest at a, slides without friction down the wire to b under gravity. Find the equation of the curve for which the particle goes from a to b in the least time. Answer. The well-known relation dT = dW

is expressed as 1

d ( mv 2 ) = mg . ds 2

and the initial conditions imply: 1 2

mv 2 = mg ∫ k . ds = mg z , ab

k being downwards directed. Thus the velocity of the particle is ds = 2g z dt

which implies 1 + (dz dx) 2

dt =

2g z

dx .

The problem consists in minimizing 1

1 + z′2

xb

∫0

2g

z

dx .

Since the integrand does not explicitly depend on x, a first integral is 1 + z ′2

z

−

z ′2 z 1 + z′2

=c

368

Chapter 5

or z 1 + z ′2 =

1

c

= k.

Thus we have: dx =

z dz . k−z

By letting z = k sin 2 θ =

k 2

(1 − cos 2θ ) ,

we have: x = 2k ∫ sin 2θ dθ =

k 2

(2θ − sin 2θ ) + K .

Since the curve passes through the origin, we must have K = 0 , so that the required equations are (setting R = k 2 ): x = R (2θ − sin 2θ )

z = R (1 − cos 2θ ) ,

the constant R being determined from the limit condition at b. The extremal is a cycloid, that is the path of a fixed point on a circle of radius R which rolls along the x-axis.

2.2

HAMILTON’S VARIATIONAL PRINCIPLE

2.2.1 Hamilton’s Postulate

We know that d’Alembert-Lagrange principle and (infinitesimal and instantaneous) virtual displacements show a differential character and let obtain the famous Lagrange equations (1788). However, these equations of motion can be deduced from a variational principle due to Hamilton (1834). If we substitute L(t , q i , q
i ) for the function f ( x, y i , y ′ i ) introduced in the variational calculus, we conclude that the Euler equations of the calculus of variations become the Lagrange equations obtained from the nonvariational principle of d’Alembert-Lagrange. This remarkable observation is at the root of Hamilton’s principle in dynamics. Let us introduce an essential definition in order to present this principle. We consider two fixed points q1 and q 2 of the configuration space Q and the following set of curves of class C 2 E = { c : [t1 , t 2 ] ⊂ R → Q ; c(t1 ) = q1 , c(t 2 ) = q 2 }.

D

The function S : E → R is defined by the action integral t2

S (c) = ∫ L(t , c(t ), c
(t )) dt . t1

(5-39a)

369

Lagrangian Dynamics, Variational Principles

The action integral is also denoted by t2

S = ∫ L(t , q i , q
i ) dt . t1

(5-39b)

Now, we can state the Hamilton’s variational principle: PR10 Among all the possible paths along which a material system can move from an instant t1 to an instant t 2 , the actual path is such that the (first) variation of the action integral vanishes:


δS =δ

t2

∫t

1

L(t , q i , q
i ) dt = 0 .

(5-40)

In other words, a motion in the configuration space Q is a mapping R → Q : t
(q 1 (t ),..., q n (t ))

making extremum the action integral. The general nature of this variational principle (viewed as a postulate) widens the field of theory compared with Newton’s laws of motion. Hamilton’s variational principle is also denoted

δS =δ

2.2.2

t2

∫t

1

( pi q
i − L∗ ) dt = 0 .

(5-41)

Hamilton’s Principle and Motion Equations

We know that dynamics can be considered in two opposite manners. In the first, a variational principle is assumed and Newton’s or Lagrange’s equations are derived as theorems. In the second, the Lagrange (or Newton) equations are taken as axioms and variational principles are deduced as theorems. We are going to prove implications between Hamilton’s principle and motion equations. PR11 The extremals relating to Hamiltonian’s variational principle verify the corresponding Newton equations. Proof. Let us consider a system of N particles of position vectors rh , of corresponding masses mh and without constraints. The extremals make the following integral extremum: t2

t2

t1

t1

S = ∫ L dt = ∫ ( t2

=∫ ( t1

3N

N

mh r
h2 − V (r1 ,..., rN )) dt ∑ 2

1

h =1

mr ( x
r ) 2 − V ( x1 ,..., x 3 N )) dt , ∑ 2

1

r =1

written with the notations of Section1.4.

370

Chapter 5

The extremals are solutions of 3N Euler’s equations: d ∂L ∂L d ∂T ∂V − i = + = 0. i dt ∂x
dt ∂x
i ∂x i ∂x

These equations, grouped together 3 by 3, are really the equations of Newton for each of particles: M 1
r
1 + ∂ r1V = 0 , . . . , M N
r
N + ∂ rN V = 0

where m1 = m2 = m3 = M 1 , . . . , m3 N −2 = m3 N −1 = m3 N = M N . PR12 The motion of the representative point of a system takes place in the configuration space iff the generalized velocities of this point satisfy the Lagrange equations. Proof. By analogy with (5-32), we have: t2

S (α ) = ∫ L(t , qαi (t ), q
αi (t )) dt t1

and also t2

S (0) = ∫ L(t , q i (t ), q
i (t )) dt . t1

Hence we have: t2 d ∂L dq i ∂L dq
αi S (α ) = ∫ ( i α + i ) dt t1 dα ∂qα dα ∂q
α dα t2

=∫ ( t1

=∫

t2

t1

i = 1,..., n

d ∂L dqαi ∂L dqαi d ∂L dqαi ( ) ( ) ) dt + − dt ∂q
αi dα ∂qαi dα dt ∂q
αi dα

d ∂L dqαi ∂L ( i − ) dt ∂qα dt ∂q
αi dα

since there is no variation at the limits. Thus we have the (first) variation: t2

∂L

t1

∂q

α S ′(α ) α =0 = ∫ (

i

−

d ∂L ) α η i (t ) dt = 0 dt ∂q
i

also written [see Eq. (5-37)]: t2

∂L

t1

∂q

δS = ∫ (

i

−

d ∂L ) δq i dt = 0 . i dt ∂q


From a previous lemma of the calculus of variations and since the various δq i are arbitrarily independent, then the last integral is zero iff ∀i ∈ { 1,..., n }: d ∂L ∂L − =0. dt ∂q
i ∂q i

Lagrangian Dynamics, Variational Principles

371

Remark. The reader will also refer to Exercise 8 where Hamilton’s principle is deduced from Lagrange’s equations and the complementary equation.

We must also mention the following PR13 Given a gauge transformation of a Lagrangian, that is L ′(t , q, q
) = L(t , q, q
) + df (t , q )

where f is a function of class C 2 of variables t and q i , then the extremals of EulerLagrange equations correspond for L and L ′ . In other words, any extremal defined by q (t ) is such that

δ L ′ ∂L ′ d ∂L ′ ≡) i − =0 dt ∂q
i ∂q δ qi

⇔

δL ∂L d ∂L ≡) i − = 0. i dt ∂q
i ∂q δq

We emphasize that f does not depend on q
i . Proof. The various variational derivatives are

δ L′ δL δ ∂ df d ∂ df ( ) )( ) = + = +( i − L i i i dt dt ∂q
i dt δq δq δq ∂q δ L d ∂f ∂f ∂ ∂f = + ( i − i ( + k q
k )) i δ q dt ∂q ∂q
∂t ∂q δL =

δ qi

⋅

So we conclude as follows:

δ L′ =0 δ qi

2.3

⇔

δL =0 δ qi

i = 1,..., n.

JACOBI’S FORM OF THE PRINCIPLE OF LEAST ACTION

The following presentation of the principle of least action shows geodesics as extremals of arc length and permits bringing closer mechanics and geometrical optics. We consider a scleronomic system S of which the potential is V (q i ) . The generalized forces associated with the n generalized coordinates q i are Qi = −

∂V ∂q i

⋅

We also recall the expression of the kinetic energy 1

T = aij q
i q
j . 2

372

Chapter 5

PR14 The generalized trajectories of the representative point of the motion of S in the configuration space and admitting the first integral1 E = T + V are the geodesics corresponding to the metric dσ 2 = 2T aij dq i dq j = 2( E − V ) aij dq i dq j

(5-42)

also written: dσ 2 = (2U + h) aij dq i dq j .

(5-43)

The representative point follows the geodesics according to the time law dσ = 2U + h . dt

(5-44)

The generalized trajectories are the extremals corresponding to t2

δ ∫ 2T dt = 0 . t1

(5-44)

Proof. The Lagrange’s equations are d ∂V 1 ∂a jk j k q
q
= − i ⋅ (aij q
j ) − i dt 2 ∂q ∂q

(5-46)

We recall that the kinetic energy defines a metric on Q: ds 2 = 2T dt 2 = aij dq i dq j . We choose for the configuration space a new metric: dσ 2 = F ds 2 = F aij dq i dq j where, a priori, F is a strictly positive function of q i so that the generalized trajectories of the representative point are geodesics and where the parameter σ is a function of t. On the one hand, if we multiply the members of the motion equations (5-46) by

dt , these dσ

equations become: d dq j dσ 1 dq j dq k dσ dt =− ∂ iV . (aij ) − ∂ i a jk dσ dσ dt 2 dσ dσ dt dσ

(5-47)

On the other hand, for this metric, the equations of geodesics of the configuration space are: j k d 2qi i dq dq + Γ =0 jk dσ dσ dσ 2

where the Christoffel symbols correspond to the line element dσ 2 = ( F aij ) dq i dq j . From this element, the equations of geodesics are written with covariant components as follows: 1

Or in an equivalent manner : h = 2(T − U ), U = −V being the force function.

Lagrangian Dynamics, Variational Principles j k d 2q h h dq dq F aih ( + Γ jk ) dσ dσ dσ 2 d 2q h dq j dq k = F aih + Γ ijk dσ dσ dσ 2 h d dq dq h d = ( F a ih )− ( F aih ) + Γ ijk dσ dσ dσ dσ d dq j dq k = ( F a ij ) + (Γ ijk − ∂ k ( F aij )) dσ dσ dσ

373

dq j dq k dσ dσ dq j = 0. dσ

But, since 1

Γ ijk = [∂ j ( F aik ) + ∂ k ( F aij ) − ∂ i ( F a jk )] 2

the previous equations become d dq j dq j dq k 1 ( F aij ) + [∂ j ( F aik ) + ∂ k ( F aij ) − ∂ i ( F a jk ) − ∂ j ( F a ik ) − ∂ k ( F aij )] =0, dσ dσ 2 d σ dσ

that is: d dq j F dq j dq k 1 dq j dq k ( F aij ) − ∂ i a jk − a jk ∂ i F =0 dσ dσ 2 d σ dσ 2 dσ dσ

or d dq j F dq j dq k 1 ∂ i F ( F aij ) − ∂ i a jk = ⋅ dσ dσ 2 d σ dσ 2 F

(5-48)

By comparing Eqs. (5-47) of the motion of the representative point with Eqs. (5-48) of geodesics, we adopt the following F=

dσ dt

and thus for this choice, we have: ∂ i F = −2 ∂ i V . By integrating, we obtain F = −2 V + C

where C is a constant. From dσ 2 = F aij dq i dq j =

and thus

dσ = 2T dt

we deduce the following expression F = 2T .

dσ 2T dt 2 dt

374

Chapter 5

Therefore, the constant is 2T + 2V = 2 E

which is often written with the function force U as follows: 2T = 2U + h .

We have so obtained the extremals as geodesics corresponding to the metric defined by dσ 2 = (2U + h) aij dq i dq j . The corresponding variational principle is written:

δ

t2

∫t

1

2T dt = 0 .

We specify that the representative point follows every corresponding generalized trajectory according to the time law defined by dσ = 2U + h dt

that is dt =

ds 2T

=

ds 2U + h

⋅

3. EULER-NOETHER THEOREM Let us introduce a general theorem which plays a fundamental role in Lagrangian mechanics and that Euler used already under another form (obviously). This so-called EulerNoether theorem lets us associate a first integral of Lagrange’s equations to any oneparameter group of diffeomorphisms (on the configuration space) which conserves the Lagrangian. Given U and U ′ opens of the configuration space Q, we recall: D

A mapping f : U → U ′ is a diffeomorphism of class C r if f is a bijection of class C r such that f

3.1

−1

is of class C r .

ONE-PARAMETER GROUP OF DIFFEOMORPHISMS We consider a material system with n degrees of freedom.

Let J be an interval of R, L : U × R n : (q, q
)  L(q, q
) be a (differentiable) Lagrangian. D

A tangent vector field X on Q is a mapping which assigns a pair (q, q
) to each point q ∈Q : X (q ) = (q, q
) .

Lagrangian Dynamics, Variational Principles D

375

An integral curve of field X on Q is a curve c : J → Q : t
c(t )

such that d c(t ) = X (c(t )) . dt

Given any field X and any point q ∈ Q , we consider: -

a neighborhood U of this point, an interval I ε = { t ∈ R : t < ε , ε ∈ R+ }, a differentiable mapping:

ϕ : I ε × U → Q : (t , q)
ϕ (t , q) = φ t (q) such that

φ 0 (q ) = q . We introduce the following definition: D

A local transformation of Q generated by a tangent vector field X is a diffeomorphism between neighborhoods of Q:

φ t : U → Q : q
φ t (q ) verifying the following differential equation d φ t (q) = X (φ t (q)) dt

(5-49)

and

φ 0 (q ) = q . Now, we can introduce the notion of one-parameter group of diffeomorphisms and prove the following PR15 Every tangent vector field on Q generates a one-parameter group of diffeomorphisms on Q. Proof. Let I = { t : t < ε , ε > 0 } and Ω be a neighborhood of {0} × Q . We consider a local transformation U → Q : q
φ t + s (q ) with reals t , s, t + s ∈ I . This local transformation satisfies, such as φ t (q) , the differential equation associated with the field X. The following integral curves associated with X t
φ t + s (q)

t
φ t (φ s (q))

having for t = 0 the same value φ s (q) [because φ 0 (φ s (q)) = φ s (q ) ], are such that:

376

Chapter 5

φ t + s (q) = (φ t
φ s )(q) , for every t , s, s + t ∈ I (see uniqueness theorem). Therefore, the local transformation of Q are such that:

φt + s = φt
φ s . In particular, we deduce:

φ − s = φ s−1 because (φ − s
φ s )(q) = φ − s + s (q ) = φ 0 (q) . In conclusion, the local transformations of Q have a group structure with composite law, φ 0 being the identity element. This group is called the one-parameter group of diffeomorphisms.

3.2

EULER-NOETHER THEOREM

Before stating the present Noether’s theorem, we say: D

A diffeomorphism φ s is admissible for Q with a Lagrangian L(q, q
) if the Lagrangian is invariant under the local transformation:

φ s : U → Q : q
φ s (q ) . Example. Given

Q 2 = { ( x , y ) : x ∈ R, y ∈ R } and L=

m 2

( x
2 + y
2 ) − V ( y ) ,

then a translation of x, namely ∀s ∈ I ⊂ R :

φ s : Q2 → Q2 : ( x , y )
( x + s , y ) is obviously admissible. Now, let us prove the theorem: PR16 If a one-parameter group of diffeomorphisms is admissible for a configuration space Q [that is: L(q, q
) invariant under various φ s ], then there is, at least, a first integral of Lagrange’s equations. Proof. Let c : R → Q : t  c(t ) be an integral curve of Lagrange’s equations. Since φ s is admissible for Q (with L), then every transformed curve φ s
c is also solution of Lagrange’s equations. Explicitly, by considering Φ : R × R → Q : ( s, t )
Φ( s, t ) = φ s (c(t )) , we know that

377

Lagrangian Dynamics, Variational Principles

φ s  c : R → Q : t
φ s (c(t )) verifies the equations of Lagrange too. We have thus: ∂L

d ∂L (Φ( s, t ),Φ
( s, t )) = (Φ( s, t ),Φ
( s, t )) i dt
∂q ∂q i

i = 1,..., n .

(5-50)

Fig. 81 Since φ s is admissible, we have [letting q i = Φi ( s, t )] : ∂L ∂Φi ∂L ∂Φ
i d
=0. + L(Φ( s, t ),Φ( s, t )) = i ds ∂q ∂s ∂q
i ∂s

By taking Eq. (5-50) into account, the last sum of 2n terms becomes: d ∂L ∂Φi ∂L d ∂Φi d ∂L ∂Φi 0 ( ). = = + dt ∂q
i ∂s ∂q
i dt ∂s dt ∂q
i ∂s

So, at point c(t ) , the first integral is denoted by

∂L ⋅ d φ (q) ∂q
ds s

s =0

;

by denoting the derivative with respect to s by q ′ , this first integral is simply

∂L ⋅ q′ . ∂q


(5-51)

Remark 1. For a nonautonomous material system with a Lagrangian L(t , q, q
) , we consider the configuration spacetime R × Q . As in Exercise 8, we view that the 2(n + 1) − dimensional space of points and directions associated with R × Q and the elements of which are ( z j (u ), z ′ j (u ))

such that - for j = 0 : - for j ≠ 0 :

dt = t′ , du dq i z j (u ) = q i (u ) , z ′ j (u ) = ⋅ du

z 0 (u ) = t (u ) , z ′ 0 (u ) =

(arbitrary parameter u)

378

Chapter 5

We know that the action integral is t2

u2

t1

u1

S = ∫ L(t , q, q
) dt = ∫ =∫

u2

u1

L (t (u ), q(u ),

dq du

dt dt ) du du du

L( z j (u ), z ′ j (u )) du

where dt du

L=L

and so this case is related to the previous study. Remark 2. We mention that if L does not explicitly depend on t (that is, the one-parameter group of time translations φ s (t , q) is admissible), then there is a first integral that is the mechanical energy. We conclude this section with two classical examples illustrating the previous essential theorem [see, e.g. Arnold (1978)]. Example 1. If the Lagrangian of a constrained material system is invariant under translations along an axis, defined by e1 for instance, then the first component of the linear momentum is a first integral. Answer. The Lagrangian of the system is L=

N

mh r
h2 − V (r1 ,..., rN ) . ∑ 2

1

h =1

The invariance under any translation, for instance:

φ s : rh
φ s (rh ) = rh + s e1 implies d φ s (rh ) ds

s =0

= e1

and the Euler-Noether theorem leads to the following first integral N

∑

h =1

N

mh r
h . e1 = ∑ mh x
h h =1

which is the first component of the linear momentum of the system. Example 2. If the Lagrangian of a material system is invariant under rotations about an axis, defined by e1 for instance, then the angular momentum about this axis (or projection of L onto e1 ) is a first integral. Answer. The invariance under the above-mentioned rotation

φ s : rh
φ s (rh ) = rh defined by

Lagrangian Dynamics, Variational Principles

379

0 0   xh   xh   1       y h  =  0 cos s sin s   y h   z   0 − sin s cos s   z   h   h  implies d rh ds

s =0

 0    =  z h  = rh ∧ e1 . − y   h

The Euler-Noether theorem leads to the well-known first integral N

∑

h =1

N

mh rh . (rh ∧ e1 ) = ∑ (mh rh ∧ rh ) . e1 = − L . e1 = − L x . h =1

4. EXERCISES First courses of theoretical mechanics deal with simple Lagrange equations; it is the reason why we will essentially consider Lagrange equations with multipliers. Exercise 1.

Two particles of same mass m move along a circle of center o and radius R. They repel each other under the action of a force inversely proportional to the cube of their distance. Friction being negligible, prove that the middle c of the arc joining positions a and b of the respective particles is moving in a circle at constant velocity. Answer. The middle c of arc ab has one degree of freedom. The chosen generalized coordinate is the angle θ locating the position vector oc. If α denotes the angle (oa , oc ) , the values of θ at a and b are respectively:

θa = θ −α ,

θb = θ + α .

The kinetic energy of the system of two particles is T=

m 2

R 2θ
a2 +

m 2

R 2θ
b2 = mR 2 (θ
2 + α
2 ).

Let us set r = ab = r 1r ,

r= r .

The force exerted on a is fa =

k k 1 = − 3 1r 3 ba r r

( k ∈ R+ ).

The work done by forces during differential displacements of a and b is

380

Chapter 5

dW = f a . d oa + f b . d ob = − =

k 1r . d (oa − ob) r3

k k 1 . d (r 1r ) = d (− 2 ) = −dV . 3 r 2r r

The potential is thus expressed as

V=

k k = ⋅ 2 2 2r 8 R sin 2 α

Since L = mR 2 (θ
2 + α
2 ) −

k , 8 R sin 2 α 2

the Lagrange’s equation: d ∂L ∂L =0 − dt ∂θ
∂θ

is reduced to d ∂L = 0. dt ∂θ


Thus the constant ∂L = 2mR 2θ
∂θ


proves that the circular motion of c is uniform.

Exercise 2.

A quarter of a homogeneous circular rod S of radius R and mass M is thrown in a vertical plane. Find Lagrange’s equations. Answer.

Let { o; e1 , e 2 } be a frame of reference. The moving rod has three degrees of

freedom, the chosen generalized coordinates are the two coordinates q 1 = x, q 2 = y of G and the inclination angle q 3 = θ of a straight line fixed in S. The kinetic energy of S is 1 1 1 T = aij q
i q
j = M ( x
2 + y
2 ) + Iθ
2 2

2

2

where I is the moment of inertia about the instantaneous rotation axis; namely:

I = MR 2 . We have thus a11 = M ,

a 22 = M ,

a33 = MR 2 ,

( aij = 0 i ≠ j ).

381

Lagrangian Dynamics, Variational Principles

The force exerted on the rod, that is the weight f = − Mg e 2 , is derivable from a potential expressed as V = Mg y . The Lagrangian is thus L=

1 aij q
i q
j − Mg q 2 . 2

From ∂L = a11 q
1 , ∂x


∂L = 0, ∂x

we deduce the following Lagrange’s equation: M
x
= 0 .

(1)

From ∂L = a 22 q
2 , ∂y


∂L = − Mg , ∂y

we deduce another Lagrange’s equation: M
y
+ M g = 0 .

(2)

From ∂L = a33 q
3 ,
∂θ

∂L = 0, ∂θ

we deduce the last Lagrange’s equation: 1 2

MR 2θ

= 0 .

(3)

Exercise 3.

A particle of mass m is constrained to move on the inside surface of a smooth paraboloid of equation z = x 2 + y 2 . By considering one multiplier, find Lagrange’s equations and give the physical meaning of the multiplier. Answer. Given primitive cylindrical coordinates u1 = r ,

u2 = θ ,

u3 = z ,

the Lagrangian of the moving particle is L=

m 2

(r
2 + r 2θ
2 + z
2 ) − mg z .

The constraint equation r2 − z = 0

implies

2r δ r − δ z = 0 .

382

Chapter 5

By referring to the general constraint equation (5-18), we have: a11 = 2r ,

a12 = 0 ,

a13 = −1 .

The Lagrange equations with one multiplier λ are thus d dt d dt d dt

∂L ∂L − = 2rλ , ∂r
∂r ∂L ∂L − = 0, ∂θ
∂θ ∂L ∂L − = −λ . ∂z
∂z

Thus we have m(
r
− rθ
2 ) = 2λ r , d 2
(r θ ) = 0 , dt m
z
= − mg − λ . m

Since in cylindrical coordinates we know that m d 2
ma = m(
r
− rθ
2 ) 1r + (r θ ) 1θ + m
z
1z , r dt

we have here: ma = 2λ r 1r − mg 1z − λ 1z .

So, from ma = mg + L ,

we deduce the following force of constraint L = λ (2r − 1z ) and the z-component is such that:

λ = − Lz . Exercise 4.

Find the law of motion of a particle of Lagrangian L=

m 2 ( x
+ y
2 ) 2

and subject to a constraint of equation y
= x

(or dy − x dt = 0 ).

Give the physical interpretation of the introduced multiplier. Answer. In this problem with one degree of freedom, x and y are primitive coordinates and we introduce one multiplier λ1 . The equation of constraint δ y = 0 corresponds to a11 = 0 and a12 = 1 in the general equation a1i δ u i = 0 . The Lagrange equations with multiplier are

Lagrangian Dynamics, Variational Principles

383

m
y
= λ1 .

m
x
= 0 ,

The motion of the particle is known by adding the equation of constraint: y
= x .

Indeed, given initial values x0 , y 0 , x
0 , y
0 we have: x = x
0 t + x0 , y = x
0

t2 2

+ x0 t + y 0 .

The Lagrange multiplier

λ1 = m
y
= m x
0 is the y-component of the force of constraint since: ∂r = L x = λ1 a11 = 0 , ∂x ∂r L . = L y = λ1 a12 = λ1 . ∂y L.

Exercise 5. A vertical hoop (with its axis always horizontal) of mass M and radius R is rolling without slipping down an inclined plane on a distance l, α denoting the inclination angle. Express the friction force from Lagrange’s equations with multiplier. Answer. The moving hoop has one degree of freedom. We choose two primitive coordinates: -

the coordinate x which measures the distance covered by the hoop moving down the plane, the coordinate θ which is the angle of rotation about the horizontal axis of the hoop.

The equation of constraint is dx = R dθ .

The kinetic energy of the moving hoop is the sum of the kinetic energy about the mass center G of the hoop and the kinetic energy of rotation about G: T = M x
2 + MR 2θ
2 . 1

1

2

2

The Lagrangian is immediately L=

M 2

( x
2 + R 2θ
2 ) − Mg (l − x) sin α .

The constraint implies:

δ x − R δθ = 0 and thus a11 = 1 ,

a12 = − R .

384

Chapter 5

Thus Lagrange’s equations with multiplier are d dt d dt

∂L ∂L − = λ, ∂x
∂x ∂L ∂L − = − Rλ , ∂θ
∂θ

that is: M
x
− Mg sin α − λ = 0 ,

(1)

MRθ

+ λ = 0 .

(2)

We add the equation of constraint x
= Rθ
,

(3)

which allows the determination of unknowns λ , x and θ . So, along the x-direction, the motion equation is immediately: M
x
− M

g 2

sin α = 0 .

From M
x
= Mg sin α + L x = M

g 2

sin α ,

where L x is the constraint force of friction, we deduce: g L x = λ = − M sin α . 2

The equation of motion shows that the hoop is rolling with half the acceleration of the corresponding frictionless motion. Exercise 6.

A small ring p of mass m frictionless slides along a circular hoop of radius R which rotates about a vertical axis at a constant angular velocity φ
= ω . This example is the one of Section 2.2.3.d of Chapter 1. (i) Find the equation of motion of p by introducing an obvious generalized coordinate. (ii) Determine the components of the force of constraint exerted on p by considering the spherical coordinates r ,θ , φ . Give the physical interpretation of Lagrange multipliers for these primitive coordinates. Answer. (i) Given the generalized coordinate θ which is the angle (k , r ) where r is the position vector of p, we have immediately: L=

m 2

R 2 (θ
2 + ω 2 sin 2 θ ) − mg R cos θ

and thus the motion equation of Lagrange is Rθ

− Rω 2 sin θ cos θ − g sin θ = 0.

385

Lagrangian Dynamics, Variational Principles (ii) Let us introduce the (primitive) spherical coordinates u 1 = r , u 2 = θ and u 3 = φ .

This problem has one degree of freedom and there are two equations of constraint. The first one is r−R=0 (⇒ δ r = 0 ), that is

a1i δ u i = 0 with a11 = 1 , a12 = a13 = 0 .

The second equation of constraint is

φ − ωt = 0

(⇒ δφ = 0 ),

that is a 2i δ u i = 0

with a 21 = a 22 = 0, a 23 = 1 . Lagrange’s equations with multipliers are: -

First: 2

m
r
− mr (θ
2 + φ
2 sin 2 θ ) + mg cos θ = ∑ λ r a r1

(1)

r =1

where the second member, equal to λ1 , is the generalized force of constraint associated with u 1 , namely:

λ1 = L ⋅ -

∂r = L . 1r = Lr . ∂r

Second: m

2 d 2
(r θ ) − mr 2 sin θ cos θ φ
2 − mgr sin θ = ∑ λr a r 2 dt r =1

(2)

= 0.

-

Third: 2 d 2 2
m (r sin θ φ ) = ∑ λ r a r 3 dt r =1

(3)

where the second member, equal to λ2 , is the generalized force of constraint associated with u 3 , namely:

λ2 = L ⋅

∂r = L . r sin θ 1φ = r sin θ Lφ . ∂φ

By taking into account of two constraints ( r = R, φ = ω t ), we notice that Eq. (2) is the equation of motion already found in (i). Given these constraints, Eq. (1) expresses the radial component of the force of constraint on p: Lr = − mR (θ
2 + ω 2 sin 2 θ ) + mg cos θ

which is λ1 .

386

Chapter 5

In the same conditions, Eq. (3) expresses the longitudinal component of the force of constraint on p: Lφ = 2mω Rθ
cos θ . The multiplier λ2 is Lφ multiplied by R sin θ . Exercise 7.

The extremity a of a homogeneous rigid rod ab is confined to move along a fixed vertical axis, when the end b is confined to move on the plane through o perpendicular to axis. The motion of the rod of mass m and length l is assumed frictionless. (i) Determine Lagrange’s equations of motion. (ii) Find first integrals and give their interpretation. Answer. (i) The extremity a has one degree of freedom, while b has two degrees of freedom a priori; but the distance between a and b is constant, which introduces a constraint. So, the rod has two degrees of freedom.

Fig. 82

Let R e = { o; e1 , e 2 , e3 } be a frame of reference such that e3 is vertical. The position of the rod is determined with respect to R e if the angle φ of position of the vertical plane oab is known with respect to the vertical plane { o; e 2 , e 3 } and if the angle θ of position of the bar in the plane oab is known with respect to the vertical axis. The angular velocity vector of the rod (with respect to R e ) is obviously ω = φ
e 3 + θ
E1

where E1 is the unit vector perpendicular to the plane { o; e ′2 , e 3 } of the rod as shown in the previous figure. We are going to calculate the kinetic energy, namely:

387

Lagrangian Dynamics, Variational Principles T =

1 2

2

m vG

1

+ I ij(G )ω iω j 2

where v G is the velocity of the mass center G with respect to R e and the various I ij(G ) are the components of the inertia tensor of the rod expressed in frame { G; E1 , E 2 , E 3 } fixed in the moving rod. First, time derivatives of oG with respect to R e and R e′ = { E1 , e ′2 , e 3 } are connected as follows: de d e′ vG = oG = oG + ω e′e ∧ oG dt dt d e′ = oG + φ
e 3 ∧ oG. dt But, from l d e′ l oG = θ
cos θ e ′2 − θ
sin θ e 3 dt 2 2 l l (since oG = sin θ e ′2 + cos θ e 3 ), 2

2

and from l

φ
e 3 ∧ oG = − φ
sin θ E1 , 2

we deduce: l l l v G = − φ
sin θ E1 + θ
cos θ e ′2 − θ
sin θ e 3 2 2 2

and thus v G2 =

l 2
2
2 (θ + φ sin 2 θ ) . 4

We mention that the previous result is also obtained from de l oa + E 3 ∧ (φ
e 3 + θ
E1 ) 2 dt l l = −lθ
sin θ e 3 − φ
sin θ E1 + θ
E 2

v G = v a + Ga ∧ ω =

2

2

since E 2 = cos θ e ′2 + sin θ e 3 . Second, since e 3 = sin θ E 2 + cos θ E 3 , we have immediately:  ml 2 12 0  1 (G ) i j 1


I ij ω ω = (θ φ sin θ φ cos θ ) 0 ml 2 12 2 2  0 0  =

ml 2
2
2 (θ + φ sin 2 θ ). 24

0   θ


   0   φ
sin θ    0   φ
cos θ  

388

Chapter 5

So, the kinetic energy of the moving rod is l 2
2
2 T = m (θ + φ sin 2 θ ) 6

and the Lagrangian is l 2
2
2 l L = m (θ + φ sin 2 θ ) − mg cos θ . 6

2

The Lagrange equations are immediately: d l2 ml 2
2 l (m θ
) − φ sin θ cos θ − mg sin θ = 0 , dt 3 3 2 2 l d (m φ
sin 2 θ ) = 0 dt 3

or explicitly:

θ

− φ
2 sin θ cos θ −

3 g 2 l

sin θ = 0 ,

φ

sin 2 θ + 2θ
φ
sin θ cosθ = 0. (ii) The constraint being frictionless, the force mg is derivable from a potential V and the mechanical energy is a first integral, namely: T +V = m

l 2
2
2 l (θ + φ sin 2 θ ) + mg cos θ . 6

2

Another first integral is obviously φ
sin 2 θ = c , that is φ
0 sin 2 θ 0 given initial angles θ 0 and φ0 . Let us express this first integral in function of angular momentum La . d From Eq. (4-23), we know that La is the moment about a of external forces plus dt mv G ∧ v a .

The external forces are mg and the forces of constraint La and Lb . We know that M a ( La ) = 0 ; M a (mg) and M a ( Lb ) are orthogonal to axis defined by e3 , so the projections of these moments on this axis are zero. In addition, the projection of mv G ∧ v a on this axis is obviously zero. Thus we conclude that d e3 ⋅ La = 0 dt which implies: e3 . La = k (constant). Let us calculate the following La = m aG ∧ v a + I ( a ) ⋅ ω . First, we have:

Lagrangian Dynamics, Variational Principles l

2

2

2

389

l m aG ∧ v a = m E 3 ∧ (−lθ
sin θ e 3 ) = − m θ
sin 2 θ E1 .

Second, from the inertia tensor I (G ) and Steiner’s theorem, we deduce, with respect to { a; E1 , E2, E3 }, the following  Ml 2 3 0 0   θ
    Ml 2 3 0   φ
sin θ  I (a) ⋅ω =  0    0 0   φ
cos θ   0   

and thus l2 θ
θ
La = ml 2 ( − sin 2 θ ) E1 + m φ
sin θ E 2 3

2

3

and finally La . e 3 = m

l2
2 φ sin θ = k . 3

So, the connection between the first integral φ
0 sin 2 θ 0 and the projection of La on the vertical axis is established. Exercise 8.

By considering the configuration spacetime of a system S, deduce Hamilton’s variational principle from Lagrange’s equations and the complementary equation; these equations correspond to the “natural trajectory” described by a material system S from one point (t1, q 1 ) to another (t 2 , q 2 ).

Answer. We know that the motion of a material system of n degrees of freedom is described by successive n-tuples q (t ) = (q1 (t ),..., q n (t )) of Q. We emphasize that the variables q i and t are treated on an equal footing and there are so expressed in function of an arbitrary parameter u. By considering the configuration spacetime R × Q , the motion of S is described by a path of points (t (u ), q i (u )) i = 1,..., n which are simply denoted by

( z j (u ))

j = 0,1,..., n

with u ∈ [u1 , u 2 ] . It is convenient to introduce the 2(n + 1) − dimensional space of points and directions associated with the timespace R × Q and of which the elements are

( z j (u ), z ′ j (u ))

j = 0,1,..., n

such that: - for j = 0 :

z 0 (u ) = t (u ) ,

z ′ 0 (u ) = t ′(u ) =

dt du

390

Chapter 5

- for j ≠ 0 :

dq i z ′ (u ) = q ′ (u ) = du

z (u ) = q (u ) , j

j

i

i

i = 1,..., n.

From the above z j = (t , q i ) ,

we deduce z
j = (1 ,

dq i du q ′i ) = (1, ) = (1, q
i ) . du dt t′

The action integral t2

S = ∫ L(t , q i , q
i ) dt t1

=∫

dq i du dt L(t (u ), q (u ), ) du dt du du

u2

i

u1

is simply written, by introducing L = L t ′ , as follows: S=∫

u2

u1

L( z j (u ), z ′ j (u )) du

where the function L is assumed of class C 2 . The (first) variation of S is u2

∂L δ z ′ j ) du j ∂z ∂z ′ ∂L ∂L d (δ z j )) du ( jδ z j + j du ′ ∂z ∂z ∂L j d ∂L d ∂L ( j δ z + ( j δ z j ) − ( j )δ z j ) du du ∂z ′ du ∂z ′ ∂z

δS =∫ ( u1

=∫

u2

=∫

u2

u1

u1

∂L j

δzj +

that is u2

∂L

u1

∂z

δS =∫ (

j

−

∂L d ∂L )δ z j du + [ j δ z j ]uu12 . j du ∂z ′ ∂z ′

Among the possible trajectories from (t1 , q1 ) to (t 2 , q 2 ) , these points corresponding to respective values u1 and u 2 , we are going to prove that the trajectory fulfilling the Lagrange’s and complementary equations is such that δ S = 0 .

First, we note the previous bracket vanishes. Second, we have for j = 1,..., n : ∂L ∂z j

and for j = 0 :

−

d ∂L ∂L d ∂ ( Lt ' ) ∂q
i ∂L d ∂L ∂ z ′ j ′ ′ ′ ′ = t − t ( ) = t − ( t ( ( )) du ∂z ′ j dt ∂q
i ∂z ′ j ∂q i ∂q i dt ∂q
i ∂z ′ j t ′ ∂L d ∂L = t ′( i − )=0 dt ∂q
i ∂q

Lagrangian Dynamics, Variational Principles ∂L dq
i ∂L ∂L d d d ∂ L ∂L (L + t ′ ) = (L + t ′ i ) t′ − − = t′ − ∂t ∂t ′ du du ∂q
dt ′ ∂z 0 du ∂z ′ 0 ∂t

∂L

∂L q ′ i ∂L ∂L d ∂L q′i d ( L + t ′ i (− 2 )) = ) t′ − t′ (L − i t′ − ∂t dt ∂t du ∂q
t ′ ∂q
t′ ∂L dL∗ ∂L ∂L d ) = 0. = t ′( − ( L − i q
i )) = t ′( + ∂t dt dt ∂q
∂t

=

We conclude that δ S = 0.

391

CHAPTER 6

HAMILTONIAN MECHANICS

In classical mechanics, the Hamiltonian theory, which dates from 19th century, does not greatly simplify the differential equations of dynamic problems in comparison with the Lagrangian version. On the contrary, the equations of motion can be often more easily obtained from the Lagrangian method. Nevertheless, the Hamiltonian formalism introduces the following advantage. The Hamilton’s equations may be simplified by transformations concerned by variables p i and q i which are independent unlike the variables q i and q
i in the Lagrangian approach. Hamiltonian variables introduce a greater freedom, notably in celestial mechanics. Of course, the Hamiltonian theory is really suited to quantum mechanics. Therefore, the area of Hamiltonian dynamics is wider than the one of Lagrangian dynamics. The Liouville’s theorem which states the invariance of the density in the phase space shows a supplementary reason to consider the independent variables p i and q i of Hamiltonian mechanics, this theorem being a foundation of statistical mechanics. Finally, the Hamilton-Jacobi equation introduces a powerful way for solving differential equations of motion.

1. N –BODY PROBLEM AND CANONICAL EQUATIONS The N-body problem consists in determining the trajectories of N particles interacting in accordance with the gravitational law of Newton. Let us show the equations. In an inertial frame of reference oxyz , we consider N particles p h of masses mh and coordinates ( xh , yh , zh ).

393

394

Chapter 6

Each particle p h is attracted by the N − 1 other particles pk with a force N

∑

fh =

mh mk

G

rhk2

 k =1 k ≠h 

(6-1)

1hk

where

rhk rhk

1hk =

( rhk = ph pk ).

We evidently have:

d 2 rh dt

2

N

∑

=

G mk

 k =1 k ≠ h 

rhk

( oph = rh ).

rhk3

The projections onto the coordinate axes lead to 3N differentials equations, namely:

mh

d 2 xh dt

2

G mh m k ( x k − x h )

N

=

∑

 k =1 k ≠ h 

(( x k − x h ) + ( y k − y h ) 2 + ( z k − z h ) 2 ) 3 2 2

and others in y and z. The force function is written U =

N

G

2

∑

 h, k k ≠ h 

mh mk (( xk − xh ) + ( yk − yh ) 2 + ( zk − zh ) 2 )1 2 2

(6-2)

where the sum is concerned with N(N-1) arrangements without repetition taken 2 by 2 (e.g. if N=4, then 12, 13, 14; 21, 23, 24; 31, 32, 34; 41, 42, 43). Since for any value l we have: ∂U =G ∂xl

m k ml ( x k − x l )

N

∑

 k =1 k ≠l 

(( x k − xl ) + ( y k − y l ) 2 + ( z k − z l ) 2 ) 3 2 2

then the 3N differential equations of motion are mh
x
h =

PR1

∂U , ∂x h

mh
y
h =

∂U , ∂y h

mh
z
h =

∂U ⋅ ∂z h

(6-3)

If the forces are conservative (that is derivable from a force function U ) then the differential equation of dynamics can be put in a canonical form.

Proof. Consider a system of N particles subject to the gravitational law in R3 and let the respective Cartesian coordinates denote ( x1 , y1 , z1 ) = ( x 1 , x 2 , x 3 ) , . . . . ( x N , y N , z N ) = ( x 3 N − 2 , x 3 N −1 , x 3 N ) .

395

Hamiltonian Mechanics

The kinetic energy of the system is 3N

T = 12 ∑ mi ( x
i ) 2 i =1

where the common coefficient m1 = m 2 = m3 represents the mass of the first particle and so on. By letting y i = mi x
i

(no summation of course!)

we obtain 3N ( yi )2 T = 12 ∑ i =1 mi

⇒

∂T ∂x i

=0

,

∂T ∂y i

=

yi mi

∂U

,

∂y i

= 0.

So, the system of motion equations (i = 1,…,3N ): ∂U ∂x i

mi
x
i =

is equivalent to the following system of 6N equations x
i =

y i ∂ (T − U ) , = mi ∂y i

y
i = −

∂ (T − U ) . ∂x i

If we introduce the function F = T −U

then the system of motion equations takes the canonical form: x
i =

∂F ∂y

i

,

y
i = −

∂F ∂x i

(i=1,…,3N )

(6-4)

by recalling that the position of indices has no significance, it is only a notation to designate 3N coordinates. Sometimes F is called a characteristic function and x i , y i are conjugate variables. We have obtained a system of 6N differential equations of first order and there are 6N unknowns: the positions and velocities of points. A priori 6N integrations are thus necessary, but first integrals can be known. By considering a problem with n degrees of freedom ( n = 3 N ), the characteristic function is generally denoted by H and the system of canonical equations is written


x
i =

∂H ∂y i

,

y
i = −

∂H ∂x i

(i=1,…,n) .

(6-5)

396

D

Chapter 6 A system of motion equations is said to be canonical if there is a function H (t , x i , y i ) fulfilling (6-5).

Before introducing such a system of 2n canonical equations in the general Hamiltonian approach of classical mechanics, let us briefly consider the problem of obtaining first integrals. PR2

A characteristic function H which does not explicitly depend on time is a first integral of canonical equations.

Proof. From dH ∂H + = dt ∂t =

∂H + ∂t

∂H ∑ ∂x i x
i + i =1 n

n

∂H ∂H

∂H

n

∑ ∂y i

y
i

i =1

∂H ∂H

∑ ( ∂x i ∂y i − ∂y i ∂x i

)

i =1

we deduce dH ∂H . = dt ∂t

So, if

∂H = 0 , we have the first integral ∂t H ( x i (t ), y i (t )) = c .

Remark. We mention that the case where H explicitly depends on t can be related to the one where H does not explicitly depend on t, but the order of the corresponding system of canonical equations is 2n+2 [e.g. Kovalevsky (1963)].

PR3

If a characteristic function does not explicitly depend on the independent variable t, then the integration of the canonical system of order 2n is reduced to the one of a system of order 2n-2, the time t being obtained from an integral.

Proof. We can view the system (6-5) as the following system of order 2n-1: ∂H 2

∂H n

2

dx ∂y = , 1 ∂H dx

...

∂y 1

dx ∂y n = , ∂H dx1 ∂y 1

(6.6) ∂H

dy1 ∂x1 , = − ∂H dx1 ∂y1

with the first integral H ( x i , y i ) = c .

∂H

...

dy n ∂x n = − ∂H dx1 ∂y1

Hamiltonian Mechanics

397

The problem facing us is the integration of a system of order 2n-2; the solution will express x 2 ,..., x n , y 1 ,... y n in function of x1 and c, time t being given by

t=∫

dx1 ∂H ∂y

+ t0 .

(6-7)

1

Example. If n = 1 , the canonical system

x
=

∂H , ∂y

y
= −

∂H ∂x

is viewed as ∂ x H dx + ∂ y H dy = 0

⇔

H ( x, y ) = c .

The integration of canonical equations is given by

t=∫

dx + t0 . ∂yH

2. CANONICAL EQUATIONS AND HAMILTONIAN

2.1

LEGENDRE TRANSFORMATION AND HAMILTONIAN

The Legendre transformation lets us establish the canonical equations of Hamilton from the Lagranges’s equations. Given a function f : R → R : x
f ( x) of class Cp ( p ≥ 2 ) and such that f ′′( x) ≠ 0 , we say: D

The Legendre transformation of f ( with respect to x) is the function L

f : R → R : x
Lf ( x) = x f ′( x) − f ( x) .

(6-8a)

By introducing the inverse function of z = f ′( x) : g : R → R : z
x = g ( z) ,

the Legendre transformation of f is viewed as L

f : R → R : z
Lf ( z ) = z g ( z ) − f ( g ( z )) .

(6-8b)

398

Chapter 6

Remark. Given the assumptions, the second derivative ( L f )′′( z ) does not vanish since we have: d

L

dz

f ( z) = g ( z) + z

dg dz

−

df dg dx dz

= g ( z)

which really implies: ( L f )′′( z ) = g ′( z ) =

1 ≠0 f ′′( x)

In addition, letting L f ( z ) = F ( z ) , we have (involution!): LL

f ( z ) = LF ( z ) = z

dF dz

( z ) − F ( z ) = z g ( z ) − ( z g ( z ) − f ( g ( z ))

= f ( g ( z )).

Example. What is the Legendre transformation of the function f ( x) =

m

2

x2 ?

Answer. We note that d f ( x) = mx dx

⇒

d2 f ( x) = m ≠ 0 . dx 2

Next, the inverse function of z = f ′(x) is x = g ( z) =

z m

and thus L

f ( z) = z

z m z 2 z2 − ( ) = . m 2 m 2m

Generalization. The definition of the Legendre transformation becomes immediately widespread to functions of vector variables x = ( x1 ,..., x r ) and y = ( y 1 ,..., y s ) .

Thus we consider a function f (x,y) of class C p ( p ≥ 2 ) for x such that det(

∂2 f ∂x i ∂y j

) ≠ 0.

The system of r equations zk =

∂f ( x, y) ∂x k

can be locally solved with respect to the r variables xi, that is xi = χ i ( z, y )

and we express:

i = 1,…,r

399

Hamiltonian Mechanics D

The Legendre transformation of f (x,y), with respect to x, is the real-valued function from R r × R s : ∂f Lx (6-9) f = i χi − f , ∂x that is explicitly Lx

f ( z, y ) =

∂f i χ ( z, y ) − f ( x ( z , y ), y ) . ∂x i

Before introducing the canonical equations, we are going to define the Hamiltonian function (or Hamiltonian) as a Legendre transformation of Lagrangian. We recall that the function L : R × R n × R n → R : (t , q i , q
i )  L(t , q i , q
i )

(6-10)

is the Lagrangian T-V (at lest of class C2 ) of a material system with n degrees of freedom. We also recall that the generalized momentum or canonically conjugate momentum to qi is pi =

∂L ∂T = i . i ∂q
∂q


(611)

∂2L We assume that the condition det( i j ) ≠ 0 is fulfilled. ∂q
∂q


If the various q
i are likened to the previous xi, if (t , q1 ,..., q n ) is likened to the previous y and if f is taken to be the Lagrangian, then we can set: D
The Hamiltonian is the Legendre transformation of L(t , q i , q
i ) with respect to q
= (q
1 ,..., q
n ) , that is

H = pi q
i − L



i = 1,…,n

(6-12)

where the various q
i are expressed in function of 2n + 1 variables t, pj, qk. Explicitly: H (t , p, q ) =

L q


L(t , p, q) = pi q
i (t , p, q ) − L(t , q, q
(t , p, q )) .

Example 1. Write down the Hamiltonian for a free particle of mass m which moves, in R3, under the action of forces derivable from a potential V.

Answer. From the Lagrangian L=

m 2

( x
2 + y
2 + z
2 ) − V ( x, y, z )

we deduce the following generalized momenta

400

Chapter 6 px =

∂L = mx
, ∂x


p y = my
,

p z = mz
.

In this example, we mention that the generalized momenta are the linear momenta. Since py p p y
= x
= x , z
= z , m m m we obtain 2 p x2 p y p z2 H= + + −L m m m

1

=

2m

( p x2 + p 2y + p z2 ) + V ( x, y, z ).

Example 2. Determine the Hamiltonian of a particle of mass m moving in a central field.

Answer. From the Lagrangian L=

m 2

(r
2 + r 2θ
2 ) − V (r ) ,

we deduce the generalized momenta pr =

∂L = mr
, ∂r


pθ =

∂L = mr 2θ
. ∂θ


θ
=

pθ

From r
=

pr , m

mr 2

we obtain

H=

p2 p r2 + θ 2 + V (r ) . 2m 2mr

Example 3. Find the Hamiltonian of a free particle of mass m which moves in a frame of axes oXYZ rotating relative to one another of axes oxyz about the common axis oz = oZ with a constant angular velocity ω = 1 . The force is derivable from a potential V ( X , Y , Z ) .

Answer. For this rheonomic problem, from  x   cos t     y  =  sin t z   0   

− sin t cos t 0

0  0 1 

X   Y  Z   

we deduce T =

and thus

m 2

( x
2 + y
2 + z
2 ) =

m
2
2
2 m ( X + Y + Z ) + m( XY
− X
Y ) + ( X 2 + Y 2 ) 2 2

401

Hamiltonian Mechanics pX =

∂L = mX
− mY , ∂X


∂L = mY
+ mX ,
∂Y ∂L pZ = = mZ
∂Z
pY =

p X
= X + Y , m p Y
= Y − X , m p Z
= Z ⋅ m

⇒

Therefore, the Hamiltonian is H=

2.2

1 2m

( p X2 + pY2 + p Z2 ) + (Y p X − X pY ) + V ( X , Y , Z ) .

CANONICAL EQUATIONS

Let us see that the Legendre transformation converts the Lagrange equations into canonical equations of the Hamiltonian formalism. PR4

A system of 2n first order differential equations called the system of canonical equations of Hamilton and following from the system of n Lagrange equations is written: q
i =

∂H , ∂pi

p
i = −

∂H . ∂q i

(6-13)

Proof. Since ∂L i ∂L i ∂L dq − i dq
− dt ∂q i ∂q
∂t ∂L ∂L = q
i dpi − i dq i − dt ∂q ∂t

dH = pi dq
i + q
i dpi −

and dH =

∂H i ∂H ∂H dq + dpi + dt , i ∂q ∂pi ∂t

we deduce: q
i =

∂H , ∂pi

∂L ∂H =− i , i ∂q ∂q

∂H ∂L . =− ∂t ∂t

The system of n Lagrange’s equations: p
i −

∂L =0 ∂q i

leads to the system of 2n canonical equations: q
i =

∂H , ∂pi

p
i = −

∂H ∂q i

describing the evolution of the material system. These equations have the canonical form of equations (6-3) where F is the Hamiltonian H.

402

Chapter 6

Remark. The equations ∂H ∂L =− ∂t ∂t

and dH ∂H ∂H ∂H ∂H = p
i + i q
i + = dt ∂pi ∂q ∂t ∂t

introduce the complementary equation: dH ∂L + =0 dt ∂t

(6-14)

which takes the following form in Lagrangian formalism: d ∗ ∂L =0 L + dt ∂t

(6-15)

where L∗ = pi q
i − L is the well-known adjoint Lagrangian. Of course, L∗ is H if we express the various q
i in function of generalized momenta. PR5

The canonical equations follow from the variational principle of Hamilton (where L = pi q
i − H ).

Proof. By recalling (5.41), Hamilton’s principle is written in this new context as t2

δ ∫ ( pi q
i − H )dt = 0 .

(6-16)

t1

We can consider again the variational problem developed in Section 2 of Chapter 5 with the function

f = pi q
i − H . The condition of extremum δ S = α S ′(0) = 0 is here written ∂f ∂f + δ q i i + δ p
i 1 ∂pi ∂q t2 ∂H = ∫ (δ pi q
i − δ pi − δ qi t1 ∂pi t2

∫t

(δ pi

∂f ∂f + δ q
i i ) dt ∂p
i ∂q
∂H + δ q
i pi ) dt = 0 . i ∂q

But we know that the last term is t2

∫t

1

pi δ q
i dt =

t2

∫t

1

pi t2

t2 d i δ q dt = [ pi δ q i ]tt12 − ∫ p
iδ qi dt t1 dt

= − ∫ p
i δ q i dt t1

since δ q i (t1 ) = δ q i (t2 ) = 0 . Thus Hamilton’s principle is written:

(sum over i)

403

Hamiltonian Mechanics t2

∫t

1

( (q
i −

∂H ∂H )δpi − ( p
i + i )δq i ) dt = 0 . ∂pi ∂q

(6-17)

Given arbitrary variations δq i and δpi of independent variables qi and pi, we deduce the canonical equations q
i =

∂H , ∂pi

p
i = −

∂H ∂q i

.

Remark. Once more, the independence of variables qi and pi let us conclude. Of course, as soon as the motion is determined, these coordinates are no longer independent since, from qi, dq i (t ) ∂L and pi = i (t , q, q
) . we obtain q
i (t ) = dt ∂q
Example. Find the canonical equations for a particle moving in a central force field.

Answer. From

H=

p2 p r2 + θ 2 + V (r ) 2m 2mr

we deduce r
=

∂H p = r , ∂pr m

p
r = −

p 3 dV ∂H = θ3 − , ∂r mr dr

θ
=

∂H p = θ2 ∂pθ mr

p
θ = −

∂H = 0. ∂θ

Phase space We know that any material system determines a point (q1 ,..., q n ) in the configuration space. But the evolution of a material system is specified in addition by the generalized velocities q
1 ,..., q
n and thus by the generalized momenta. Given an open U ⊂ R n , we note that pairs (p,q) are points of a vector space U × R n and we say: D

The space of 2n-tuples ( pi , q i ) of independent conjugate variables is called the phase space associated with a material system.

We denote this 2n-dimensional space by P and thus ( p1 ,..., p n , q1 ,..., q n ) = ( p, q) ∈ P . The motion of a given material system is represented by a phase point (p,q) which describes a curve in the phase space in accordance with the corresponding Hamilton’s canonical equations.

404

2.3

Chapter 6 FIRST INTEGRALS AND CYCLIC COORDINATES

We have already defined the first integral notion in the Lagrangian context and it is immediately transposable to the Hamiltonian formalism. PR6

The Hamiltonian of a scleronomic material system is constant during the motion.

Proof. By considering the complementary equation, we have 0=

∂L dH =− dt ∂t

⇒

H =C.

So, in the case of scleronomic and conservative systems (with potential V ), PR6 of Chapter 5 becomes: PR7

Every scleronomic and conservative system has the Hamiltonian as a first integral called the Jacobi integral, which is nothing else than the mechanical energy H = E .

The proof is like the one of the recalled proposition; namely, from the Euler theorem about the homogeneous functions, we deduce H = pi q
i − L =

∂T i q
− L = 2T − L = T + V . ∂q
i

(6-18)

We recall the following D

A generalized coordinate qk is cyclic if it does not explicitly occur in the Lagrangian or Hamiltonian: ∂L ∂q

k

=0

⇔

∂H ∂q k

= 0.

In this case, we find again PR7 of Chapter5; that is, the generalized momentum conjugate to qk is a first integral: ∂H ∂q k

=0

⇒

p
k = 0

⇒

pk = c .

Relevance of cyclic coordinates If a coordinate (e.g. qn) is cyclic, then the conjugate momentum is a first integral ( p n = c ). So the Hamiltonian is a function of other qi, of other pi, of an arbitrary constant c (determined by the initial conditions) and of t, namely:

H (t , p1 ,..., p n−1 , q1 ,..., q n −1 , c) . So the problem comprises n − 1 generalized coordinates. Its solution will let us determine the ∂H cyclic coordinate by integrating q
n = . ∂c

405

Hamiltonian Mechanics

Remark 1. To integrate the canonical equations is to search the independent first integrals. If all these were known, the problem (and thus its solution) would be reduced to a system 2n finite equations. Remark 2. The first integrals are deduced from the canonical equations. So, in the example of a central force field, we have p
θ = −

which implies that

∂H =0 ∂θ

pθ = mr 2θ


is constant during the motion. We find again a well-known result since the coordinate θ is cyclic. Remark 3. It turns out to be necessary to choose a coordinate system that gives the maximum of cyclic coordinates, this choice making the most of symmetries of the problem. In the previous example, no Cartesian coordinate would be cyclic. The Routh’s method.

If a cyclic coordinate exists, for instance qn, we know that the Hamiltonian H (t , p1 ,..., p n−1 , q1 ,..., q n −1 , c) depends on an integration constant. In this context, the Routh’s method consists in combining at once the Lagrangian and Hamiltonian formalisms. Given r cyclic coordinates q 1 ,..., q r , we say: D

The Routhian is the function R defined by r

R(t , q1,..., q n , p1,..., pr , q
r +1,..., q
n ) = ∑ pi q
i − L(t , q, q
) . i =1

The differential r

dR = ∑ q
i dpi + i =1

=

r

∑ q
i dpi − i =1

n

r

n

pi dq
i − ∂L dt − ∑ ∂Li dqi − ∑ ∂Li dq
i ∑
∂t i =1 i =1 ∂q i =1 ∂q n

n

pi dq
i − ∑ ∂Li dqi − ∂L dt ∑ ∂t i = r +1 i =1 ∂q

implies ∀i ∈ {1,…,r}: ∀i ∈ { r + 1 ,…,n}:

∂R = q
i , ∂p i ∂R ∂L =− i i ∂q
∂q


∂R

,

∂q i ∂R ∂q

i

= − p
i =−

∂L ∂q i

.

(6-19)

406

Chapter 6

The first collection of 2r equations is the one of canonical equations for the r cyclic coordinates with the Hamiltonian R, the second collection of equations fulfills the n − r Lagrange’s equations: ∂R d ∂R ( i)− i =0 dt ∂q
∂q

i = r + 1,..., n .

But, for the cyclic coordinates, we have: ∀i ∈ {1,…,r}:

∂L ∂q

i

=0

⇒

∂R ∂q i

=0

and the corresponding momenta pi are arbitrary constants ci to be determined from the initial conditions. So, the n − r Lagrange’s equations are solved with the Routhian R (t , q r +1 ,..., q n , q
r +1 ,..., q
n , c1 ,..., c r ) .

2.4

LIOUVILLE’S THEOREM IN STATISTICAL MECHANICS

We know that any point of the phase space, called phase point, determines a possible state of a given material system. At the initial time, the state of the system is determined by a i 2n-tuple (pi,q ) in the phase space and the motion of the system is followed by a representative point which describes a unique path in the phase space according to the 2n Hamilton’s equations. Of course, if the number of degrees of freedom of the material system is very large and thus the number of canonical equations too, then obtaining the (unique) solution of the system of these equations can be impossible in practice. In addition, in the case of systems containing a great number of particles, it will be impossible to know the initial conditions of each of them. In order to overcome such difficulties, we are going to introduce a fundamental theorem of statistical mechanics due to Liouville. Whereas a well-determined phase point cannot be specified to represent the state of a system, we consider a set of neighboring phase points. These phase points represent neighboring states of the one of the material system and determine a cloud of possible states of the given system. So, we imagine a collection of phase points such that any of which could potentially represent the material system; in other words, we imagine a great number of systems corresponding to a given material system but with slightly different initial conditions.

Fig. 83

407

Hamiltonian Mechanics

The motion of each phase point represents the (independent) motion of a particular system. Let Ω be the phase volume element occupied by the neighboring phase points at the initial time t0 and let Ω′ denote the volume element occupied by the previous phase points at a later date t 0 + dt. We note that the number of representative phase points of a material system is the same in Ω and in Ω′ since any phase point initially within Ω can never leave this volume. Indeed, if a representative phase point of a possible system escaped at a given time, that would mean this point would occupy the same position as a phase point belonging to the boundary of the volume; but, since the (unique) motion is exactly determined from a given position of the phase space, then two corresponding material systems would move in the phase space along the same path. We emphasize that any phase path representing the (independent) motion of any system of Ω, that is a particular solution of the Hamilton’s equations, cannot intersect another and the number of representative phase points is invariable! Now, we can prove the following Liouville’s theorem. PR8

The density of representative points in phase space corresponding to the motion of a given material system is invariable during the motion.

Proof. According to the preliminaries, we must prove that the volumes Ω and Ω′ are equal, which will imply that the number of representative phase points per unit volume remains constant during the motion. During a time variation dt all the phase points of Ω move and their coordinates vary from pi, qi to the following

pi′ = pi + dpi ,

q ′ i = q i + dq i .

(6-20)

The relation between corresponding volumes Ω and Ω′ is obviously Ω′ = where J =

∂ ( p' , q ' ) Ω ∂ ( p, q )

∂ ( p' , q ' ) is the Jacobian of the coordinate transformation defined by (6-20). ∂ ( p, q )

From

p'i = pi + p
i dt ,

q ′ i = q i + q
i dt

∂pi′ ∂p
= δ ij + i dt , ∂p j ∂p j

∂pi′

we deduce

∂q ′ i ∂q
i dt , = ∂p j ∂p j and therefore the Jacobian is expressed as

∂q

j

=

∂p
i ∂q j ∂q ′ i ∂q j

dt

= δ ij +

∂q
i ∂q j

dt

408

Chapter 6

1+

∂p
1 dt ∂p1

∂p
1 dt ∂p 2

...

∂p
1 dt ∂p n

...

∂p
1 ∂q n

dt

. J=

. . .

.

∂q
n dt ∂p1

∂q
n dt ∂p 2

...

∂q
n ∂q
n dt ... 1 + n dt ∂p n ∂q

By limiting the expansion of J to the first order in dt, we obtain: ∂p
i ∂q
i + i ) = 0, J =1 + ∑( ∂q i =1 ∂p i n

but from n

∑( i =1

∂p
i ∂q
i + )=0 ∂pi ∂q i

we deduce that J = 1 and we have really obtained the expected equality Ω' = Ω .

From this result, we are going to prove that the density ρ (t , p, q ) of points in the phase space, such that the number of systems (whose representative points lie within Ω) is equal to ρ Ω , remains constant during the motion. Indeed, by expressing that this number is constant: d ( ρ Ω) = 0 dt

and knowing that d Ω = 0, dt

we deduce:

and thus

dρ Ω=0 dt dρ = 0, dt

which means that the density of possible systems or the number of phase points per unit phase volume is invariable during the motion: dρ ∂ρ ∂ρ ∂ρ = + p
i + i q
i = 0 . dt ∂t ∂pi ∂q

(6-21)

Remark. In differential geometry, the phase volume Ω is denoted by dp ∧ dq. Since the number of representative points of Ω (moving in the phase space) is invariable, that is d d dρ ( ρ dp ∧ dq ) = 0 , and since (dp ∧ dq ) = 0 , we deduce = 0. dt dt dt

Hamiltonian Mechanics

409

3. CANONICAL TRANSFORMATIONS The choice of generalized coordinates and generalized momenta is not unique; if we know that the Lagrange equations are invariable under changes of generalized coordinates, on the contrary we will have to search for transformations preserving the canonical form of Hamilton’s equations. Transformations called canonical have this property and such particular transformations play a fundamental role in Hamiltonian mechanics, in symplectic geometry and in celestial mechanics for instance. Such transformations can be very helpful insofar as new coordinates are cyclic. We are going to explain that.

3.1

LAGRANGE AND POISSON BRACKETS We have already defined the phase space P and we say:

D

The phase spacetime or state space is R×P .

3.1.1

Preservation of Canonical Form and Poisson Bracket First, we note the following

PR9

Any diffeomorphism1 on the state space does not necessarily preserve the form of Hamilton’s canonical equations.

Proof. Let us consider the canonical system q
i =

∂H , ∂pi

p
i = −

∂H ∂q i

and the following diffeomorphism

Q I = Q I (t , pi , q i ) ,

PI = PI (t , pi , q i )

I , i = 1,..., n.

In the “new” coordinates, the Hamiltonian H is written as follows: K (t , P , Q ) = H (t , p(t , P , Q ), q (t , P , Q )) .

Before continuing, let us introduce the definition of the Poisson bracket. Given an open I of R, opens Ω and Ω′ of Rn, we consider two functions at least of class C1, namely f and g: I × Ω × Ω ′ → R respectively defined by f (t , p, q ) and g (t , p, q ) . D
The Poisson bracket 2 of functions f and g is the function defined by 

{f,g} = ∑ ( i

1 2

∂f ∂g ∂f ∂g − ). ∂q i ∂pi ∂pi ∂q i

Bijection which is differentiable as well as its inverse (both of class C2 for instance). For instance, the Liouville’s theorem is conveniently expressed as dρ ∂ρ + {ρ,H} = 0 = ∂t dt

(6-22)

410

Chapter 6

Under the diffeomorphism the equations of Hamilton take the following form: ∂Q I ∂Q I ∂H ∂Q I ∂H Q
I = +∑ ( i − ) ∂t ∂q ∂p i ∂pi ∂q i i =

∂Q I ∂Q I ∂K ∂Q J ∂K ∂PJ ∂Q I ∂K ∂Q J ∂K ∂PJ +∑ + − ( ) ∑i ∂p ( ∂Q J ∂q i + ∂P ∂q i ) ∂t ∂PJ ∂p i ∂q i ∂Q J ∂pi i i J

=

∂K ∂Q I ∂K + { QI ,QJ } + { Q I , PJ } ⋅ J ∂PJ ∂t ∂Q

In the same manner we obtain: ∂P ∂K ∂K + { PI , PJ } . P
I = I + { PI , Q J } J ∂t ∂Q ∂PJ

However the canonical form of Hamilton’s equations is preserved after transformation iff there is a function H (t , PI , Q I ) such that ∂H Q
I = , ∂PI

∂H P
I = − I , ∂Q

that is not always the case, which proves the proposition. Now, the question is to know under what conditions the canonical form is preserved; namely: The diffeomorphism defined by

PI = PI (t , p, q) ,

Q I = Q I (t , p, q )

preserves the canonical form of Hamilton’s equations if { Q I , Q J }={ PI , PJ } = 0 { Q I , PJ } = δ JI

( ⇔ { PI , Q J } = −δ IJ )

and if there is a function F (t , PI , Q I ) such that QI =

∂F , ∂PI

PI = −

∂F . ∂Q I

Indeed, we have: ∂2F ∂K ∂ ∂F Q
I = + = ( + K), ∂t ∂PI ∂PI ∂PI ∂t

∂2F ∂K ∂ ∂F − P
I = − =− I( + K) , I I ∂t ∂Q ∂Q ∂Q ∂t with the following transformed Hamiltonian H =

∂F +K. ∂t

Hamiltonian Mechanics 3.1.2

-

411

Poisson Bracket and Symplectic Matrix

Among other things, the Poisson brackets let us: verify if a function f (t , p, q ) is a first integral, find new first integrals from known first integrals, express the motion equations under a very simple form, and so on.

Let us denote the coordinates of any point x of the phase space by x A = ( pi , q i ) where A∈{1,…,2n} and i∈{1,…,n}. We introduce the following (2n × 2n) matrix the importance of which will be later underlined. D
The matrix  0 J = (ω AB ) =  − I



I , 0 

(6-23)

where I is the ( n × n ) unit matrix, is called the symplectic matrix. The inverse matrix is

(

J −1 = ω AB

)

such that

ω AB ω BC = δ AC . Of course, we have J 2 = −I and 0 − I   . J −1 = t J = − J =  I 0  

So, the symplectic matrix is antisymmetric and of determinant +1. The Poisson bracket of functions f and g is expressed as 

{ f , g } = ω AB

∂f ∂g . ∂x A ∂x B

(6-24)

In matrix notation, this is written:

{ f , g } =  ∂f

 ∂p

∂f   ∂q 

0   I 

− I   0 

 ∂g     ∂p   ∂g     ∂q 

412

Chapter 6 t

 ∂f    ∂p = −   ∂f    ∂q 

 ∂g     ∂p  .  ∂g     ∂q 

J

PR10 The 2n Hamilton’s equations are denoted by ∂H ∂x B

x
A = ω AB

(6-25)

with A, B ∈ {1,...,2n } . Proof. For the n first indices these equations are

p
i = ω iB

∂H

= ω i n+ j

∂x B and for the n last indices they are q
i = ω n+i B

∂H ∂x

B

∂H ∂x n + j

= ω n+i j

∂H ∂x

j

=−

=

∂H ∂q i

∂H . ∂p i

Canonical equations and Poisson bracket The expression of the Poisson bracket of functions f (t , p, q ) and H (t , p, q ) is

{ f , H } = ∑ ( ∂f i n

i =1

∂H ∂f ∂H − ) ∂q ∂p i ∂p i ∂q i

that is


{ f , H } = ω AB

∂f ∂H . ∂x A ∂x B

(6-26)

PR11 The canonical equations of Hamilton have the following elegant expression:

x
A = { x A , H }

(6-27)

Proof. We have

{

∂x A ∂H ∂H x ,H =ω = ω BC δ BA C B C ∂x ∂x ∂x H ∂ = ω AC C ∂x A = x
. A

}

BC

So, for A = 1,..., n and A = n + 1,...,2n respectively, we have:

p
i = { pi , H },

q
i = { q i , H }.

(6-28)

413

Hamiltonian Mechanics First integral existence PR12 The existence condition of a first integral f of canonical equations of Hamilton is ∂f + { f , H }=0. ∂t

(6-29)

Proof. This is obvious since n ∂f dp i ∂f dq i df ∂f = + ∑( + ) dt ∂t i =1 ∂pi dt ∂q i dt ∂f = + {f , H } = 0. ∂t

Lie algebra First, we recall that, given a commutative field K, an algebra on K or K-algebra is a vector space E on K provided with a bilinear mapping *: E × E → K . In other words, we say: D

A K-algebra is a vector space K , E ,+ provided with an (inner) law * such that: ∀x, y, z ∈ E : ( x + y ) ∗ z = ( x ∗ z ) + ( y ∗ z ) , ∀x, y, z ∈ E : x ∗ ( y + z ) = ( x ∗ y ) + ( x ∗ z ) , ∀k1, k2 ∈ K , ∀x , y ∈ E : (k1 x ) ∗ (k2 y ) = k1k2 ( x ∗ y ) .

In addition, we say: D

A Lie algebra is an algebra for which the inner law is anticommutative and satisfies the Jacobi’s identity, that is: ∀x , y ∈ E : x ∗ y = − y ∗ x , ∀x , y, z ∈ E : ( x ∗ y ) ∗ z + ( y ∗ z ) ∗ x + ( z ∗ x ) ∗ y = 0.

Let C ( R × P ) denote the space of real-valued functions on a state space. PR13 The space C ( R × P ) together with the Poisson bracket forms a Lie algebra.

Proof. From the Poisson bracket definition, we have immediately,

∀k1, k2 ∈ R, ∀f , g , h ∈ C ( R × P ) : (i) The bilinearity:

{ f , g + h} = { f , g} + { f , h} , { f + g , h} = { f , h} + {g , h} , {k1 f , k 2 g} = k1k 2 { f , g} .

(6 − 30)

414

Chapter 6

(ii) The anticommutativity:

{ f , g} = −{g , f } . (iii) Jacobi’s identity:

{{ f , g }, h} + {{g , h}, f } + {{h, f }, g} = 0 .

(6-30’)

Let us prove this last with the help of ω AB : ∂ ∂f ∂g ∂h ∂ ∂g ∂h ∂f (ω CD C ) B + ω AB A (ω CD C ) A D ∂x ∂x ∂x ∂x ∂x ∂x ∂x D ∂x B ∂ ∂h ∂f ∂g + ω AB A (ω CD C ) ∂x ∂x ∂x D ∂x B

ω AB

∂2 f ∂g ∂f ∂2g ∂h ∂ 2 g ∂h ∂g ∂ 2 h ∂f AB CD =ω ω ( A C + C + C ) B +ω ω ( A C ) B D A D D A D ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x 2 2 ∂ h ∂f ∂h ∂ f ∂g + ω ABω CD ( A C + C ) B . D A D ∂x ∂x ∂x ∂x ∂x ∂x ∂x AB

CD

The second and third terms cancel and so on. Indeed, by changing the names of summation indexes A → D, B → C , C → A, D → B in the third term, this last becomes

ω DC ω AB

∂ 2 g ∂h ∂f , ∂x D ∂x A ∂x B ∂x C

that is the second term opposite. In the same manner, the fourth term cancels the fifth and the sixth term cancels the first. The two additional properties: { f , gh } = g { f , h } + h { f , g },

(6-31)

∂ { f , g } = { ∂f , g } + { f , ∂g } ∂t ∂t ∂t

(6-32)

are immediately proved from the Lagrange bracket definition.

Poisson theorem

PR14 If f and g are first integrals of Hamilton’s equations, then their Poisson bracket is a first integral. Proof. From and we deduce:

∂t f + {f , H} = 0 ,

∂ t g + {g , H } = 0

{H , { f , g }} + { f , {g , H }} + {g , {H , f }} = 0 , {H , { f , g}} − { f , ∂ t g} + {g , ∂ t f } = 0

415

Hamiltonian Mechanics

and thus ∂ { f , g } + {{ f , g }, H } = 0 . ∂t

From this theorem of Poisson, we can prove the following PR15 Given a conservative system ( H = E ), the existence of a first integral f (t , p, q ) implies that the successive derivatives

∂i f are first integrals. ∂t i

Proof. The theorem of Poisson lets us assert that integral iff ∂t f + {f , H} = 0

{ f , H } is

a first integral. But f is a first

and thus ∂ t f is a first integral. Then, Poisson’s theorem means that {∂ t f , H } is a first integral. But ∂ t f is a first integral iff ∂2 f + {∂ t f , H } = 0 , ∂t 2 ∂2 f is a first integral, and so on. thus we conclude ∂t 2 D

Any two functions f and g are in involution or involutive if their Poisson bracket is zero.

PR16 If a function f is involutive with other functions g and h, then it is involutive with their Poisson bracket {g , h}.

Proof. Since we have { f , g } = { f , h} = 0 , the identity of Jacobi implies

{ f , {g , h}} = −{g , {h, f }} − {h, { f , g}} = 0 . 3.1.3

Lagrange and Poisson Brackets

Given a diffeomorphism on the phase space:

Q I = Q I ( pi , q i ) ,

PI = PI ( pi , q i ) ,

then for any P and Q of { P1 ,..., Pn , Q1 ,..., Q n } we say: D


The Lagrange bracket of coordinates P and Q is the real-valued function: n

( P, Q ) = ∑ ( i =1

∂pi ∂q i ∂q i ∂pi ). − ∂P ∂Q ∂P ∂Q

(6-33a)

416

Chapter 6

The diffeomorphism being denoted by y A = y A ( x B ) with A, B ∈ {1,...,2n}, the previous definition is expressed as follows: D

The Lagrange bracket of any two coordinates yC and yD is the real-valued function ( y C , y D ) = ω AB

∂x A ∂x B

.

∂y C ∂y D

(6-33b)

PR17 There is the following relation between Poisson and Lagrange brackets: 2n

∑(y

{

}

C

, y D ) y D , y B = δ cB .

C

, y D ) { y D , y B } =ω AE

D =1

(6-34)

Proof. We have: 2n

∑(y D =1

= ω AE δ RE ω RS

3.2

∂x A ∂x E RS ∂y D ∂y B ω ∂y C ∂y D ∂x R ∂x S

A B A B ∂x A ∂y B ES ∂x ∂y S ∂x ∂y = ω ω = δ = δ CB . AE A C S C S C S ∂y ∂x ∂y ∂x ∂y ∂x

CANONICAL TRANSFORMATION

We are going to consider canonical transformations (which enrich Hamiltonian mechanics); for instance, such transformations of generalized coordinates and momenta preserve the canonical structure of the motion equations. Let us consider a neighborhood in a configuration space and the corresponding domain in the phase space. D
A diffeomorphism

Pi = Pi ( p, q ) ,

Q i = Q i ( p, q )

is a canonical transformation on the phase space if the differential pi dq i − Pi dQ i is exact; in other words, if there is a real-valued function S on the phase space such that 

pi dq i − Pi dQ i = dS

(6-35)

dpi ∧ dq i = dPi ∧ dQ i .

(6-36)

or if 

417

Hamiltonian mechanics Example. Given Pi = f (q i ) cos pi ,

i = 1,..., n ,

Q i = f (q i ) sin pi

let us specify f (q i ) such that this transformation be canonical. Answer. Since ∂f ∂f cos pi dq i − f sin pi dpi ) ∧ ( i sin pi dq i + f cos pi dpi ) i ∂q ∂q ∂f = − i f dpi ∧ dq i , ∂q

dPi ∧ dQ i = (

the transformation is canonical iff ∀i ∈ { 1,..., n }: −

∂f f =1 ∂q i

⇔

1 2

∂( f )2 = −1 ∂q i

that is

( f (q i )) 2 = −2q i + c i

( ci ∈ R ) .

3.2.1 Canonical Transformations and Brackets PR18 A diffeomorphism Pi = Pi ( p, q ) ,

Q i = Q i ( p, q )

is a canonical transformation on the phase space iff ( Pi , Pj ) = 0 ,

Proof. We consider

(Q i , Q j ) = 0 ,

( Pi , Q j ) = δ i j .

(6-37)

∂( P , Q ) ≠ 0. ∂ ( p, q)

The differential form p r dq r − Ps dQ s is written in function of variables Pi and Qi as follows: ∂q r ∂q r dPs + ( p r ( P , Q ) − Ps ) dQ s pr ( P , Q) s ∂Ps ∂Q (summation over r and s).

It is exact iff ∂Ps ∂p r ∂q r ∂p r ∂q r ∂ 2qr ∂ 2q r + p − + p = r r ∂Pk ∂Q s ∂Q s ∂Pk ∂Q s ∂Pk ∂Pk ∂Pk ∂Q s

∂P and, since s = δ sk , it is exact iff ∂Pk ∂p r ∂q r ∂q r ∂p r = ( Pk , Q s ) = δ sk ; − s s ∂Pk ∂Q ∂Pk ∂Q

(∑) r

418

Chapter 6

in addition, since

∂Ps ∂Q

m

=

∂Pm

= 0 , the condition

∂Q s

∂p r ∂q r ∂Q

m

∂Q

s

−

∂Ps ∂Q

m

∂p r ∂q r

=

∂Q ∂Q s

m

−

∂Pm ∂Q s

becomes ∂p r ∂q r ∂Q ∂Q s

m

−

∂q r ∂p r ∂Q ∂Q s

m

= (Q s , Q m ) = 0 ;

finally, the condition ∂p r ∂q r ∂p r ∂q r ∂ 2q r ∂ 2q r = + pr + pr ∂Pm ∂Ps ∂Ps ∂Pm ∂Ps ∂Pm ∂Pm ∂Ps

is ∂p r ∂q r ∂q r ∂p r = ( Pm , Ps ) = 0 . − ∂Pm ∂Ps ∂Pm ∂Ps

By recalling the zero square property of the exterior differentiation ( d
d = 0 ), we conclude that a canonical transformation on the phase space is a differentiable mapping which preserves the 2-form

ω = dpi ∧ dq i = dPi ∧ dQ i . Important remark.

We denote a general 2-form ω ∈ Ω 2 (P ) as follows:

ω = 12 ω AB dx A ∧ dx B .

(6-38)

In the following explicit expression

ω = 12 (ω ij dx i ∧ dx j + ω i n+ j dx i ∧ dx n + j + ω n+i j dx n+i ∧ dx j + ω n +i n + j dx n +i ∧ dx n+ j ) , we notice that the various ω ij are zero since there is no term of type dpi ∧ dp j , the various

ω n+i n+ j are zero since there is no term of type dq i ∧ dq j and the only nonzero terms are ω i n+i = −ω n+i i = 1 . Of course, the symplectic matrix J = (ω AB ) appears.

The previous skew-symmetric invariant bilinear form plays a fundamental role in symplectic geometry, this form being necessary to study the natural symplectic structure of the phase space. Such a study is outside the field of this book, but it is introduced, for instance, in Talpaert [2000]. PR19 A diffeomorphism on P is a canonical transformation iff the matrix associated to the Lagrange bracket is the symplectic matrix J, iff the matrix associated to the Poisson bracket is J −1 .

419

Hamiltonian mechanics

Proof. A diffeomorphism y A = y A ( x B ) defines a canonical transformation iff ∂x A ∂x B ω CD dy C ∧ dy D = ω AB dx A ∧ dx B = ω AB C D dy C ∧ dy D ∂y ∂y = ( y C , y D ) dy C ∧ dy D

(sum over C and D )

iff ( y C , y D ) = ω CD

iff

{y

C

, y D } = ω CD

[since (6-34)].

Remark. On the phase space, the existence criterion of a canonical transformation is

⇔

( PI , PJ ) = 0 ,

(Q I , Q J ) = 0 ,

( PI , Q J ) = δ IJ

{PI , PJ } = 0 ,

{Q , Q } = 0 ,

{Q

I

J

J

, PI } = δ .

(6-39)

J I

PR20 A transformation is canonical iff it preserves the Poisson bracket. Proof. Given a transformation y A = y A ( x B ) , we have: ∂f ∂g ∂g ∂y A ∂y B CD ∂f =ω { f , g }x = ω ∂x C ∂x D ∂y A ∂y B ∂x C ∂x D ∂f ∂g = { y A, yB } A B , ∂y ∂y CD

but { f , g }y = ω AB

∂f ∂g ∂y A ∂y B

and thus { f , g }y = { f , g }x

⇔

ω AB = { y A , y B }

that is iff the transformation is canonical.

3.2.2

Canonical Transformations and Generating Functions

Let us consider a neighborhood in a configuration space and the corresponding domain in the state space R × P . D

A diffeomorphism defined by Pi = Pi (t , p, q ) ,

Q i = Q i (t , p, q ) ,

t =t

is a canonical transformation on the state space R × P if the differential ( pi dq i − Hdt ) − ( Pi dQ i − Hdt ) is exact;

420

Chapter 6

in other words, if there is a real-valued function S on the state space such that pi dq i − Hdt = Pi dQ i − Hdt + dS

(6-40)

dpi ∧ dq i − dH ∧ dt = dPi ∧ dQ i − dH ∧ dt .

(6-41)

or if

By recalling PR5, we know that the variational principle holds for two collections of canonical variables {pi,qi} and {Pi,Qi} at once, that is t2

δ ∫ ( pi q
i − H ) dt = 0 , t1

δ ∫ ( PI Q
I − H )dt = 0 . t2

t1

By referring to PR13 of Chapter 5, it is certain that the two previous integrands do not differ dS of an arbitrary function S which does not depend on from more than the total derivative dt variables q
i and Q
I too. So, the variation of t2

∫t

1

t2 t 2 dS ( pi q
i − H )dt − ∫ ( PI Q
I − H )dt = ∫ dt = S ( 2) − S (1) t1 t1 dt

vanishes at limits (1) and (2) whatever the described function S. D

The previous function S (t , pi , q i , PI , Q I ) such that pi dq i − PI dQ I = ( H − H )dt + dS

is called the generating function of the canonical transformation. The generating function S is a priori dependent on 4n+1 variables. However 2n+1 variables are independent in view of the following 2n relations: PI = PI (t , p, q) ,

Q I = Q I (t , p, q ) .

Consequently, any generating function is one of the following types: S1 (t , q, Q ) , S 2 (t , P , q) , S 3 (t , p, Q ) , S 4 (t , p, P ) .

(i)

Generating function S1 (t , q, Q ) .

∂(qi , Q i ) We suppose det ≠ 0 in such a way that qi and Q i are independent coordinates. i ∂ ( pi , q ) We have: ∂S ∂S ∂S pi dq i − H dt = Pi dQ i − H dt + 1 dt + 1i dq i + 1i dQ i ∂t ∂q ∂Q

421

Hamiltonian mechanics ⇒

pi =

∂S1 , ∂q i

∂S1 , ∂Q i

Pi = −

H =

∂S1 +H . ∂t

(6-42)

 ∂2S  These relations define the canonical transformation (non singular if det  i 1 j  ≠ 0 ).  ∂Q ∂q  Example. The generating function

S1 = ∑ q i Q i i

generates the following canonical transformation pi = Q i ,

( H =H ).

Pi = − q i

This transformation is assuredly canonical because dp i ∧ dq i = dPi ∧ dQ i is fulfilled. By considering the generating function S1 , let us prove the following PR21 Canonical transformations preserve the canonical form of Hamilton’s equations.

Proof. Given the canonical equations p
i = −

∂H , ∂q i

q
i =

∂H ∂pi

and the canonical transformation Pi = Pi (t , p, q ) ,

Q i = Q i (t , p, q ) ,

we have: ∂P ∂P ∂P ∂H ∂PI ∂H P
I = ∂ t PI + I p
i + Ij q
j = ∂ t PI − I + ∂pi ∂pi ∂q i ∂q j ∂p j ∂q =−

∂2S + {PI , H } ∂Q I ∂t

[ since (6-42) ] .

But, the Poisson brackets are invariant under the canonical transformations, therefore we have: {PI , H } = − ∂HI ∂Q and thus ∂ 2 S1 ∂H P
I = − − I ∂Q ∂t ∂Q I ∂H [ since (6-42) ]. =− ∂Q I The invariance of the other Hamilton equations is proved in the same way.

422

(ii)

Chapter 6 Generating function S 2 (t , P , q)

The generating function S2 is defined from the Legendre transformation of S1 with respect to Q: L

S2 (t , P , q ) = − Q S1 (t , q, Q ) = −(

∂S1 i Q − S1 (t , q, Q )) ∂Q i

= PiQi + S1 (t , q, Q ). From

pi dq i − Hdt = Pi dQ i − H dt + dS1 (t , q, Q ) we deduce dS 2 (t , P , q) = pi dq i + Q i dPi + ( H − H ) dt . and thus pi =

∂S 2 , ∂q i

Qi =

∂S 2 , ∂Pi

H =

∂S 2 +H ∂t

(6-43)

with the condition  ∂2S det  i 2  ∂q ∂Pj 

(iii)

  ≠ 0.  

Generating function S 3 (t , p, Q )

The generating function S3 is defined from the Legendre transformation of S1 with respect to q: ∂S L S 3 (t , p, Q ) = − q S1 (t , q, Q ) = − ( 1i q i − S1 (t , q, Q )) ∂q = − pi q i + S1 (t , q, Q ) . Differentiation leads to Pi = −

∂S 3 , ∂Q i

qi = −

∂S 3 , ∂pi

H =H+

∂S3 ∂t

(6-44)

with

 ∂ 2 S3   ≠ 0. det  j  p Q ∂ ∂   i

(iv)

Generating function S 4 (t , p, P )

The generating function S4 is defined from the Legendre transformation of S2 with respect to q:

423

Hamiltonian mechanics L

S 4 (t , p, P ) =− q S 2 (t , P , q) = −(

∂S 2 ∂q

i

q i − S 2 (t , P , q))

.

= S1 (t , q, Q ) + Pi Q − pi q . i

i

Differentiation leads to qi = −

∂S 4 , ∂pi

∂S 4 , ∂Pi

Qi =

H =

∂S 4 +H. ∂t

(6-45)

with  ∂ 2S4 det   ∂pi ∂Pj 

  ≠ 0.  

Real symplectic group

We are going to see that the generating functions for canonical transformations are obviously related and we are going to consider an essential matrix of second derivatives of generating functions. Given a diffeomorphism on the phase space defined by y A = y A (x B )

A, B ∈ {1,...,2n}

Pi = Pi ( p, q ) ,

Q i = Q i ( p, q )

that is explicitly: we say: PR22 A canonical transformation exists iff there is a matrix X such that t

X J −1 X = J −1

( or t X J X = J )

where  ∂p   ∂x   ∂P X =  B  =   ∂y   ∂q  ∂P

∂p   ∂Q  . ∂q   ∂Q 

A

Proof. First, we deduce successively From (i):

∂pi ∂Q j

=

∂ 2 S1 ∂Q j ∂q i

=−

∂Pj ∂q i

,

from (ii):

∂pi ∂ 2 S2 ∂Q j , = = ∂Pj ∂Pj ∂q i ∂q i

from (iii):

∂Pj ∂ 2 S3 ∂q i , = − = ∂Q j ∂Q j ∂pi ∂pi

(6-46)

424

from (iv):

Chapter 6

∂ 2 S4 ∂q i ∂Q j =− =− . ∂Pj ∂Pj ∂pi ∂pi

 ∂x A  Second, we denote the matrix  B  by X BA .  ∂y 

( )

We have obviously: ∂x A ∂y B ∂y ∂x B

C

= X BA ( X −1 ) CB = δ AC .

For instance, ∂p1 ∂y B ∂p1 ∂y B = 1, … , = 0, … ∂y B ∂p1 ∂y B ∂q 3

We emphasize that

(X )

−1 A B

 ∂y A  =  B  .  ∂x 

Now let us prove that t

X −1 J −1 = J −1 X .

Indeed, we have: t

X −1

 ∂P  ∂p =  ∂P   ∂q

∂Q   ∂p  ∂Q   ∂q 

and t

X −1 J −1

 ∂Q  ∂p =  ∂Q   ∂q

∂P   ∂q  − ∂p   ∂P = ∂P   ∂p   ∂q   ∂P

−

∂q   ∂Q  . ∂p   ∂Q 

Since  ∂q − ∂P −1 J X =  ∂p   ∂P

−

∂q   ∂Q  , ∂p   ∂Q 

we have proved that J −1 X = t X −1 J −1

and thus t

X J −1 X = J −1 .

We note that such a fundamental condition of canonical transformation exists whatever the generating function we choose.

Hamiltonian mechanics

425

In addition, there exists an essential structure of group, namely: PR23 The set of matrices of canonical transformations on the phase space has a structure of group. Proof. (i) The composition of any two matrices of the set (matrix multiplication) is a matrix of the set. Indeed, given matrices X and Y such that t

XJX =J,

t

( XY ) J ( XY )= t Y t X J X Y = t Y J Y = J .

t

Y JY = J ,

we have

(ii) The associativity results from the associative law of matrix multiplication. (iii) The unit element is the unit matrix I. It is such that t IJI=J. (iv) Every element X of the set has an inverse, namely: X −1 = J −1 t X J .

Indeed, we have: X −1 X = J −1 t X J X = J −1 J = I

and X −1 is an element of the set because t

( J −1 t X J ) J ( J −1 t X J ) = t J ( X t J −1 t X ) J = t J ( X J J X −1 J −1 ) J

= − t J J −1 J = J .

4. HAMILTON-JACOBI EQUATION

We know that the solution of the canonical system of motion equations is obvious when all the coordinates are cyclic. The analytical integration method of Jacobi is famous for solving insoluble problems by the Lagrangian or Hamiltonian formalisms. This method based on Hamilton works is very effective in celestial mechanics, in perturbation problems and other areas. 4.1

HAMILTON-JACOBI EQUATION AND JACOBI THEOREM

The canonical transformations are particularly useful insofar as the transformed Hamiltonian H is simpler than the original Hamiltonian H. A clever man as Jacobi thought to

426

Chapter 6

choose a zero transformed Hamiltonian in order that the solutions P i and Q i of “new” canonical equations

∂H P
i = − i ∂Q

∂H Q
i = ∂Pi

be constants. Therefore there are 2n (independent) first integrals of motion of the representative point in the phase space. The Hamilton-Jacobi equation is obtained from such a judicious canonical transformation. If the transformed Hamiltonian H is zero then the existence criterion of a canonical transformation for a generating function of type S (t , q, Q ) is pi dq i − Pi dQ i = H dt + dS .

Identification implies the 2n equations pi =

∂S ∂q i

Pi = −

∂S ∂Q i

and the famous nonlinear Hamilton-Jacobi equation


∂S ∂S + H (t , i , q i ) = 0 . ∂t ∂q

(6-47)

In the Jacobian theory, the 2n equations of Hamilton being P
i = 0

Q
i = 0 ,

then the 2n (independent) first integrals of motion of the representative point in the phase space are

Pi (t , pi , q i ) = ai

Qi (t , pi , q i ) = bi

where ai and b i are 2n arbitrary constants, the position of the index of bi doing not matter. Given an orbit of a descriptive point in the phase space, the values of parameters ai and b i are the motion constants. In mechanics, only the search of a complete integral of the Hamilton-Jacobi partial differential equation is interesting (and not the general solution). D

A complete integral of the Hamilton-Jacobi equation is a solution which depends on as many arbitrary independent integration constants as coordinates q i and t, on q i and t.

However, the Hamilton-Jacobi equation shows S through partial derivatives. Therefore, one of the arbitrary constants is necessarily additive (since it doesn’t alter the partial derivatives).This additive constant plays no role in the canonical transformation (only the partial derivatives of S are present) and we express precisely:

Hamiltonian mechanics D

427

A complete integral of the Hamilton-Jacobi equation is a solution of the equation depending on as many nonadditive independent constants as coordinates q i , on q i and t.

It is denoted by S (t , q i , b i )

by taking the “new” coordinates Q i as arbitrary constants. Now, the n relations pi =

∂

S (t , q i , b i )

∂q at the initial instant are n equations linking the n constants b i and initial values of pi and q i together. So, they allow to calculate the integration constants from particular initial conditions. i

Next, the n following equations ai = Pi = −

∂ S (t , q i , b i ) i ∂b

provide the various ai from the initial conditions since the right-hand member is known from initial values of q i . Therefore, the problem is solved because the last equations let us obtain the coordinates q i from initial conditions and time: q i = q i (t , ai , b i ) .

It is the powerful integration method of Jacobi. Remark 1. Don’t forget the existence condition of every canonical transformation:

det (

∂2S ) ≠ 0. ∂q i ∂b j

Remark 2. The notation of a generating function by S (t , q i , Q i ) is fully justified because we have ( Q i being constants): dS = pi dq i − H dt = L dt .

Thus S (corresponding to a complete integral of the Hamilton-Jacobi equation) is nothing else than the action integral S = ∫ L (t , q, q
)dt , where this integration is “along the solution q(t).” With each complete integral is associated a class of orbits of a representative point in the phase space fitting the least action principle.

428

Chapter 6

Jacobi’s theorem

PR24 If

S (t , q i , b i ) is a complete integral of the Hamilton-Jacobi equation, then the

functions pi (t , ai , b i ) and q i (t , ai , b i ) obtained by solving the 2n equations pi =

∂ ∂q

i

ai = −

S (t , q i , b i ) ,

∂ S (t , q i , b i ) i ∂b

constitute the general integral of Hamilton’s canonical equations. Proof. On the one hand, by considering pi =

∂ ∂q

i

S (t , q i , b i ) , the derivative of

∂S ∂S + H (t , q i , i ) = 0 ∂t ∂q

with respect to parameters b i leads to ∂2S ∂H ∂ 2 S + = 0. ∂t ∂bi ∂p j ∂q j ∂bi

(1)

On the other hand, the total derivative of ai with respect to time, namely a
i = 0 , is written: ∂2S ∂ 2S + q
j = 0 ∂b i ∂t ∂bi ∂q j

(2)

(since b
i = 0 ).

Subtracting (1) and (2), we obtain: ∂2S ∂H (q
j − ) = 0 j i ∂q ∂b ∂p j and the canonical transformation condition det (

∂ 2S ) ≠ 0 leads to n Hamilton’s canonical ∂q j ∂bi

equations: q
j =

∂H . ∂p j

Now, on the one hand, the derivative of the Hamilton-Jacobi equation with respect to coordinates q i leads to ∂2S ∂H ∂H ∂ 2 S + + = 0. ∂t ∂q i ∂q i ∂p j ∂q j ∂q i On the other hand, the total derivative of pi = p
i =

(3)

∂S with respect to time is ∂q i

∂ 2S ∂ 2S + q
j . i i j ∂q ∂t ∂q ∂q

(4)

429

Hamiltonian mechanics

Subtracting (4) and (3), we obtain: p
i = −

∂H ∂2S ∂H + (q
j − ) i i j ∂q ∂q ∂q ∂p j

which implies the n other canonical equations: p
i = −

∂H . ∂q i

Remark. We have shown a canonical transformation (ai , b i ) ↔ ( pi , q i ) and the general integral of the canonical equations of Hamilton:

pi = pi (t , ai , b i ) ,

q i = q i (t , ai , b i ) .

Any other general integral pi = pi (t , ci , d i ) ,

q i = q i (t , ci , d i )

(5)

doesn’t define a canonical transformation and doesn’t allow obtaining a complete integral of the Hamilton-Jacobi equation. However, there is an exception, namely: If the constants ci and d i are respectively the various initial values pi0 and q0i , then the previous general integral shows a canonical transformation (ci = pi0 , d i = q0i ) ↔ ( pi , q i ) : pi = pi (t , pi0 , q0i ) , Indeed, the general integral (5) with

q i = q i (t , pi0 , q0i ) .

D( pi , q i ) ≠ 0 leads to the system D(ci , d i )

ci = ci (t , pi , q i ) ,

d i = d i (t , pi , q i )

of 2n first integrals of Hamilton canonical equations and thus the following first integrals:{ ci , c j }, { ci , d j } and { d i , d j }. We immediately have: (ci , d j )t 0 = δ i j ,

(ci , c j )t 0 = 0 ,

( d i , d j )t 0 = 0 ,

{ ci , d j } t 0 = − δ i j ,

{ ci , c j } t 0 = 0 ,

{ d i , d j } t0 = 0

and since the Poisson brackets are first integrals we deduce (for every time): { ci , d j } = − δ i j ,

{ ci , c j } = 0 ,

{ di,d j } = 0 .

Therefore, the general integral (5), where ci and d i are respectively the initial values of various pi and q i , is a canonical transformation.

430

4.2

Chapter 6 SEPARABILITY

Obtaining a complete integral of the Hamilton-Jacobi equation is often a fastidious even unsolved problem; nevertheless, searching this integral is sometimes made easy. We restrain our study to the (frequent) case of conservative material systems with scleronomic Hamiltonian. In the case of a scleronomic Hamiltonian i = 1,..., n

H ( pi , q i ) = E

the Hamilton-Jacobi equation ∂t S + H (

∂S i ,q ) = 0 ∂q i

has a complete integral of type S = − E t + S (q 1 ,..., q n , E , b 2 ,..., b n )

(6-48)

where ∂ t S = 0 . Therefore, the Hamilton-Jacobi equation becomes: H(

∂S i ,q ) = E . ∂q i

(6-49)

The theorem of Jacobi immediately leads to the general solution of the canonical equations of Hamilton: pi =

∂S ∂q i

a1 = −

∂S ∂S =t− ∂E ∂E

aj = −

∂S ∂b j

i = 1,..., n , ( or t − t 0 =

∂S ), ∂E

j = 2,..., n .

Example. In the case of the simple harmonic oscillator, the Hamilton-Jacobi equation is ∂S obtained by replacing p by in the Hamiltonian, that is: ∂q ∂S 1 ∂S 2 k 2 + ( ) + q =0 2 ∂t 2m ∂q

where q is the displacement and k the stiffness of the spring. This Hamilton-Jacobi equation is easily solved by considering S = S (q, b) − b t where b is the energy of the simple harmonic oscillator.

Hamiltonian Mechanics

431

The Hamilton-Jacobi equation H(

∂S , q) = b ∂q

is here 1 ∂S 2 k 2 ( ) + q = b. 2m ∂q 2 Thus the complete integral is

S = km

2b

∫

k

− q 2 dq − b t .

We have: a=−

km ∂S =− k ∂b

⇒

t−a = −

⇒

q=

∫

dq 2b k − q 2

+t

k m cos −1 ( q) k 2b

2b k cos( (t − a)) k m

The constants a and b are linked to initial conditions. At initial instant (t = 0) the oscillating point of mass m is relaxed at equilibrium position q0 with p0 = 0 . Given the initial conditions, the Hamilton-Jacobi equation

(

∂S 2 ) = 2m b − km q 2 ∂q

implies 0 = p0 = (

∂S )0 = 2m ∂q

b−

k 2

q02

and we find again an obviousness: the parameter b is the (initial) mechanical energy b = k q02 = m ω 2 q02 = E . 2

2

Therefore, we have: q = q0 cos(ω (t − a )) and the initial condition q (0) = q0 implies the constant a must vanish. In conclusion, S is the generating function of a canonical transformation and is expressed by S = mω (given the initial conditions).

∫

q02 − q 2 dq − m ω 2 q02 t 2

432

Chapter 6

PR25 A scleronomic mechanical system with a cyclic generalized coordinate, e.g. q1 , shows a complete integral of the Hamilton-Jacobi equation of type S = − E t + a1 q 1 + S (q 2 ,..., q n ; a1 , E , b 2 ,..., b n −1 )

(6-50)

where a1 is an arbitrary constant.

Proof. The scleronomic character explains the first term. In addition, if q1 is a cyclic coordinate we have: ∂L =0 ∂q1

⇒

∂H = 0 = − p
1 . ∂q1

∂S conjugate to q1 is an arbitrary constant a1 . 1 ∂q The complete integral is thus of type (6-50). Since it must verify the Hamilton-Jacobi equation, then the function S must satisfy:

The generalized momentum p1 =

H ( q 2 ,..., q n , a1 ,

∂S ∂S ,..., n ) = E . 2 ∂q ∂q

In the complete integral (6-50) the variables t and q1 are separated from the other variables. The reader will generalize if t , q1 ,..., q k are cyclic coordinates, with H (q k +1 ,..., q n , p1 ,..., pn ) . D

A scleronomic conservative mechanical system is called separable if the corresponding Hamilton-Jacobi equation shows a complete integral S of type:

S = −E t +

n

∑ S i ( q i ; b)

(6-51a)

i =1

where b denotes the set of arbitrary constants. This complete integral will be also denoted simply by: n

S = − E t + ∑ Si ( q i ) .

(6-51b)

i =1

where only the corresponding variable qi which appears in each Si is indicated. Make clear that the separability existence is dependent on the choice of generalized coordinates. Remark. Solving the Hamilton-Jacobi partial differential equation is not a priori more simple than solving Hamilton canonical equations, but separability leads to quadratures! Conditions for separability

There are cases where a Hamiltonian system may be quickly solved by separation of variables; they are the most simple Hamiltonian cases.

433

Hamiltonian Mechanics

Stäckel1 showed a general form of the Hamiltonian for which we can have separation of variables. Historically, conditions for the separability of variables were obtained by Morera and Dall’Aqua, and works of Levi-Civita were essential. Later, Weinacht2 found all the coordinate systems of two and three dimensions for which separation is possible. At present, the study of separable systems has numerous applications, as for instance in astronomy. A necessary and sufficient condition for the separability can be shown. Let n

S = − Et + ∑ S i (q i ) i =1

be a complete integral of the Hamilton-Jacobi equation. It is such that

H (q i ,

(where the notation

dS i

dS i dq i

)=E

i = 1,..., n

emphasizes the derivation with respect to the only concerned

dq i generalized coordinate), and thus, ∀k ∈ {1,..., n} : dH dq k

=

∂H ∂q k

+

∂H d 2 S k =0 ∂p k (dq k ) 2

(no summation).

⇔ d 2Sk k 2

(dq )

=−

∂H ∂q

k

∂H ∂p k

∂H ≠ o. ∂p k

by assuming ∀k :

For every r ≠ k , we have: d dq

⇔

[

r

(

∂H ∂q

k

∂H )=0 ∂p k

∂ ∂H ∂ ∂H dp r ∂H ∂H ∂ ∂H ∂ ∂H dp r − k[ r( ( k )+ ( k) r] )+ ( ) ]=0 r ∂p r ∂q dq ∂p k ∂q ∂q ∂p k ∂p r ∂p k dq r ∂q ∂q

and since dp r dq

r

=−

∂H ∂q r

∂H , ∂p r

we obtain the conditions for separability:

( 1

∂ 2 H ∂H ∂ 2 H ∂H ∂H ∂ 2 H ∂H ∂ 2 H ∂H ∂H − − ) ( − ) = 0. ∂q r ∂q k ∂p r ∂p r ∂q k ∂q r ∂p k ∂q r ∂p k ∂p r ∂p r ∂p k ∂q r ∂q k

Stäckel, P. 1890, Math. Ann. 35, 91.  1893, ibid. 42, 537. 2 Weinacht, J. 1923, Math. Ann. 90, 279.

(6-52)

434

Chapter 6

We consider Hamiltonians H = T + V such that T = 12 a ij (q ) pi p j ,

V = V (q ) .

n(n − 1) necessary and sufficient conditions for the separability (6-52) are 2 polynomials of fourth degree in p, where the coefficients of the terms of fourth, second and zero-th degrees are equal to zero:

So, the

∂ 2T ∂T ∂T ∂ 2T ∂T ∂T ∂ 2T ∂T ∂T ∂ 2T ∂T ∂T − − + = 0, ∂q r ∂q k ∂p r ∂p k ∂q r ∂q k ∂p r ∂p k ∂p r ∂q k ∂q r ∂p k ∂p r ∂p k ∂q r ∂q k

(1)

∂ 2V ∂T ∂T ∂ 2T ∂T ∂V ∂ 2T ∂T ∂V − − ∂q r ∂q k ∂pr ∂pk ∂pr ∂q k ∂pk ∂q r ∂q r ∂pk ∂pr ∂q k

(2)

+

∂ 2T ∂T ∂V ∂T ∂V ( r k + k r)=0, ∂pr ∂pk ∂q ∂q ∂q ∂q

∂ 2T ∂V ∂V ∂V ∂V = a kr k =0 k r ∂p r ∂p k ∂q ∂q ∂q ∂q r

(3)

( r ≠ k ).

Equation (1) determines the form of T , equations (2) and (3) determine V.

Stäckel’s case

The important case where the various akr vanish for k ≠ r (the Hamiltonian containing a sum of n momenta squared) was studied by Stäckel. We say: PR26 A material system such that H=

n

1 2

∑ f i (q) pi2 + V (q) i =1

is separable iff there are:

( )

(i) a nonsingular matrix uij of which each element depends on the only variable qi and such that n

∑ f i uij = δ 1j , i =1

(ii) functions vi of the corresponding variable qi such that n

∑ f i vi = V

(potential).

i =1

Proof. First, from the separability, we are going to deduce (i) and (ii).

435

Hamiltonian Mechanics n

A complete integral S = ∑ S i (q i ; b) verifies the Hamilton-Jacobi equation H = E ( = b1 ) , i =1

that is: n

1 2

dS

∑ f i ( dq ii ) 2 = b1 − V . i =1

Derivations with respect to successive parameters bi lead to

fi

dS i ∂ dS i ( ) =1 , dq i ∂b1 dq i

fi

dS i ∂ dS i ( )=0 dq i ∂b j dq i

n

∑ i =1 n

∑ i =1

j = 2,..., n .

We note that each fi depends on the only corresponding variable qi and the system of the previous equations can be written: n

∑ i =1

f i u ij = δ 1j

where each u ij depends on the only corresponding generalized coordinate qi. In addition, the expression n

V = b1 − 12 ∑ f i ( i =1

n

dq

) 2 = δ 1j b j − 12 ∑ f i ( i

= ∑ f i (u ij b j − 12 ( i =1

n

is really of type

∑

n

dS i

i =1

dS i dq

i

dS i dq

i

)2

)2 )

f i vi where each vi only depends on the corresponding variable qi.

i =1

Second, if by hypothesis we have: n

∑ i =1

f i u ij = δ 1j ,

n

∑

f i vi = V

i =1

then the Hamilton-Jacobi equation H = b1 , that is n

1 2

∑ i =1

fi (

∂S 2 ) + V = b1 , i ∂q

is written n

∑ i =1

f i ( 12 (

∂S 2 ) + vi − u ij b j ) = 0 . i ∂q

436

Chapter 6

Therefore, the Hamilton-Jacobi equation has solutions of the following type:

S = S1 (q1 ; b) + ... + S n (q n ; b) by choosing, for each i:

(

∂S 2 ) = 2 (u ij b j − vi ) i ∂q = φ i (q i ; b).

A complete integral is thus written: n

S = −E t + ∑

∫ i =1

φ i dq i .

We note that the Jacobi’s theorem leads to the general solution of Hamilton’s canonical equations, that is pi = φ i

(1)

and n u1 a1 = − ∂S = t − ∑ 22 ∫ i dqi ∂E i =1 φi

also written: n

t − t0 = ∑

∫ i =1

ui1

φi

dq i ,

(2)

and n

a j = −∑

∫ i =1

u ij

φi

dq i

j = 2,..., n,

(3)

these last equations (3) determining the trajectories.

Example. We consider a system with two degrees of freedom of which the Hamiltonian has the form H = 12 [ A( x, y ) p x2 + B( x, y ) p 2y ] + V ( x, y )

where x, y are generalized coordinates and px, py the conjugate momenta.

(i) If the system is separable, characterize the functions A, B, V and find the expression of a complete integral. (ii) Deduce the equations of motion. Answer. (i) A complete integral of the separable system is of type S = S1 ( x , E , b ) + S 2 ( y , E , b )

and thus 1 2

[ A( x, y ) (

dS1 2 dS ) + B ( x, y ) ( 2 ) 2 ] = E − V . dx dy

437

Hamiltonian Mechanics

By letting f ( x, E , b) = 12 (

dS1 2 ) , dx

g ( y, E , b) = 12 (

dS 2 2 ) , dy

we obtain A( x, y ) f ( x, E , b) + B( x, y ) g ( y, E , b) = E − V .

(1)

So, we have the system of equations A

∂f ∂g +B =1 , ∂E ∂E

A

∂f ∂g +B = 0. ∂b ∂b

Since S is a complete integral, we have: ∂2S ∂x∂E D= 2 ∂ S ∂y∂E

∂2S ∂x∂b ∂2S ∂y∂b

and thus dS1 ∂ dS1 ∂g ( ) dx ∂E dx ∂E = dS1 ∂ dS1 ∂g ( ) ∂b dx ∂b dx

∂f ∂E ∆= ∂f ∂b =

dS 2 ∂ dS 2 ( ) dy ∂E dy dS 2 ∂ dS 2 ( ) dy ∂b dy

dS1 dS 2 D ≠ 0. dx dy

So, the previous system of equations has one solution 1 0 A =

B =

∆ ∂f ∂E ∂f ∂b

∂g ∂E ∂g ∂b

=

1 , ∂f ∂f ∂E ∂g ∂E ( ) − ∂b ∂f ∂b ∂g ∂b

1 0 ∆

=

−1 . ∂g ∂f ∂E ∂g ∂E ( ) − ∂b ∂f ∂b ∂g ∂b

By letting P=

1 , ∂f ∂b

Q=

−1 , ∂g ∂b

X =

∂f ∂E , ∂f ∂b

Y =−

∂g ∂E ∂g ∂b

438

Chapter 6

where P and X do not depend on y and Q and Y do not depend on x, we have so obtained: A=

P , X +Y

B=

Q . X +Y

Equation (1) is thus written: V = E − Af − Bg =

EX + EY − Pf − Qg . X +Y

By letting the function (independent on y)

ξ = EX − Pf , and the function (independent on x)

η = EY − Qg , we have: V=

ξ +η X +Y

and H=

ξ +η 1 ( P p x2 + Q p 2y ) + . X +Y 2( X + Y )

Conversely, given such a Hamiltonian there is separability, the corresponding HamiltonJacobi equation is 1 P ∂S Q ∂S [ ( )2 + ( )2 + ξ + η ]= E . 2 ∂y X + Y 2 ∂x A complete integral of type S = S1 ( x, E , b) + S 2 ( y, E , b) is such that S1 is obtained from P dS1 2 ( ) = EX − ξ + b 2 dx

and S2 is obtained from Q dS 2 2 ( ) = EY − η − b ; 2 dy

that is: 2 ( EX − ξ + b dx + P

S=∫

(ii)

We have successively:

px =

∂S = ∂x

2 ( EX − ξ + b) P

py =

∂S = ∂y

2 ( EY − η − b) Q

∫

2 ( EY − η − b) dy . Q

439

Hamiltonian Mechanics

t − t0 = a=−

X dx Y dy ∂S , =∫ +∫ ∂E 2 P( EX − ξ + b) 2Q( EY − η − b)

∂S dx = −∫ + ∂b 2 P( EX − ξ + b)

∫

dy 2Q( EY − η − b)

.

From the two previous results, we deduce:

dx 2 P( EX − ξ + b)

dy

=

2Q( EY − η − b)

and

dt =

X

dx

+

2 P( EX − ξ + b)

Y

dy

2Q( EY − η − b)

=

( X + Y ) dx 2 P( EX − ξ + b)

.

So, the trajectory and the motion are determined by the following equations:

dx 2 P( EX − ξ + b)

=

dy 2Q( EY − η − b)

=

dt . X +Y

5. EXERCISES Exercise 1.

Write down the Hamiltonian for a free particle in cylindrical coordinates. Write the canonical equations of Hamilton and the Liouville-Boltzmann equation. Show that this last equation leads to Hamilton’s equations.

Answer. By considering the generalized coordinates q1 = r ,

q2 = θ ,

q3 = z ,

the Lagrangian for the free particle is

L = T − V = 12 (r
2 + r 2θ
2 + z
2 ) − V (t , r ,θ , z ) . Since the generalized momenta are pr =

∂L = r
, ∂r


the Hamiltonian per unit mass is written

pθ =

∂L = r 2θ
, ∂θ


pz =

∂L = z
, ∂z


440

Chapter 6

H = pi q
i − L = 12 ( p r2 +

pθ2 r2

+ p z2 ) + V (t , r ,θ , z )

.

The canonical equations of motion are

θ
=

r
= p r , p
r = −

∂H pθ2 ∂V , = 3 − ∂r ∂r r

pθ r2

p
θ = −

,

z
= p z ,

∂V , ∂θ

p
z = −

∂V . ∂z

The Liouville-Boltzmann equation is written: df ∂f ∂f ∂f pθ ∂f ∂f pθ2 ∂V ∂f ∂V ∂f ∂V = + pr + + pz + ( 3 − )− − = 0. 2 dt ∂t ∂r ∂θ r ∂z ∂pr r ∂r ∂pθ ∂θ ∂p z ∂z

This linear homogeneous equation in f leads to the system of canonical equations: dpθ dp dp z dt dr dθ dz = = = = 2 r = = . pθ 1 pr pz pθ ∂V − ∂V − ∂V − ∂θ ∂z r2 r 3 ∂r Exercise 2. From the well-known equations of motion of a particle (of a system) written in generalized coordinates: q

i + Γ ijk q
j q
k = a i where a i = − g ik

∂V , ∂q k

(i) express the motion equations in cylindrical coordinates and the Liouville-Boltzmann equation, (ii) express the motion equations in spherical coordinates and the Liouville-Boltzmann equation, (iii) find again the corresponding expressions of Lagrangian, Hamiltonian and canonical equations in spherical coordinates. Answer. (i) Given ds 2 = dr 2 + r 2 dθ 2 + dz 2 ,

the only nonzero Christoffel symbols are 1 . r

r Γθθ = −r ,

Γ rθθ = Γθθ r =

q
1 = r
= R ,

Θ q
2 = θ
= , r

We denote q
3 = z
= Z ,

where R is the velocity component in the r-direction, Θ is the velocity component perpendicular to the (r , z ) − plane and Z is the velocity component along the z-axis.

441

Hamiltonian Mechanics The motion equations are the following: 1 q

1 + Γ 22 q
2 q
2 = − g 11

⇔

∂V ∂q 1

Θ 2 ∂V
R= − , r ∂r

and 2 1 2 2 q

2 + Γ 12 q
q
+ Γ 21 q
2 q
1 = − g 22

⇔

1 ∂V d Θ 2 Θ ( ) + r
= − 2 dt r r r r ∂θ

⇔


= − RΘ − 1 ∂V , Θ r r ∂θ

∂V ∂q 2

and q

3 = − g 33

⇔

∂V ∂q 3

∂V . Z
= − ∂z

With the six variables r ,θ , z , R, Θ, Z , which are not canonically conjugate, the equations of the motion of the particle are simplified. For a phase density f (t , r ,θ , z , R, Θ, Z ) , the Liouville-Boltzmann equation is ∂f ∂f Θ ∂f ∂f Θ 2 ∂V ∂f RΘ 1 ∂V ∂f ∂V ∂f +R + +Z +( − ) −( + ) − = 0. ∂t ∂r r ∂θ ∂z ∂r ∂R r r r ∂θ ∂Θ ∂z ∂Z

(ii) Given ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2 ,

the only nonzero Christoffel symbols are r Γ φφ = −r sin 2 θ ,

r Γθθ = −r ,

Γ rθθ = Γ θθ r =

1

Γ rφφ = Γ φφ r =

1

r r

,

θ Γ φφ = − sin θ cos θ ,

,

φ φ Γ θφ = Γ φθ = cot θ ,

We denote q
1 = r
= R ,

Θ q
2 = θ
= , r

q
3 = φ
=

Φ r sin θ

442

Chapter 6

where R is the velocity component in the direction of the radius vector simply called the radial velocity, Θ is the velocity component along the meridian and Φ is the velocity component along the circle of colatitude. The motion equations are the following: 1 q
3 q
3 = − g 11 q

1 + Γ 122 q
2 q
2 + Γ 33

⇔

∂V ∂q 1

Θ 2 + Φ 2 ∂V R
= , − r ∂r

and 2 1 2 2 2 q
3 q
3 = − g 22 q

2 + Γ 12 q
q
+ Γ 21 q
2 q
1 + Γ 33

⇔

∂V ∂q 2


= − RΘ + Φ cot θ − 1 ∂V , Θ r r r ∂θ 2

and 3 1 3 3 3 3 q

3 + Γ13 q
q
+ Γ 31 q
3 q
1 + Γ 23 q
2 q
3 + Γ 32 q
3 q
2 = − g 33

⇔

∂V ∂q 3


= − RΦ − ΘΦ cot θ − 1 ∂V . Φ r r r sin θ ∂φ

With the six variables r ,θ , φ , R, Θ, Φ , which are not canonically conjugate, the equations of the motion of the particle are simplified. Given the phase density f (t , r ,θ , φ , R, Θ, Φ ) , the Liouville-Boltzmann equation is thus written:

Θ 2 + Φ 2 ∂V ∂f Φ ∂f ∂f Θ ∂f ∂f ) +( − + +R + ∂r ∂R ∂r r ∂θ r sin θ ∂φ ∂t r 1 ∂V ∂f 1 ∂V ∂f RΘ Φ 2 RΦ ΘΦ ) = 0. cot θ − ) + (− − cot θ − + (− + r r r ∂θ ∂Θ r r r sin θ ∂φ ∂Φ (iii)

Given a mass m, we have immediately: L=

m 2 (r
+ r 2θ
2 + r 2 sin 2 θ φ
2 ) − V (t , r ,θ , φ ) , 2

H = pi q
= i

1 2m

( p r2

+

pθ2 r2

In generalized coordinates, we note that L=m g q
i q
j − V (t , q ) , 2 ij

H = 21m g ij pi p j + V (t , q) .

+

pφ2 r 2 sin 2 θ

) + V (t , r ,θ , φ ).

443

Hamiltonian Mechanics

The canonical equations

{

}

p
i = { pi , H }

q
i = q i , H ,

are written as it follows:

r
=

p ∂H = r , ∂p r m

θ
=

pθ mr 2

,

φ
=

pφ mr 2 sin 2 θ

,

pφ2 pθ2 ∂H ∂V p
r = − = + − , 3 3 2 ∂r mr mr sin θ ∂r p
θ =

pφ2 mr sin θ

p
φ = −

2

2

cot θ −

∂V , ∂θ

∂V . ∂φ

We note that the generalized momenta are written with the variables R, Θ, Φ as follows:

p r = mr
= mR , pθ = mr 2θ
= mr Θ , pφ = mr 2 sin 2 θ φ
= mr sin θ Φ .

Exercise 3. Is the diffeomorphism P = q cot p ,

Q = ln(q −1 sin p)

a canonical transformation on the 2-dimensional phase space ? Answer. Yes it is, because q cos p dp + cot p dq ) ∧ ( dp − q −1 dq ) 2 sin p sin p 1 = ( 2 − cot 2 p ) dp ∧ dq sin p

dP ∧ dQ = (−

= dp ∧ dq .

Exercise 4. Find the motion equation of a simple harmonic oscillator by introducing the generating function S1 = 12 m ω (q ) 2 cot Q ,

where q is the displacement and ω 2 = k m .

444

Chapter 6

Answer. We have: p=

∂S1 = m ω q cot Q , ∂q

P=−

∂S1 1 ω (q) 2 = m . ∂Q 2 sin 2 Q

This last relation leads to 2P sin Q mω

q= and thus

p = 2m ω P cos Q .

The Hamiltonian is invariable under the transformation because the time doesn’t explicitly appear in the generating function. Since p=

∂T = m q
∂q


and V =

k 2

(q) 2 ,

the Hamiltonian is written: H = T +V =

( p) 2 k + (q) 2 = ω P cos 2 Q + ω P sin 2 Q = ω P . 2m 2

Since the coordinate Q is cyclic, we conclude that the conjugate momentum P is constant, it is E ω . The motion equation is ∂H =ω Q
= ∂P

which implies Q = ωt + c ,

where c is an integration constant (fixed by the initial conditions). We find the well-known displacement q=

2E mω 2

sin (ω t + c) .

445

Hamiltonian Mechanics Exercise 5.

In the central force problem, find a complete integral of the Hamilton-Jacobi equation and the general solution of motion equations. Answer. The Hamiltonian is well-known:

H = 21m ( pr2 + pθ2 r 2 ) + kr . The Hamilton-Jacobi equation is

k ∂S 2 1 ∂S 2 ∂ t S + 1 [( ) + 2 ( ) ] + = 0 . 2 m ∂r r ∂θ r We search a complete integral of type S = − E t + S1 (r ) + S 2 (θ )

which implies 1 [( ∂S ) 2 + 1 ( ∂S ) 2 ] + k = E 2m ∂r r 2 ∂θ r

that is

(

dS 2 2 k dS 2 ) = r 2 [ 2m E − 2 m − ( 1 ) ] . dθ r dr

Since the two members are functions of respective θ and r, we obtain: dS 2 = A. dθ

This expected result, S2 = Aθ , implies: r 2 [ 2m E − 2 m

⇒

k dS 2 − ( 1 ) ] = A2 r dr

S = − E t + Aθ ±

∫

2mE − 2m

k r

−

A2 dr . r2

The general solution is such that − b1 =

∂S = −t ± m ∫ ∂E

− b2 =

∂S =θ ± ∂A

∫r

dr 2mE − 2m k r − A2 r 2

A dr 2

2mE − 2m k r − A2 r 2

;

thus, for a trajectory through (r0 ,θ 0 ) at instant t0 , the previous equations are well-known: t − t0 = ±

m 2

∫

r

r0

dr E − k r − A2 2mr 2

446

Chapter 6 r

dr

r0

r 2 2mE − 2m k r − A2 r 2

θ − θ0 = ± A ∫

.

These equations give the positions ( by θ (r ) ) and position instants.

Exercise 6.

A particle moves in a vertical plane under the action of its weight mg without friction. This plane is constrained to rotate about a vertical axis with constant angular velocity ω. (i) Calculate the Hamiltonian. (ii) Find the Hamilton-Jacobi equation and a complete integral. (iii) Determine the general solution of Hamilton canonical equations. Answer. (i) By introducing the cylindrical coordinates r ,θ , z and knowing that θ
= ω , the Lagrangian is written: L=

m 2 m (r
+ r 2 θ
2 + z
2 ) − mg z = (r
2 + z
2 + r 2 ω 2 ) − mg z . 2 2

There are two degrees of freedom, the generalized coordinates are r and z. Since the generalized momenta are pr =

∂L = m r
, ∂r


pz =

∂L = m z
, ∂z


the Hamiltonian is immediately: H = pi q
i − L =

m 1 ( pr2 + p z2 ) − r 2 ω 2 + mg z . 2m 2

(ii) The Hamilton-Jacobi equation is ∂S ∂S 2 ∂S 2 1 + ( ( ) + ( ) ) − m r 2 ω 2 + mg z = 0 . 2 ∂t 2m ∂r ∂z

By separating the variables t , r , z : S = − E t + S1 (r ) + S2 ( z ) ,

the Hamilton-Jacobi equation becomes: dS 2 dS 2 1 ( ( 1 ) + ( 2 ) ) − m r 2 ω 2 + mg z = E 2m dr 2 dz

or

(

1 dS1 2 m 2 2 1 dS 2 ( ) − r ω ) + ( ( 2 ) + mgz ) = E . 2m dr 2m dz 2

So, the first term only depends on variable r and the second term only on variable z; they are thus constants.

447

Hamiltonian Mechanics

By putting 1 dS1 2 m 2 2 ( ) − r ω = E1 , 2m dr 2 1 dS 2 2 ( ) + mg z = E2 2m dz

with E1 + E2 = E , we have a complete integral immediately: S = −( E1 + E2 ) t ± ∫ 2mE1 + m 2 r 2ω 2 dr ± ∫ 2mE2 − 2m 2 g z dz . (iii) The general solution of motion equations follows from a1 = −

∂S , ∂E1

a2 = −

∂S ∂E2

that is a1 = t ±

∫

m dr 2mE1 + m r ω 2 2

2

,

a2 = t ±

∫

m dz 2mE2 − 2m 2 g z

.

Exercise 7.

The Hamiltonian of a system with two degrees of freedom is H=

1 ( 12 ( g p12 + f p 22 ) + gϕ + fψ ) f +g

where f and ϕ are functions of the generalized coordinate q1, and g and ψ are functions of the coordinate q2. Find a complete integral of the Hamilton-Jacobi equation. Answer. It is a Stäckel’s case of separation of variables and we have: S = − Et + S1 (q 1 ) + S 2 (q 2 ) .

The Hamilton-Jacobi equation is written: dS dS 1 1 [ 2 ( g ( 11 ) 2 + f ( 22 ) 2 ) + gϕ + fψ ] = E f +g dq dq

that is 1 ( 1 ( dS1 ) 2 + 1 ( dS 2 ) 2 ) + ϕ + ψ − E − E = 0 . 2 f dq1 g dq 2 f g g f

By introducing the constant C, the separation is expressed as 1 dS1 2 ϕ E ( ) + − =C, 2 f dq1 f f 1 dS 2 2 ψ E ( ) + − = −C . 2 g dq 2 g g

448

Chapter 6

So, we have S1 = ± ∫ 2Cf − 2ϕ + 2 E dq1 , S 2 = ± ∫ − 2Cg − 2ψ + 2 E dq 2

and thus

S = − Et + S1 + S 2 .

Exercise 8. Find a complete integral of the Hamilton-Jacobi equation of a spherical pendulum of length R. Establish the general solution of canonical equations of Hamilton. Answer. In this problem of two degrees of freedom the Lagrangian is written L=

m 2

R 2 (θ
2 + φ
2 sin 2 θ ) + mg R cosθ

where the two generalized coordinates are the colatitude θ and the longitude φ ,and the Hamiltonian is pφ2 1 2 H ( pφ , pθ ,θ ) = ( p + ) − mg R cosθ . θ 2 2mR sin 2 θ The Hamiltonian-Jacobi equation is written: ∂t S +

1 2 mR

2

((

∂S 2 1 ∂S 2 ) + ( ) ) − mgR cosθ = 0 . 2 ∂θ sin θ ∂φ

In this scleronomic problem where φ is a cyclic coordinate, we search a complete integral of type: S = − E t + cφ + Θ(θ ) .

The Hamilton-Jacobi equation that is 1 2mR

(( 2

dΘ 2 c2 ) + ) − mgR cosθ = E dθ sin 2 θ

lets us obtain θ. Thus a complete integral is S = − E t + cφ ± ∫ 2mR 2 ( E + mgR cosθ ) − c 2 sin 2 θ dθ . The general solution of motion equations follows from m R 2 dθ

a1 = −

∂S =t ± ∂E

a2 = −

∂S = −φ ± c ∫ ∂c sin 2 θ

∫

2mR 2 ( E + mgR coθ ) − c 2 sin 2 θ

= t − t0

dθ 2mR ( E + mgR cosθ ) − c 2 sin 2 θ 2

.

449

Hamiltonian Mechanics

Remark 1. The first part of the general solution shows the “horary law”, the second leads to φ (θ , E , c, a2 ) . Remark 2. We can immediately verify that: ∂ 2S ∂E ∂φ

∂ S )= ∂ai ∂q j ∂2S ∂c ∂φ 2

det(

∂2S ∂E ∂θ

∂ dΘ ∂E dθ ≠ 0. ∂ dΘ ∂c dθ

0 =

∂2S ∂c ∂θ

1

Exercise 9.

A particle of mass m and position vector ox moves in a vertical plane under the action of its weight mg and a central force inversely proportional to the square of the distance from the center r = ox . (i) Express the Hamiltonian in function of variables X =r+z ,

Y =r−z

where z is the height of the particle relative to o. (ii) Find the general solution of the canonical equations of Hamilton. Answer. (i) From the expression of the potential mk 2 V = mgz − r we deduce L = T −V =

m 2

k∈R

( x
2 + z
2 ) − mgz + m

k2 r

and thus H = p x x
+ p z z
−

m

x = r2 − z2 =

XY

2

( x
2 + z
2 ) + mgz −

mk 2 x +z 2

2

.

Since

we deduce L=

z=

X −Y 2

( X − Y ) 2mk 2 m ( X
Y + XY
) 2 X
− Y
2 +( + ( ) ) − mg 2 4 XY 2 2 X +Y mg

( X 2 − Y 2 ) − 2mk 2 m X
2 Y
2 )− 2 = ( X + Y )( + 2 4 X 4Y X +Y The generalized momenta pX =

∂L , ∂X


pY =

∂L ∂Y


450

Chapter 6

imply 4X pX , X
= m( X + Y )

Y
=

4Y pY . m( X + Y )

Therefore, the Hamiltonian is written: H = p X X
+ pY Y
− L mg

=

2 m( X + Y )

( X p X2 + Y pY2 ) + 2

( X 2 − Y 2 ) − 2mk 2 X +Y

.

In this scleronomic case, where X and Y are separable variables, the complete integral is of type S = − Et + S1 ( X ) + S 2 (Y ) and thus

mg ( X 2 − Y 2 ) − 2mk 2 dS1 2 dS 2 2 (X ( ) + Y ( ) ) + 2 =E dX X +Y m( X + Y ) dY 2

that is: dS dS 2 2 mg 2 mg 2 X ( 1 )2 + X − EX + Y ( 2 ) 2 − Y − EY = 2mk 2 . m dX 2 m dY 2

(ii) By letting 2 m

X (

dS 1 2 mg ) + X dX 2

2

− EX = C

and 2 m

Y(

dS 2 2 mg 2 ) − Y − EY = 2mk 2 − C , dY 2

we obtain: m C m 2 gX + − dX = (1) 2 2 X 4

S1 = ± ∫

mE

S2 = ±∫

mE

and 2

+

m 2 k 2 mC m 2 gY − + dY = (2) . 2Y 4 Y

Therefore, we obtain: S = − Et + (1) + (2)

and the general solution is given by a1 = −

∂S , ∂E

a2 = −

∂S . ∂C

451

Hamiltonian Mechanics

Exercise 10.

In spherical coordinates r ,θ ,φ , let 2

2

pφ p H = 1 ( pr2 + θ2 + 2 2 2m r r sin θ

e2 b2 2 2 eb r sin θ − )+ pφ + eV (r ) 2 8mc 2mc

be the Hamiltonian of a particle of charge e subject to a central electric field [ potential V (r ) ] and to a constant magnetic field B = b 1z (along pole axis), c being the velocity of light. (i) Deduce there is no separable solution of the Hamilton-Jacobi equation. (ii) If the term (eb c) 2 is neglected, show the Hamilton-Jacobi is separable. Find a complete integral and the general solution of Hamilton canonical equations. Answer. (i) The coordinate φ being cyclic, let us try a separable solution of type S = − E t + R (r ) + Θ(θ ) + Aφ

where the constant A is the generalized momentum associated with φ. The Hamilton-Jacobi equation is written: dR 2 1 dΘ 2 A eb 2 2 eb 1 r sin θ − (( ) + 2 ( ) + 2 2 ) + A + eV (r ) = E 2 2m dr r dθ r sin θ 8mc 2mc 2

2 2

or dR dΘ A2 eb e 2b 2 4 2 1 A r 2 + e r 2 V (r ) − E r 2 + (r 2 ( ) 2 + ( ) 2 + 2 ) − r sin θ = 0 . 2m dr dθ sin θ 2mc 8mc 2 Two successive partial derivatives lead to the following absurd result: ∂2 c 2b 2 4 2 ( r sin θ ) = 0 . ∂θ ∂r 8mc 2 There is thus no separable solution. (ii) If (

eb 2 ) = 0 , then the Hamilton-Jacobi equation is written: c r 2 dR 2 eb dΘ 2 A2 1 2 2 2 ( ( ) − (( ) + 2 ) = 0 , A r + e r V (r ) − E r ) + 2m dr 2mc 2m dθ sin θ

where the first term depends only on r and the second term only on θ; there are two (constant) opposite terms: r 2 dR 2 eb ( ) − A r 2 + e r 2 V (r ) − E r 2 = B , 2m dr 2mc A 1 dΘ 2 (( ) + ) = −B . 2m dθ sin 2 θ 2

Therefore, a complete integral is S = −E t ±

∫

eb 2mB A − 2meV (r ) + 2mE + 2 dr ± c r

∫

− 2mB −

A2 dθ + Aφ . sin 2 θ

452

Chapter 6

The general solution of motion equations is − a1 =

∂S = −t ± ∂E

− a2 =

∂S = ±∫ ∂A

− a3 =

∫

m dr eb 2 mB A − 2 meV ( r ) + 2 + 2mE c r

e b dr 2c

∂S = ±∫ ∂B

eb

2mB A − 2me V (r ) + 2 + 2mE c r

m dr r

2

eb c

A − 2 meV ( r ) +

2mB

r2

,

± + 2mE

− A sin 2θ dθ

±∫

− 2mB − A2 sin 2 θ

∫

− dθ − 2mB − A2 sin 2 θ

,

+φ.

BIBLIOGRAPHY

The books in question are essentially at the root of the theoretical mechanics teachings of the author for undergraduate and third year engineering students.

Arnold, V., 1978, Mathematical methods of classical mechanics, MIR (Moscow, 1974), Springer Graduate Texts in Math. N°60 Springer-Verlag, New York. Brousse, P., 1981, Mécanique analytique, Vuibert, Paris. Chevalier, L., 1996, Mécanique des systèmes et des milieux déformables, Ellipses-Marketing, Paris. Flanders, H., 1963, Differential forms with applications to the physical sciences, Academic Press. Goldstein, H., 1956, Classical mechanics, Addison-Wesley, Reading, Mass. Kovalevsky, J., 1963, Introduction à la mécanique céleste, Armand Colin, Paris. Landau, L. and Lifshitz, E., 1960, Mechanics, Addison-Wesley, Reading, Mass. Lichnerowicz, A., 1964, Eléments de calcul tensoriel, A. Colin, Paris. Mantion, M., 1981, Problèmes de mécanique analytique, Vuibert, Paris. Meriam, J., 1980, Dynamics, SI version, John Wiley & Sons. Meriam, J., 1975, Statics, SI version, John Wiley & Sons. Pars, L., 1965, A treatise on analytical dynamics, Heinemann, London. Poincaré, H., 1957, Méthodes nouvelles de la mécanique céleste, 3 vol.,Dover Publications. Scheck, F., 1994, Mechanics from Newton’s laws to deterministic chaos, Springer-Verlag. Simon, K., 1979, Mechanics, Addison-Wesley, Reading, Mass. Talpaert, Y., 1982, Mécanique analytique, vol. 2, Talpaert (Ed.). Talpaert, Y., 1987, Mécanique générale et analytique, Ellipses-Marketing, Paris. Talpaert, Y., 2000, Differential geometry with applications to mechanics and physics, Dekker, New York. Whittaker, E. T., 1970, A treatise on the analytical dynamics of particles and rigid bodies, Cambridge University Press. Wintner, A., 1947, The analytical foundations of celestial mechanics, Princeton University Press.

453

INDEX

A

C

absolute acceleration, 28, 304 absolute derivative, 215

calculus of variations, 212, 360 canonical equations, 395, 401, 412 canonical isomorphism, 166 canonical system, 396 canonical transformation, 416, 419-420, 423425 canonically conjugate momentum, 399 center of reduction, 14 central axis, 15, 41 central axis of inertia, 288 central ellipsoid of inertia, 288 central force field, 403, 405 change of basis, 140, 191, 312 change of cobasis, 140 change of natural basis, 198 characteristic equation, 282 characteristic function, 395 Christoffel formulae, 206 Christoffel symbols, 204 Christoffel symbols of first kind, 205 class of inertial frames, 306 cloud of possible states, 406 compatible with constraint(s), 81, 88, 90-91 complementary equation, 359, 402 complete integral, 426, 427 complete variation, 363 completely antisymmetric p-linear form, 178 completely symmetric tensor, 154, 155 component, 3 components of dynam, 5 compression, 48 concurrent (spatial forces), 44 cone of kinetic friction, 54 cone of static friction, 54 configuration space, 72, 74, 354 configuration spacetime, 358 conjugate tensor, 168 conjugate variables, 395 conservation of angular momentum, 318 conservation of mass (theorem), 311 conservative force field, 108 consistent with constraints, 81, 88, 90-91 constraint force, 34 constraint (particle subject to ), 86

( ) components, 208 absolute differential of ( ) components, 208 absolute differential of

1 0 0 1

absolute differential of a vector field, 207, 214 absolute frame of reference, 303-304 absolute velocity, 22 acceleration, 215 action and reaction (principle of), 39 action integral, 368, 378, 427 addition in L( E , F ) , 136 addition of dynams, 8 addition of p-forms, 185 addition of tensors, 156 adjoint Lagrangian, 359, 402 adjoint of a p-form, 219 adjoint of a q-vector, 218 admissible diffeomorphism, 376 affine space, 2 algebra, 413 alternation mapping, 178 angle of kinetic friction, 54 angle of static friction, 54 angular momentum, 307-308, 319-322 angular momentum theorem, 316 angular velocity, 18 angular velocity tensor, 312-314 angular velocity vector, 314 annular-linear constraint, 58, 67 antisymmetric tensor, 154 antisymmetrization, 178 axis of sliding dynam, 14

B ball-and-socket joint, 59, 67 Bernoulli-Lagrange vector, 81 bilateral constraint, 75 body force, 49 brachistochrone problem, 367 bras, 137

455

456

constraint (system subject to a ), 75 constraints (system subject to ), 77 contact dynam, 48 continuous vector field, 92 contracted multiplication, 159 contracted product, 159, 172 contraction, 158 contravariant components, 171 contravariant-contravariant representation, 173 contravariant-covariant representation, 173 contravariant vector, 146 coordinate basis, 195 coordinate line, 193 coordinate system, 193 coordinates, 3, 192 coplanar forces, 43-44 Coriolis acceleration, 23, 39 Coriolis force, 323 Coulomb friction, 50 couple, 12, 13, 14, 16 covariant component, 170 covariant-contravariant representation, 173 covariant-covariant representation, 174

( ) -covariant derivative, 208 ( ) -covariant derivative, 209 1 1

0 2

covariant derivative of t,, 211 covariant representation, 170 covariant representation of vector differential, 203 covariant vector, 146 covector, 136, 144, 219 curl of covector field, 228-229 curl of second order tensors, 229 curvilinear coordinates, 194, 197-199, 202 cyclic coordinate, 360, 404

D d’Alembert’s principle, 342 d’Alembert-Lagrange principle, 342 decomposable p-form, 183 density, 264 determinant, 217 diffeomorphism, 374 differential of function, 220 differential of vector, 195 dimension, 2 directional derivative of function, 221 directional derivative of tensor field, 224 divergence of tensor field, 227 divergence of vector field, 225

Index domain of a material system, 263 double contraction, 160 double pendulum, 73 dry friction, 50 dual basis, 137 dual space, 137 dummy index, 139 dynam, 3-5 dynam of forces, 36, 95, 317 dynam of internal forces, 39 dynam of mechanical actions, 35 dynam of velocities, 21 dynam of virtual displacements, 93 dynam of virtual velocities, 93 dynamic dynam, 318

E eigenvalue, 281 eigenvector of tensor, 281 Einstein summation convention, 318 elements of reduction, 5, 35 embedding, 67 equal tensors, 156 equality of dynams, 6 equilibrium, 38 equilibrium conditions, 40-42 equilibrium (particle in ), 98, 107, 109 equilibrium (rigid body in ), 100 equilibrium (system of particles in ), 107, 109 equilibrium (system of rigid bodies in ), 100 equiprojective field, 10, 20 equivalent systems of vectors, 4, 40 Euclidean space, 3 Euclidean vector space, 174 Euler angles, 82, 85 Euler’s equation 212, 364, 365 Euler-Noether theorem, 374 Euler pendulum, 23 exterior algebra, 188 exterior calculus, 178 exterior form of degree p, 178 exterior multiplication, 186 exterior product of 1-form, 180 exterior product of p-form, 184, 186 exterior product of q-vectors, 188 exterior product space, 185 external constraint, 37 external forces of constraint, 37 external mechanical action, 34 extremal 365, 369-371 extremum principles, 360

457

Index F Fermat’s principle, 360 field of moments of dynam, 91 field of virtual velocities, 85 first integral, 310, 359, 366, 396, 413, 415 first variation, 364 flat mapping, 166 force, 300 force associated with q i , 357 p-form, 178 frame (of reference), 2, 192 free bodies (system of), 86 free-body diagram, 68-70 free index, 139 free particle, 81 friction force, 50 friction laws of Coulomb, 52 frictionless constraint, 55 functional, 362 fundamental formula of acceleration, 22 fundamental formula of rigid body kinematics, 19 fundamental principle of dynamics, 302, 303 fundamental tensor, 164

G Galilean principle of inertia, 302 generalized coordinates, 72, 78-79, 340 generalized force, 106, 107, 357 generalized momentum, 359, 399 generalized trajectory, 73, 355, 372 generalized velocity, 73, 340, 355 generating function, 420, 427 geodesic, 212-213, 215, 371-374 given external forces, 36 given forces, 37 gradient of tensor field, 224-225 graduation, 188 Guldin (or Pappus) theorems, 271

H hat, 180 Hamilton-Jacobi equation, 426, 430 Hamilton’s variational principle, 369 Hamiltonian, 399 harmonic oscillator, 430 Hero of Alexandria, 360 hoist, 119

holonomic constraint, 76 holonomic system, 340 hyperstatic (collection of forces), 45 hypostatic (collection of forces)

I ideal constraint, 96-97 ignorable coordinate, 360 impulse, 311 indefinite signature, 176 inertia cylinder, 280 inertia ellipse, 280 inertia ellipsoid, 279, 286-287 inertia ellipsoid equation, 278 inertia force, 342 inertia law, 302 inertial frame of reference, 304 integral (with respect to a mass distribution), 266 integral of f along a curve, 362 integral curve of field, 375 interaction forces, 37 internal forces of constraint, 37 internal mechanical action, 34 internal potential, 325 invariant Lagrangian, 376-379 invariant of dynam, 11 involutive function, 415 isostatic ( collection of foces), 45

J Jacobi’s identity, 413, 414 Jacobi’s integral, 359, 404 Jacobi’s theorem, 428

K kets, 137 kinematic dynam, 21 kinetic dynam, 307, 317 kinetic energy, 308, 326-328, 356 kinetic energy theorem, 323, 328-330 kinetic friction (coefficient of ), 53 kinetic friction force, 51 kinetic momentum, 307 Koenig expression of angular momentum, 308 Koenig expression of kinetic energy, 309, 324 Kronecker tensor, 153

458

L Lagrange bracket, 415, 416 Lagrange’s equations, 347, 349 Lagrange multiplier, 351 Lagrangian, 349 Laplacian of function, 230 Laplacian of vector field, 231 laws of friction, 52 least time (principle of ), 360 Legendre transformation, 397, 399 length of arc, 211 Levi-Civita symbol, 216 Lie algebra, 413 line element, 201 line of action, 35 linear mapping, 135 linear momentum, 307 linear momentum theorem, 310 Liouville’s theorem, 407, 409 local reference frame, 304 local transformation, 375 lowering mapping, 166

M mass of material system, 264 mass per unit area, 264 mass per unit length, 265 mass per unit volume, 264 material symmetry, 287 material system, 33 Maupertuis, principle, 361 mechanical action, 34 mechanical system, 33 method of isolation, 68 metric element, 201 metric tensor, 176 mixed tensor, 150, 152 modeled material system, 34 moment of dynam, 5 moment of inertia, 272-273 moment of inertia about origin, 273 moment of inertia about coordinate axes, 273 moment of inertia about coordinate planes, 273 moment of inertia about any axis, 277 moment of order k, 272 momentum canonically conjugate to q i , 359 motion, 300 motion law, 302 motion of mass center (theorem), 311 multilinear form, 306

Index multiplication of dynam by scalar, 9 multiplication of dynams, 10 multiplication of a p-form by a scalar, 185 multiplication of a linear mapping by a scalar, 136 multiplication of a tensor by a scalar, 157

N natural basis, 195 natural frame, 195 natural trajectory, 389 N-body problem, 393 neutral equilibrium, 110 Newton’s first law, 302 Newton’s second law, 302 Newton’s third law, 303 nonholonomic system, 340 norm 176, normal force, 49 (normal) friction force, 51 normal stress, 47 number of degrees of freedom of a constraint, 55 number of degrees of freedom of a system, 71 nutation angle, 87

O one-form, 136 one-parameter group of diffeomorphisms, 375 orthogonal dynams, 10 orthogonal vectors, 165 orthonormal basis, 176

P parallel forces, 44 parallel translation, 213 parallel transport, 213 parallelogram of forces (law of ), 303 partially antisymmetric tensor, 155 partially symmetric tensor, 154 phase point, 406 phase space, 403 phase spacetime, 409 pivot, 65, 68 pivoting moment, 49 plane support, 60, 68 point, 2

459

Index point space, 2, 192 Poisson bracket, 409, 411 Poisson theorem, 414, 415 position vector, 3 postulate of initial conditions, 301 potential, 108 power of external forces, 323-330 power of internal forces, 323-330 precession angle, 87 pre-Euclidean space, 3 pre-Euclidean vector space, 164 primitive coordinates, 72 principal axes, 282, 286-287 principal axes of inertia, 283 principal moment of inertia, 283 principle of Galilean relativity, 305 principle of least action, 361 product of a dynam by a scalar, 8 product of a p-form by a scalar, 185 product of a tensor by a scalar, 157 product of dynams, 9 product of inertia, 272 proper motion (angle of ), 87 pseudo-Euclidean vector space, 176 pseudo-norm, 176 punctual constraint, 55

R radius of gyration, 280 raising mapping, 167 reaction, 34 reciprocal basis, 169 reciprocal vector, 169 rectilinear constraint, 56 rectilinear coordinates, 194, 202 reduction of a vector system, 12 redundant constraint, 45 relative acceleration, 23 relative minimum, 362 relative velocity, 22 representation of a dynam, 6 representative point, 355 resultant of a dynam, 5 rheonomic constraint, 76, 341 Ricci identities, 206 Ricci theorem, 211 rigid body, 74 rigid body motion, 92 rolling moment, 49 rolling without slipping, 26 rotational kinetic energy, 328 Routh’s method, 405-406 Routhian, 405

S scalar, 158, 219 scalar invariant, 12 scalar multiplication, 164, 170 scalar product, 164, 172 scleronomic constraint, 76, 341 scleronomic system, 340 screw joint, 64, 68 second order contravariant tensor, 150 second order covariant tensor, 148 separable mechanical system, 432 separability (condition for ), 433 sharp mapping, 167 shear stress, 47 short path (principle of ), 360 signature, 176 skew-symmetric p-linear form, 178 slider-crank, 122 sliding dynam, 14, 16 sliding guide, 63, 68 sliding hinge, 61 sliding pivot,, 61, 68 sliding velocity, 26 smooth constraint, 55 space connected with a rigid body, 17 space of points and directions, 377, 389 spherical pendulum, 73 square of dynam, 10 stable equilibrium, 110 Stäckel’s case of separability, 434-439 state space, 409 static friction (coefficient of ), 53 static friction force, 51 statically determinate structure, 45 Steiner’s theorem, 288-290 stiffness, 300 stress, 47 strict components, 183, 189, 216 subdivision, 269 sum of dynams, 7 sum of p-forms, 185 sum of tensors, 156 surface force, 49 symmetric tensor, 153-154 symplectic group, 425 symplectic matrix, 411 system of canonical equations (of Hamilton), 401 system of Euler’s equations, 367 system of vectors, 3

460

Index

T Tangent vector field, 374 Tangential force, 49 (tangential) friction force, 51 tension, 48 tensor algebra, 158 tensor criterion, 162-163 tensor density, 216 tensor field, 201 tensor inertia, 274 tensor multiplication, 157

( tensor of type ( tensor of type ( tensor of type ( tensor of type (

tensor of type

tensor of type

0 1 1 0 0 2 2 0 1 1

) , 144 ) , 145 ) , 146 ) , 149 ) , 150

( ) , 152 q p

tensor on point space, 199 tensor product, 143, 157 tensor product space, 147, 149, 151, 152 tensor space, 152 Torricelli theorem, 102 trace, 236 transformation of tensor components, 199 transport acceleration, 23 transport velocity, 22,87 transposed tensor, 151 translational kinetic energy, 328 trihedron of Copernicus, 304

uniform rectilinear translation, 306 unilateral constraint, 75 unstable equilibrium, 110

V variational calculus, 368 variational derivative, 365, 371 vector 145, 219 q-vector, 188 vector invariant, 11 velocity, 214 velocity field, 19 velocity vector … relative to … 18, 19 virtual angular displacement, 93 virtual angular velocity vector, 93 virtual displacement, 33, 80 virtual kinematic dynam, 93 virtual rotation (vector of ), 85 virtual velocities (field of ), 85-86 virtual velocity, 33, 82, 84 virtual power, 94 virtual work, 94 virtual work principle, 98-102

W wedge, 180 welded connection, 35 welded joint, 67

Z U unconstrained rigid body, 82 undetermined multipliers, 350-354

zero dynam, 7, 12 zero form, 184 zero tensor, 153