This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Physics Editor: Chris Hall Development Editor: Ed Dodd Assistant Editor: Brandi Kirksey Editorial Assistant: Stefanie Beeck
ALL RIGHTS RESERVED. No part of this work covered by the copyright hereon may be reproduced or used in any form or by any means-graphic, electronic, or mechanical, including but not limited to photocopying, recording, taping, Web distribution, information networks, or information storage and retrieval systems-without the written permission of the publisher.
Technology Project Manager: Sam Subity Marketing Manager: Mark Santee Marketing Assistant: Melissa Wong Marketing Communications Manager: Darlene Amidon Brent Project Manager, Editorial Production:
Teri Hyde Creative Director: Rob Hugel Art Director: John Walker Print/Media Buyer: Rebecca Cross Permissions Editor: Roberta Brayer Production Service: Lachina Publishing
Services
For product information and technology assistance, contact us at Cengage learning Academic Resource Center 1-800-423-0563 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions. Further permissions questions can be e·mailed to [email protected].
Library of Congress Control Number: 2007937232 ISBN-13: 978'0-495-38693-3
Text Designer: Patrick Devine Design
Photo Researcher: Jane Sanders Miller Copy Editor: Kathleen Lafferty Illustrator: Lachina Publishing Services,
Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA
(engage Learning is a leading prOVider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil and Japan. Locate your local office at international. cengage.com/region
fxamView~ and ExamView Pro@are trademarks of FSCreations, Inc. Windows is
a registered trademark of the Microsoft Corporation used herein under Iicense_ Macintosh and Power Macintosh are registered trademarks of Apple Computer, Inc. Used herein under license.
Cengage Learning products are represented in Canada by Nelson Education, ltd.
For your course and learning solutions, visit academic.cengage.com
Purchase any of our products at your local college store or at our preferred online store www.ichapters.com.
Printed in Canada 123-15671211100908
We dedicate this book to our colleagu e Jerry S. Faughn, whose dedicati on to all aspects of the project and tireless eHorts through the years are deeply appreci ated.
CONTENTS OVERVIEW
~1echanics
PART 1 CHAPTER I CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9
Introduction 1 Motion in One Dimension 24
Vectors and Two~Dimen5ionQIMotion 54 The Laws of Motion 83 Energy 119 Momentum and Collisions 161
Rotational Motion and the Low of Gravity 190 Rotational Equilibrium and Rotational Dynamics 228
Solids and Fluids 268
PART 2
The -modynamics
CHAPTER 10 CHAPTER 11 CHAPTER 12
Thermol Physics 322 The Lows of Thermodynamics 385
PART 3
Vibrations and Waves
CHAPTER 13 CHAPTER 14
Vibrations and Waves 425
Sound 459
Electricity and Magnetism
PART 4 CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER
Energy in Thermol Processes 352
15 16 17 18 19
20 21
Electric Forces and Electric Fields
497
Electrical Energy and Capacitance
531
Current and Resistance
570
Direct-Current Circuits
594
Magnetism 626 Induced Voltages and Inductance 663 Alternating-Current Circuits and Electromagnetic Waves
PARTS
Light and Optics
CHAPTER 22 CHAPTER 23 CHAPTER 24 CHAPTER 25
Reftectian and Refraction of Light Mirrars and Lenses 759 Wave Optics 790 Optical Instruments 823
PART 6
tv odem Physics
CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER
26 27 28 29 30
APPENDIX A APPENDIX B APPENDIXC APPENDIX D APPENDIX E
Relativity 847 Quantum Physics 870 Atomic Physics 891 Nuclear Physics 913 Nuclear Energy and Elementary Particles Mathematics Review A.l An Abbreviated Table af Isotopes A.14 Some Useful Tables A.19 SI Units A.21
MeAT Skill Builder Study Guide A.22
Answers to Quick Quizzes, Example Questions, Odd~Numbered Multiple-Choice Questions, Conceptual Questions, and Problems A.52 Index 1.1
Standards of Length, Mass, and Time 1.2 The Building Blocks of Matter 4 1.3 Dimensional Analysis 5 1.4 Uncertainty in Measurement and Significant Figures 1.5 Conversion of Units 9 1.6 Estimates and Order-of-Magnitude Calculations 11 1.7 Coordinate Systems 13 1.8 Trigonometry 14 1.9 Problem-Solving Strategy 16 Summary 18
7
Rotational Equilibrium and Rotational Dynamics 228 8.1
Torque 228 Torque and the Two Conditions for Equilibrium 8.3 The Center of Gravity 234 8.4 Examples of Objects in Equilibrium 236 8.5 Relationship Between Torque and Angular Acceleration 239 8.6 Rotational Kinetic Energy 246 8.7 Angular Momentum 249 Summary 254
Vectors and Their Properties 54 3.2 Components of a Vector 57 3.3 Displacement, Velocity, and Acceleration in Two Dimensions 60 3.4 Motion in Two Dimensions 62 3.5 Relative Velocity 70 Summary 74
CHAPTER 4
The Laws of Motion
232
CHAPTER 9
Solids and Fluids
CHAPTER 3
196
CHAPTER 8
CHAPTER 2
83 Forces 83 4.2 Newton's First Law 85 4.3 Newton's Second Law 86 4.4 Newton's Third Law 92 4.5 Applications of Newton's Lows 4.6 Forces of Friction 101 Summary 108
Angular Speed and Angular Acceleration \90 7.2 Rotational Motion Under Constant Angular Acceleration 194 7.3 Relations Between Angular and linear Quantities 7.4 Centripetal Acceleration 199 7.5 Newtonian Gravitation 207 7.6 Kepler's Laws 215 Summary 218
190
7.1
268
9.1 9.2 9.3
States of Matter 268 The Deformation af Solids 270 Density and Pressure 276 9.4 Variation of Pressure with Depth 279 9.5 Pressure Measurements 283 9.6 Buoyant Forces and Archimedes' Principle 284 9.7 Fluids in Motion 290 9.8 Other Applications of Fluid Dynamics 296 9.9 Surface Tension, Capillary Action, and Viscous Fluid Flow 299 9.10 Transport Phenomena 307 Summary 311
4.1
Part 2: Thermodynamics CHAPTER 10
94
Thermal Physics
CHAPTER 5
Energy
119
5.1 Work 119 5.2 Kinetic Energy and the Work-Energy Theorem 5.3 Gravitational Potential Energy 127 5.4 Spring Potential Energy 135 5.5 Systems and Energy Conservation 141 5.6 Power 143 5.7 Work Done by Q Varying Force 147 Summary 150
CHAPTER 6
Momentum and Collisions 6.1 6.2
Momentum and Impulse 161 Conservation of Momentum 166
161
124
322 ) 0.1 Temperature and the Zeroth Law of Thermodynamics 10.2 Thermometers and Temperature Scoles 324 10.3 Thermal Expansion of Solids and Liquids 328 10.4 Macroscopic Description of an Ideal Gas 335 10.5 The Kinetic Theory of Gases 340 Summary 345
322
CHAPTER 11
Energy in Thermal Processes
352 Heat and Internal Energy 352 11.2 Specific Heat 355 11.3 Calorimetry 357 11.4 Latent Heat and Phose Change 359 11.5 Energy Transfer 366 11.6 Global Warming and Greenhouse Gases 375 Summary 377
11.1
y
vi
Contents
CHAPTER 12
The Laws of Thermodynamics
16.7 The Parallel-Plate Capacitor 547 16.8 Combinations of Capacitors 549 16.9 Energy Stored in a Charged Capacitor 16.IOCapacitors with Dielectrics 557 Summary 562
385
12.1 Work in Thermodynamic Processes 385 12.2 The First Low of Thermodynamics 388 12.3 Thermal Processes 390 12.4 Heat Engines and the Second Low of Thermodynamics 12.5 Entropy 408 12.6 Human Metabolism 413 Summary 416
399
Part 3: Vibrations and Waves CHAPTER 13
Vibrations and Waves
425
13.1 Hooke's Law 425 13.2 Elastic Potential Energy 428 13.3 Comparing Simple Harmonic Motion with Uniform Circular Motion 432 13.4 Position, Velocity, ond Acceleration as a Function of Time 436 13.5 Motion of a Pendulum 439 13.6 Damped Oscillations 442 13.7 Waves 443 13.8 Frequency, Amplitude, and Wavelength 445 J3.9 The Speed of Waves on Strings 447 13.10 Interference of Waves 448 13.11 Reflection of Waves 449 Summary 450
17.1 Electric Current 570 17.2 A Microscopic View: Current and Drift Speed 572 17.3 Current and Voltage Measurements in Circuits 574 17.4 Resistance, Resistivity, and Ohm's Low 575 17.5 Temperature Variation of Resistance 579 17.6 Electrical Energy and Power 580 17.7 Superconductors 584 17.8 Electrical Activity in the Heart 585 Summary 588
CHAPTER 18
Direct-Current Circuits
594
18.1 Sources of emf 594 18.2 Resistors in Series 595 18.3 Resistors in Parallel 598 18.4 Kirchhoff's Rules and Complex DC Circuits 603 18.5 RC Circuits 607 18.6 Household Circuits 611 18.7 Electrical Safety 612 18.8 Conduction of Electrical Signals by Neurons 613 Summary 615
626
19.1 19.2 19.3 19.4 19.5 19.6 19.7
Magnets 626 Earth's Magnetic Field 628 Magnetic Fields 630 Magnetic Force on a Current-Carrying Conductor 633 Torque on a Current Loop and Electric Motors 636 Motion of a Charged Particle in a Magnetic Field 639 Magnetic Field of a Long, Straight Wire and Ampere's Low 642 19.8 Magnetic Force Between Two Parallel Conductors 645 19.9 Magnetic Fields of Current Loops and Solenoids 646 19.10 Magnetic Domains 650 Summary 652
14.1 Producing a Sound Wave 459 14.2 Characteristics of Sound Waves 460 14.3 The Speed of Sound 461 14.4 Energy and Intensity of Sound Waves 463 14.5 Spherical and Plane Waves 466 14.6 The Doppler Effect 468 14.7 Interference of Sound Waves 473 14.8 Standing Waves 475 14.9 Forced Vibrations and Resonance 479 14.lOStanding Waves in Air Columns 480 14.11Beats 484 14.12 Quality of Sound 486 14.13The Ear 487 Summary 489
CHAPTER 20
Induced Voltages and Inductance
Part 4: Electricity and Magnetism CHAPTER 15
497
15.' Properties of Electric Charges 497 15.2 Insulators and Conductors 499 15.3 Coulomb's Law 500 15.4 The Electric Field 505 15.5 Electric Field Lines 510 15.6 Conductors in Electrostatic Equilibrium 513 15.7 The Millikan Oil-Drop Experiment 515 15.8 The Van de Graaff Generator 516 15.9 Electric Flux and Gauss's Law 517 Summary 523
663
20.1 20.2 20.3 20.4
Induced emf and Magnetic Flux 663 Faraday's Law of Induction 666 Motional emf 670 Lenz's Low Revisited (The Minus Sign in Faraday's Law) 674 20.5 Generators 676 20.6 Self-Inductance 680 20.7 RL Circuits 683 20.8 Energy Stored in a Magnetic Field 686 Summary 687
CHAPTER 21
Alternating-Current Circuits and Electromagnetic Waves 696
CHAPTER 16
Electrical Energy and Capacitance
Current and Resistance 570
Magnetism
459
Electric Forces and Electric Fields
CHAPTER 17
CHAPTER 19
CHAPTER 14
Sound
555
531
16.1 Potentiol Difference and Electric Potential 531 16.2 Electric Potential and Potential Energy Due to Point Charges 538 16.3 Potentiols and Charged Conductors 542 16.4 Equipotential Surfaces 543 16.5 Applications 544 16.6 Capocitance 546
21,1 Resistors in an AC Circuit 696 21.2 Capacitors in an AC Circuit 699 21.3 Inductors in on AC Circuit 701 21.4 The RLC Series Circuit 702 21.5 Power in on AC Circuit 707 21.6 Resonance in a Series RLC Circuit 708 21.7 The Transformer 710 21.8 Maxwell's Predictions 712 21.9 Hertz's Confirmation of Maxwell's Predictions 21.10 Production of Electromagnetic Waves by on Antenna 714
713
Contents
21.11 Properties of Electromagnetic Waves 715 21.12 The Spectrum of Electromagnetic Waves 720 21.13 The Doppler Effect for Electromagnetic Waves 722 Summary 723
CHAPTER 22
732
22.1 The Nature of Light 732 22.2 Reflection and Refraction 733 22.3 The Low of Refraction 737 22.4 Dispersion ond Prisms 742 22.5 The Rainbow 745 22.6 Huygens's Principle 746 22.7 Total Internal Reflection 748 Summary 751
Nuclear Physics
790
913
847
Galilean Relativity 847 The Speed of Light 848 Einstein's Principle of Relativity 850 Consequences of Special Relativity 851 Relativistic Momentum 858 Relativistic Energy and the Equivalence of Mass and Energy 859 26.7 General Relativity 863 Summary 865
Nuclear Fission 937 Nuclear Fusion 941 Elementary Particles and the Fundamental Forces Positrons and Other Antiparticles 944 Classification of Particles 945 Conservation Laws 947
30.7 The Eightfold Way 949 30.8 Quarks and Color 950 30.9 Eleetroweak Theory and the Standard Model 30.lOThe Cosmic Connection 954 30.11 Problems and Perspectives 955 Summary 956
823
CHAPTER 26
929
Nuclear Energy and Elementary Particles 30.1 30.2 30.3 30.4 30.5 30.6
823
Part 6: Modern Physics 26.1 26.2 26.3 26.4 26.5 26.6
899 902
CHAPTER 30
800
25.2 The Eye 824 25.3 The Simple Magnifier 829 25.4 The Compound Microscope 830 25.5 The Telescope 832 25.6 Resolution of Single-Slit and Circular Apertures 25.7 The Michelson Interferometer 840 Summary 841
Relativity
891
29.1 Some Properties of Nuclei 913 29.2 Binding Energy 916 29.3 Radioactivity 918 29.4 The Decay Processes 921 29.5 Natural Radioactivity 926 29.6 Nuclear Reactions 927 29.7 Medical Applications of Radiation Summary 931
CHAPTER 25 25.1 The Camera
891
CHAPTER 29
CHAPTER 24
Optical Instruments
27.4 Diffroction of X-Rays by Crystals 876 27.5 The Compton Effect 879 27.6 The Dual Nature of Light and Motter 880 27.7 The Wove Function 883 27.8 The Uncertainty Principle 884 Summary 886
28.3 The Bohr Madel 894 28.4 Quantum Mechanics and the Hydrogen Atom 28.5 The Exclusion Principle and the Periodic Table 28.6 Characteristic X-Rays 905 28.7 Atomic Transitions and Lasers 906 Summary 908
759
24.1 Conditions for Interference 790 24.2 Young's Double-Slit Experiment 791 24.3 Change of Phase Due to Reflection 795 24.4 Interference in Thin Films 796 24.5 Using Interference to Read CDs and DVDs 24.6 Diffraction 802 24.7 Single-Slit Diffraction 803 24.8 The Diffraction Grating 805 24.9 Polarization of Light Waves 808 Summary 815
870
27.1 Blackbody Radiation and Planck's Hypothesis 870 27.2 The Photoelectric Effect and the Particle Theory
28.1 Early Models of the Atom 28.2 Atomic Spectra 892
23.1 Flat Mirrors 759 23.2 Images Farmed by Concave Mirrors 762 23.3 Convex Mirrors and Sign Conventions 764 23.4 Images Formed by Refraction 769 23.5 Atmospheric Refraction 772 23.6 Thin Lenses 773 23.7 Lens and Mirror Aberrations 781 Summary 782
Wave Optics
Quantum Physics
CHAPTER 28 Atomic Physics
CHAPTER 23
Mirrors and Lenses
CHAPTER 27
of Ught 872 27.3 X-Rays 875
Part 5: Light and Optics Reflection and Refraction af Light
vii
952
835 Appendix Appendix Appendix Appendix Appendix
A: B: C: D: E:
Mathematics Review A.l An Abbreviated Table af Isotopes Some Useful Tables A.19 SI Units A.21
MeAT Skill Builder Study Guide
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems A.52 Index 1.1
A.14
A.22
937
943
_~AC'BOUT THE AUTHORS
Raymond A. Serway received his doctorate at Illinois Instimte of Technology and is Professor Emeritus aLJames rvladison University. In 1990 he received the Madison Scholar Award atJames Madison University, ,..· here he taught for 17 years. Dr. Serway began his teaching career at Clarkson University, where he conducted research and taught from 1967 to 1980. He was the recipient. of the Distinguished Teaching Award at Clarkson University in 1977 and of the Alumni Achievement Award from Utica College in 1985. As Guest Scientist at the IBM Research Laboratory in Zurich, S",!itzeriand, he worked ,·,'ith K. Alex t\'lliller, 1987 Nobel Prize recipient. Dr. Serway also was a visiting scientist at Argonne National Laboratory, where he collaborated with his mentor and friend, Sam Marshall. In addition to earlier editions of this textbook, Dr. Sen,,'a)' is the coauthor of Principles of Ph)lSics, fourth edition; Physics for Scient.ists and Engineers, seventh edition; Essentials o/College Physics; and Alodenz Ph)'sics, third edition. He also is the coauthor of the high school t.extbook Physics, published by Holt, Rinehart and 'Vinston. In addition, Dr. Senvay has published more than 40 research papers in the field of condensed matter physics and has given more than 70 presentations at professional meetings. Dr. Serway and his wife, Elizabeth, enjoy traveling, golf, gardening, singing in a church choir, and spending time with their four children and eight grandchildren. Chris Yuille is an associate professor of physics at Embry-Riddle Aeronautical University (ERAU), Daytona Beach, Florida, the world's premier institution for aviation higher education. T-Te received his doctorate in physics from the University of Florida in 1989 and moved to Daytona after a year at ERAU's Frescau, Arizona, campus..t\1though he has taught courses at all levels, including postgraduate, his primary interest has been the delivery of introductory physics. He has received several awards for teaching excellence, including the Senior Class Appreciation Award (three times). He conducts research in general relativity and quantum theory, and was a participant in the JOVE program, a special three-year NASA grant program during \vhich he studied neutron stars. His work has appeared in a number of scientific journals, and he has been a featured science writer in Analog Science Fiction/Science Fact magazi ne. 1n add ition to th is textbook, he is coauthor of Esse'ntials of College Physics. Dr. Vuille el~oys tennis, swimming, and playing classical piano, and he is a fanner chess champion of SL Petersburg and Atlanta. In his spare time he writ.es fiction and goes to the beach. His '\life, Dianne Kowing, is an optometrisl for a local Veterans' Administration clinic. His daughter, Kira VllilleKml,:ing, is a meteorology/communications double m~jor at ERAU and a graduate of her father's first-year physics course. He has two sons, Christopher, a cellist and fisherman, and James, avid reader of Disney comics. Jerry S. Faughn earned his doctorat.e at the University of\rfississippi. I-Ie is Profes-
sor Emeritus and fonner chair of the Department of Physics and Astronomy at Eastern Kenmcky University. Dr. Faughn has also ,...' ritten a microprocessor interfacing text [or upper-division physics students. He is coauthor ofa nonmathematical physics text and a physical science text for general education students, and (with Or. Scrway) the high-school textbook Physics, published by Holt, Reinhart and Winston. He has taught courses ranging from the lower division to the graduate level, but his primary interest is in studentsjusr beginning 1.0 learn physics. Dr. Faughn has a wide variety of hobbies, among \vhich are reading, travel, genealogy, and old-time radio. His 'tvife, Mary Ann, is an avid gardener, and he contributes to her efforts by staying out of the way. His daughter, Laura, is in family practice, and his son, David, is an attorney.
viii
PREFACE
Coltege PhJsi.cs is written for a one-year course in introductory physics usually taken by students majoring in biology, the health professions, and other disciplines including environmental, earth, and social sciences, and technical fields such as architecture. The mathematical techniques used in this book include algebra, geometry, and trigonometry, but not calculus. This textbook, which covers the standard topics in classical physics and 20thcentury physics, is divided into six pans. Part I (Chapters 1-9) deals with Ne,..,,~ tonian mechanics and the physics of fluids; Part 2 (Chapters ]0-L2) is concerned 'with heat and thermodynamics; Pan 3 (Chapters 13 and 14) covers wave motion and sound; Pan 4 (Chapters 15-21) develops the concepts of electricity and magnetism; Part 5 (Chapters 22-25) treats the propen.ies or light and the field of geometric and \vave optics; and Part 6 (Chapters 26-30) provides an introduction to special relativity, quantum physics, atomic physics, and nuclear physics.
OBJECTIVES The main o~jectives of this introductory textbook are twofold: to provide the student with a clear and logical presentation of the basic concepts and principles of physics, and to strengthen an understanding of the concepts and principles through a broad range of interesting applications to the real world. To meet those objectives, we have emphasized sound physical arguments and problem-solving methodology. At the same time, \-\le have attempted to motivate the student through practical examples that demonstrate the role of physics in other disciplines.
CHANGES TO THE EIGHTH EDITION A number of changes and improvements have been made to this edition. Based on comments [rom users of the sevel1lh edition and reviewers' suggestions, a m~~jor c1Tort \vas made to increase the emphasis on conceptual understanding, to add new end-of-chapter questions and problems that are informed by research, and to improve the clarity of the presentation. The new pedagogical features added to this edition are based on current trends in science education. The following represent the major changes in the eighth edition.
Questions and Problems We have substantially revised the end-of-chapter questions and problems [or this edition. Three new types of quest.ions and problems have been added: •
Multiple-Choice Questions have been introduced with several purposes III mind. Some require calculations designed 1.0 facilitate students' familiarity with the equations, the variables used, the concepts the variables represent, and the relationships benveen the concepts. The rest are conceptual and are designed to encourage conceptual thinking. Finally, many student.s are required to take multiple-choice tests, so some practice with that form of question is desirable. Here is an example ofa multiple-choice question: 12. A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. "Vhat happens to the acceleration or the trLlck as its trailer leaks sand al a constant rate through a hole in its bot.tom? (a) It decreases at a steady ratc. (h) It increases at a steady rate. (c) It increases and then decreases. (d) It decreases and then increases. (e) 11. remains constant.
ix
x
Preface
The illstrtlCLOf may select multiple-choice questions to assign as homework or usc them in the classroom, possibly with -peer instruction" methods or in conjunction with "'clicker" s}'stcms. More than 350 multiple-choice questions are included in this edition. Answers LO odd-numbered Illultiple-choice questions dl-C included in the Answers ~eLliulI at tile end uf the book, and answers to all questions are found in the l1JslnLCtor~'i Solutions Afmlllal and on the instrucLOr's POllJPrl-Prlurp CD-ROlvI.
•
Enhanced Content problems require symbolic or conceplllal responses from the slllden t. A s)'l1Ibo(;( Enhanced Conlent problem requires the sludent to obtain an answer in terms of symbols_ In general, some g·uidance is built into the problem stateIllent. The Roal is to belter train the studenlto deal with mathematics at a level appropriate to this course. Most studcnts at this Icvel are uncomfortable \,..-ith symbolic equations, which is unfortunate because symbolic equations are the most efficient vehicle (or presenting relationships between physics concepts. Once students understand the physical concepts, their ability [Q solve problems is greatly en hallCed. As soon as the numbers ,lre substilUted into an equation, however, allthe concepts and their relationships to one another are lost. melded together in the student's calculator. The symbolic Enhanced Content problems train students LO postpone substitution of values, facilitating their ability to think conceptually lIsing the equations. An example of a symbolic Enhanced Content prOblem is provided here: 14'- An objeCl of mass
In is dropper! from the roof of a building of heighl 11_ \·Vhile the objecl is falling, a wind hlowing parallel to the face of the building cxcns a conslane horizollLal force Fon the ol~jccl. (a) Ilow long does it take lhe objcclto slrike the grolllld? Express lhe time I in terms of g and h. (b) Find an cxprcssiUll in lcrms of "/Ii and Ffar the accc1cralion fl) of tllC ohjcCl in the horizontal direction (taken as lhe positive x-direclion). (c) How far is the object displaced horil.olltall}' before hilling lhc ground? Answer in terms of HI, g, I': and II. (d) Find lhe magnilude of the object's accclcralioll while il is falling, using [he variables F, 111, and g.
A cOllcrlJlUal Enhanced Content ,noblem encourages the student to think \'erballr and conceptually about a gi\·en physics problem rather than rely solely on compUl 1n.) are placed on a fJ"ictionless table in contace \"jth each other. A horizontal force of magnitude Fis applied to lhe block ofm,jacksonville U'lliversitJ; Ted Eltzroth, Elgin Commul/it)' Coffege; i\·lanin Epstci II, California Slate U1!iversit)', Los Angelr's: Florence [top, Virginia Slate UnivI'T.I'il)'; ~'I ike [ydenbeq~, /liew iVII'xi,/) Stall' Universit), at Alamogordo; Davene Eyrcs, North Sealtle Community College; BreI! Fadem, MllhLenbF.rg Collegl'". Greg: Falabella, Wagner College; 1\'lichaeJ Falcski, Delta Co/ll'ge; .Jacqueline Faridani. Shiptll'1lSburg UniT'ersily; Abu Fasihuddin, Vnivl"f.\ity oleO/mediCl/t; Scott Fedorchak. CfllJI/Jb(,1t Uniwniry; Frank Ferronc, Drexl'/
xxi
xxii
Preface
UnivN.I'II)'; l.~arland Fish, KaLall!azoo Valley COl/wnwily Collf'gt:; KCIll Fisher, CollIlI/bus S/nfe Communily Col~ lege; Allen I-lora, !/ood Collegr.Jlllles Friedrichsen. Austill Community Collegf'", Cylllhi,l Galo\'ich. UniT'ersify U/Ai01 -lhern Cohmu/o: Ticu G,lmalie, ArJwllSu.\· Sif/le Ulliversil)'-LRAF!~: And)' Gavrin, Indiana UniverSil)" Purdue Univn\'ity hl([irtnapoli~; Michael Giangnlildc, OaJdulld Com1llullily Collegl': Wdls Gordon, Ohio Fa/lfy UlIiVl'r.litJ; Charlt:S Grabowski, CaITol! CommulIit)' College; Roben Cramer, [.aile (;il} Commullity ColllKI'", Janusl Grcbowicz, Ulltun,lily oj Ilou~lulI-J)uwnluwn;~:loITis Greellll'ood, Sa 1/ jarinlo College Cen/ml; David Grah, Cal/lloll. UTlitlersily; Fred Crossc, SuSqlll'lWlI11fl Univnsil)': 1-I; Brooke M. Pridmore, Clayloll State Universily; joseph Pries1, Mia'flli University; james Purcell, GeOl"{{ia Siale [Jniversity; "V. Sieve Quon, Feulura Coffege; Michael Ram, Stall! Uninasity OliVeT/} YOI'll at BuJ/alo; Kl-'J"l R_cihd, The Ohio Slnle UlIillersil)'; }.'I. AnLhony Reynolds, Embry-Riddle Aeronautical University; Barry Robertson, Queen's University; Virginia Roundy, Califinnia ,\'tate University, Flllh'rlon; Larry Rowan, UnivrrsiiJ ofNo'rtlt CaroliHa, CllllfJd /-lill; Dubravka Rupnik, I.ollisiana Slale UniV/:rsit)'; William R. Savage:, Universil)' of Iowa; Reinhard A, Schumacher, Carneg'; Donald D. Snyder, Indimw Univenit)" at SolttMend; George Strobel, Univenit)' of Cnll"/:;rin; Carey E. Strotl<Jeh, Virginia Stnle Universit)'; Thomas \V. Taylor, Clevdalld Stale UlIiversit~'1; Perry A. Tompkins, Samford
xxiii
xxiv
Preface
l..hriversit): L. L. Van Zandt. P/t1dllt' University; Howard G. Voss, Al'izol/a Slate UTliTlf.rsil),;James Wanliss, Embry-niddlR Aeronll/ltical University; Larry Weaver, Kansas Slale Ulliucrsit),; Donald II. \"'hite. Western Oregon State Collegt"; Bernard Whiting, Ulliversi(v ojFlon-da; George A. Williams, VlIivm·ity oj l./tllh;Jerry H. Wilson. Metropolitan Statp C.ollege; Roben M. Wood, UnivprsitJ' ojGporgin: and Clyde A. Zaidins, UlliversitJ ojColnmrlo af Df1Iver.
Gerd Konemeyer and Randall Jones contributed several end-oF-chapter problems. especially those or interest to the life sciences. Ed\'.:ard F. Redish of the Cniversity of Maryland graciously allowed us to list some of his problems rrom the Activity Based Physics Project. ',Ve are extremely grateful to the publishing team at the Brooks/Cole Publishing Company for their expertise and outstanding work in all aspects orthis project. In particular, we thank Ed Dodd, who tirelessly coordinated and directed our efforts in preparing the manuscript in its \'arious stages, and Sylvia Krick, who transmitted all the print ancillaries. Jane Sanders Miller, the photo researcher, did a great job finding phoLOs or physical phenomena, Sam Subity coordinated the media program for the text, and Rob Hugel helped translate our rough sketches into accurate, compelling art. Katherine '''Tilson of Lachina Publishing Services managed the difficult task of keeping produClion moving and on schedule. :\tfark Santee, Teri Hyde, and Chris Hall also made numerous valuable comribmions. Mark, the book's marketing manager, was a tireless advocate for the text. Tcri coordinated the entire production and manufacturing of the text, in all its various incarnations, from start to finish. Chris provided justtbe right amoum of guidance and vision throughout the project. \·Ve also thank David Harris, a great team builder and motivator with loads of enthusiasm and an infectious sense of humor. Finally, we are deeply indebted to our wives and children [or their love, support, and langterm sacrifices.
Raymond A. Serway Sl. Petersburg, Florida Chris Yuille Daytona Beach, Florida
ENGAGINii.ApPLICATIONS
:\ll1lOugh pll\sics is relc\-ant to so much in Ollr modem lives. it Ill.n not be obvious to sluclclllS in an imroducLOry course. In this eighth t'diLion of Colhgl' Physirs, h(' continue a d{'~ilin feature beg-ull in the ~t'vt'nlh edition. This feature makes the relevance of physics LO everrday life more obvious by pointing oul specific applications in tILe form of a mal'ginal notc. Some of these .-.pplicalions pertain to lhe life sciencc:-. and are marked with Ihe DNA icon The lisl below is not illlended to be a complete listing of all the applications of the principles of physics found in this textbook. Man~' other applic:niolls are LO be found within the text and especially in the worked l'xamples, conn:pwal qlle~lions, and cnd-ol~chaplerproblems .
II.
Chapter 3 The longjump. p. 66 Chapter 4 Se.-'al belts. p. HI. Iidicopler Hight. p. 93 Colliding \~hiclt:s. p. 94 Sl..ydi\ing, p. lOR
II
a III
II II
• "
II
Chapter 5 Fl:lgellar llloWllIelll: bioluminescence, p. 142 I"-I('roid impac!' p. 142 Dict versus ("Xl'ITi~(' in weighl-Ios~ programs, p.147
Chapter 6 Boxing and brain ir~jury, p. 11';3 11~jllry to passcng:c..:rs in car collisions. p. Ili5 (;Iaucoma testillg:, p. 169 Professor Goclcl:rrd \'Jas righl all along: Rockeb \\'clrk in spacer p. 178 IVlullislage rod..l'I:.. p. 179
pp.330-331
II III
II III •
198
ArtifIcial gr;wil)'. p. 203 I\anked madw:ns. p. 205 Wh), is the Sun hot? p. 213 Geosynchronolls orbit and telecommunications salellitcs. p. 21 i
111 III
\\'ol'king oITbn;akra~t. p. 354 Physiolllg-y of cxercise, p. ~-\54 Sea breaesand Ihcl'lllOlb. p. 355 Home il1SlIlatioll, pp. :-\liH-369 Staying warlll in the arnie, p. 370 Cooling automobile clIgilll·S. p. 37\ Algal hlooms in ponds ancllakes. p. 371 Hod)' tClltperalllrc. p. :n2 Light-culored Sllllllllt:l dothing, p. :{7:l Thennog:rapln·. p. 3i~ Radi;lIioll IhermOllletl'rs for mcasuring boch lemperatllre. p. 3;3 Thermal radiation :tnd Ili~ht vision, p. 374 Thermo:-. huttles. p. 37:, Global w,lIllling and g-n:t:nhollse gases. p. 375
"
Refriger;uurs and he.-·;u I}umps, PI). 102-·103 ~Peqwlllal lIlolion~ machines, p. 409 The di I"('(tion of tillle. p. ·112 Humall metabolism. pp. 113-416 Fighting- fat, p. 41f}
Chapter 13
Archery. p. '129 Pi.t;
qU0 X 103 kg/m:\ to g/cm 1 .
4.50 g/cm i
Answer
1.6
ESTIMATES AND ORDER·Of·MAGNITUDE CALCULATIONS
Celling an exact answer to a calculation may One!1 be difficult or impossible, either for mathematical reasons or because limited information is .n-ailable. In these GISeS, estimates can yield lIseful approximate answers thaI can determine whether a mor(> precise calculation is nccessalT Estimmes also sen'e as a partial check if the exact calculations are actuall), carried oul. If a large answer is expected bllt a small exact answer is obtained, there's an error ~omc\'iherc. For many problems, knowing the approximme value of a quantity-within a factor of 10 or so-is sufficient. This approximate valuc is called an order-ofmagnitude estimate. and requires finding' the power of to that is closest to the actual value of the quantity. For example, 75 kp; - 10 2 kg, where the symbol means "is on the order of" or "is approximately." Increasing a quan[itr by three orders ofmagniwde means that its value increases by £1 [actor of 10~ = 1 000. Occasionally the process or making such estimates results in fairly crudc answers, bill answers tcn times or morc too large or small are still useful. For example, suppose you're interested ill how mallY people have contracled a certain disease. Any estimates under ten thousand arc small compared with Eanh's tOlal population, bUl a million or more would be alarming. So even relatively imprecise inrorl1laLion can provide valuable guidance. In developing these estimates, you can take considerable liberties WiLh the numhers. For example, '1T - 1, 27 ~ 10, alld 6.:) -- 100. To gel. it less crude estima[c, it's permissiblc [Q use slightly more accurate numbers (e.g., '1T - 3, 27 - 30, 65 -- 70). Hettel' accuracy can also be obtained by systematically underestimating as man>' numbers as YOli o\,crcstim;He. Some quantities may be completely unknown, but ii's slandard to make reasonable guesses, as the examples show.
EXAMPLE 1.6 Goal
Brain Cells Estimate:--
_
Develop a simple estimate.
Problem
Estirnate the nurllber of cells in the human
brain. Strategy Estima[e the volume of a human brain and divide by the estimated volume of one cell. The brain is located in the tipper portion of the head, with a \'olumc lhaLcould be approximated bya cube = 20 cm 011 a side_
..
Brain cells, consisting of about 10% neurons and 90% glia, \'ary greatly in size. with dimensions ranging from a few microns to a meter or so. As a guess, take d = 10 miCl-ons as a typical dimension and consider a cell to be a cubc with each side having [hat length.
e
.........,
'
Solution Estimate of the volume ora human brain:
~"'"'' = f'
ESlimaLe the volume ora cell:
\~."" =
Divide Lhe volume ora brain by the volume ora cell:
number of cells
~,
= (0,2 Ill)" =
,l" =
(10 X 10
6
8 X 10-3 111"
m)"
\~)I'l1fi)(lOS HI) = 10' bills
t\lultiply this value by the approximate lunar distance:
# of" dollar bills - (4 X 10"_ kill) (10' billS) lkm
That's the same order of magniwde as the U.S. national debt!
QUESTION 1.7
Based on the answer, about how many stacked pennies would reach the Moon? EXERCISE 1.7
lTow man)' pieces of cardboard, typically found at the back of a bound pad of paper, would match thc height or the \Vashington monument, about 170 m laB? Answer
havc
[Q
stack up
LO
....... 10:· (Answers may vary.)
EXAMPLE 1.8 Goal
yOll
Estimale a
Number of Galaxies in the Universe \"oIUIllC
and a number density, and combine.
Problem GiH'n that astronomers can sec about 10 billion light years into space and that there are 14 galaxies in our local group, 2 million light years from the next local group, estimate the number of galaxies in thc observable universe. (NOIc: One light )'ear is the distance traveled by light in one ~·ear. ~lbollt9.5 X 10 15 m.) (See Fig. 1.3.)
Strategy From thc known information, we can estimate the number of galaxies per unit volume. The local group of J4 galaxies is contained in a sphere a million light years in radius, with the Andromeda group in a similar sphere. so there are about JO g£llaxics within a volume of radius Imillion light years. Multiply that number density by the volume of the observable universe.
~
....
"
... , ..
,
Solution Compute lhe approximale volume of galaxies:
FIGURE 1.3 In tllisdcep-spacc photugraph. theH' are few stars-jml galaxies w;thout ('Itt!.
l'fg
of the local group
v" = ~1TTS -
,
(10" ly):I
~
=
10'" Iy'
1.7
E.stimate the density of galaxies:
Coordinate Systems
13
# of' galaxies density of galaxies = -----''-:--V", 10 galaxies
Iy'
I OIB
Compute t.he approxi mate volume or the observable universe: Multiply the density or galaxies by \I,:
# of galaxies
~
(density of galaxies)V;,
(
_ aalaXieS) . (lo,oly')
IO-I'~
10
-
~
I)~
galaxies
.
Notice the approximate nature of Iht: computal ion, which uses 41T/3 -- 1 on two occasions and l'l 10 for the number of galaxies in the local group. This is completely justified: Using the actual numbers would be pointless, because the other assumptions in the problem-the size of the observable universe and the idea that the local galaxy density is representative of the density everyv,,'here-are also vcry rough approximations. Further, there was nothing in the problem that refJuired using volumes of spheres rather than volumes of cubes. Despite all these arbitrary choices, the ans\ver still gives useful informatlon, becallse it rules out a lot of reasonable possible answers. Before doing the calculation, a guess ofa billion galaxies might have seemed plausible. Remarks
1.7
13
............. ~
QUESTION 1.8 Of the fourteen galaxies in the local group, only one, the Milky Way, is not a dwarf gablxy. Estimate the number of galaxies in the universe that are not dwarfs. EXERCISE 1.8 Given that the nearest star is about 4 light years "I\o\'ay and thatlhe galaxy is roughly a disk 100 000 light years across and a thousand tight years thick, estimate the number of stars in the Milky \·Vay galaxy. ~ 10 12 stars (Estimates will vary_ The actual answer is prob:lbly close 1.0 4 X lOll stars.)
Answer
COORDINATE SYSTEMS
Many aspects of physics deal \\lith locations in space, \vhich require the dcllnition of a coordinate system. A poilU on a line can be located with one coordinate, a point in a plane with two coordinates, and a point in space \vith three. A coordinate system used to specify locations in space consists of the following: • A fl xed reference point 0, called the origi1J • A sct of specified axes, or directions, with an appropriate scale and labels all the axes • rnsl:l'uctions on labeling a point in space relative to the origin and axes
One convenient and commonly used coordinat.e system is the Cartesian coordi· nate system, sometimes called the rectangular coordinate system. Such a system in two dimensions is illust.rated in Figure 104. An arbitrary point in this system is labeled with the coordinates (.~, y). For example, the point P in the figure has coordinates (5, 3). If we start at the origin 0, we can reach P by moving 5 meters horizontally 1.0 the right and then 3mel.ers vertically upv..' ards. In t.he same vvay, the point Qhas coordinates (-3, 4), which corresponds to going 3 meters horizontally to t he left of the origin and 4 meters vertically upwards frolll there. Positive x is usually selected as right. of the origin and positive y upward frolll the origin, but in two dimensions this choice is largely a Illallcr of tasle. (In t.hree dimensions, however, there are "right-handed" and "left-handed" coordinates, which lead 10 minus sign differenccs in certain operalions. These will be addressed as necded.)
>'
(m)
10
Q
• (-3.4)
3 p
.(5,3) :r(m)
()
5
10
FIGURE 1.4 Dcsigll;:J1jon of poi illS in a Iwo-dimcllsional Cartesian coordinale syslem. Ever}' poim is Iabc/cd witli roordinalcs (x. y).
14
Chapter 1
Introduction
Sometimes it's more convenient to locate a point in space by its plane polar coordinates (1; 0), as in Figure ].5. In this coordinate system, an origin Ganci a reference line are scleClcd as shown. A poin1 is then specified by the distance r from the origin to the point and by the angle fJ between the reference line and a line dnlwn from the origin to the point. The standard rcference line is usually selected to be the positive x-axis of a Cartesian coordinate system. The angle 8 is considered positive when measured counterclockwise from the reference line and negative when mcasured clockwise. For example, if a point is specified by the polar coordinates 3 m and 60 we locate this point by moving alit 3 m from the origin 0 at an angle of 60 above (coullterclocbvise from) the reference line. A point specined by polar coordinates 3 m and -60 is located 3 m out ['rom the origin and 60 belm\' (c1ocbvise [rom) the reference line. 0
Refi.:rcllcc line
,
0
FIGURE 1.S
A poJarcllordillate
S"~[f'm.
0
1.8 TRIGONOMETRY Consider the riglll triangle shown in Active Figure 1.6, where side)' is opposite the angle 8, side;r is adjacent to the angle 8, and side ris the hypotenuse orthe triangle. The basic trigonometric functions de filled by such a triangle are the ratios of the lengths of the sides or the triangle. These relationships are called the sine (sin), cosine (cos), and tangent (tan) funClions. fn terms 01'8, the basic trigonometric functions arc as follows: 1
)'
. Sill () '=
,
"'r
,
cox
= .'1- .
The average speed of an object over a given time interval is lhe LOlal dislance traveled divided by the total time elapsed: Average speed
total distance total time
81 unit: meter per second (m/s) symbols, this equation might be written 11 = dll, ",,,ith the letter rJ understood in context to be the average speed, not a velocity. Because total distance and total time arc ahlo,'ays positive, the average speed will be positive, also. The definition of average speed completely ignores 'what may happen between the beginning and the end of the motion. For example, you might drive from Atlanta, Georgia, to St. Petersburg, Florida, a distance of about 500 miles, in 10 hours. Your average speed is 500 mi/l0 h = 50 mi/h. It doesn't matter if you spent two hours in a traffic jam traveling only 5 mi/h and another hour at a rest stop. For average speed, only the total distance traveled and total elapsed time are important. ]11
EXAMPLE 2.1 Goal
The Tortoise and the Hare
Apply the concept of average speed.
Problem A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops [or a gO.O-min nap. Upon awakening, he remembers the race and runs twice as fast. Finishing the course in a lotal time of 1.75 h, the rabbit wins the race. (a) Calculate the average speed of the rabbit. (b) "Vhat was his average speed before he stopped for a nap? Strategy Finding the overall average speed in part (a) isjust a matter of dividing the total distance by the total time. Part (b) requires lWO equations and two unknowns, the latter turning out to be the two different average speeds: VI before the nap and v2 after the nap. One equation is given in the statement of the problem (v2 = 2v l ), whereas the other comes from the factlhe rabbit ran for only 15 minutes because he napped for gO minutes. ".
~
Solution (a) Find the rabbit's overall average speed.
Apply the equation for average speed:
Average speed
total distance total time 2.29 km/h
4.00 km 1.75 h
2.2
Velocity
27
(b) Find the rabbit's average speed before his nap. Surn the running times, and seL the Slim equal to 0.25 h:
II
+
= 0.250 h
12
(I)
Substitute V2 = 2v, and the values of til and d';!. inLO Equation (1):
Remark As seen in this example, average speed can be calculated regardless of any variation in speed over the given time interval. QUESTION 2.1 Does a doubling of an object's average speed always double the magnitude of its displacement in a given amount of time? Explain. EXERCISE 2.1 Estimate the average speed of the Apollo spacecraft in meters per second, given thal the craft LOok five days to reach the Moon from Earth. (The Moon is 3.8 x 1Qli rn from Earth.)
Answer
- 900 m/s
Unlike average speed, average velocity is a veeLOr quantity. having both a magnitude and a direction. Consider again the car of Figure 2.2, moving along the road (the x-axis). Let the car's position be xj at some time and x/at a later time 'I' In the Lime interval at = t.r - ti , the displacerncl1l of the car is.6.x = -'i - Xi"
'j
The average velocity divided by tJ./:
v during
a tirne interval
at
is the displacement
ax
TABLE 2.1
51 unit: meter per second (m/s)
Unlike the average speed, which is always positive. the average velocity of an object in one dimension can be either positive or negative. depending on the sign of the displacement. (The lime interval ~I is always positive.) In Figure 2.241, for example, the average ,'clocil)' of the car is positive in the upper illustration, a positive sign indicating motion to the right along the x-axis. Similarly, a negative average velocity for the car in the lower illustration of rhe figure indicates that it moves to the left along the x-axis. As an example, we can use the data in Table 2.1 to find the average velocity in the time interval from point ® to point@ (assume two digits are significant):
ax
52
m -
Definition of average velocity
[2.2]
v==:tJ.I
V ~ -tJ.-I
+-
30
Position of the Car at Various Times Position I (s) @ @
Aside froll) meters per second, other common units for average velocity are feet per second (ft/s) in the U.S. customary system and centimeters per second (cm/s) in the cgs system. To further illustrate the distinction between speed and velocity, suppose we're watching a drag race from the Goodyear blimp. In one run we see a car follow the straight-line path from ® to @ shown in Figure 2.3 during the time interval 1J.t,
®c:>Xi
@
~l
x
FIGURE 2.3 A drag race viewed from a blimp. One car follows the red slraighl-line palh from ® [email protected] :l second car follows lhe h[lJe curved p:llh.
28
Chapter 2
Motion in One Dimension
and in a second run a car follm.. . s the curved path during the same interval. From the definition in Equation 2.2, the two cars had the same average velocity because they had the same displacement = XI ~ Xi during the same time interval dt. The car taking the curved route, however, traveled a greater distance and had the higher average speed.
ax
( 100 yd
FIGURE 2.4 (Quick Quiz 2.1) The path followed by a confused football player.
(TIP 2.3 Slopes of Graphs '\ The word slope is often lIsed in reference to the graphs of physical data. Regardless of the type of data, the stolle is given by
change in vertical axis
Slope
=
change
.. ttl
honzontal
. aXIS
Slope carries units.
TIP 2.4 Average Velocity vs. Average Speed Average velocity is nollhe same as average speed. If you nm from x= 0 III 10 x= 25 malld back to your starting point in a Lime
int.erva\ of 5 s, the average velocity is zero, whereas the average
speed is 10 mho
QUICK QUIZ 2.1 Figure 2.4 shows the unusual path of a confused football player. AFter receiving a kickoff at his own goal, he runs down field to within inches of a touchdown, then reverses direction and races back until he's tackled at the exacllocation where he first caught the ball. During this run, which took 25 s, what is (a) the total distance he travels, (b) his displacement, and (c) his average velocity in the x-direction? (d) Whal is his average speed?
Graphical Interpretatian of Velocity If a car moves along the x-axis from ® to @
Instantaneous Velocity Average velocity doesn't take into account the details of what happens during an interval of time. On a car trip, for example, you may speed up or slmv down a number of times in response to the traffic and the condition of the road, and on rare occasions even pull over to chat with a police officer about your speed. 't
~
Ih ) 4 mi.S( 10 -h- 3600 s
=
6.7 mi
(d) Find the average acceleration from 200 s to 300 s, and fll1d the displacement..
The slope of the green line is the average acceleration from 200 ,to 300 s (Fig. 2.19b):
a= ~
(1.0 X 10' - 1.7 X 10') llli/h LOX 10',
-1.6 (mi/h)/s
2.5
One-Dimensional Motion with Constant Acceleration
41
200
ISO _ 100
~
5,
50 0 -50 -50
t"
-100 (h)
(a) !'")
200
~tJu
150
ElO
_ 100 ~
::, 50
S,
0
o
t
-50 -50
50
100
0 -100
-100
(d)
(e)
FIGURE 2.19
(Example 2.7) (a) Velocil}'\". time graph Ii)! the Aceb. (b) Thl.:
~lopl.: ofthe"lt::epe~l
1:111genl blue litH: g-ivcs the peak a,cdt.:r:ttioll. and lhe slope 01 lhe greell lim: i~ Ill(' average acn.·lcl",uitlll between 200 sand :H10s. (1.:) The area llllder the vclocityv.,. lillle gr<Jph ill some lilllC illlen'"~ 1("', = :(5.0 X 10' s)( -5.0 X 10' mi/h) ~
-1.3 X 103(mi/h)(s)
= (2.4 X 10"
+ 9.0 X JO" + 2.5 X 10'
-1.3 X 10')(mi/h)(s) = lrrrr..- • • • • •
8.9 mi •
• • • • • • • ...011I
Remarks There are a number of ways to find the approximate area uncleI' a graph. Choice of technique is a sonal pre[ercnce.
per~
QUESTION 2.7 According tu the graph in Figure 2.19a, at whal different times is the acceleration zero? EXERCISE 2.7 Suppose the velocity vs. time graph of another train is given in Figure 2.l9d. Find (a) the maximum instantaneous acceleration and (b) the total displacement in the imer\'al ['1"0111 0 S [Q 4.00 X IO't s.
Answers
(a) 1.0 (mi/h)/s
(b) 4.7 mi
42
Chapter 2
Motion in One Dimension
2.6 FREELY FALLING OBJECTS
GALILEO GAl/LEI ftolian Physicist and Astronomer (1564-1642) Galileo formulated the lows that
govern the motion of objects in free fall. He olso investigated the motion
of on object on on inclined plane, established the concept of relative motion, invented the thermometer, and discovered that the motion of a
swinging pendulum could be used to measure time intervals. After designing and constructing his own
telescope, he discovered four of
Jupiter's moons, found that our own Moon's surface is rough, discovered sunspots and the phases of Venus, and showed that the Milky Way consists of an enormous number of stars. Galilee publicly defended Nicolaus
Copernicus's assertion that the Sun is at the center of the Universe (the heliocentric system). He published Dialogue Concerning Two New World Systems to support the Copernican model, a view the Church declared to be heretical. After being taken to Rome in 1633 on a charge of heresy, he was sentenced ta life imprisonment and loter was confined to his villa ot Arcetri, near Florence, where he died in 1642.
'Vhen air resistance is negligible, all objects dropped under the influence of gravity near Earth's surface fall toward Earth with the same constant acceleration. This idea may seem obvious mclay, but it wasn't until about 1600 that it \'vas accepted. Prior to that time, the teachings of the great philosopher Aristotle (384-322 H.C.) had held that heavier objects fell faster than lighter ones. According to legend, Galileo discovered the law of falling objects by observing that two different weights dropped simultaneously from the Leaning Tower of Pisa hit the ground at approximately the same time. Although it's unlikely that this particular experiment was carried out, we know that Galileo performed Illany systematic experiments with objects moving on inclined planes. In his experimenLS he rolled balls dO\;,.'n a slight incline and measured the distances they covered in successive time intervals. The purpose of the incline ,\o'as to reduce the acceleration and enable Galileo to make accurate measurements of the intervals. (Some people refer to this experiment as "diluting gravity.") By gradually increasing the slope of the incline he was finally able to draw mathematical conclusions about freely falling objects, because a falling ball is equivalent to a ball going down a vertical incline. Galileo's achievements in the science of mechanics paved the \-va)' for Newton in his development of the Im.. . s of motion, \vhich 've will study in Chapter 4. Try the following experiment: Drop a hammer and a feather simultaneously from the same height. The hammer hits the floor first because air drag has a greater effect. on the much lighter feather. On August 2, 1971, this same experiment was conducted on the Moon by astronaut David Scott, and the hammer and feather fell with exactly the same acceleration, as expected, hitting the lunar surface at the same time. Tn the idealized case ,vhere air resistance is negligible, such marion is called free fall. The expression freely faMing objer' doesn't necessarily refer to an object dropped from rest. A freely falling o~ject is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all considered freely falling. ',Ve denote the magnitude of tile free-fall acceleration by the symbol g. The value of g decreases with increasing altitude, and varies slightly 'with latitude as well. At Earth's surface, the value of gis approximately 9_80 l11/s2 . Unless stated otherwise, we willllse this value for gin doing calculations. For quick estimates, use g= 10 111/52 . If \-\fe neglect air resistance and assume that the free-fall acceleration doesn't vary \.,..ith altitude over short vertical dislances, then the motion of a [reely falling object is the same as motjol1 in one dimension under constam acceleration. This means that the kinematics equations developed in Section 2.6 can be applied. It's conventional to define "up" as the + )I-direction and to use )' as the position variable. In that case the acceleration is a = -g = -9.80 m/s2 . In Chapter 7, we study the variation in gwith altitude. QUICK QUIZ 2.6 A tennis player on serve tosses a ball straighlup. While the ball is in free fall, does its acceleration (a) increase, (b) decrease, (c) increase and then decrease, (d) decrease and then increase, or (e) remain constant? QUICK QUIZ 2.7 As the tennis ball of Quick Quiz 2.6 travels through the air, does its speed (a) increase, (b) decrease, (c) decrease and then increase, (d) increase and then decrease, or (e) remain the same? QUICK QUIZ 2.8 A skydiver jumps out of a hovering helicopter. A few seconds later, anolher skydiverjumps out, so they both fall along the same vertical Jine relative to the helicopter. Both skydivers fall vlo'ith the same acceleration. Does the vertical distance bet,.,..een them (a) increase, (b) decrease, or (c) stay the same? Does the difference in their velocities (d) increase, (e) decrease, or (f) stay the same? (Assume g is constant.)
2.6
EXAMPLE 2.8
Not
0
43
Bad Throw for a Rookie!
Goal Apply the kinematic equ
5) -
,=
118 m/s
9.80 m/s
,=12.0s
(4.90 m/s')(L2.0 5)'
945111
(c) Phase 2: Find the velocity of the rocket just prior to impact.
+ (118 Ill/S) , - (4.90 m/,)I'
Find the tin1e to impacl by selling)' = 0 in Equation (6) and using the quadratic formula:
0= 235
Substitute this value of t into Equat.ion (5):
v = (-9.80 m/s')(25.9 s)
t =
III
25.9 s
+ 118 Ill/S = -136m/s
~
~
Remarks You may think thal it is more natural to break this problem into three phases, with the second phase ending at the maximum height and the third phase a free fall from maximum heig·hlLO the ground. Although t.his approach gives the correct answer, it's all unnecessary complication. Two phases arc sufficient, one lor each different acceleration. QUESTION 2.10 [f, instead, some fuel remains, at what height should the engines be fired again to bri'lkc the rocket's fall and allO\v a
pcrfcClly soft landing? (Assume the samc acceleration as during t.he initial ascent.) EXERCISE 2.10
An experimental rocket designed to land upright falls freely from a height of 2.00 X 10~ 111, starting at rest. At a height of 80.0 m, the rocket's engines start and pro\'ide constant upward accelcration until the rockellands. Vlhat acceleralion is reguired if the speed on touchdowil is to be zero? (Neglect air resislance.) Answer
14.7 m/s~
Multiple-Choice Questions
2.1
_ a
Displacement
The displacement of all object moving along the defined as the change in position or the object, "'" ~ xJ - xi
x-aX1S IS
[2.1]
where Xi is the initial position of the object and Xi is its final position. A vector quantity is characterized by both a magnitude and a direction. Ascalar quantity has a magniLUde only.
2.2
o~icct
Average speed ==
[2.4]
Ii
2.5 One-Dimensional Motion with Constant Acceleration
is given by total distance
u=vo+a./
total time
v
"'x
[2.2]
v==:-~
tJ - l,
The average velocity is equal to the slope of the straight line .joining the initial and final points on a graph of the position of the object versus time. The slope of the line tangent to the position \IS. time curve at some point is equal to the instantaneous velocity at that time. The instantaneous speed of an objecl is defined as the magnitude of the instantaneous velocity.
2.3
IJ -
The instantaneous acceleration of an object at a certain time equals the slope of a velocity vs. time graph at that instant.
The average velocity during a time int.erva\ 6.l is the displacement 6.x divided by D.t.
"'t
"'t
l'j- v, --
The most useful equations that describe the motion of an object moving \\"ith constant acceleration along the x-aXIS are as follows:
Velocity
The average speed or an
=-tiu =
47
Acceleration
The average acceleration a. of an object undergoing a change in \'elocity Av during a time interval tit is
[2.6]
ti.x = vol.
+ ~aL2
[2.9]
v2 =
+ 2atix
[2.10]
uo'.!
All problems can be solved with the first two equations alone, the last being convenient when time doesn't explic· itly enter the problem. After the constants are properly identified, most problems reduce [0 one or two equations in as many unknov,,·ns.
2.6
Freely Falling Objects
An object falling in the presence of Earth's gravity exhibits a free-fall acceleration directed toward Earth's center. If air friction is neglected and if the altitude of the falling o~iect is small compared with Earth's radius, then we can assume that the free-fall acceleration g = 9.8 111/s2 is constant over the range of Illation. Equations 2.6, 2.9, and 2.10 apply, with a. = -g.
FOR ADDITIONAL STUDENT RESOURCES, GO TO WWW.SERWAYPHYSICS.COM
•
•
1. An arrow is shot straight up in the air at an init.ial spced of 15.0 m/s. After how much t.ime is the arrow heading clowl1\vard at a speed of 8.00 m/s? (a) 0.714 s (b) 1.24 s (c) 1.87 s (d) 2.355 (e) 3.22 s 2. Acannon shell is fired straight up in the air at an initial speed of 225 m/s. After how much timc is the shell at a height of 6.20 X ]02 III and heading down? (a) 2.96 s (b) 17.3 s (c) 25.4 5 (d) 33.6 s (e) 43.05 3. ''''Then applying the equations of kinematics for an object moving in one dimension, which of the following statements must be true? (a) The velocity of the object must remain constant. (b) The acceleration of the object must remain constant. (c) The velocity of the object must increase with time. (d) The position of the o~ject must increase with time. (e) The velocity of the object must ahvays be in the same direction as its acceleration. 4.Ajuggler throws a bmvling pin straight up in the air. After the pin leaves his hand and while it is in the air, \vhich statcmCIll is tfue? (a) The velocity of the pin is always in the samc direction as its acceleration. (b) The velocity of the pin is ncver in the same direct.ion as its
acceleration. (c) The acceleration of the pin is zero. (d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity oftlle pin is in the same direction as its acceleration on the way up. 5. A racing car starts from rest and reaches a final speed v in a time t. If the acceleration of the car is constant during this rime, which of the roJlowing statements must be true? (a) The car travels a distance vt. (b) The average speed of the car is v12. (c) The acceleration of the car is viI. (d) The velocity of the car remains constant. (e) None of these 6. A pebble is dropped from rest from lhe top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 seconds? (a) 9.8 In (b) 19.6 m (c) 39 m (d) 44 m (ej 27 m 7. An object moves along the x-axis, its position measured at each instant of time. The data are organized intO an accurate graph of x vs. t. ·Which of the following quantities cannol be obtained from this graph? (a) the velocity at any instant (b) the acceleration at any instant (c) the displacement during some time interval (d) the average
48
Chapter 2
Motion in One Dimension
\"clocity during some time interval (e) the speed panicle at any instant
or the
8. People become ul1collllOl"Lable in an elevator if" it accelerates fmm rest al a rate such that it attains a speed of abOlU 6 rn/s after descending ten stories (abo11130 111).
\\'hat is the approximate magnitude of its acceleration? (Choose the closest answer.) (a) 10 111/52 (b) 0.3 m/s:! (e) 0.6 mis' (el) I m/s' (e) 0.8 mis'
neglect air rl·iction.) (a) The speed of the red ball is less than thai of Ihe blue ball. (b) The speed of the red ball is greatcr than that of the blue ball. (c) Their velocities are equal. (d) The speed of each ball is greater than 1/0. (e) The acceleration of the blue Inll is greater than that or the red b well as il~ initial Spl'l'.x = =
Substitute v). = 0 at maximum height, and the fact that v"J = (l1.0 ,;,/s) sin 20.0°:
67
Motion in Two Dimensions
0.722 m
-2(9.80 m/s')
~
_
Remarks Although modeling the long jumper's motion as that of a projectile is an oversimplification, the values obtained are reasonable.
QUESTION 3.6 True or False: Because the x-component of the displacement doesn't depend explicitly on g, the horizontal distance traveled doesn't depend on the acceleration of gravity. EXERCISE 3.6 A grasshopper jumps a horizontal dist.ance of ] .00 III from rest, ",lith an initial velocity at a 45.0" angle with respect to the horizontal. Find (a) the initial speed of the grasshoppcr and (b) the maximum height reached.
Answers
(a) 3.13 m/s (b) 0.250 m
EXAMPLE 3.7 Goal
The Range Equation
Find an equation for the maximum horizolllal displacemcm ofa projectile nred from ground level.
Problem An athletc participates in a long:jump compctition, leaping into the air with a velocity V o at an angle with t.he horizonLal. Obtain an exprcssion for the length of 1he jump in tcrms of vo, 00 , and g. Strategy
eo
Usc the results of Example 3.6, eliminating the time l from Equations (I) and (2).
.
.
,
~
~
Solution Use Equation C1) of Exarnple 3.6 to nnd the time of night, l:
Substitute t.his expression for t into Equation (2) of
~x =
() en
ax =
9 1/- V"
Vo
cos
l
=
(110 cos 0 0
Example 3.6:
Simplify:
Remarks
sin 00) g
0I) cos O' j) SIJ)
g
Substitute the identity 2 cos On sin 00 = sin 20 0 to reduce t he foregoing expression to a single t.rigonomctric function: ~
l( 2Vo
,
,
-,
,
-
(I)
.
.
,
~
The usc of a trigonometric identity in t.he nnal step isn't necessary, but it makcs Question 3.7 easier to
ans,·\'cr. QUESTION 3.7 VVhal. angle On produces the longesljurnp? EXERCISE 3.7 Obtain an expression for the athlete's maximum displacement in the vertical direction, and g.
Answer
6.)'111.11\ =
vo2 sin 2 0 0
2g
6.Ym;n;
in terms of
Vn,
00 ,
68
Chapter 3
Vectors and Two-Dimensional Motion
EXAMPLE 3.8 That's Quite an Arm Goal Solve a two-dimensional kinematics problem with a non horizontal initial \'clocity, starting and ending at different heights.
110'::
20.0 m/s ......
Problem A stone is thrm..·!1 upward from the LOp of a building at an angle of30.0° to the horizontal and with .0 m, the )'-displacement ""hen the stone hits the ground. Using the quadratic formula, solve for the time. To solve pan (b), subsl.iwte the time [rom pan (a) into the componems of the \'e1ocity, and substitute the same time into the equation for the x-displacement to solve pan (c).
,
\
\ I \ \ \
\ I
(>:.-45.0 Ill) I
-, FIGURE 3.19
(Example .3.8)
.,
".
Solution (a) Find the time of flight. Find the initial x- and J-components of'the velocity:
Find the y-displacemcnt, taking Yo vo)' = 10.0 m/s:
=
= vo,. =
V o•
0,)' = -45.0111, and
Reorganize the equation illLo standard form and use the quadratic formula (see Appendix A) 1.0 find the posit.ive
vosin
= (20.0 m/s) (cos 30.0°) = +17.3 m/s eo = (20.0 m/s)(sin 30.0°) ~ +10.0 m/s
6} =
J -
eo
Vo cos
)'0
= vo./ - ~gt'2
-45.0 m = (10.0 m/s)/- (4.90 m/s')/' 4.22s
/=
l'OOt:
(b) Find the speed
at.
impact.
=
Substitute the value of I found in part (a) into Equation 3.14a to find the y~col1lponentof the velocity at impact:
v,
Use this value of vJ ' the Pythagorean theorem, and the fact that V x = vO.t = 17.3 m/s to find t.he speed of t.he stone at impact.:
v = V V, 2
v nJ -
g/
= -31.4
~
=
10.0 m/s - (9.80 m/s') (4.22 5)
m/s
v,' = V( 17.3 m/s)' +
+
(-31.4 m/s)'
35.9 m/s
(c) Find the horizontal range of the stone. Substitute the time of" night. into the range equation:
.:'»- = x =
Xo
=
(v o cos
ell = (20.0 m/s)(eos 30.0°)(4.22 s)
73.1 m
~
~
Remarks The angle at which the b~lll is lhrown affects the vclocity vect.or throughout it.s subsequent. motion. but doesn't affect the speed at a given height. This is a consequence of the conservat.ion of energy, described in Chapter 5. QUESTION 3.8 True or False: All other things being equal, if the ball is thrmvn at hal r the given speed it will t.ravel half as far.
3.4
Motion in Two Dimensions
69
EXERCISE 3.8 Suppose the stone is thrown from the same heighl as in the example at an angle of 30.0 degrees below the horizontal. If it strikes the ground 57.0 m away, find (a) the time of flight, (b) the initial speed, and (c) the speed and the angle of the velocity vector with respcct to the horizontal at impact. (Hinl: For part (a), use the equation for the xdisplacement to eliminate vol [,"om the equation [or the J-displacemcnt.) 0
Answers
(a) 1.57 s (b) 41.9 m/s (c) 51.3 mis, -45.0"
So far we have studied only problems in which an object with an initial velocity (ollmvs a trajectory del.errnilled by the acceleration or gravity alone. Tn the more general case, other agents, such as air drag, surface friction. or engines, can cause accelerations. These accelerations. taken together, form a vector quantity with components ax and fl),. \·Vhen both components are constant, we can lise Equations ~.ll and 3.12 to study the motion, as in the next example.
EXAMPLE 3,9 Gool
The Rocket
Solve a problem involving accelerations in two directions.
Problem Ajet plane traveling horizontally at 1.00 X 102 m/s drops a rocket from a considerable height. (See Fig. 3.20.) The rocket immediately fires its engines, accelenlting at 20.0 m/s:! in the x-direction while falling under the influence of gravity in the y-direct:ion. When the rockct has fallen 1.00 kill, lind (a) its velocity in the y-direcl.ion, (b) it.s velocity in thc x-direction, and (c) t.he magnitude and direction of its velocity. Neglect air drag and aerodynamic lin.
,,
,,
,,
!1y= -1.OOX
,, \, 103m\
,,
Strategy Because the rocket maintains a horizontal orientation (say, through ~ gyroscopes), the x- and )'-components of acceleration are independent of" each FIGURE 3.20 (Example 3.9) other. Usc the time-independent equation for [he velocily in the y-d i rection LO find the y-component. of the velocity after the rocket falls 1.00 kill. Then calculate t.he time of the rail and use that time to find the velocity in the x-direction .
,..
.
. . . .... . .
.....
....
"
Solution (a) Find the velOCity in the )I-direction. Use Equation 3.14c:
vv' -
Substilllt:e VOl' = 0, g = -9.80 m/s2 , Clnc! ily = ~ 1.00 X 1O.'l m, and solve for V.I.:
0 = 2(-9.8 m/s')(-1.00
v,. =
X
10' m)
-lAO X 10' m/s
(b) Find the velocity in the x-direCLion. Find thc time il takes the rocket 1.0 drop LOO X -10 3 m, using the y-componcnl of the velocity:
SllbslitUl.e I, vox, and a.~ inl.O Equation 3.11a velocity in the x-direction:
10
find the
-lAO X lQ-,> m/s = 0 Vx
= v o, ~
+
(
9.80 m/s-' ) I
1/) ~ 1.00 X 10' m/s
I =
14.3 s
+ (20.0 m/s')(14.3 s)
386 m/s
(c) Find the magnitude and direction or the velocity. Find the magnitude lIsing the Pythagorean theorem and the results of paris (a) and (b):
v ~ Vv,' + v,' ~
411 m/s
=
V( -lAO
X 10'm/s)'
+ (386 m/,)'
70
Chapter 3
Vectors and Two-Dimensional Motion
e=
Use the inverse tangent function to find the angle:
tan-
l
( lJ,) = V
tan
x
.
,
~
Remarks
_,(-1.40X10'm/s) 386 m/s
=
-19. go
.............................~
Notice the symmetry: The kinematic equations for the x- and y-c1il-eclions are handled in exacLiy the same
way. Having a nonzero acceleration in the x-eli rcctioll doesn't greatly increase the difficulty of the problem. QUESTION 3.9
True or False: Neglecting air friction, a projectile ,,\·jth a horizontal acceleration slays in the air longer than a projectile thaL is freely railing. EXERCISE 3.9
Suppose a rockel-propelled motorcycle is fired from rest horizontally across a canyon l.00 km wide. (a) ',"hat minimum constam acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 kill lower than the starting point? (b) At ""hat speed docs the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction. Answers
(a) 13.1 mis' (b) 202 m/s
In a stunt similar to that described in Exercise 3.9, motorcycle daredevil Evel Knievel tried to vault across Hells Canyon, part of the Snake River system in Idaho, on his rocket-powered Harley-Davidson X-2 "Skycycle." (See the chapter-opening photo on page 54). He lost consciousness at takeoff and released a lever, premaLUrel)' deploying his parachute and falling short of the other side. He landed safely in the canyon.
3.5
)'
RELATIVE VELOCITY
Relative velocit)' is all about relating the measurements of two different observers, one moving '.vith respect to the other. The measured velocity of an object depends on t.he vclocity of the observer with respect to the object. On higlnvays, for example, cars moving in the saine direct.ion are often rnoving at high speed relative to Earth, but. relative to each other they hardly move at all. To an observer at rest at the side of the road, a car might be traveling at 60 mi/h, but. to an observer in a truck travcling in the same direction at 50 mi/h, the car would appear to be traveling only]O mi/h. So measurements of velocity depend on the reference frame of' the observer. Reference frames are just coordinate systems. Most of lhe time, we use a stationary frame of reference relative to Earth, but occasionally 1ve usc a moving frame of reference associated with a bus, car, or plane moving with constant velocity relative to Earth. In t,.".o dimensions relative velocity calculations can be confusing, so a systematic approach is important and useful. Let E be an observer, assumed stationary with respect to Earth. Let l.\VO cars be labeled A and B, and int.roduce the following notation (see Fig. 3.21): rAI~ = the position of Car A as measured by E (in a coordinate system fixed with
respect to Earth).
-::k:-:-----jf-----x E
FIGURE 3.21 The position orCar A rehtliw: to Car B can be round by vector subl raction. The rate or change or the rCSll1tant \'ector with respen to lime is the rdative vdncilr equation.
rBE
= the position orCaI' B as measured by E.
fAll
= t.he position orCaI' A as measured by an observer in Car B.
According to the preceding notation, the nrst letter tells us "..· hat the veclor IS pointing at and the second lett.er tells us where the position vector starts. The posiand rBE, are given in the Agure. tion vectors of Car A and Car B relati\'e to E, How do \ove find 1,,1\' the position of Car A as measured by an observer in Car B? 'Ve simply draw an arrO\'v pointing from Car B to Car A, which can be obtained b)' subtracting THE from r.'\E:
rr\E
3.5 ~
I"AI\
=
~
rAE -
Relative Velocity
71
~
rBE
[3.15]
Now, the rale of change of these quantities ,.. . ith lime gives us the rel0, ;1 girl call throw the balt a maximullI hori/omal distance H. on a level field. How far can she throw lhe same ball vertically upward? Assume her musclcs give the ball the same speed in each case. (Is lhis (Issumplion valid?)
67. _ A student decides to measure the muzzle vclocity ora pellet shot from his gun. He poilll~ the gun horizontally. He places a target 011 a n~rtical wall a distance Xa\e1)' from the gun. The pellet hits the target ,1 venical distance )' below the glill. (a) Show thalthe position of the pellet whcn tl'aveling thmllgh the air is given by)' = Ax:!, where A is a UHlstaJl(. (b) Express the constanL A in terms of the inilial (muzzle) velocityand the frce-f;lll acceleration. (c) If x = 3.00 III ami )' = 0.21O m, \."hat is the initial speed of the pellet?
62. _ The equation ora parabob is y = ax'! + bx + r, \dlere a, U. and r are constants. The x- and )'-coordinl"~ "I light
7.70 X 10' N 1':Il~1 {/=--= 3.50 X 10' kg
11/
=
2.20 m/s2
m!s')t
= 12.0 m!s
(b) Find the time necessary to reach a speed of 12.0 m/s. Apply the kinematics "dacit)' equation:
+ v"
v=
(I{
o-
(12.0 m!s)'
= (2.20
-->
1=
5.45 s
(c) Find the resistance force after the engine is turned off. Using kincmatic!i. find the net acceleration due to rC"lislance forces: Substitmc the acceleration into NeWLon's second law, finding the resistance force:
F«"" =
11/(1
= 20(50.0 m)
-->
(I
=
-1.44 mis'
= (3.50 X 10' kg)( - 1.44 m!s') =
-5041\
....
....
Remarks The propeller excns a force 011 the air, pushing it b~ICk\''''
J~ =
~
T,-Fg=O
1.00 X 10' :-I
Force
x-Component
J.Component
T,
-'/; cos 37.0°
'/; sin 37.0
T
'1"; cos 53.0°
T2 sin 53.0"
°
-1.00 X 102:-.!
~
2
~
T,
L F.~ = 2: F, =
Apply the conditions for equilibrium to the knot, using (1) the components in the table:
(2)
-7'1 COS
37.0°
7; sin 37.0"
+ J; cos 53.0" =
c
0
+ 7; sin 53.0" - 1.00 X
10':-1 = 0
0
37.0 There are two equations and n,,'o remaining unknowns. T, = T J (cos cos 53.00 Solve Equalion (I) (0,7;: Substitute the result for 1; into Equation (2):
'Ii
sin 37.0°
7;
=
7; =
)
= T (0.799) = 133T '0.602 .,
+ (l.33T,)(sin 53.0°) - 1.00 X 10' 1\ =
60.1 N 1.337;
1.33(60.0:-1) =
...... -
....
°
79.9 N ~
Remarks It's very easy to make sign errors in this kind of' problem. One way to avoid them is to ahvays measure the angle of a "ector from the positive x-direction. The ~igonomctric functions of the angle will then automatically give the correct signs for the components. For example, T 1 makes an angle of L80° - 37" = 143 0 with respect to the positi"e )"-axis, and its x-component, 7~ cos 143°, is negati,'c, as it should be.
QUESTION 4.5 How would the answers change if a second traffic light were attached beneath the first? EXERCISE 4.5 Suppose the traffic light is hung so that the tensions 71 and 7; are both equal La 80.0 ~. Find the new angles they make with respect to t.he x-axis. (By symmetry, these angles will be the same.) Answer
Both angles arc 38.7°.
EXAMPLE 4.6
4.5
Applications of Newton's Lows
(a)
(b)
97
Sled on a Frictionless Hill
Goal Use the second law and the normal force in an equilibrium problem. Problem A sled is tied to a tree on a [riclionless, sno\\,covered hill, as shmvl1 in Figure 4.l3a. If the sled weighs 77.0 N, find the force exerted by the rope on the sled and the magniwde of the force Ii exerted by the hill on the sled. Strategy \·Vhen an object is on a slope, it's conven iell! to use tilted coordinates, as in Figure 4.13b, so that the nannal force n is in the y-direction and the tension force T is in the x-direction. In thc absence of friction, the hill exerts no force on the slcd in the x-direction. Because the sled is at rest, the conditions for equilibrium, 2.J~V; = 0 and "iF), = 0, apply, giving two equations for the two unknm.. . ns-the tension and the normal force. ~
r···························
FIGURE 4.13 (£xamplo.: 4.6) (a) A sled tied to a tree on a frictionless hill. (h) A free-hody diagram ror the ~led.
.
.
~
Solution Apply Newton's second law to the sled, with -; = 0: Extract the x-component from this equation to find T The x-component of the normal force is zero, and the sled's \veight is given by mg = 77.0 :"!. Writc the y-component of )..Icwton's second law. The y-component of the tension is zero, so this equation wil1 give the normal force.
2: I~=
38.5 N
T=
2: I~ = 11,
=
T - (77.0 N) sin 30.0° = 0
T+ 0 - mgsinfJ
0
+
11, -
mgcos 0 =
11, -
(77.0 N)(cos 30.0°) = 0
6fi.7 N
.
lrrr.... • • • • • • • • • •
....
Remarks Unlike its value on a horizontal surface, n is less than the weight 0(' the sled when the sled is on the slope. This is because only pan of the force of gravity (the X~cOll1pOnellt) is acting to pull the sled down the slope. The y-component of the force of gravity balances the normal force. QUESTION 4.6
Consider the same scenario on a hill with a steeper slope. ''''ould the magnitude of the tension in the rope get larger, smaller, or rcmain the same as before? How would thc normal force be affected? EXERCISE 4.6
Sllppase a child of weight w climbs onto Lhe sled. If Lhe tension force is measured child and the magnitude of the normal force acting on the sled. Answers
ill
= 43.0 ~,
II
= 104
1O
be 60.0 N, find the weight of the
~
QUICK QUIZ 4.5 For the woman being pulled forwarcl on the toboggan in Figure -:1-.14, is the magnitude of the normal force exerted by the ground on the toboggan (a) equallo the LOtal weight of the woman plus the toboggan, (b) greater than the total \-"eight, (c) less than the total weight, or (d) possibly greater than or less than the Lotal weight., depending on the size of the '.. .· eight relative to the tension in the rope?
Accelerating Objects and Newton's Second Law "Vhen a net force acts on an object, Lhe object accelerates, and we lise NeWLOn's second law to analyze the motion.
FIGURE 4.14
(Quick Quit ·\.5)
98
Chapter 4
EXAMPLE 4.7
The
Lows 01 Motion
Moving
0
Crote
Goal Apply the second law of motIon for a system nOt in equilibrium, LOgether with a kinerllalics equation. The combined weight of" the cr"'lte and dolly in Figure 4.15 is 3.00 X 10 9 N. If" the man pulls on the rope with a constant force or 20.0 N, \vhat is thc acccleration of thc
Problem
system (crate plus dolly), and how far will it move in 2.00 s? Assume lhe system starts from rest and that there are no friclion forccs opposing the motion. Strategy \Ve can find the acceleration of the system from NCWlon's second law. Because the force exerted on the system is constant, its accelcration is constant. Therefore, we can apply a kinemal ics equation LO find the distance travcled in 2.00 s.
FIGURE 4.15
(Example 4.7)
,...-
~
Solution Find the mass of'lhe system frol11 the definition of\...' eight, tu =
W
11/=-= g
mg:
Find [he acceleration of the system from the second law: Use kinemalics to find the distance moved in 2.00 s, v.lith V o = 0:
{l
x
=
1':
3.00 X 10' N = 30.6 kg 9.80 Ill/s2 20.0 N = 30.6 kg
m
, 0.654 m/s'
~x = ~ a,I' = & (0.654 m/s')(2.00 sf =
1.31
In
.....
....................
~
Remarks Note thal the constant app! ied force of' 20.0 N is assumed to act on the system at all times during its motion. If the force were removed at some instant, the system would continue to move with constant velocity and hence zero acceleration. The rollers have an effeClthat was neglected, here.
QUESTION 4.7 What effect docs doubling the weight have on the acceleration and thc displacement?
EXERCISE 4.7 A man pulls a 50.0-kg box horizont.ally from rest \,,'hile exerting a constant horizontal force, displacing the box 3.00 min 2.00 s. Find the force the man exerts on the box. (Ignore friClion.)
Answer
75.0 N
EXAMPLE 4.8 The Runaway Car Goal
Apply the secondlav.. and kinemat.ic eqUal ions to a problem involving an object moving 011 an incline.
Problem (a) A car or mass '!II is on an icy driveway inclined at an angle tJ = 20.0°, as in Figure 4.16£1. Determine the acceleration of the car, assuming the incline is frictionless. (b) If the length of the drireway is 25.0 In and the car starts from rest at the top, how long does it take to travel to the bottom? (c) What is the car's speed at the botLOm? Strategy Choose tilted coordinates as in Figure 4.16b so thal the normal force Ii is in the positive y-dircction, perpendicular to the drive\vay, and the positive ~'-axis is down the slope. The force of gravity Fg then has all x-component, mgsin e, and ay-component, -mKcos e. The components of Newton's second law form a system of two equations and two unknowns for the accel- FIGURE 4.16 eration down thc slope, {Ix, and the normal force. Parts (b) and (c) can be solved with the kinematics equations.
(a)
(Ex;unplc 4.8)
4.5
Applications of Newton's Laws
.
.
~
99
...,
Solution (a) Find the acceleration of the car. -'>
,,~
Appl)' Newton's second law:
ma
L..;F
Extract the x- and y-components from the second law:
(1)
max =
(2)
0 =
Divide Equation (J) by
1n
and substitute the given values:
a.., = gsin
---+
=
Fg
+
L F.l: = L 1~ =
e=
-'>
n
tllgsin
(J
-1Ilgcos
e + II
(9.80 m/s') sin 20.0° = 3.35 mis'
(b) Find the time taken for the car to reach the boltom.
Use Equation 3.L1b (or displacement, with
v(h
= 0: 1=
3.86 s
(c) Find the speed of the car at the bottom of the driveway. Use Equation 3.11a for velocity, again with vox = 0:
'\ = a) =
(3.35 m/s')(3.86 s) =
12.9 m/s
.
~
•••• ...A
)Jotice that the final answer for the acceleration depends only on gand the angle (J, not the mass. Equation (2), which gives the normal force, isn't useful here, but is essential \.. . hen friction plays a role.
Remarks
QUESTION 4.8 H the car is parked on a more gentle slope, how will the time required for it to slide to the bottom of the hill be affected? Explain. EXERCISE 4.8 (a) Suppose a hockey puck slides clown a frictionless ramp with an acceleration of 5.00 mjs2. What angle docs the ramp make \vith respect to the horizontal? (b) If the ramp has a length of 6.00 m, how long does it take the puck to reach the bottom? (c) Now suppose the mass of the puck is doubled. ',\'hat's the puck's new acceleration down the ramp? Answer
(a) 30.7° (b) J .55 s (c) unchanged, 5.00 mis'
EXAMPLE 4,9
Weighing a Fish in an Elevator
Goal Explore the eHect of acceleration on the apparent weight oran object.
A man \veighs a fish with a spring scale auached to the ceiling oran elevator, as shmvn in Figure 4.17 15, Two objecr.5 are cunnected by a string that passes over a frictionless pulley as in Active Figure 4.18, where IfI j < 1/1'2. and tll and at are the respective magnitudes
of" the accelerations. ·Which mathematical statement is true concerning the magnitude of the acceleration a':!. 01" lllass m,} (a) fl'2 < g (b) fl'";l > g (c) tt.1 = g (d) a? < a l (c)
a'l
>
til
16. An object or mass m undergoes an acceleration a down a rough incline. \-Vhich of the following forces should ltol appear in t.he free-bod)' diagram for the object? Choose all correct answers. (a) the force of gravity (b) (c) rhe normal force of the incline on the objcct (d) the force of friction down the incline (c) the force of friction up the incline (f) lhe force of the object on the incline
ma
• 1. A ball is held in a person's hand. (a) Identify all the external forces acting on the ball and rhe reaclion to each. (b) lfthe ball is dropped, ,,"·hat force is exerted on it while it is l~llling? Identify the reaction forcl: in this case. (:-.Jeglect air resistance.) 2.lf gold were suld by weight, would ),ou rather buy it in Denver or in Death Valle)'? If it were sold by mass, in which of the I\VO locations would you prefer to buy it? '",'hy?
8. Analyi'.e the motion of a rock dropped in water in terms of its speed and acceleration as it falls. Assume a resistive force is acting on the rock that increases as the vclocil}' of the rock increases. 9. In the moL ion picture It I-JajJjJenn! Unp Night (Columbia Pictures, ]934), Clark Gable is st
SECTION 4.1 FORCES SECTION 4.2 NEWTON'S FIRST LAW SECTION 4.3 NEWTON'S SECOND LAW SECTION 4.4 NEWTON'S THIRD LAW 1. The heaviest invertebrate is the giant squid. \."hich is estimated to have a weight of aboul 2 lons spread oul over its length of70 fecI. What is its weight in newtons? 2. A foot hall pUlller accelerales a football from rest to a speed of ]() m/s during the lime in which his toe is in contact with the ball (abOllt 0.20 s). If the football has a mass of 0.50 kg, whal average force does the punter exert on the ball? 3. A 6.0-kg object undergoes an acceleration of 2.0 l11/s:1. (a) What is the magnitude of the resilltant force acting on it? (b) If th is same force is applied to a ~1.0-kg object, whal acceleration is produced? 4.
10. A 5.0-g bullet IC FIGURE P4.31
32.
113
object with mass 1111 = 5.00 kg rests on a frictionless hQl-i/.olltal table and is connected to a cable that passes O\'er a pulley and is then fastened to a hanging object with mass Ill',!, = 10.0 kg, as showll in Figure P4.36. Find the acceleration of each ol~ject and the t.ension in the Glblc. All
FIGURE P4.36 36, 82. A flre helicopter carries a 620-kg bucket or water at the end of a 20.0-m-long cable. Flying back from a fire at a cOnstanl speed of 40.0 m/s, the cable makes an angle of 40.0° with respect to the venical. Determine the force exerted by air resist;ame in all four case~. -
EXAMPLE 5.1 Goal
Sledding Through the Yukon
Apply (he basic definitions of work done by a constant force.
Problem An Eskimo returning [rom a successful nshing trip pulls a slcd loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of' 1.20 X I02N on the sled by pulling on the rope. (a) How much ,...' ork does he clo on the sled if the rope is horizontal 1.0 the ground (8 = 0° in Fig. 5.6) and he pulls the sled 5.00 Ill? (b) Ho\\' much work docs he do on the sled if 0 = 30.0° and he pulls the sled (he same distance? (Treat the sled as a point particle, so details such as the point of all.achmenr of the rope make no difference.) Strategy Substitute the g'iven values of Fand dx into the basic equations for work, Equations 5.\ and 5.2.
FIGURE 5.6 (Examples 5.1 anel 5.2) An Eskimo pulling a sled with a rope 011 an angle 0 to the horizolll;J1,
122
Chapter 5
Energy
,.. ..
,
~
Solution (a) Find the work clone \","hen the force is horizolllal.
W = F!lx = (1.20 X 10 2 N)(5.00
Use Equation 5.1, substituting the given values: (b) Find the work done \-vhen the force is excncd 30 0 angle.
at
Ill) =
6.00 X lO'J
a
Use Equalion 5.2, again substituting the given values:
Remarks The normal force fl, the gravitational force mg, and the upward cOl1lponcm of the applied force do no work on the sled because they're perpendicular La the clisplaccmcm. The mass of the sled didn't come into play here, but it is imporlant when the effects of friClion must be calculated and in the next section, where we introduce [he work-energy theorem.
EXERCISE 5.1 Suppose the Eskimo is pushing the same 50.0-kg sled across level tcrrain with a force of 50.0 N. (a) If he does 4.00 X 10 2 J of work on the sled while exerting the force horizontally, through what distance must he ha\'e pushed it? (b) If he exerts the same force at an angle 0("45.0° with respect to the horizontal and moves the sled through the same distancc, hO\\' much \\lork does he do on the sled?
QUESTION 5.1 How does the answer [or the \v'ork done by the applied force change if the load is doubled? Explain.
Answers
(a) 8.00
III
(b) 283 J
Work and Dissipative Forces :.;. .~
:5
(h)
The edge of,
= -5.88
0
= -5.88
X
10"J
Chapter 5
130
Energy
(c) Repeal the problem, if y = 0 two meters above @. Find PF.i, the potential energy at ®: Find PEl' the potential energy at
PE;
®:
= mgy, =
(60.0 kg)(9.80 111/5')(8.00
Ill) ~
4.70 X 10 3 J
PEf = "'{Di = (60.0 kg)(9.8 l11/s')(-2.00 m) = -1.18 X 10 3 J
Compute the change in potential energy:
PEf - PEi = -1.18 X J03J - 4.70 X 10'J
-5.88 X 10'J .
~
......... -
~
Remarks These calculations show that the change in the gravitational potential energy when the skier goes from the top of the slope LO the bottom is -5.88 X 10 3 .1, regardless oj 111,,, zeTO Itrvel sdecled. QUESTION 5.4 If the angle of the slope is increased, docs the change of gravitational potential energy between L\...·o heights (a) increase, (b) decrease, (c) remain lhe same? EXERCISE 5.4 1f the zero level for gravitational potelltial energy is selected to be midway dO\vn the slope, 5.00 m above point ®, find the initial potential energy. the final potential energy, and the change in potential energy as the skier goes rrom point ® to ® ill Figure 5.]4·.
Answer
2.94 kj, -2.94 kJ, -5.88 kJ
Gravity and the Conservation of Mechanical Energy
TIP 5.4 Conservotion Principles There arc many conservation laws like the conservation of mechanical energy ill isohtl.ed systems, as in Equation 5.13. For example, tllOmt'lllllm, angular momentum, and ckClric charge arc all conser\'ed qU spct.'c1 at the bottom depends Dill}' on til t.he mass 11/, accclcriltion a, and time I. Strategy The power is provided by t.he engine. which creates a nonconscrvative force. L:se the \~'ork-energ-)' ,... .
theorem together wilh Ihe work done by the engine, WFIlj.;illC' and the work done by the drag force, \r\~lr//--T------
SECTION 5.4 SPRING POTENTIAL ENERGY 19. Find the height from which you would have
a ball so that it would have a speed of9.0 m/sjust bcfore it hits the ground.
h
l
to drop
20. \-\,'hen a 2.50-kg oqjcct is hllng vCl'lically all a ccrlain light spring described by Hooke's law, lhc spring st.retches 2.76 cnl. (a) What is the force constant of the spring? (b) If the 2.50-kg object is removed, how far will the spring streIch if a 1.25-kg block is hung on it? (c) I-low much work must an external agellt do LO stretch the same spring 8.00 crn from its tlllslrelchcd position? 21. An accelerometer ill a control system consists of a 3.65-g objeCl sliding on a horizontal rail. A low-mass spring is connected between the o~jecl and a llange at one end of the rail. Grease on the rail makes SLllic rriclion ncgligible, bm rapidl}' d'imps out vibrations of the sliding
26. Truck suspcnsions oft.en have "helper springs" that engage at high loads. One such arrangement is a leafspring with a helper coil spring mounled on the axle, as shown in Figure P5.26. When the main leaf spring is compresscd by distance )'0' the helper spring engages and then helps 10 support any additional load. Suppose the leaf spring constant is 5.25 X 10" N/m, the helper spring constant is 3.60 X 105 N/m, and Yo = 0.500 m. (a) \\'hat is the compression of the leaf' spring ror a load of !).OO X ]O~, N? (b) How much work is done in comprc~sing the springs?
Problems
155
the spring and oscillaLes freely on a horizontal, frictionless surface as in Active Figure 5.20. The initial goal of this problem is to find the \'clocit), at the equilibrium poinl after the block is released. (a) What objects constitute the system, and through what forces do they interaCL? (b) What are the t\.. .o points of interest? (c) Find the energy stored in the spring when the mass is stretched 6.00 cm from equilibriulll and again when the mass passes through equilibrium after being released from rest. (d) \Vritc the conservation of energy equation for this situation and solve il for the speed of the mass as it passes equilibrium. Substitme to obtain a numerical value. (e)\,\fhat is the speed at the halfway point? \'\'hy isn't it half the speed at equilibrium? 1.0
FIGURE PS.26
27.11 The chin-up is one exercise tbat can be Llsed Lo strengthen the biceps muscle. This muscle can exert a force of approximately 800 N as it comraClS a distance of 7.5 em in a 75-kg malc. 3 I-Jow much work can the biceps muscles (one in each arm) perform in a single contraclion? Compare this amount or work '\lith the cncl"!D' required to lift a 75-kg person 40cm in performing a chin-up. Do you think the biceps muscle is the only muscle involved in pedorming a chin-up? 28. " A nea is able to jurnp about 0.5 m. It has been said th;:H if a nea ,vere as big as a human, it would be able Lo jump over a lOO-sLOry building! When an animal jumps, it convens lVork done in contracting llluscles into gravitational potential energy (with sotTle steps in beLween). The maximum force exerted by a muscle is proportional LO its cross~secLional area, and the work clone by the Illuscle is Ihis force times the length ofcomraction.lrwc m:lgnificd a nea by a factor of I 000, the cross section of its muscle would increase by I 000 2 and the length 01" contraClion would increase by I 000. How high would this ~supernea be able to jump? (Don't forget thm the 111,-155 of the "superflea" increases as well.) 129. )A 50.0-kg projectile is fired al an angle of 30.0° above the horizontal with an initial speed of 1.20 X 10 2 m/s fwm the LOp of a cliff 142 III above level ground, where Lhe Rround is taken to be)' = O. (a) What is the initial LOlal mechanical energy of the projectile? (b) Suppose the projectile is traveling 85.0 m/s at its maximum height of ')' = 427 m. How much work has been dOllc on I he projec\ ile by air friClion? (c) What is the speed of t he projectilc immediately before it hits the ground if air friction does one and a half times as much work on lhc projectile when it is going down as it did when it was going up?
30.
31.
~G.
mil
A prOjectile of mass '11/ is fired horizontally with an initial speed of lJ u from a height of 11 ;Lbove a l1at, desert surface. Neglecting air friction, at the installt befc)I"c the prqjeclile hits the ground, find the following in terms of 111, vo, 11., and g: (a) the work clone by the force of gravily on lhe prqjeClile, (b) the change in kinetic energy of the projectile since it. was fired, and (c) the nnal kinetic cllergy of the projectile. (d) Are any of the answers changed if Lhe iniLial angle is changed?
m
A horizontal spring attached to a wall has (\ force constant of 850 N/m. A block of mass 1.00 kg is attached
P. I'app;\s cl. aI., "Nolluniform shortcnillg: in lht: biccp~ brachii duringdb,)\\' llexi{ll1," .l0llnwl uf A1J/Jfird J'ily~-101{)gJ 92, 2:l81. :!OO;l.
SECTION 5.5 SYSTEMS AND ENERGY CONSERVATION 32. A50-kg pole val/her running at 10 mf,s vaults over the bar. Her speed when she is abovc the bar is 1.0 ntis. NeglecL air resistance, as well as any energy absorbed by the pole, and determine her ,dtilUdc as she crosses the bar. 33. A child and a sled with a combined mass of 50.0 kg slide down a l'riClionless slope. If the sled starts from reSL and has a speed of 3.00 m/s at the boltom, what is the height of the hill?
34. Hooke's law describes a certain lighL spring of un Siretched 1cnRth 35.0 em. When one end is attached to the top ora door frame and a 7.50-kg object is hung from the other end, the length of the spril1g is 41.5 em. (a) Find its spring constant. (b) The load and the spring are takcn down. Two people pull in opposite dircctions on the ends of the spring, each with a force of 190 N. Find the length of the spring in this situation.
35.
am A
0.250-kg block along a horil.ontal lrack has a speed of 1.50 m/5 immediately before colliding with a light spring of force const.ant 4.60 N/m located at the end of' the track. (a) What is the spring's maximum COJll~ pression if the track is frictionless? (b) If the Lrack is 1101 frictionless, ,,'auld the spring's maximum compression be greater than, less tholl, or equal 10 the value obtained in pan (a)?
137.ITarzan swings on a 30.0-m-long vine initially inclined at N/m 2 (28 Ib/in. 2 ). Death results in 50% of cases in which the \"hole-body impact pressure reaches 3.4 X 105 N/m 2 (50 Ib/in. 2 ). Armed with the data above, we can estimate the forces and accelerations in a typical car crash ~lI1d see how scat belts, air bags, and padded intcriors can reduce thc chance of death or seriolls injury in a collision. Consider a typical collision
APPLICATION "
Injury to Passengers in Car Collisions
166
Chapter 6
Momentum and Collisions
involving a 75-kg passenger not wearing a seal belt, lraveling at 27 m/s (60 mi/h) who comes La rest in about O.OJO s after striking an unpadded dashboard. Using r~,\.llt = 11/.Vj - 'fII,V j , we find that F
_ mv; -
o-
!/lUI
(75 kg)(27 m/,)
ill'
0.010 s
-2.0 X 10' N
and
nvl
a= -
Int
=
27
m/s .
0.010 s
_
.>
= 2 100 m/s- =
2700 mis' 2 g = 280r! 9.8 m/s b
Ifwe assume the passenger crashes into the dashboard and windshield so [hat the head and chest, with a combined surface area of 0.5 m 2 , experience the force, we find a whole-body pressure of
2.0 X 10' N _
.>
0.::' mF (in units of
IO~
N)
10
FIGURE 6.5
Force 011 a car versus time for a typical collision.
(a)
:;
4
X
,.>
10'> N/m-
'V'e see that the force, the acceleration, and the ,""hole-body pressure all exceed the threshold for hltality or broken bones and that an unprotectcd collision at60 mi/h is almost certainly fatal. ''''hat can be done to reduce or eliminate the chance of dying in a car crash? The most important factor is the collision time, or the time it takes the person to come to rest. If this time can be increased by 10 to 100 times the value of 0.01 s for a hard collision, the chances of survival in a car crash are much higher because the increase in 8.l makes the contact force 10 to 100 times smaller. Seat belts restrain people so that they come LO rest in about the same amOUlll of time it takes to SLOp tIle car, typically about 0.15 s. This increases the effective collision time by an order of magnitude. Figure 6 ..1 shows the measured force on a car versus time for a car crash. Air bags also increase the collision time, absorb energy from the body as they rapidly deflate, and spread the contact force over an area of the body of about 0.5 m 2 , preventing penetration \vounds and fractures. Air bags must deploy \'ery rapidly (in less than 10 ms) in order to SLOp a human traveling at 27 m/s before he or she comes LO rest against the steering column about 0.3 III away. To achi.eve lhis rapid deployrncnl, accelerometers send a signal LO discharge a bank of capacitors (devices that store electric charge), which then ignites an explosive, thereby filling the air bag with gas very quickly. The electrical charge for ignition is stored in capacitors to ensure that the air bag continues LO operate in the event of damage to the battery or the car's electrical system in a severe collision. The important reduction in potentially fatal forces, accelerations, and pressures to tolerable levels by the simultaneous use of seat belts and air bags is summarized as follows: H a 75-kg person traveling at 27 m/s is SLOpped by a seat belt in 0.15 s, the person experiences an average force org.s kN, an average acceleration of 18g, and a whole-budy pressure of 2.8 X 10" N/m 2 [or a contact. area of 0.5 m2 . These values are about one order of magnitude less than the values estimated earlier for an unprotected person and well below the thresholds for life-threatening injuries.
p
6.2 ++
(b)
FIGURE 6.6
(a) A coilisiull between two objects rt:sulting from direct
contacl. (b) A collision bt:twcCIl two
charged objects (in this case, a protoll and a hdiulil lluclcLI1;).
CONSERVATION OF MOMENTUM
"Vhen a collision occurs in an isolated system, the total momentum of the system docsn't change 'with the passage of time. Instead, it remains constant both in magniwde and in direction. The momenta or the individual objects in the system may change, but the \'eCLOr sum of all the momenta will not change. The tOtal momenWill is therefore said to be conserved. In this sectioll, Vle will see how the laws of motion lead LIS to this important conservation law. A collision may be the result of physical contact between two objects, as illustrated in figure 6.6a. This is a common macroscopic event, as when a pair of bil-
6.2
Conservation of Momentum
liard balls or a baseball clJ1d a bal strike each other. By contrast, because contact on a submicroscopic scale is hard 10 define accurately, the 1l00ioll of rollision must be generalized to that scale. Forccs betv..'ecll t\'..' o ol~jects arise from the electrostatic interaction of the electrons in the surface atoms of" the objects. As will be
Before collision
(al
disclissed in Chapter 15, electric charges arc eirhcT posilive or negalive. Charges
with the same sign repel each oLher, \v'hilc charges \vith opposite sign attract each other. To understand the distinction bctween macroscopic and microscopic collisions, consider the collision between two positive charges, as shown in Figure 6.6b. Because the two particles in the figure arc both positiyely chargcd, they repel each Olher. During such a microscopic collision, particles need not wllch in the normal sense in order to intcract and transfer 1TI01llCIllUIll. Activc figure 6.7 shm.. . s an isolated s)"stem of t\\-·o panicles before and a ["tel' they collide. B)' aisolaLecl," wc mean that no external forces, such as Ihe gravitational force or friction, act on the system. Before the collision. the n:locities of the two panicles are Vii and V~j; after the collision, the velocities are Vlj and v~f' The impulse-momenulm thcorem applied LO 1lI 1 bccomes ~
F~1ti{=
Like\.. . ise, for
1fI'!.,
we
167
After colli.'\ion
(hi ACTIVE FIGURE 6.7 Ikfore and af'll:r it head-Oil colli:.ion belween two objccts. Tht: momentum or each 1)1~{~n changes as a resull of lht: collision, bill the lOtal momentum of the system I"t'mains conSlant.
~
1lI l V lj -
1JI1V li
h~lve
Fj~til. = 1fI~V2f-
nl':!.V'l.I
F
where j\n is the 'cr receives a shot wilh the ball (0.0600 kg)
2
14. A O.SOO-kg football is thrown toward the cast with a speed of 15.0 m/s. A stalionary receiver catches the ball and brings it 10 rest in 0.0200 s. (a) What is thc impulse delivcred to the ball as it's caught? (b) What is the average force exerted 011 the receiver?
4
(,j
5
17. The forces shown in the force \IS. time diagram in Figure P6.17 act on a 1.5-kg particle. Find (a) the impulse for the interval from l = 01.0 l = 3.0 s p
Reference linc
(,,)
p f'
e
',s
o
Refel'CllCC !illl'
(hl FIGURE 7.2 (a) Tht' poilll POll a rutal illg compact d i1>C at I = O. (I» As the disc rotau.:s. Pmovcs thruugh all arc lcnglll s.
SI unit: Tadian (rad) For example, if a point on a disc is at 0; = 4 rad and rolates to angular posillon Of= 7 rad, the angular displacement is AO = Of- 0; = 7 rad - 4 rad =.:1 rad. Note that we use angular variables to describe 1he rotating disc because each point on the disc undergoes lhe same angular displacement in any given time interval. Using lhe definition in Equation 7.2, Equation 7.1 can be ,,,,rilten more generally as 6.0 = 6.s/I; where Lis is a displacement along the circular arc subtended by lhc angular displacement. I laving dclined angular displacemellts, it's nawral to de/ine an angular speed: The average angular speed W;\I' of a rotating rigid ol~jeCl during lhe time interv,ll Lil is the angular displacemelllLifJ divided by 6.(:
w"
er -(), '" - - = l; -
I,
u/'
,.
'. ",
- - - t ' - - ' - - - ' - - - - - - ., ()
M ~
J
[7.3]
FIGURE 7.3
I\>. a point all the
(ompan disc mon·.~ frOIJl @ 10 @.
Sf unit: radian per second (rad/s)
lIlt:: diK rotates through the angk fiO = OJ- 0...
192
Chapter 7
Rotational Motion and the Law of Gravity
Tip 7.1 Remember the Radian Equation 7.1 uses angles measured in radians. Angles expressed in terms of degrees must first be cQlwerLed to
radians. Also, be sure to check whether your calcillator is in degree or radian mode when soh"ing prohlems in\'oh'ing rotation.
For very short time intervals, the average angular speed approaches the instantaneous angular speed,just as in the linear casco
The instantaneous angular speed w of a rotating rigid object is the limit of the average speed 6.8/lJ.l as the time interval fj.f, approaches zero: w
= lim .:)./ -Jo
t;e
0
[7.4]
I:1l
SI unit: radian per second (rad/s) \·Ve take w LO be positive \ovhen 8 is increasing (couTlLCrclock\.,rise motion) and negative ,..,hen e is decreasing (clockwise Illotion). \'Vhcn the angular speed is constant, the instantaneous angular speed is equal to the average angular speed.
EXAMPLE 7.1 Goal
Whirlybirds
Convert an angular speed in revolutions per minute to radians per second.
Problem The rotor on a helicopter llIrns at an angular speed of3.20 X 10 2 revolutions per minute. (In this book, we sometimes use the abbreviation rprn, but in most cases we use rev/min.) (a) Express this angular speed in radians per second. (b) If the rotor has a radius of 2.00 Ill, what arclength does the tip or the blade trace out in 3.00 X 102 s?
Strategy During one revolution, the rotor turns through an angle or 21T radians. Use this relationship as a conversion factor.
... ...,
'-' ...
Solution (a) Express this angular speed in radians per second.
Apply the conversion factors] rev = 21T rae! and 60 s = I min:
rev w = 3.20 X 10'-min = 3.20 X 10' loeJ;eed v. Even though the car moves at a constant speed, it still has an acceleration. To understand this, consider the defining equation for ;wcrage acceleration: ~
---t
aav
=
vr
-
if
~
(a)
~
Vi
[7.12] /i
v,
vI
The numerator represents the difference bCl\\'een the velocity vectors and Vi' These vectors may have the same magnilude, corresponding to the same speed, but if t.hey have different direrlimu, their dirference can't equal zero. The direc[joll or the car's velocity as il moves in I he cil-culal' path is continually changing, as shown in Figure 7.6b. For circular motion at constant speed, the acceleration vector always points toward the center of the circle. Such an acceleration is called a centripetal (centcr-seeking) acceleration. Its magnil.llde is givcn by
., ,r .,.
(f=(
[7.13]
To derive Equmion 7.\3, consider Figure 7.7', the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure 8.5b illustrates the fact that the component of the force perpendicular LO lhe lever arm causes the torque. QUESTION 8.2 To make the wedge more cf'fective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? EXERCISE 8.2 A man lies one end ora strong rope 8.00 m long to the bumper of his truck, 0.500 m from the ground, and the other end to a vertical tree trunk at a height of 3.00 m. He lIses the truck to create a tension of 8.00 X 101 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point.
Answer
2.28 X 10 3 N· In
8.2
TORQUE AND THE TWO CONDITIONS FOR EQUILIBRIUM
An object in mechanical equilibrium must satisfy the follmving t\-vo conditions: 1. The net external force must be zero:
2. The net external torque must be zero:
2: F =
0
L -:; =
°
The fi rst condition is a statement of lranslaLional equilibrium: The sum of all forces acting on the object rnust be zero, so the object has no translational acceleration, ~ 1" = O. The second condition is a statement of rotational equilibrium: The sum of --'W'tl ~ all torques on the object must be zero, so the object has no angular acceleration, ~._..::~, ~ = O. For an object to be in equilibriulll, it must both translate and rotate at a constant rate. This large balanced rock :lllhe Because we can choose any location for calculating tQHlues, it's usually best to Garden of the Gods in Colorado Springs. Colorado, is in mechanical select an axis thal will make at least one torque equal to zero,justLO simplify the equilibriulll. net torque equation.
a
EXAMPLE 8.3
Balancing Act
Goal Apply the conditions of equilibrium and illustrate the use of different axes for calculating the netLOrque on an object. Problem A woman of mass m. = 55.0 kg sits on the left end of a seesaw-a plank of length L = 4.00 Ill, pivoted in the middle as in Figure 8.6. (a) first compute the torques on the seesmv abom an axis that passes through the pivol. point. ''''here should a man of mass 1\1 = 75.0 kg sit if the system (seesaw pillS man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of 1n = 12,0 kg. (c) Repeat part (b), but1.his tirnc compute the p1 torques ahout an axis through the left end of the plank. Strategy Refer La Figure 8.6b. In parI. (a), apply the second condition of equilibrium, :£7 = 0, computing tOrques around the pivot point. The mass of the plank forming the seesav·,' is distribUled evenly on either side of the pivot point, so t.he torque exerted by gravity on the plan k, 7Kravil}" can be computed as if all the plank's mass is concentrated at. the piVOl point. Then Tgr,wily is zero, as is the torque exerted by the pivot. because their lever arms are zero. In part (b) the first
(a)
I.
-x
2.00 rn
(b)
FIGURE 8.6 (a) (Example R,3) T\,·o people on a seesaw. (b) Frcc body diagram for the plank.
8.2
Torque and the Two Conditions for Equilibrium
233
condition of equilibrium, IF = 0, must be applied. Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer. .
~
Solution (a) Where should the man sit
[Q
• • • • • • • • • • • • • • • • • • • • '''11
balance the seesaw?
Apply the second condition ofequilibriulll to the plank by sClling the sum of the torques equalLO zero: Thc [-Irst two torques are zero. Let x represent. the man's distance from the pivol. The woman is at a distance L/2 from the pivot.
o+ 0 -
Solve this equation for x and evaluate it:
(b) Find thc normal force seeS
50.0:\ = 0
I=
P
->
11
= 50.0 N
Apply the second condition of equilibrium, computing torques around the base of the ladder, with Tgr'l\' standing for the torque due to the ladder's 50.0-0.' \-veight: The torques due to friction and the normal force are zero about 0 because their moment arms are zero. (tv!oment arms can be found from Fig. 8']3c.)
0+ 0 -(50.0 N)(5.00 m) sin 40.0 0
From Equation (l), we now have I= P= 21.0 N. The ladder is on the verge of slipping, so write an cxpressian for the maximum force of static friction and solve for fL,:
21.0 N =
P~
21.0
+ P(IO.O m)
sin 50.0 0 = 0
~
I= I..",,, =
1'-,11
= 1'-,(50.0 N)
21.0N 1'-, = -.- - = 0.420 ~O.O N
~
~
_
P
Remarks Note that torques \-"ere computcd around an axis through the bottom of the ladder so that only and the force of gravity contributcd nonzero torques. This choice of axis reduces the complexity of the torque equation, often resulting in an equation with only onc unknown. QUESTION 8.8 If a 50.0 N monkey hangs from the middle rung, \'.,'Quld the coefficient of static friction be (a) doubled, (b) halved, or (c) ullchangcd? EXERCISE 8.8 If thc coefficient of static friction is 0.360, and the same ladder makes a 60.0° angle \vith respect to the horizontal, how far along the length of the ladder can a 70.0-kg painter climb before the ladder begins to slip?
Answer
8.5
6.33 m
RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION
\,Vhen a rigid object is subject to a net torque, it undergoes an angular acceleration that is directly proportional to the net torque. This result, \..: hich is analogous to NeWlon's sccond law, is derived as follows.
240
Chapter 8
Rotational Equilibrium and Rotational Dynamics
F,
"
,
,... ... ----
--"
The system shown in Figure 8.14 consists of an object of mass 111 connected to a very light rod of length r. The rod is pivoled at the point 0, and its movement is confined to rotation 011 a frictionless horizort{rd table. Assume that a force J~ acts perpendicular La the rod and hence is tangent La the circular path of the object. Because there is 110 force tu up pose this tangential force, the ubject ulHlergocs a tangential acceleration {l, in accordance with Nevy·ton's second law:
m.
'\ O~r,
,".. ......... __ .- ... ..."
"
FI = rna l
Multiply both sides or th is equation by r: FIGURE 8.14 An object of Illass 11/ atlached 10 a 11ghl rod ofkngth r moves ill a circular path 011 a frictionless horizontal surface while "langential force F, acts 011 it.
Fir =
"UTl t
Substituting the equation a, = -ra relating tangential and angular acceleration into the above expression gives [8.4]
The lert side of Equation 8.4 is the torque acting on the object about its axis of rotation, so we can rewrite it as T=
" mr-a
[8.5]
Equation 8.5 shows that the torque on the object is proportional to the angular acceleration of the object, \vhere the constant of proportionality mr 2 is called the moment of inertia of the object of mass m. (Because the rod is very light, its moment of inertia can be neglected.)
QUICK QUIZ 8.1
Using a screwdriver, you try to remove a sere'.. .· from a piece of furniture, but can't get it to turn. To increase the chances of success, you should use a screwdriver that (a) is longer, (b) is shorter, (c) has a narrower handle, or (d) has a wider handle.
Torque on a Rotating Object Consider a solid disk rotating about its axis as in Figure 8.15a. The disk consists of many particles at various distances from the axis of rotation. (See Fig. 8.15b.) The torque on each onc of these panicles is given by Equation 8.5. The net torque on the disk is given by the sum of the individual torques on all the particles:
2;, = (2;",)"2)",
[8.6)
Because the disk is rigid, all of its panicles have the same angular acceleration, so a is not involved in [he SUIll. If the masses and distances of the panicles are labeled with subscripts as in Figure 8.15b, then 2:1nr 2
= In(r]2
+
1n 2T2'2
+
1I/.:3r3 2
+ ..
This quantity is the moment of inertia, I, of lhe whole body: Moment of inertia
[8.7]
~
FIGURE 8.15 (n) A solid disk rotating abolll its axis. (b) The disk consists of man~' panicles. all with the same angular acceleration,
'0 =
[8.8]
Equation 8.8 says that the angular acceleration of an extended rigid object is proportional to the net torque acting on it. This equation is the rotational analog of Newton's second Ia,..· of motion, with torque replacing force, moment of inertia replacing mass, and angular accelerat.ion replacing linear acceleration. Although the moment of inertia of an object is related to its mass, there is an important dif~ ference bet,veen thern. The mass m depends only 011 the quantity of matter in an object, ,vhercas the moment of inertia, I. depends on both the quantity of rnatter and its distribulion (through the /1. term in 1= 17111';1) in the rigid object.
QUICK QUIZ 8.2
A COIlS[antnet torque is applied to an object. vVhich one of the following will not be constant? (a) angular acceleration, (b) angular velocity, (c) moment of inertia, or (d) center of gravity. QUICK QUIZ 8.3 The two rigid objects shown in Figure 8.16 have the same mass, radius, and angular speed. If the same braking torque is applied to each, lvhich takes longer 1.0 stop? (a) A (b) B (c) more information is needed The gear system on a bicycle provides an easily visible example of I.he relalionship between torque and angular acceleration. Consider first a five-speed gear system in which the drive chain can be adjusted 1O wrap around any of five gears attached lo the back wheel (Fig. 8.17). The gears, vvilh different radii, arc concentric with the wheel hub. ''''Then the cyclist begins pedaling from rest, the chain is auachedto the largest gear. Because il has the largest radius, lhis gear provides lhe largestlorquc 1.0 the drive ~vheel. A large torque is required initially, because the bicycle starts from rest. As the bicycle rolls faster, the tangential speed of the chain increases, eventually becolTling too fast for the cyclist to maintain by pushing the pedals. The chain is then moved 1O a gear with a smaller radius, so the chain has a smaller tan~ genl.ial speed lhat the cyclist can more easily maintain. This gear doesn't provide as much torque as the first, but the cyclist needs to accelerate only [Q a somewhal higher speed. This process continues as the bicycle moves faster and faster and the cyciistshiflS through all five gears. The fifth gear supplies the Im...est torque, bUlllm.. · the main function of that torque is to counter the frictional torque from the rolling tires, v.'hich tends to reduce lhe speed or the bicycle. The small radius of the fifth gear allows dle cyciistLO keep lip with the chain's movement by pushing the pedals. A 15~speed bicycle has the same gear structure on the drive ""heel, but has three gears on the sprocket. connected to the pedals. By combining differenl positions of the chain 011 the rear gears and the sprockct gears, 15 different lorques are available.
~ Rotational analog af Newton's
sp.cono low
A
B
FIGURE 8.16
(Quick Quiz 8.3)
APPLICATION Bicycle Gears
FIGURE 8.17 The drive wheel and gears of a bicyc:1e.
More on the Moment of Inertia As wc havc seen. a small object (or a particle) has a moment of inenia equal to mr 2 about some axis. The moment of inenia of a com/Josile ol~ject about some axis is just the sum or the moments of inenia of thc object·s components. For example, suppose a Illno. =
mg
=
~
fa to the cylinder in Figure 8.23c:
Notice the angular acceleration is c1ocbvise, so the torque is negative. The normal and gravity forces have zero moment arm and don't contribute any torque. Solve for T and substitute a = a/ R (notice that both and a are negati\"e):
(2)
I,"
-TR = ~!VIR'a
Ct'
Substitute the expression for Tin Equmion (3) inw Equation (1), and solve for the acceleration:
rna = - mg - ~NTa
Substitute the values for m, l\JI, and g, gelling substitute a inLO EquaLion (3) to get T:
a = -5.60 mis'
(l,
(solid cylinder)
then
a=
T=
mg
m + ttv!
8.40 N
(b) Find the distance the bucket falls in 3.00 s. - H5.60 m/s')(3.00 sF
Apply the disphlcemellL kinematic equation for constant acceleration, \-"ith 1= 3.00 sand V o = 0: trrr... • • • • • • • •
Remarks Proper handling of signs is very important in these problems. All such signs should be chosen ini~ tially and checked mathemaLically and physically. In this problem, for example, both the angular acceleration (l' and the acceleration a are negative, so a = tt/I?. applies. If the rope had been wound the other wayan the cylinder, causing coul1lerclock\'.:ise rotation, the wrque woule! have been positive, and the relationship would have been a = -a/ R, with the double negative making the righthand side positive, just like the len-hand side.
EXERCISE 8.11 A hollow cylinder of Illass 0.100 kg and radius 4.00 cm has a string wrapped several times around it, as in Figure 8.23d. J f the string is attached to a rigid support and the cylinder allowed La drop from rest, find (a) the acceleration of the cylinder and (b) the speed of the cylinder ,"vhen a meter of string has ullwound off of it. Answers
(a) -4.90 mis'
(b) 3.13 mls
QUESTION 8.11 How would the acceleration and tension change if most of the reel's mass were at its rim?
8.6
ROTATIONAL KINETIC ENERGY
In Chapter 5 we defined the kinetic energy of a particle moving through space with a speed vas the quantity !mv 2. Analogollsly, an o~iect rotating about some axis with an angular speed w has rotational kinetic energy given bytlw2 . To prove this, consider an object in the shape of a thin rigid plate rotating around some axis perpendicular to its plane, as in Figure 8.24. The plate consists ormany small particles, each of Illass m. All these particles rotate in circular paths around the
8.6
Rotational Kinetic Energy
axis. If r is the distance of one of the particles from Ihe axis of rotation, the speed of that particle is v = rw. Because the total kinetic energy of the plate's rotation i., the slim of all the kinetic energies Clssociated with it'i particles, we have £..J ra . "'(' ~,-=
') = "'('
~mw-
~
'1 ") ~ml·w-
247
wt
t-=o
[8.16]
Not.e that conservation of angular momentum applies to macroscopic objects such as planets and people, as '.. .ell as to atoms and molecules. Thcre arc many examples or conservation of angular momentum; one of the most dramatic is that of a flgure skater spinning in the finale of her act. In Figure 8.28a, the skater has pulled her arms and legs close to her body, reducing their distance from her axis of rotation and hence also reducing her moment of inertia. By conservation of angular momentum, a reduction in her moment of inerlia must increase her angular velocity. Coming out of the spin in Figure 8.28b, she needs to reduce her ang·ular velocity, so she extends her arms and legs again, increasing her moment of inertia and t.hereby slowi ng her rotation. Similarly, \-"hen a diver or an acrobat 'wishes to make several somersaults, she pulls her hands and feet close to the trunk of her body in order LO rotate at a greater angular speed. 1n this case, the cxternal force clue to gravit)' acts through her center of gravity and hence exerts no torque about her axis or rotation, so the angular momentum about her center of gravity is conserved. For example, when a diver wishes to double her angular speed, she musl reduce her mOfnellt of inertia to half its initial value. An interesting astrophysical example of conservation of angular momentum occurs when a massive star, at the end of its lifetime, uses up all its fuel and collapses under the influence of gravitational forces, causing a gigantic outburst of energy called a supernova. The best-studied example of a remnant. of a supernova explosion is the Crab Nebula, a chaotic, expanding mass of gas (Fig. 8.29). In a
FIGURE 8.28 Michelle Kwau controls her momcnt orincnia. (a) By pulling in her arms anrl legs, she reduces her momellt ofillcrtia and increases her angular velocity (rate of spi 11). (b) Upon lallding, eXlending her arms and legs increases hel 1ll0rnenl orillerlia and helps sial\' her spill
TighTly curling her body, a diver decreast:s her moment of inertia, increasing her <Jllgular vc1ocilY.
(bl
8.7
"
~
] .::
£ ~ ~
(a)
(b)
Angular Momentum
251
FIGURE 8.29 (a) The Crab Nebula in lhe constellation Tam·us. This neblda is llw remnant ora supernova secn on Earlh in A.D. 1054. Ir is 10cHed some 6 300 liKht-years away and is approximately 6lighl-ycars in diameter, still expanding outward. A pubarclccp inside lhe nebula (]asllf>s 30 times every second. (b) Pulsar orr (c) Pulsar on.
(c)
supernova, part of the star's mass is ejected into space, where il even wally condenses into new stars and planets. Most of what is left behind typiGdly collapses into a neutron star-an extremely dense sphere ofmal.ler with a diameter of about 10 k m, greatly reduced from the 10"-km diameter of the origi nal slar and containing a large fraction of the star's original mass. [n a neutron star, pressures become so great that atomic electrons combine with proLOns, becoming neutrons. As the moment of inertia of the system decreases during the collapse, the star's rotational speed increases. More than 700 rapidly rotating neutron stars have been identified sillce their first discovery in 1967, with periods of rotation ranging from a millisecond to several seconds. The neutron star is an amazing- system-an o~ieCl with a mass greater than the Sun, fitting com forI ably ,vithin the space of a small county and rotating so fast that the tangential speed of the surface approaches a sizable fraction of the speed of light!
APPLICATION Rotating Neutron Stars
QUICK QUIZ 8.5 A horizontal disk with moment of" inertia I[ rotates with angular speed w[ about a venical frictionless axle. A second horizontal disk having moment of inertia 12 drops onl0 the first, initially not rotating but sharing the same axis as the first disk. Because their surfaces are rough, the two disks evenLUall>' reach the same:.: angular speed UJ. The ratio w/w l is eqllal to (a) '/1, (b) J,lI, (c) 1,/(/, + '2) (d) 12/(/, + '2) QUICK QUIZ 8.6 [[global \-I.'arming continues, it's likely that some ice from the polar ice caps of the Earth will melt. and the water will be distributed closer to the equator. If th is occurs, would the length of I he day (one revolution) (a) increase, (b) decrease, or (c) remain the same?
EXAMPLE 8.14 The S innin Stool Goal
Apply conservation oranguhtl-momentlll11 to a simple system.
Problem A student sits on a pivoted stool while holding a pair of weights. (See Fig. 8.30.) The sLOol is free to rOlate about. a vertical axis with negligible friction. The moment of inertia of sludent, weights, and stool is 2.25 kg ·m 2 . The student is set in rotation with arms out.stretched, making one completc turn every 1.26 s, arms outstretchcd. (a) \Vhat is the initial angular speed of the system? (b) As he rotates, he pulls the weights inward so that the new moment of inertia of the system (sLudent, objects, and stool) becomes 1.80 kg' m 4 . What is the new angular speed of Ihe system? (c) Find the work done by the swclent on the system while pulling in the \veights. (Ignore energy lost through dissipation in his muscles.)
Wr
I~ I
(a)
(b)
FIGURE 8.30 (Example H.14) (a) The sllldenl is g-ivcn all inilial angular spced while holding lWO
Strategy (a) The angular speed can be obtained frolll the frequency, weights OUl. (b) The ;lIlgU!;lf specrl increases as which is the inverse of the period. (b) There are no external lorques act- the student draws the weights inwards. ing on the system, so the new angular speed can be found with the principle of conservation of angular mOlllenlUm. (c) The work done on the system during this process is the same as the systern's change in rotational kinetic energy.
252
Chapter 8
Rototional Equilibrium and Rotational Dynamics
,.. ..
..................................
~
Solution (a) Find the initial angular speed of the system. Invert the period to gCllhe frequency, and multiply by 27T:
W, ~
2"'f= 2.,,/1'= 4.99 rad/s
(2)
(2.25 kg' m")(4.99 rad/s) = (1.80 kg' m')wj
(b) After he pulls the weights in, what's the system's new angular speed? Equate the initial and final angular momenta of the system:
Substitute and solve {or the Filla I angular speed
w/
w! =
6.24 rad/>
(c) Find the work the student docs on the system.
I'VSllIf!t:1H -"K-llw'-!lw" U r - 2 f j 2' I
Apply the ''lork-energy theorem:
= H1.80 kgm 2)(6.24 rad/s)2
- H2.25 kg· m 2 )( 4.99 rad/5)'
.....
.....
. . .. .. . . . . . . . . . . . . . . . . . .
Remarks Although the angular momentum the student does work on the system.
,
.
.....................
-.
- .......
or the syst.em is conserved, mechanical energy is not conserved because
QUESTION 8.14 If the student suddenly releases the weights, ,...·ill his angular speed increase, decrease, or remain the same? EXERCISE 8.14 A SLar "vith an initial radius of l.0 X l08 m ancl period of 30.0 days collapses suddenly to a radius or l.0 x 10 4 m. (a) Find the period ofrmation after collapse. (b) Find the work done by gravity during the collapse if the mass of the SLar is 2.0 X 10:{O kg. (c) "Vhat is the speed oran indestructible person standing on the equator of the collapsed star? (Neglect any rclativistic or thermal effects, and assume the star is spherical before and after it collapses.) Answers
(a) 2.6 X 10- 2 S
EXAMPLE 8.15
(b) 2.3 X 10.12 J
(c) 2.4 X 10' m/s
The Merry-Go-Round
Goal Apply conscl'vmian of angular momentum ,.. . hilc combining two moments ofinenia. Problem
A mcrry-go-round modeled as a disk of mass
J\t1 = 1.00 X 10 2 kg and radius R = 2.00 m is rotating in a
horizontal plane about a frictionless vertical axle (Fig. 8.31). (a) Arter a student with mass m = 60.0 kg jumps on the rim of the merry-go-round, the system's angular speed decreases La 2.00 rad/s. If Lhe studenL 'walks slowly from the edge toward the center, find the angultJr speed of the systern when she reaches a poim 0.500 m from the center. (b) Find the change in the system's rotational kinetic energy caused by her movement to the center. (c) find the work done on the slUdent as she walks to ,. = 0.500 111.
FIGURE 8.31 (Example 8.15) As the student walks tuward the cen1cr of the rotating platform, the 1fI01llCllLOf inertia of the ~~'stern (stlldcl'll plus plalform) decrcases. Because angular momentum is consen'cd, 1hc 'lng-IlIaI' speed of 111(; system must increase.
Strategy This problem can be solved with conservation of angular momentum by equating the system's initial angular momentum ,,,,,hen the student stands aLthe rim to the angular momentum when (he student has reached r = 0.500 m. The key is to fi nd the different moments of inertia.
8.7
,..-'
.
Angular Momentum
253
..........................................
~
Solution (a) Find the angular speed when the student reaches a point 0.500 111 from the centcr.
Iv = ~MR' = ~(1.00 X 10' kg)(2.00 m)'
Calculate the moment of inertia of the disk, Tf):
= 2.00 X 10' kg· m'
Calculate the initial momcnt of inertia of the s1.lldenl. This is the same as the moment ofincnia ora mass a distance R from the axis:
Is = mR' = (60.0 kg)(2.00 m)' = 2.40 X 10' kg· m'
Surn the t\VO moments of inertia and multiply by the initial angular speed to nnd L i , the initial angular momenturn of' the system:
Li = (I"
+
{,)w i
= (2.00 X 10' kg· m' = 8.80 X 10' kg·
mr/ =
Calculate the student's final moment of inertia, f v ' when she is 0.500 m from the center:
1.\/=
Thc mornent of inertia of the platform is unchanged. Add it to the student's Hnal moment of inertia, and multiply by the unknown final angular spced [Q nnd Li
/1= (If)
Equate the initial and final angular momenta and solve for the final angular speed of the system:
+
+ 2.40 X 10' kg· m')(2.00 rad/s)
m'js
(60.0 kg)(0.50 m)2 = J5.0 kg· m 2
ISJ}w/= (2.00 X 10 2 kg·m'!
+
]5.0 kg·m 2)wJ
L, = Lf
(8.80 X 10' kg· m'/s) ~ (2.15 X 10' kg· m')wf WI =
4.09 rad/s
(b) Find the change in the rotational kinetic energy of the system.
Calculate the initial kinetic energy of thc system:
KE,
=
~/,w,'
= H 4.40 X 10' kg
111
2
)(2.00 rad/s)'
= 8.80 X lO' J Calculate the final kinetic energy of the system:
KI~= ~/fw/ ~ H215 kg·m')(4.09 rad/s)' ~ 1.80 X 10'J
Calculate the change in kinetic energy of the system:
faj -
KE, =
920)
(c) Find the work done on the student. The sLUdentunclergoes a change in kinetic energy thal equals the work done on hcr. Apply thc work-l'nergy theorem:
HI - A K" -.u. L:.,tutal ionary stand, and a resistive force of 120 N is applie(1 tangent to the rim of the tire. (a) What force Illust be appliefl by a chain passing O\'er a 9.00-cl11diameter sprocket in order to give the wheel an acccleration of4.50 rad/s'2? (b) \",'hm force is rcquired if you shift to a 5.60-cm-diamctcr sprocket? 139,1..-\ 150-kg mcrq·g-o-round in the shape ofa uniform, solid. horizontal disk of radius 1.50 III is set in motion by wl'apping a rope abom the rim of the disk and pulling on the rope. \\'hat constant force must be exerted 011 the rope to bring the mcrq-go·round from resL to an angular speed of 0.500 rcv/s ill 2.00 s?
-to.
mil
An Atwood's machine consists of blocks of masses == 10.0 kg and t11 2 == 20.0 kg llttached b)' a cord run· ning- over a pulley as in Figure P8.40. The pulley is a solid cylinder wiLh mass M == 8.00 kg and radius r::: 0.200 m. The block of Illass 1/1../, is allowed 10 drop, and the cOHI 111 1
FIGURE .8,40
262
Chapter 8
Rotational Equilibrium and Rotational Dynamics
{lIrns the plllley Wilholll slipping. (a) Why must the len· sion T';! be greater than the tension T 1? (b) What is the
147.IA solid, uniform
disk of radius 0.250 m and mass 55.0 kg rolls down a ramp of lenglh 4.50 m that makes an angle of 15.0" with the horizontal. The disk stans from rest from the top of the ramp. Find (a) the speed of the disk's center of lIlass when it reaches the hOllom of [he I"amp and (b) the angular speed of the disk at the bOllom of the ramp.
acceleration of Ihe system, assuming the pulle)' axis is frictionless? (c) Find the tensions T 1 and 7"'1' 41. An airiinCl'lands \\,jth a speed of50.0 Ill/S. Each wheel of
the plane has a radius of 1.25 III and a moment of inertia of 110 kg' Ill:!, At LOllchdown, the wheels begin to spin under the action offriclion. Each wheel supports a weigh I of 1.10 X 10-1 N, and the wheels alLain their angulal' speed in 0.480 s while rolling without slipping. What is the coefficielll of kinetic friction between the wheels and the nlll\\'a~'? Assume thaI the speed orthe plane is constant.
SECTION 8.6 ROTATIONAL KINETIC ENERGY 42. A GIl" is designed LO get its energy from a rotating flywheel with a radius of 2.00 III and a mass of 500 kg. Before a trip. the flywheel is allached to an eleClrie mOLor, which brings the nrwhcers rotational speed lip 10 5 000 rc\/min. (a) Fine! U1C kinelic energy stored in the flywheel. (b) I[the fI)'wheel is to supply energy to the car ;:IS a 1O.0-l1p mowr would, find the lenglh of time the car could rUIl before the flywheel would h,wc to be broughr back up to speed. 43. A hori/.onLal 800-N merry-go-round of radius 1.30 m is starLcc! from rest b)' a constant horizontal force of
50.0 N applied tangcl1lially
(0
the merry-go-round. Find
the kinetic energy of the merrY'-go-roulld after (Assllllle il is a solid cylinder.)
44.
~.OO
s.
Four o~jcCl VI. This result is often expressed by the statement that swiftly moving fluids exert less pressure than do slowly moving fluids. This irnportant i~lCt enables us to understand a wide range of everyday phenomena. QUICK QUIZ 9.7 You observe \.wo helium balloons floating next to each other aL Lhe ends of stri ngs secured to a t.able. The facing surfaces of the balloons are separated by 1-2 em. You blow Lhrough the opening between the balloons. '''That happens to the balloons? (a) They move toward each otiler. (b) They move a,\'ay from each other. (c) They are unallected.
EXAMPLE 9.13 Goal
Shoot-Out at the Old Water Tank
Apply Bernoulli's equation to fllld the speed ora fluid.
Problem A nearsighted sheriff fires at a caule rustler with his trusty six-shooter. Fortunately for the rustler, the bullet misses him and peneLrates the town water tank. causing- a ]c,lk (Fig. 9.3]). (a) lfthe top of the tss-scctional "rea is large compared lo the hole's (A:.! » AI), so the water level drops vcry slowl>' and V 1 = 0. Apply Bernoulli's cquation to points Q) and @ in Figurc 9.:~ I, notin~ lhat PI equals atmospheric pressure PI' at the hole and is approximately the samc al thc top of the waler tank. Part (h) can be soh·cd with kinematics,just as if the waleI' wcre a ball thrown horil.ontally.
gr;", il = o.
-3.00
III
= -(4.90 m/5')I'
1= 0.782 s Compute the horizontal distancc thc slrearn travels in this tin1e:
....
,
x = vo~/ = (?.13 l11/s)(0.782
s)
2.45 m
.
... ~
Remarks As the analysis or pan (a) shows, the speed of the water emerging from the hole is equal to thc speed acquired by an object falling freely through the vertical distance It. This is known as Torricelli's law. QUESTION 9.,13
As time passes, what happens to the speed of the water lca\·ing the hole? EXERCISE 9.13
Suppose, in a similar situation, the watcr hit~ the ground 4.20 m from the hole in the tank. If the hole is 2.00 m above the ground, how far above the hole is the watcr le\·el? Answer
2.21 m above the hole
EXAMPLE 9.14
Fluid Flow in a Pip.e
Gool Solve a problem combining Bernoulli's equation and the equation of continuity. Problem A large pipe with a cross-sectional area of 1.00 m'l descends 5.00 m and narrows to 0.500 m~, where it terminatcs in a valve at point Q) (Fig. 9.32). If the pressure at point ® is atmospheric pressure. and the \·alve is opened \vide and water allowed to flow freely, [md thc speed of the water leaving the pipe. Strategy The equation of conlinuit), together with Bernoulli's equation, constilUte twO equations in two unknowns: the speeds v and v 2 . Eliminate v'2. from I.kr· noulli's equation with the equation ofconlinllily, and solve for VI. J
® "11 vt
"
,
",---0)
-----------~J~l
VI FIGURE
9.32
(F.xample9.14)
296 ~
Chapter 9
...
Solids and Fluids
.
~
Solution ',Vrite Bernoulli's equation: Solve the equalion of continuity for v2: (2)
In Equation (1), sel PI = P2 = Po, and substitute the expression for v 2' Then solve for v j ,
(3)
SubstlLUtc the given values:
V, =
11.4 m/s
.
.
~
,
...
Remarks This speed is slightly higher than the speed predicted by Torricelli's law because the narrowing pipe squeezes the fluid.
QUESTION 9.14 Why does setting AI = A 2 give an undefined answer for the speed VI? Hinl: Substitute A 1 = verifywhethcr or not a contradiction is obtained.
1\2
into Equation (3) and
EXERCISE 9.14 Water flowing in a horizontal pipe is at a pressure of 1.40 X 105 Pa at a point where its cross-sectional area is 1.00 m 2 . "\Then the pipe narrows to OAOO m 2 , the pressure drops [Q 1.16 X 105 Pa. Find the water's speed (a) in the wider pipe and (b) in the narrmvcr pipe. Answer
(a) 3.02 m/s
(b) 7.56 m/s
9.8
OTHER APPLICATIONS OF FLUID DYNAMICS
In this section we describe some common phenomena that can be explained, at least in pan, by Bernoulli's equation. In general, an object moving through a fluid is acted upon by a net upward force as the result of any effeu that causes the fluid to change its direction as it flows past the object. For example, a golf ball struck with a club is given a rapid backspin, as shown in Figure 9.33. The dimples on the ball help entrain the air along the curve of the ball's sufi-ace. The figure shows a thin layer of air wrapping parnvay around the ball and being deflected downward as a result. Because the ball pushes the air down, by Newton's third 1£1\\1 the air must push up on the ball and cause it to rise. \Vithout the dimples, the air isn't as well entrained, so the golf ball doesn't travel as far. A tennis ball's fuzz performs a similar funct.ion, though the desired result is ball placement rather than greater distance.
=
FIGURE 9.33 A spinning go[fbal[ is
m = 5.66 X 10' kg
. .....
..
Remarks Note the factor of two in the last equation, needed because the airplane has two wings. The density of the (
This equation shows that the exhaust speed is reduced in the atmosphere, so rockets arc actually more dTective in the vacuum of spacc. Also of interest is the appearance of the density p in the denominator. A lower density working nuid or gas will give a higher exhaust speed, which partly explains why liquid hydrogen, which has a very low density, is a fuel of choice.
9.9
299
HOME PLUMBING
Consider the portion ora home plumbing syslem ShO\'.'I1 in Figure 9.38. The water trap in the pipf' below the sink captures a plug of water that prevents sewcr gas from finding its way from the sewer pipe, lip the sink drain, and inlO the home. Suppose the dishwasher is draining and the \V betv./een the solld surface and a line drawn tangent to the liquid at the surf"ce is called the contact angle (Fig. 9.44). The angle is less than 90 0 for
¢, "
,,
'\
\,Vater
,, Glas...; ,
w
~
W
FIGURE 9.43 A liquid ill C0l11aCl wilh where PJ is the densit)' of the fluid. Al the instant the sphere begins to Fall, the force of friction is zero because the speed of the sphere is zero. As the sphere accelerates, its speed increases and so does Fr. Finally, at a speed called the terminal speed v" the net force goes to zero. This occurs 'when the net upward force balances the downward force of gravity. ThclTfore, the sphere reaches lcrrninal speed ,,,",hen ~
1~+B=w
or
When this equaLion is solved for Terminal speed
VI'
we get 2r:!g
-+
lI,
= - - (p - p)
97)
[9.29]
Sedimentation and Centrifugation If an object isn't spherical, we can Slill use the basic approach just described LO dClcrmine its terminal speed. The only difference is that we can't usc Stokes's law for the resistive forcc. Instead, "...e assume that the resistive force has a magnitude given by F r = kv, where II. is a coelTicient that must be determined experimentally. As discussed previollsly, the object reaches its terminal speed when the downward force of gravity is balanced by the nClupward force, or
w=B+Fr
[9.30]
where B = PJgV is the buoyant force. The volullle Vol" the displaced r1uid is related La the density p of the falling o~jccl by V= 'In/p. Hence, we can express the buoyanI. force as
PI B= -mg P We substitute this expression for 13 and F,. condition):
PI
ml{ = -
P
=
mg
kv/ into Equation 9.30 (terminal speed
+
IW I
or v = II!g(l _ PI) I k P
[9.31]
The terminal speed for panicles in biological samples is usually quite small. For example. the terminal speed for blood cells falling through plasma is about 5 cm/h in the gravitational field of the Earth. The terminal speeds [or the molecules that make up a cell arc many orders of magnitude smaller than this because of their much smaller mass. The speed at which materials fall through a fluid is called the sedimentation rate and is important in clinical analysis. The sedimentation rate in a fluid can be increased by increasing the effective acceleration g th"t appears in Equation 9.31. A fluid containing various biologi-
Summary
cal molecules is placed in a centrifuge and whirled at very high angular speeds (Fig. 9.53). Under these conditions, the panicles gain a large radial acceleration a r = '(I2/ r = w2.,. that is much greater than the free-fall acceleration, so we can replace gin Equmioll 9.31 by w~.,.and obtain
311
w
[9.32) This equation indicates thatlhe sedimentmion rate is enormously speeded up in a centrifuge (w 2r» K) and that lhose panicles with the greatest mass will have the largest tcrminal specd. Consequently the most massive particles will settle out on the buttolll ofa testlllbe first. FIGURE 9.53 a ccmrifllg:c
Silllplifkd diaJ..j:rarll of
(top view).
SUMMARY 9.1
States of Matter
9.4
Malter is normally classi fled as bei ng in one of t.h ree states: solid, liquid, or gaseolls. The fourth state oflnatter is called (l plasma, which consists ofa neulral syst.em of charged particles interacting electromagnetically.
9.2
The Deformation of Solids
The clastic properties ofa solid can be described using the concepts of" stress and strain. Stress is related to lhe force per unit area producing a deformation; strain is a measure of the amount of deformation. Stress is proportional lO strain, and the constant of proponionality is the elastic modulus: Stress
= clastic modulus
X strain
[9.1]
Three COlllmon types or deformation are (I) lhe ITS1Slance ()f a solid to elongation or compression, characterized by Young's modulus Y; (2) the l'csistance to displacement or rhe h\Ces of a solid sliding past each oUler, characterized by the shear mocltilus S; and (3) lhe resistance of a solid OJ" liquid to a change in volume, characterized by the bulk modulus B. All three types or deformatioll ohey laws similar to Ilookc's law for springs. Solving problems is usually a matter of identif)~ng the given physical variables and solving- for the ullknown variable.
9.3
Variation of Pressure with Depth
The pressure in an incompressible fluid val-ies with depth It according to the expression p
=
Po + pgh
[9.11]
where PI) is atmospheric jJl"essllre (1.013 X 10-') Pa) and pis the density of the fluicl. Pascal's principle states that when pressure is e in length if [he o~jecl is accelerated upwards at a rate of 3.0 m/s 2 ? (c) WhIll
FIGURE P9.11
[]J\r\7hen water freezes, it expands abollt 9.00%. 'What would
be the pressure increase inside your automobile engine block if the water in it frozc? The bulk modulus of ice is 2.00 X lO~t :-.11m 2 .
Aluminum
fIJI
The lOlal cross-sectional area of the load-hearing calcified portion of the two forearm bones (radius and ulna) is approximately 2.4 cm~, During (l car crash, lhe forearm is slammed ag-ainst the dashboard. The arm comes to rest from an initial speed of 80 km/h in !i.0 ms, '[the arm has an effective milSS 01"3.0 kg and bone material can withstand a maximum compressional Slress of 16 X L0 7 Pa, is lhe arm likely to wilhs[and lhe crash?
SECTION 9.3 DENSITY AND PRESSURE 13.
mJ
Suppose two worlds, each having mass M and radius N, coalesce into a single world. Due to gravitational contraction, the combined world has a radius of only ~R. What is the t point in the ocean is in the Mariana Trench, aholll J I km deep. The pressure at the ocean noor is hug-e, ahollt 1.13 X IO~ :\'/m 2 . (a) Calculate the change in volume of 1.00 Ill' of water carried from the surface to the bottom of the Pacil-ic. (h) The densitr of water at the surfacc is 1.03 X 10 3 kg/Ill:I , Find its densit~ at the bot· tom. (c) 1nip.nl and rc~rro ducible reference temperature for the Kelvin scale; it occurs at a temperature of O.OloC and a pressure of 4.58 mm of mercury. The temperCllure althe triple point of waleI' on the Kelvin scale occurs at 273.16 K. Therefore, the 51 unit of tempera~ ture, the kelvin, is defined as 1/273.16 of the temperature of the triple point of water. Figure 10.6 shows the Kelvin temperatures for various physical processes and structures. Absolute zero has been closely approached but never achieved. \'Vhal would happen to a substance ifits temperature could reach 0 K? As Figure 10.5 indicates, the substance \·,:ould exert zero pressure on the ' . . alls of its container (assuming the gas doesn't liquefy or solidify on the ,...'ay to absolute zero). rn Section 10.5 ..." e show thal the pressure of a gas is proportional to Lhe kinetic energy of the molecules of that gas. According La classiG11 physics, therefore, the kinetic energy of the gas would go to zero and there ,...,ould be no Illation al all of the individual cOlllponents of" the gas. According to quantulll theory, however (to be discussed in Chapter 27), the gas would £llways retain some residual energy, called the zero-jJoinl pnergy, at [hat low temperallire.
J05
The Celsius, Kelvin, and Fahrenheit Temperature Scales 10 1
J O~
J>-~t~
Surface of rhe Sun Copper melts \,Va1(,1" frccles Liquid nitrogen
10 1oI\~
PI \/:
'Ii 7;
7"
P =]' ~
-->
J
I
'I;
" 372 K Pr = ( 1.0 I X 1O"Pa)--=
303 K
1.24 X W', Pa
(b) Find the magnitude or the friction force
acting all the cork. Apply Newton's second law to the cork just before it leaves the boule. Pin is the pn:ssure inside the bOILle, and J~,Ul is the pressure Olllside.
2: F= 0
PillA -
~IUIA
1'~'lilti()1I = PillA - ~)lIIA =
-
(Pill -
l~riLlion
=
0
j~HlI)A
(1.24 X 10' I'a - 1.01 X 10' l'a)(2.30 X 10- 1 m') j'lliai"ll
.
---?
=
5.29 N .~
Remark Notice the use, once again, of the ideal gas la\\' in Equation (I). \,yhencver compi1ring the state ofa gas at different points, this is the best ,vay to do the mat h. One other poi Ill: Heating the gas blasted [he cork out of the boule, ,,'hich meant the gas did ,..:ork Oil the cork. The work done by an expanding gas-dri\'ing pistons and generators-is one of the foundations of modern technology and ,.. . ill be studied extensively in Chapter 12.
(\"'0
QUESTION 10.7
As the cork begins to move, what happells to the pressure inside tile bottle? EXERCISE 10.7
A tire contains air at a gauge pressure of 5.00 X 10 1 Pa at ,l temperature of 30.0°C. After nightfall, the temperalllre drops to -I a,Q°c. Find the new gauge pressure in the lire. (Recall lhat gauge pressure is absolut.e pressure minus aunospheric pressure. Assume constant volume.) Answer
3.01 X 10.1 Pa
EXAMPLE 10.8 Submer ing a Balloon Goal
Combine the ideal gas law with the equation of hydrostatic equilibrium and buoyancy.
Problem A sturdy balloon with volume 0.500 m:~ is attached to a 2.50 X 102_kg iron weight and tossed overboard into a freshwater lake. The balloon is made of a light material of negligible mass and elasticity (although j[ can be compressed). The air in the balloon is initially at atmospheric pressure. The system fails to sink and there aiT no more '.. .·eights, so a skin diver decides to drag it deep enough so that the balloon will rcmain submerged. (a) Find the volumc oflhe balloon at the point where the system will remain submerged, in equilibrium. (b) vVhat's U1C balloon's pressure at that point? (c) Assuming constant tempcralUre, 1.0 ,·",hat minimum dept.h must the balloon be dragged?
Strategy As the balloon and weight are dragged deeper into t.he lake, the air in the balloon is compressed and the volume is reduced along with the buoyancy. At SOllle depth 11 lhe total buoyant force acLing on the balloon and weight, 8 ba1 + 13 Fe , will equal the total weight, W haJ + w F,., and the balloon will remain at that depth. SubstiLLHe these forces into Newton's second law and solve for the unknowtl volume of the balloon, answering pan (a). Then use the ideal gas law to find the pressme, and the equation of hydrostatic equilibrium to nnd the dept.h.
10.4
Macroscopic Description of on Ideal Gas
339
..
,.. ... Solution (a) Find the ,'olume of the balloon at. dlC equilibrium poinL. mho
Find the \"olume of the iron, VFc : Find the maijS of the balloon, \vhich is equal to the mass of the air if\\'e neglect thc mass of the balloon's material:
\
~< = -
Pr,·
/II'M'
-
....
2.50 X I O~ kg = -::-,--,----cc;-:--"---;;- = 0.031 8 m'
Apply ~ewLOn's second law to the systcm when it"s in equilibrium: Substitute the appropriate expression for cach term: Cancel the g's and solve lor the volume of the balloon, Vb;! I: 0.6-15 kg
+ 2.50
X
10' kg - (1.00 X 10" kg/m')(0.031 8111') 1.00 X 10' kg/ Ill'
(b) \o\'hat's the balloon's pressure at the equilibrium point?
Now use the ideal gas 1m\-' LO lind the pressurc, assuming constanttemper(lture, so that 71 = 7 . J
/',,"J II/Oj --=--= J->/\'; nR1~, V,
p =-1',=
Ij
I
=
0500 m 3 ( , 1.01 0.219111'
2.31 X 10' Pa
(c) To what minimum depth musl the balloon be dragged'
Use the equation of hydrostatic equilibrium to find the depth:
P, = I~ilm
+
pgh
2.31 X 10" Pa - 1.01 X 10" Pa pg
=
....
(1.00
X
lO'kgfl11')(9.80 m/5')
13.3111
Remark Once again, the ideal gas law was used to good effect. This problem shows how e\'en answering a fairly simple question can require Ihe application of ~e\'cral different physical concepts: density. buoyancy. the ideal gas law, and hydrostatic equilibrium.
QUESTION 10.8 Ira glass i'i turned upside down and then submerged in water. what happens to the \'olume of the trapped air as the glass is pushed deeper under \\Iatcr? EXERCISE 10.8 A boy takes a 30.0-cl11:\ balloon holding air at 1.00
£lUll al the surface or a freshwater lake down to a depth of 4.00 Ill. Find the volume orthe balloon atthi'i depth. Assume the balloon is made orJight material of little elasticity (although it can be compressed) and Ule temperature of the tr~lpp('d air remains constant.
Answer
21.6 cm:1
340
Chapter 10
Thermal Physics
As previously slated, tile flU !TIber of molecules contained in one mole of any gas is Avogadro's number, tV..I = 6.02 X 10 23 particles/mol, so N
n=-
[10.10]
l\/A
,",,"here n is the number of moles and N is t.he number of molecules in the gas. With Equation ]0.10, we can rewrite the ideal gas law in terms of the total number of molecules as
PV = nRT =
li RT N...
or Ideal gas law
-+
[10.11]
PV= NIi"T where
Boltzmann's constant
'J" Jill = - R = 138 X 10-·" j/K
~
N"
.
(10.12]
is Boltzmann's constant. This reformulation of the ideal gas law will be used in the next seClion to relate the temperature of a gas to the average kinetic cllcrg}' of panicles in the gas.
10.5
THE KINETIC THEORY OF GASES
In Section 10.4 we discussed the macroscopic properties of an ideal gas, including pressure, volume, number of moles, and temperature. In this section we consider the ideal gas model from the microscopic point o[view. \11it..roy;cll IllOleCllle~
120
80
40
200
'100
liOO
!:IOO
v (lll/S)
I 000
I 20U
! 400
I GOO
ACTIVE FIGURE 10.15 The Maxwell speed distribulloll fIX JO" nilrogell molecules at 300 K and yon K. The IOlal :H"e~l under either CllIYC l..'quals the totaillumbe." 01 11101el"ldc.~. The most prc,babh.: speed urnI" th(: l\\'('r:age spel'd 1',1\' alld the rool11le;ll\-~qLlare speed 11,,,,. .In:- illdi("ilted for I he ~IO(J-" Cllr\'c.
344
Chapter 10
Thermol Physics
TABLE 10.2 Some rms Speeds Gas
Molar Mass (kg/mol)
1-1,
2.02 X 10-:1
He
4.0 X 10-'
1-1,0
18 X 10-:1 20.2 X 10-"
Ne
APPLYING PHYSICS 10.2
1902 I 352 637 602 51l
N:! and CO
28.0 X 10-5\
NO
30.0 X 10-'
0,
32.0 X 10-'
CO,
44.0 X 10-:1
SO,
64.1 X 10-"
494 478 408 338
EXPANSION AND TEMPERATURE
Imagine a gas in an insulated cylinder with a movable piSLOIl. The pistoll has been pushed imvard, compressing the gas, and is now reJcased. As the molecules of the gas strike the piston, they move it outward. E.xplain, from the point of vic\\' of the kinetic theory, how the expansion of this gas causes its lernperature Lo drop.
piston to move \vilh some velocity. According to the conservation of momentulll, the molecule must rebound with less speed than it had before the collision. As these collisions occur, the average speed of the collection of molecules is therefore reduced. Because temperature is related to the averagc speed of the molecules, the tcmperaturc of the gas drops.
Explanation From the point of view of kinetic theory, a molecule colliding with the piston causes the
EXAMPLE 10.9 Goal
A C linder of Helium
Calculate the illlernal energy ora system and the average kinetic energy per molecule.
Problem A cylinder contains 2.00 mol o[ helium gas at 20.0°C. Assume the helium behaves like an ideal gas. (a) Find the to[c the clock to run J~lst ur sIO\...-? pair of eyeglass frames are made or epoxy plastic (cocf.l"iciCIll or linear expansion = 1.~O X 10- 7~ (Fig. 11.5). The slab allows energy to transfer rrom Lhe region of higher temperature to t.he region of lower temperanlre by thermal conduction. The rate of energy transfer, qp = Q/tJ.I, is prop0rlional to the cross-sectional area of the slab and the temperaLUre differ· CJ1ce and is inversely proportional to the thickness of the slab:
a\vot)' from )our bod)' br heaL The
pdmary insulaling mediulll is the air trapped in small pod.els within the material.
).late that qp has units of wallS when Qis injoules and D.l is in seconds.
11.5
Suppose (\ substance is in the shape oCa long, uniform rod orlcngth I~. as in Figure 11.6. We assume the rod is insult.:s the gao,;.
385
Chapter 12
386
The Laws of Thermodynamics
where we have set the magnitude F of the external force equal to /.)1\, possible because the pressure is the same everywhere in the system (by the assumption of equilibrium). Note thal if the piSLOil is pushed downward, 6.y = Yf-)" is negati\'c. so we need an explicit negative sign in the expression for l\1to make the work posi[ive. The change in \'olume of the gas is 6V= A 6.)', which leads to the following definition: The work W done on a gas at constant pressure is given by
w=
-p~v
[12.1]
\\'here Pis the pressure throughout the gas and 6.\'is the change in volume of the gas during the process. If the gas is compressed as in Active Figure 12.lb. aVis negaliYc and the work done on the gas is posiLivc. If the gas expands, LlVis positive and the work done on the gas is negative. The work done by the gas on its environmem, Well'" is simply the negative of lhe work done on lhe gas. In the absence of a change in volume, the work is zero.
EXAMPLE 12.1 Goal
Work Done by' on EXJ1anding Gas
Apply the definition of work
rH.
constant pressure.
Problem In a system similar to that shown in Active Figure 12.1, the gas in the C)'lincier is at a pressure equal to 1.01 X I(f' Pa and the piston has an area ofO.100 111~. As encrgy is slowly added to the gas by heat, the piston is pushed up a disl;lnce of 4.00 em. Calculate Ihe \\lark done by the expanding gas on the surroundings, V\~II\' assuming the pressu rc rcma i I1S constant. Strategy The \\Iork done on the environment is thc negative or the work done on the gas given in Equation 12.1. Compute the chang-c in vohunc and lTluhiply by the pressure. . ..., r'
Solution
= A ~y =
Find the change ill volume of the gas, ~V; which is the cross-sectional area times the displacement:
~v
Multiply this result by the pressure, getting the work the gas docs on the clwironment, \1~0I:
We",
(0.100111')(4.00 X 10-2 Ill)
= 4.00 X 10- 3 Ill'
= =
P ~\I = (1.01 X 10; l'a)(4.00 X 10 -'Ill')
-l04J •••••
..rI
Remark The volume of the gas increases, so the work done on the ellYironment is positi\"e. The work done on the system during this process is lr= -404j. The energ)" required to perform positive work on the em'ironment must come from the energy of the gas. (See the next section for more details.) QUESTION 12.1 If 110 energy were added to the gas during the expansion, could the pressure remain constant? EXERCISE 12.1 Gas in a cylinder similar to Figure 12.1 moves a piston with area 0.200 m:? as energy is slowl) added to the system. If 2.00 X 103 J of work is done on the environment and the pressure of the gas in the cylinder remains constam at 1.01 X 10!> Pa, find the displacement or the piston.
Answer 9.90 X 10-2
In
Equation 12.1 can bc used to calculate the work done on the system only when rhe pressure of the gas remains constant during the expansion or compression. A pl"OCeSS in ,·vhich t.he pressure remains constant is called an isobaric process.
12.1
Work in Thermodynamic Processes
The pressure \IS. volume graph, or PV diagram, of an isobaric process is shmvn in Figure ]2.2. The curve on such
V{ (950 X 10- 3 111') T ~ T - ~ (293 K) -;--_.:-:----ccco-~
{
, II,
3
111 ')
733 K (2.00 X 10·' Pa)( 1.00 X 10-3 111 3 )
Again using til
Q
~ /';u -
W
Q= 4.50 X IO'J - (-3.00 X 10'./) al
~
7.50 X 10'./
constant pressure
Substitute values into Equation 12.6:
Q=
nC,,/';T = ~I1R/,;T
= ;(8.21 X W-' 11101)(8.3] JlK· 11101)(733 K - 293 K) =
7.50 X 10']
~
12.3
(e)
1-f0\\'
Thermal Processes
393
would the answers changc [or a diatomic gas?
tJ.u= nC"tJ.T= (~+ l)nRtJ.T
Obtain the new change in illternal energy, DoLl, noting that C" = ~R for a diatomic gas:
= i(8.21 X 10-' rnol)(8.31 j/ K· mol)(733 K - 293 K)
tJ.U = Obtain the new energy transferred by heat. Q:
7.50 X 10'.1
Q = "CI,tJ.T =
(i + l)nRtJ.T
=;(8.21 X 10 'mol)(8.3Ij/K·mol)(733K-293K)
Q. = 1.05 X 10'.1
......
.
~
Remarks Notice that problems involving diatomic gases are no harder than those with monatomic gases. It'sjust a malleI' of" ;')cUu.'iting the molar specific heats. QUESTION 12A True or False: During a constant pressure compressioll, the tcmperallire of an ideal gas must always decrease, and the gas mllst always exhaust thermal energy (Q < 0). EXERCISE 12A Supposc an ideal monatoIT1ic gas at an initial temperature of 475 K is compressed frolll 3.00 L to 2.00 L while its pressure remains constant at 1.00 X JOJ Pa. Find (a) the work done on the gas, (b) the change in internal energy, and (c) the energy transrerred by heat, Q.
Answers
(a) 1.00 X 10'j
(1)) -1.'iOI
(c) -250.1
Adiabatic Processes In an adi;:lbatic process, no energy enters or leavcs the system by heat. Such a system is insulated, thermally isolated from its environment. In general, hm,.. . evcr, the system isn't mechanically isolated, so it can still do work. A sufficiently rapid process m~lY be considered approximatel}' adiabatic because there isn't time [or any sign ificant transfer of energy by hcat. For adiabatic processes Q = 0, so the flrst 1m·\' becomes
tJ.U= HI
(adiabatic processes)
The work done during an adiabalic process Girl be calculated by finding the change in the internal encl-gy. Alternately. the work can bc computed from a P\I diagram. For an ideal gas undergoing an adiabatic pl-ocess, it can be shown that. p\f"t = constant
[I2.8a]
where [12.8b] is called the (uliabfltir i'11df'x of the gas. Values or the adiabatic index for several different gases are given in Table 12.J. After computing the COllstant on the righthand side of ECJll(llion 12.8a and solving for the pressure P, [he area under the curve in thcPFdiagral1l call be found by cOllnting boxes, yielding the work. If a hot gas is allowed to expand so quickly that there is no time for energy to enter or leave the systern by heal, rhe work done on the gas is negative and the il1\ernal energy decreases. This decrease occurs because kinetic energy is transferred from the gas molecules to the moving piston. Such an adiabatic expansion is of praClical importance and is nearly n:alil.cd in all internal combustion engine when a gasoline-air mixture is ignited and expands rapidly against a piston. The following example illust.rates this process.
394
Chapter 12
EXAMPLE 12.5 Goal
The Lows of Thermodynamics
Work and an Engine Cylinder
Use the firsllaw
1.0
find the work done in an adiabatic expansion.
Problem In a car engine operating at a frequency or 1.80 X l03 rev/min, the expansion of hot, high-pressure gas againsla piswn occurs in abollt]O inS. Because energy transfer by heal typically takes a time on the order of minutes or hours, iCs safe La assume little energy leaves the hot gas during the expansion. Find the work done by the gas on the piston during this adiab~llic expansion by assuming the engine cylinder contains 0.100 moles or
an ideal monatomic gas that goes from J.200 X 10 3 K to 4.00 X 102 K, typical engine telllperalllres, during the expansion. Strategy Find the change in internal energy using the given temperatures. For an adiabatic process, this equals the work done on the gas, which is the negative of the work done on the environment-in this case, the piston. .
~
Solution Start with the first law, laking Q = 0: Find 6.Ufrol11 the expression for the internal energy oran ideal monatomic gas:
Remarks The work clone on the piston comes at the expense of the internal energy of the gas. In an ideal adiabatic expansion, the loss of internal energy is completely converted into useful work. In a real engine, there are ahva}!s losses.
QUESTION 12.5 In an adiabat.ic expansion of an ideal gas, why must the change in temperature always be negative? EXERCISE 12.5 A monatomic ideal gas ,,,·it.h volume 0.200 L is rapidly' compressed, so the process can be considered adiabatic. If the gas is in itially at 1.0 I X lO,i Pa and 3.00 X 10 2 K and the final temperature is 477 K, find the work done by the gas on the envi ron ment, H~'m'.
Answer
-17.9 J
EXAMPLE 12.6
An Adiabatic EXJ>ansion
Goal Use the adiabatic pressure vs. volume relation to find a change in pressure and the work done on a gas. Problem A monatomic ideal gas at an iniLial pressure of 1.01 X 10 5 Pa expands adiabatically from an initial volume of 1.50 m3 , doubling its volume. (a) Find the nev.' pressure. (b) Sketch the PV diagram and estimate the work done on the gas. Strategy There isn't enough information to solve this problem with the ideal gas law. Instead, usc Equation 12.8 and t.he given information to find the adiabatic index and the constam C for the process. For part (b), sket.ch the P\! diagram and count boxes to estimate the area under t.he graph, which gives the work.
1.00 0.80
0.60 0.40 0.20
, 1.00
2.00
3.00
If (m )
FIGURE 12.4 (Example 12.6) The jJVdiagram of an adiabatic expansion: the graph of p~ CV.,., where Cis a COllstMlt ;lIld y ~ C,/C".
12.3
Thermal Processes
395
. ..
~.
.,
Solution (a) Find the ncw pressure.
e'i
First, calculate the adiabatic index:
~R
5
y=-=-~-
c=
Use Equation 12.8a to find the constant C:
CIt
ttR
P,II,"
= (loOI
~ lo99
The constant C is fixed for the cntire process and can be used to rind IJ'1:
3 X 10' Pa)(l.50 111')5/"
X 10" Pa·m'
C = 1',11," = P,(3.00
111")5/:;
1.99 X 10"Pa·I11" = P,(6.24 111')
P, =
3.19 X 10' Pa
(b) Estimate the work done un the gas [rom a PV diagram.
Count the boxes betwcen VI = 1.50 Ill:~ and V2 = 3.00 m:l in the graph of P= (1.99 X 10!'> Pa' m!"')V-5/:I in the IJV diagram shmvn in Figure 12.4:
Number or boxes = 17
Each box has an 'area' 0["5.00 X 1O:~.J.
W= -]7'5.00 X
....
IO"J = -8.5 X lO"J .
~
Remarks The exact answer, obtaincd with calculus, is -8.43 X 10'lJ, so our result is a very good estimate. The answer is negative becausc thc gas is expanding, doing positive work on the environment, thereby reducing its own internal energy. QUESTION 12.6 For an adiabatic expansion benveen two given volumes and an initial pressure, which gas does more work, a monalomic gas or a diatomic gas? EXERCISE 12.6 Repeal the preceding calculations for an ideal diatomic gas expanding adiabatically from an initial volume of 0 ..500 m~~ LO a final volume of 1.25 m'\ starting at a pressure of PI = 1.01 X 10 5 Pa. Use the same techniques as in the example.
Answers
P, = 2.80 X 10' Pa, W= -4 X 10 '.I
Isovolumetric Processes An isovolumetric process, sometimes called an isorhoTic process (\vhich is harder to remember), proceeds at conSlant volurne, corrcsponding to vcrtical lines in a PVdiagram. If1.he volume doesn't change, no work is done on or by the systcm, so H'= 0 and the nrstla\\' of thermodynamics reads
t1U= Q
(isovolumclric process)
This rcsult tells LIS that in an isovolumetric process, the change in internal energy of a system equals the energy transferred to the system by heat. From Equation 12.3, the energy I ransfcrred by heal in constant volume processes is given by
Q= "C"f}."!"
EXAMPLE 12.7 Goal
[12.9]
An Isovolumetric Process
Apply thc firstlmv
La
a constant-volulllc proccss.
Problem How much t.hermal encrgy must be added to 5.00 moles ofmonal.Omic ideal gas al3.00 X 10'lKand with a constant. volume of 1.50 L in ordcr 1.0 raise thc tcmperallll"C of the gas (Q ~.80 X 102 K?
396
Chapter 12
The Laws of Thermodynamics
Strategy The energy transferred by heat is equal to the change in the internal energy of the gas, which can be calculated by substitution imo Equation 12.9. ,-
~
Solution ApplyEquT
tJ.U- W
- nR'fln
399
11m reser....oir.at Ii,
\t; CJ)
r_~
Engillt:
(PV Area)
HEAT ENGINES AND THE SECOND LAW OF THERMODYNAMICS
A heat engine takes in energy by heal. and partially convcrts it to other forms, such (IS electrical and mechanical energy. In a typical process for producing electricity in a power plant, for instancc, coal or some other fuel is burned, and the resulting internal energy is used to convert V'later to steam. The steam is thcn directed althe blades of a turbine, selling it rOlating. Finally, the mechanical energy associated with Ihis rotation is used to drive an electric generator. In another heat engincthe internal combustion engine in an automobile-encrgy enters the engine as flle) is injected inLO the cylinder and combusted, and a fraction of this energy is converted to rnechanical energy. In general, a heat engine carries somc working substance through a cyclic process! during which (1) energy is transferred by heat [rom a source at a high temperature, (2) work is done by the engine, and (3) energy is expelled hy the engine by heat to a source atlo".·cr temperature. As an example, considcr the operation of a sleam engine in which the working substance is water. The water in the engine is carried through a cycle in which it first evaporatcs into steam in a boiler and then expands against a piston. After the steam is condensed with cooling water, it rt;turns to the boiler, and the process is repe~l[ed. It's useful to draw a heat engine schcmatically, as in Active Figure 12.9. The engine absorbs energy Qh from the hot reservoir, does \,,'ork ltVltlll.
p
' - - - - - - - - - - If
FIGURE 12.10 The PVdiagram lor an arhitrary cyclic process. The arca enclosed by the curve c.:quals the 11Ct work done.
Chapter 12
400
The Laws of Thermodynamics \~'e can think of thermal efficienc)' as the ralio of the benefit received (work) to the cost incurred (energy lransfer at the higher temperature). Equation 12.12 shows that a heal engine has 100% efficienc), (p = I) onl}' if Q,- = 0, meaning no energy is expelled to the cold reservoir. In other words, a heat engine with perfect cflicicncy would have lU lise all Lile illjJlIl clleq;y rUl duing IllcclJ A.
WCI\
for the isobaric process
Compute the work on [he SyStCI11, with pressure constant:
I'(L)
.
=
(b) Find 1:1u,.\II' cess A ---7 B.
"I~(
1.00
I'(L)
1.-,
10
:")
IJ
2.00
.
Solution (a) Find nand T fj \ ...·ith the ideal gas law:
401
HrCA = -P.6\!= -(1.01 X l05 Pa)(5.00 X ]0-3 m :\
- l.50 X 10-2 m") We" =
1.01 X 10'.1
.~
402
Chapter 12
The Lows of Thermodynamics
Find the change in internal energy,
aUe/!:
!:iU,,, = ~TlR!:iT = ~(0.203 mol)(S.31 ]lK·lI1ol)
X (3.00 X 10' K - 9.00 X 10' K) !:iUcA =
Compute the thermal energy,
QGI'
[rom the Hrsllaw:
QGl
~ !:iU", - WCA = -1.52 X 10 3 .1 -
= (e) Find the net change in internal energy, 6.UI1l "I' for the cycle:
-1.52 X 10'J l.01 X 10'.1
-253 X lO'J
= l.52 X L03j
+0
- 1.52 X 10'.1 = 0
(f) Find the energ)' input, Q,.; the energy rejected, Qr; the thermal efficiency; and the net work performed by the engine: Sum all the positive contributions to find Qh:
Q" = Q,1f< + Que ~ 1.52
X
10'.1
+
1.67 X 10'.1
3.19 X 103 J Sum any negative cOlltributions (in this case, there is only one):
Find the eng"inc efficiency ancllhe net work done by the engine:
Q, =
-2.53 X 10'.1
253 X 10'J
1'=1_IQ"=l IQ"I WCIIg- = - (\t\~1B
+
H/nc
3.19
+
lO'J
0.207
WCA )
= -(0 - 1.67 X 10 3 .1 ~
X
+ 1.01
X
10'J)
6.60 X JO'J
~
~
Remarks Cyclic problems are rather lengthy, but the individual steps are often shon substitutions. Notice that the change in internal energy for the cycle is zero and that the net work done on the environmenL is identical to the net [hennal energy transferred, both as they should be. QUESTION 12.11 If BC\'~'ere C. (a) Find Q, !:iU, and Wforlhe constant volume process A ~ B. (b) Do the samc for the isothermal proccss B ---? C. (c) Repeat, for the constanL pressure process C -> A. (d) Find Q", Q" and the efficiency. (e) Find We",. Answers (a) Qcli1 = !:iUAB = 151J, W;lIl = 0 (b) !:iUilC = 0, QBC = -Wile = 1.40 X 10'/ -151.1, WC,\~ 101.1 (d) Q,,~29IJ,Q,= -252.1,e=0.13'1 (e) We", = 39.1
(c) QC,'
= -252 J,
!:iUC..' =
Refrigerators and Heat Pumps Heat engines can operate in reverse. In this case, energy is ir~jecled into the engine, modeled as \",ork IV in Acti\'e Figure 12.12, resulting in energy being extracted from the cold reservoir and transferred to the hot reservoir. The system now operates as a heat pump, a comrnon example being a refrigerator (Fig. ]2.13). Energy Q{ is extracted from the interior of the refrigerator and del ivered as energy Q" to the warmer air in the kitchen. The work is done in the compressor unit of the refrigerator, compressing a refrigerant such as freon, causing its temperature to increase.
12.4
Heat Engines and the Second Low of Thermodynamics
A household air conditioner is another example of a heat pump. Some homes are both heated and cooled by heat pUlllpS. In \'linter, the heat pump extracts energy Qr from the cool outside air and delivcrs energy QI! to the warmer air inside. ln summcr, encrgy Qr is removed from tile cool inside air, while energy Q/i is ejected (Q the \",arm air olltside. For a refrigerator or an air conditioner-a heat pump operaling in cooling rnode-work Wis what you pay for, in terms 01" electrical energy running the compressor, whereas Qr is the desired benefit. The most efficient refrigerator or air conditioner is onc that removes the greatest amount or energy from the cold rescrvoir in exchange for the least amount of work.
403
Ikal pump
The coefficient of performance (COP) for a refrigerator or an air cOllclitiOiler is the magniLUde of thc energy extractcd from thc cold reservoir, !Q,.I, divided by the work Wpcrfonned by the device:
.
COP (cooling mode
) =IQ,I -W
Cold reservoir at T£
[12.13]
51 unit: dimensionless The larger this ratio, the bettcr the performance, because morc energy is being removcd for a given amount of '.. .' ork. A good refrigerator or air conditioner \vill have a COP of 5 or 6. A heat pump operating in healing mode warms the inside ora house in wintcr by extracting energy from the colder outdoor air. This statemelll may seem paradoxical, but. recall that. this process is equivalent to a refrigerator rcmoving energy from its interior and ejecting it int.o the kitchen.
ACTIVE FIGURE 12.12 Sc!lemalic diagram ora heal pump, which takes in energy Q; > () from ,s, brain, ~nd skeletal mllscles.!\.'lore intense activit), increases the metabolic nlte to a maximum of about 1 600 W for a superb racing cyclist, although such a high raLe can only be maintained for periods ofa few seconds. \'Vhcn we sit watching a riveting film, \ve give off about as much energy by heal as a briglH (250-W) lightbllib. Regardless of level of activity, the daily food intake shouJdjust balance the loss in internal energy if" a person is not 1.0 gain weight. Further, exercise is a poor substitute [01' dieting as a method of losing weight. For example, the Joss of I pound of body f~u requires the muscles La expend 4 100 kCIIrr(': Ann/lin,
Huol..~.
Q
t1/
-=-+-
K. II. Cuop"r. l\alll;ll11 :-.!("I' Yt,rk. 1~1l;8.
[12.21]
t1/ In t.his definition, absoltlle value signs arc used to show that e is a positive number and to avoid explicitly using minus signs required by our definitions of Wand Q in the first law. Table 12.6 shows the efflcienc}' of" workers engaged in different activities for several hours. These values were obtained by rneasuring the power output and simultaneolls oxygen consurnption of mine workers and calculating the metabolic rale from their oxygen consumption. The table shows that a person can steadily supply mechanical power for several hours at about 100 \V with an efficiency of about 17%. It also shows the dependence of efficiency on activity, and that (' can drop to values as low as 3% for hig'hly inefficient activities like shoveling, which involves lrIallY slans and stops. Finally, it is inlercsting in comparison 10 the average results of Table 12.6 that: a superbly conditioned athlete, efficiently coupled to a mechanical device for extracting power (a bike!), can supply a pO\'..-er of around 300 \'V for about 30 minutes at a peak efficiency of22%.
TABLE 12.6 Melabolic Rale, Power OutpUl, and Efficiency for Differenl AClivilies a Metabolic Rate
Power Output
I1U 11/
TV 11/
(watts)
(waUs)
Efficiency e
Cycling
50S
96
0.19
Pushing loaded coal GIl'S in a mine
525
90
0.17
Shoveling
570
17.5
0.(13
Activity
"Sol/rrl': "Inter- anrl Intra-lndh-idll;11 niITt:n:nCl'~ In Energ"~ Expcndiwn' and Mechanic-al Efficil'l1C\"~ C. H. Wy lid h,lIIl cl ~d .. 1:'r~(IIlIJllIifS !). 17 (1 ~)(ili).
~~MARY 12.2
12.1 Work in Thermodynamic Processes The work clone on a gas at a constant pressure is w~
-Pt1\1
[12.1]
The work done on the gas is positive if the gas is compressed (J.Vis negative) and negative if the gas expands (liVis positive). In general. the work done on a gas that lakes it from some initial state to some final state is the negative of the area under the curve all a PVdiagram.
The First Law of Thermodynamics
According to Ihe first law of thermodynamics. when a system undergoes a chllnge from one state (0 anal her. the change in ilS intel-nal energy liUis
[12.2] where Q is the energy transferred into lhe system by heat and W is the work done on the system. Q is positive when energy enters the system by heat and negative when the sys·
Multiple-Choice Questions lelll loses energy. W is positivt> when work is done on thc system (for example, by compression) and ncgati\'c when the system does positive work on its environment. The change of the intcrnal energy, 6.U, of an ideal gas is givcn by
D.U=
11
[12.51
G',,6.T
where Ct , is the molar specific hem at constant volume.
12.3
IQ,I IQ"I
W;·lIg I!==--~
IQ"I
[12.12]
Ileat pumps are heat engines in reverse. In a refrigerator the heat pump removes t.hermal eller~n' [rom inside the refrigerator. I-leal pumps operaling in cooling mode have coefficient of performance given by COP (cooling mode)
Thermal Processes
An isobaric process is one thai occurs at constant pressure. Tile work done 011 the system in such a process is -p 6.\~ whereas the thermal energy transferred by heat is given by
417
=
IQ,I
W
[12.13]
A heat pump in heating mode has coefficient of perfor-
mance
[12.6]
COP(hcating mode)
=
IQ"I W
[12.14]
with the molar heat capacity at constanl pressure given by C,,~C"+n.
In an adiabatic process no energy is transferred by heat bet.vl.'een the systcm and its surroundings (Q = 0). in this case the f-lrst law gives !i.U = \'11; which means the internal energy changes solely as a consequence of work being done on t.he systelll. Thc pressure and volume in adiabatic processes are related by
pvr =
constant
112.801
where'Y = epiC" is the adiabatic index. In an isovolumetric process the volume doesn'l change and no work is done. For such processt:s, the first law gives /1U= Q All isothermal process occurs:ill constant lempcraLurc. The work done by an ideal gas on the environment is \;11V. 12.
A
2
+--;F V (I i t (; rs) o'------!---:---!c---c2 3 ,1 FIGURE P12.S
SECTION 12,2 THE FIRST LAW OF THERMODYNAMICS
(Problems:l and 13)
m
A cylinder of volume 0.300 m:1 contains 10.0 mol of neon gas at 20.0° C. Assume neon behaves as an ideal gas. (a) What is the pressure of the gas? (b) Find the il1lernal energy of the gas. (c) Suppose the gas expands at constant pressu,'c to a volume of 1.000 m 3. f-low much work is done on the gas? (d) What is the tcmperature of the gas at the new volume? (e) Find the internal energy of the gas when its volume is 1.000 m 3 . (f) Compute the change in the internal energy during the expansion. (g) CompUle AU - \IE (h) Must thermal energy be transferred to the gas during the constant pressure expansion or be taken away? (i) Compute Q, [J1C thermal energy transfer. (j) 'What symbolic relmionship between Q. I1U, and Wis suggested by the values obtained?
6. Sketch a PV diagram of the following processes: (a) A gas expands at constant pressure PI from volume VI to volume \12- It is then kepI al constant volume while the pressure is reduced La P,;!, (Il) A gas is reduced in pressure [rom PI to P2 while its volume is held const rotale~ in a circle \\,;th uniform angular spcecl, its projection Qalong the x~axis moves with simple harmullic moti()I1.
13.4
(c) 0.735 s
POSITION, VELOCITY, AND ACCELERATION AS A FUNCTION OF TIME
\·Ve can obtain an expression for the position of an ol~ject moving with simple harmonic Illotion as a fUI1Clion of time by returning to the relationship between simple harmonic motion and uniform circular Illotion. Again, consider a ball on the rim of a rotating turntable of radius A, as in Active Figure 13.12. 'We refer {Q the circle made by the ball as the referena! cirde for the Illotion. \Ve assume the lUrntablc revolves at a (;O'nsl.al1t angular speed w. As the ball rotates on the reference circle, the angle 0 made by the line OP with the x-a.xis changes with time. Meam·vhile, the prqjection of P on the x-~lXis, labeled point Q, moves back and forth along the axis with simple harmonic motion. From the right triangle OPQ, we see that cos 8 = xl A. Therefore, the x-coordinate ofthe ball is
x= AcosO Because the ball rOlateS with constant angular speed, it follows that 8 = wt (see Chapler 7), so we have
x
~
A cos (wi)
[13.12]
In one complete revolution, the ball rotates through an angle of21T fad in a time equal to the period T In other '.. .ords, the motion repeats itself every T seconds. Therefore, [13.13]
INhere fis the frequency of the motion. The angular speed of the ball as it moves ,-lround the reference circle is the same as the angular frequency of the projected simple harmonic mOlion. Consequently, Equation ]3.12 can be written x = A cos (2",{/)
[13.]4a]
13.4
Position, Velocity, and Acceleration as a Function of Time
This cosine function represents the position of an object moving with simple harmonic mOL ion as a (uncl.ion of time, and is graphed in Active Figure 13.13a. Because the cosine function varies between 1 and -I, x varies between A and - A. The shape of the graph is called sinusoidal. Active Figul'cs I3.13h and 1~{.13c represent curves for velocity and acceleration as a function of time. To find the equation for the velocity, usc Equations ]3.6 and 13.I4a l.Ogether with the identity cos2 e + sin 2 e = 1, obtaining
v= -Awsin(21Tjl)
[13.14b]
where we have used the fact that w = V!(/ nt. The ::t sign is no longer needed, because sine can Lake both positive and negative values. Deriving an expression for the acceleration involves substituting Equation 13.1421 into Equation 13.2, Newton's second law for springs:
,,= - Aw' cos (21T{I)
x=
x
437
II cosw/
(a)
v
(h)
n
«)
[13.14c] (/=-w'lAcoswt
The detailed steps of these derivations are left as an exercise [or the studelll. Notice that ,\'hen the displaccment x is at a maximum, at x = A 01" X = -A, the velocity is zero, and when x is zero, the magnitude of the velocity is a maximum. Further, when x = + A, its most positive value, the acceleration is a maximum but in the negative x-direction, and when x is at its most negative position, ". = -A, the acceleration has its maximum value in the positive x·dircction. These facts are consistent with our earlier discussion of the points at which TJ and a reach their m::Lximum, minimum, and zero values. The maximum values of the position, velocity, and acceleration are always equal to the magnitude of the expression in front of the trigonometric function in each equation because the largest value or either cosine or sine is 1. Figure 13.14 illustrates one experimental arrang-emellt that demonstrates the si nusoidal nature of simple harmonic motion. An object. connected to a spring has a marking pen attached to it. While the object vibrates venically, a sheet of paper is moved horizontally with constant speed. The pen traces out a sinusoidal pattern.
ACTIVE FIGURE 13.13 (a) Displacement, (b) \'elocil)'. and (c) acceleration \"ersus time far an object 1ll0\crimClll:(em)
36. A "seconds" pendulum is one that moves through its equilibriulll position once each second. (The period of the pendulum is 2.000 s.) The length of a seconds pendululll is 0.9927 m at Tokyo and 0.994 2 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations? 137.IA pendululll clock that works perfenly on the Eanh is taken to the ~'Ioon. (a) Does it run f~lst or slow there? (b) If the clock is started at 12:00 midnight, wh,ell will it read after one Eanh day (24.0 h)? Assume the free-fall acceleration on lhe 1'\'100n is 1.63 Ill/S2. :lR. An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0"C. (a) When placcd in a room at a temperature of -S.U"C, will it gain lime or lose time? (b) HO\.. . milch lime will it gain or lose eve I)' hour? Hint: See Chaptcr 10. 39. The free-fall acceleration on Mars is 3.7111/52. (a) \·Vhat length of pendulum has a period of I s on Earth? \'Vhal length of pendulum would ha"e a 1-s period all Mars? (b) An object is suspended from a spring with force constant 10 N/m. Find the mass suspended from this spring that would rcsult in a period of I s on Earth and on Mars. 40. A simple pendululll is S.OO m long. (a) What is thc period of simple harmonic motion for this pendulum if il is located in an elcvator accelcrating upward at 5.00 mjs2?
f----'--+--!;--f-----:--+- I
(,)
FIGURE P13.42
43. A certain I'M radio station broadcasts jazz music at a frequcncy of 101.9 MHz. Find (a) the wave's period and (b) its wavelength. (Radio waves are electromagnetic Wilves lhilt travel at the speed of light, 3.00 x 10 8 m/s.) 44. The dist.ance between twO successive minima of a transverse wave is 2.76 m. Five crests of the wave pass a given point. along- the direction of travel e\'ery 14.0 s. Find (a) the frequency of the \'i·ave and (b) the wm'e speed. 45. A harmonic wave is travcling along a rope. It is observed that the oscillalor Ihal gencrates the wave complctes 40.0 vibrations in 30.0 s. Also, a given maximum travels 425 cm along thc ropc ill 10.0 s. What. is the wavelength? 46.
II
A bat can detect small objects, such as an insect, whose siJ;c is approximately equal to one wavelength of the sound the bat makes. Ifbats emita chirp ata frequcnc)' of 60.0 kHz and the speed of sound in air is 340 mis, what is the smallest insect a bat Gill detect?
4S6
Chapter 13
Vibrations and Waves
47. A cork on t.he surface of a pond bobs up and down two times per second on ripples having a wavelength of 8.50 em. lfthe cork is 10.0 III from shore, how long does it take a ripple passing the cork to reach the shore? 11.8.IOcean waves an: traveling to the east at ~I.O m/s with a dis-
tance 0[20 III between crests. With whal frequency do the waves hillhe from of a boat (a) when the boat is at anchor and (b) when the boat is moving \.. .csnvarcl at 1.0 m/s? ~.O
kg
FIGURE P13.57
SECTION 13.9 THE SPEED OF WAVES ON STRINGS 49. A phone cord is 4.00 In long and has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the laut corel. The pulse makes [our trips down and back along lhe cord in 0.800 s. What is the tension in the cord? 50. A circLis performer ~tretchcs a tightrope bem'cell tWO towers. He strikes one end of Lhe rope and sends a wave along it toward the other tower. He notes thaI it takes the wave 0.800 s to reach the opposite [Ower, 20.0 Tn away. If a I-m length of the rope has a massorO.350 kg, find the tension in the tightrope.
5!. A transverse pulse moves along a stretched cord of length 6.30 II'l having a mass of 0.150 kg. If Lhe tension in the cord is 12.0 N, find (a) the wave speed and (b) the time it takes the pulse to trave! the length of the core!. 52. A taUl clothesline is 12.0 III long and has a mass of 0.375 kg. A transverse pulse is produced by plucking one end of the clothesline. If the pulse takes 2.96 s to make six round trips along the clothesline, find (a) the speed of the pulse and (b) thc tension in the clothesline. 153.ITransversc w;wcs with a speed of 50.0 mls arc to be produced on a sll'elched string. A 5.00-m length orstring ""'ith a tOI.itiw x-direction \\·jth a speed 1'. The wa\l· frolllS are planes parallel La Ihe J-:-plane.
No\\'" consider a small portion of a ""we from t hat is at a great distance (relati\"c to A) from the source, as in Figure 14.6. fn this case the rays are nearly parallel to
each other and the wave fronts are very close to being planes. At distances from the source that are great relative lO the wan~length, lherefore, we can approximate the wave front with parallel planes, called plane waves. Any small portion ofa spherical wave that is far from the source can be considered a plane w~we. Figure 14.7 illustrates a plane wave propagating along the x-axis. If the positive x-direction is taken 1.0 be the direction of the 'Nave Illation (or ray) in this figure, then the wave fronts are parallcl to the plane containing the y- and z-axes.
EXAMPLE 14.3 Goal
Intensity Variations of a Point Source
Relatc sound imensities and lheir di.lilances frolll a point source.
Problem A small source emits sound waves with a power outpul of 80.0 \,\t. (a) Find the intensity 3.00 In from the source. (b) At whaL distance would the intcnsity be one-fourth as much as it is at r = 3.00 m? (c) Find the distance at which the sound level is 40.0 dB. Strategy The source is small, so the emitted wa\'es are spherical and the intensity in pan (a) can be found by substituting values into Equation 14.8. Pan (b) invol\'es solving for rin Equalion 14.8 followed by substitution (although Eq. ILl.9 can be used instead). In part (c), con\"crt from the sound intensity level 1.0 the intensity in W/m'!., using Equation 14.7. Then subslitute into Equation 14.9 (although Eq. 14.8 could be used instead) and soh'e for r2 . ~
. . . . ..
.
. ....
Solution (a) Find the inlensil)' 3.00 m from the source. Substitute ?P'l\ = 80.0 \,,- and r = 3.00 minto Equalion 14.8:
1= '!P.,
+rr,.'
= _8::.:0:.:.'0=--.:.:\'\_',-, = 41T(3.00 m)'
0.707W/m'
(b) At what distance would the intensity be one-fourth as much as it is at r::= 3.00 Ill? Take 1= {O.707 \'\'/m'!.)/4, and sol\'e for rin Equation 14.8:
,.=
(::'J'
80.0W ]" [ 4'11 ( 0.707 W/m')/ ~.O
(c) Find the distance at. which the sound level is 40.0 dB. Convert the intensity level of 40.0 dB to an intc:nsit>1 in W/m' by solving Equation 14.7 fa,. I:
40.0 = 10 log
(.!.-) II)
-->
4.00
= log (.!.-) II)
6.00 m
Chapter 14
468
Sound
Solve Equation 14.9 [or J}, substitute the intensity and the result of part (a), and take the square rooL:
"
I,
r~-
I,
I"r
2
--. "2-"
(3.00m)'(
1'2
......
=
=
, I,
1'1- -
1'2,
0.707W/m' ) 1.00 X 10- 8 W/m'
2.52 X 10"
Jl)
..........................~
Remarks Once the intensity is known at one position a certain distance away from the source, it's easier to use Equation 14.9 rather than Equation 14.8 to find the intensity at any other location. This is particul~1rly true for pan (b), where, llsing Equation 14.9. we can see right away that doubling the distance reduces the imcnsiLy to oncfourth its previous value.
QUESTION 14.3 The power aUlpLlL of a sound system is increased by a [aeLOr of 25. By what factor should from the speakers so lhe sound inten~ity is the same?
yOll
adjust your distance
EXERCISE 14.3 Suppose a cenainjet plane creates an intensity level of 125 dB ala distance 0[5.00 m. "\That intensity level does it creale on the gTound directly underneath it when Oying at. an altitude of 2.00 km? Answer
73.0 dB
14.6
THE DOPPLER EFFECT
I f a car or truck is moving while its horn is blmving, the frequency of the sound yOll hear is higher as rhe vehicle approaches you and lower as it moves away from you. This phenomenon is one example of the DopJJler effect, named for Austrian physicist Christian Doppler (1803-1853), who discovered iL. The same effect is heard if you're on a motorcycle and the horn is stationary: the frequency is higher as you approach the source and Im,,'cr as yOll move 3\vay. Although the Doppler efTect is most often associated "vith sound, it's common to all waves, including light. In deriving the Doppler effect, we assume the air is stationary and that all speed measurements are made relative to this stationary medium. The speed Vo is the speed of the observer, V s is the speed of' the source. and v is rhe speed of sound.
Case 1; The Observer Is Moving Relative to a Stationary Source tn Active Figure 14.8 an observer is moving with a speed of IJO w"iard the source (considered a point source), which is at rest. (vs = 0). '/lie take the frequency of the source to be Is, the ,,,'avelength of the source to be As, and the speed of sound in air to be v. If both obsen'er and source are stationary, the obsen·cr detcclS Is wave fronlS per second. (That is, ,,,hen V o = 0 and V s = 0, the obsen'cd frequency fo equ1 ring or length L lixetl at both C'l1d~. TIl(' ch~lrach;rislic freqllt:ncies clr\,illl'al ion llll"m ;L harlllonic series: (b) the flindalllenta] fl"equcllCr. or first harmonic; (c} the second h:ll"l'n,)nic; and (d) the thirrl harmonic. I\'ole lh:1l N (a)
v
Ji
A;" =
=
v 2L
[14.15]
Recall that the speed ofa wave on a string is v = \/F/J.L, where Fis the tension in the string and J..L is its mass per unit length (Chapter 13). Substituting imo Equation 14.15, we obtain [14.16] (c)
Multillash phntographs of standillgwa\e pancrns in a cord driven by a \"ibralOr ai, the left end, The singleloop pattern in (a) represcliis the
fundamemal frequency (n
=
I). the
two-loop paHern in (b) the second
harmonic (n:= 2), and Ihe Ihree-loop paHern in (e) the lhird h,lnTlonic (II = 3).
This lowest frequency or vibration is called the fundamental frequency of the vibrating string, or the first harmonic. The first harmonic has nodes only at the ends: the points of attachment, with node·anlinode paLLerll ofN-A-N. The next harmonic, called the second harmonic (also called the first overtone), can be constructed by inserting an additional nodeanlinode segment between the endpoints. This makes the pattern N-A-N-A-f\T, as in Active Figure 14.18c. We count the node-antinode pairs: N-A, A-N, N-A, and A-N, four segments in all, each representing a quarter ' I. . avclength. \'Ve then have L = 4(A:/4) = A2 , and the second harmonic (first overtone) is
h
= -
v Lv 2 (9_Lv)
A2
= -
=
=
2Ji
This frequency is equal to lwice the fundamental frequency. The third harmonic (second overtone) is constructed similarly. Inserting one Illore J\'-A segment, "ie obtain Active Figure 14.18d, the patt.ern of' nodes reading K-A-N-A-N-A-N. There are six node-anti node segments, so L = 6(A 3 /4) = 3(A:/2), which means that A~ = 2L/3, giving
fl
v A,
3v 2L
=-=-=
3J;
All the higher harmonics, it turns out, are positive imeg·er multiples of the fundamental:
1" The frequencies!l>
2/1,
=
nJi
n = 2L
(F-
V;
11.
= 1,2,3, ...
[14.17]
3.il, and so on form a harmoll.icseries.
QUICK QUIZ 14.4 Which of the following frequencies are higher harmonics of a slring wilh fundamental frequency of 150 Hz? (a) 200 Hz (b) 300 Hz (c) 400 Hz (d) 500 I-Iz (e) 600 Hz \Alhen a stretched string is distorted to a shape that corresponds to anyone orits harmonics, after being released it vibrates only at the frequency of that harmonic. If the string is struck or bm.. .ed, however, the resulting vibration includes different amounts of' various hannonics, including the fundamental frequency. Waves not in the harmonic series are quickly damped out on a string fixed at both ends. In effect, when disturbed, the string "selects" the standing-wave rrequencies. As "ie'll
14.8
see later, the presence of several harmonics on a string gives stringed instruments thcir characteristic sound. which enables LIS to distinguish one from anothcr cven when they are producing identical fundamental frequencies. The frequcncy of a string on a musical instrument can be changed b>1 varying either the Icn>iion or the Ipllgth. TIl(' tf'nsion in glJit:.:l:r ::Inc! violin slrings i,1; V::lJ"ied by turning pegs on the neck of the instrumclll. As the tension is increased, the frequency of the harmonic series increases according to Equation 14.17. Once the instrument is tuned, the musician varies the rrequency by pressing the strings againslthe neck at a variely of positions, thereby changing the effective lengths of the vibrating portions of the strings. As the length is reduced, the frequency again increases, as [ollo\'.'s from Equation 14.17. Finally, Equation 14.17 shows that a string of fixed lcngth can be made to vibrate at a lower fundamental frequency by increasing its mass per unit length. This increase is achieved in the bass strings of guitars and pianos by wrapping the strings with metal windings.
EXAMPLE 14.7 Goal
Standing Waves
477
APPLICATION Tuning a Musical Instrument
Guitar Fundamentals
Apply standing-wave concepts to a stringed instrumcnt.
Problem The high E string on a certain guitar measures 64.0 em In length and has a fundamental frequcncy of 329lh. \Vhen a gllitar~ ist presses down so that the sU-ing is in contact with the first fret (Fig. 14.19£1), the string is shortencd so that it plays an F note that has a frequency 0[349 Hz. (a) How far is the fret [rom the nut? (b) Overtones can be produced on a guitar string by gently placing the index finger in the location of a node of a higher harmonic. The string should be touched, but. not depressed against. a fret. (Given the width of a finger, pressing l.00 hard v,,Iill damp out higher harmonics as well.) The fundamental Frequency is thercby suppressed, making it possible to hear overtones. Where on the guitar string relative LO the nut shoulclthe finger be lightly placed so as to hear the second harmonic? The fourth harmonic? (This is equivalent to finding the location of the nodes in each case.)
(a)
bridge
~nd
frel
Strategy For pan (a) lise Equation ]4.15, corresponding LO the funda(b) mcntClJ frequency, to find the speed of waves on the string. Shortening lhe string by playing a higher note doesn't affect the wave speed, ,...· hich FIGURE 14.19 (Example 1'1.7) (a) Plaring an F note all a guitar. (b) Some pans ora guilar. depends only on the tension and linear density of the su-ing (vl.'hich are unchanged). Solve Equat.ion 14.15 for the ncw length L, using the new Fundamental frequency, and subtract this length [1'0111 the original length to find the distance frol11 the nut to the first frer. In part (b) remember that. the distance from node to node is haifa v,:avelength. Calculate the wavelength, divide it in two, andlocC\te the nodes, "vhich are integral numbers of half-,vavelengths [rom the nuL ""oip: The nut is a small piece of wood or ebony at the LOp of the fret hoard. The distance from the nut 10 the bridge (belm\' the sound hole) is the length of the string. (See Fig. 14.19b.)
Solution (a) Find the distance from the nut to the first frct. Substiwtc L(( = 0.640 m and}; = 329 Hz into Equalion 14.15, lInding the wave speed on the string:
It ;
v 9L -
0
v; 2 Lof, ; 2(0.640 m)(329 Hz) ; 421 m/s 421 m/s 2 ( 349 Hz )
Solve Equation 14.-15 for the length L, and substitute the wave speed and the frequency of an F note. Sllbtractthis length from t.he originallengt.h L o to find lhe distance from the nul to the firSl fret.:
D.X;
Ln - L
~
64.0
~
. 0.603
elll -
III ;
60.3
60.3
CIll;
CIll
3.7
CIll
478
Chapter 14
Sound
(b) Find the locations of nodes for the second and fourth harmonics.
The second harmonic has a \'\'avelength A:? = 1.... 0 = 64.0 em. The distance from nullo node corresponds to haIfa wavelength. The fourth harmonic, of wavelength ;\4 = ~Lo = 32.0 CIll, has three nodes bet\vecn the endpoints:
Remarks Placing a finger at the position Ax = 32.0 em damps out the fundamental and odd harmonics, but not all the higher even harmonics. The second harmonic dominates, however, because the rest of the string is free to vibrate. Placing the finger at .1.x = 16.0 (Ill or 48.0 em damps Ollt the first through third harmonics, allowing the founh harmonic 1O be heard.
QUESTION 14.7 True or False: If a guitar string has length L, gently placing a thin object at the position (~)II twill always result in the sounding a higher harmonic, where 11 is a positive intcger. EXERCISE 14.7 Pressing the E string dm.. .' n on the fret board just above the second fret pinches the string firmly against the fret, gi\'· ing an F-sharp, which has frequency 3.70 X 10' Hz. (a) Where should the second frel be localed' (b) Find lwo locations \,'here you could tol/ch thc open E string and hear the third harmonic. Answers (a) 7.l cm from the nut and 3,4 cm from the first fret. )Jote that the distance from the first to the second fret isn't the same as from the nut to the first fTeL (b) 2].3 Clll and 42.7 cm from the nut
EXAMPLE 14.8 Goal
Harmonics of a Stretched Wire
Calculate string harmonics, relate them to sound, and combine them ' . .· ith tensile stress.
Problem (a) Find the frequencies of the fundamental, second, and third harmonics of a steel wire 1.00 III long ,,,,,ith a mass per unit length of 2.00 X 10-3 kg/Ill and under a tension ofSO.O N. (b) Find the wavcJengths of the sound 'Naves created by the vibrating wire for all three modes. Assume the speed of sound in air is 34-5 m/s. (c) Suppose the wire is carbon steel with a density of 7.80 X 10 3 kg/nl"'I, a cross-sectional area A = 2.56 X 10-7 m 2 , and an elastic limit or2.S0 X 108 Pa. Find the fundamental frequency if the wire is tightened to the elastic limit. Neglect an)' stretching or the wire (which would slightly reduce the mass per unit length).
Strategy (a) It's easiest to find the speed of waves on the wire then substitute into Equation H.15 to find the first harmonic. The next tVI'O are Illultiples of the first, given by Equation 14.J7. (b) The frequencies of the sound waves are the same as the frequencies of the vibrating wire, but t.he wavelengths al-e different. Use 11£ = fA, ,.,.'here u£ is the speed of sound in air, LO find the wavelengths in air. (c) Find the force corresponding to the elastic limit and substitute it into Equation 14.16.
Solution (a) Find the first three harmonics at the given tension. Use Equation l3.ISLO calculate the speed orthe \\lave on the wire:
v=
!?:' I -~
/.L
v 2L
Find the \-"ire's fundamental frequency from Equation 14.15: Find the next two harmonics by multiplication:
\
I,
=
2I,
\
80.0 N , =2.00XIO-m!s 2.00 X 10 3 kg! m
2.00 X 10' mls 2(1.00 m)
= 2.00 X 10' Hz ,j; =
1.00 X 10' Hz
3I,
= 3.00 X 10' Hz
14.9
Forced Vibrations and Resonance
479
(b) Find the wavelength of the sound waves produced. Solve v" = JA for the wavelength and substitute 1he frequencies:
~
vif,
= (345 m/s)/(1.00 X 10' Hz) =
3.45 m
A, =
v/};
= (345 m/sl/(2.00 X 10' Hz)
1.73 m
A" =
vi;;
~ (345 m/sl/(3.00 X 10' Hz)
l.l5 m
A,
(c) Find the fundamental frequency corresponding LO the elastic limit. Calculate the tension in the wire from the elastic limit:
Suhstitute the values of F, f.l, and L into Equation 14.16:
F - = clastic limit
A F
=
Ii
= 2L \ / ;
Ii
= 2(1.00 m) \
---+
F = (elastic limit)"'\'
(2.80 X 108 Pa)(2.56 X 10-7 m')
= 71.7 N
IfF I
!
71.71\ 200 X 10 'kg/m
94.7 Hz • _ •• _.A
.
~
Remarks From the ans\ver to pan (c), it appears we need to choose a thicker wire or use a better grade of steel with a higher elastic limit. The frequency corresponding to the elast.ic limit is smaller than the fundamental! QUESTION 14.8 A string on a guitar is replaced with one of Im·ver linear densil)'. To obtain the samc frequency sound as previously, must the tension of the ne\v string be (a) grcater than, (b) less than, or (c) equalLO the tcnsion in the old string? EXERCISE 14.8 (a) Find the fundarnental frequcncy and second harmonic if the tension in dle wire is increased to 115 N. (Assume the wire doesn't stretch or break.) (b) Using a sound speed 0[345 mis, find the \v;welengl.hs of the sound waves produced.
Answers
14.9
(a) 1.20 X 10' Hz, 2.40 X 10' Hz
(bl 2.88
Ill,
1.44 m
FORCED VIBRATIONS AND RESONANCE
In Chapter 13 we learned that the energy of a damped oscillator decreases over time because of friction. It's possible to compensate for this energ'y loss by apply~ ing an external force that. does positive work on the system. For cxample, suppose an objccr-spring systcm having some naUiral frequency of vibration.fo is pushed back and fonh by a periodic force \\~th frequency! The system vibrates althe frequency/of the driving force. This type or motion is referrcd to as a forced vibration. ll.s amplitude reaches a maximum when the frequency of the driving force eqllals the natural frequcnC)' of tJ1e system ic"l' called the resonant frequency of the system. Under this condition, t.he systcm is said to be in resonance. In Section 14.8 we learned that a stretched string can vibrate in one or more of its natural modes. Here again, if a periodic force is applied to t.he string, the amplitude or vibration increases as the frequency of the applied force approaches one of the su-ing's nault'al frequencies of vi braLion. Resonance vibrations occur in a wide variety of circumstances. Figure 14-.20 illustrates one experiment tJlat demonsu·ates a resonance condition. Se\'eral pendulums of different lengths are suspended from a flexible beam. If one of them, such as A, is set in mOlion, the others begin to oscillate because ofvibrmions in the flexible beam. Pendulum C, the same length as A, oscillates with the greatest amplitude because its natural frequency matches that of pendulum A (the driving force). Another simple example of resonance is a child being pushed on a s\ving, which is essentially a pendulum with a natural frequency that depends on its length. The swing is kept in motion by a series of appropriately limed pushes. For its amplitude 10 increase, the swing 111ust be pushed each time it returns to the person's hands.
AV B FIGURE 14.20 I{csOIl,l11ct-·.lfpendulum 1\ is SCI. ill oscillation, ani)' pcndullim C, wilh a lenglll matching lhal of 1\, will e,cmualh' oscillate with ,I large ;ImpliLud..:. or resonale. The :!lTOh'S indicl\c motion perpcnrliClllar 10
the page.
480
Chapter 14
Sound
APPLICATION Shattering Goblets with the Voice
FIGURE 14.21 Standing-wave pattern in a vibrating wineglass. The glass will shatter if the :l.Inplilude of vibration becomes too large.
This corresponds to a frequency equal to the natural frequency of the swing. If the energy put into the system per cycle of mOlion equals the energy lost due LO friction, the amplitude remains constanl. Opera singers have been known to set crystal goblets in audible \'ibration with lhr.ir pnwe.rflll voices, as shown in Figure 14.21. This is yet another example of resonance: The sOllnd waves emill.ed by the singer can set up large-amplitude vibrations in the glass. If a highly amplified sound wave has the right frequency, the amplilUde of forced vibrations in the glass increases to the point where the glass becomes heavily su"ained and shaLters. The classic example of structural resonance occurred in 1940, \.. . hcn the Tacoma Narrm.. . s bridge in the slate of \VashingLOn \.. .as set in oscillation by the wind (Fig. 14.22). The ampliulde of the oscillations increased rapidly and reached a high value until [he bridge ultimately collapsed (probably beG1USe of met,...-
14,11
Beats
485
mon' out or phase, then illlO phase again. and so on. As a conscClucncc, a listener al some fixed point hears an aliernation in loudness. knowil as beats, The IlllmbCT or beats per second, or the bmt jiHjlil'Jlf)', equells the difference in rrequency between the two sources: };, =
II, - I,l
~
[14.20]
where J" is the beal frequency and I\ and j~ are the t,,'o frequencies. The absolute ,'alue is used because the beat rref]uenc), is a pusiti,'c quantity and will occur regardless of the order or subt raction. A stringed instrument stich as a piano can be tUlled by beating a note on the instrument against a note or knm"'11 frequency. The string can then be wned to the desired frequenc), by a(~jll.stinp; llle tension LllllilllO healS ' between the onset ora loud ~ollnd and the ear's protecti\'(.' reaction, howevcr, so a vcry sudden IOILd sOllnd can still damage the car. The complex stnlclure or the human cal" is believed to be related to the Ltct then mammals evol\"ed from seagoing creatures. In comparison, insect cal"s are considerably simpler in design bccause insects hm'(' always heell bnd residents, A typical insect ear consists of an City Icvel of these sounds can reach 103 dB. measured a distance of 5.0 III from the source. Determine the intensity le\-e! of the infrasound 10 km from the source, assuming the souncl cllerg)' rildi;:lles ulliformly ill illl directiolls. lB. A family ice show is held at an enclosed arena. The skaters perform lo music playing at :llevcl of80.0 dB. This intcnsit)' level is too loud for your baby, who yells at 75.0 dB. (a) What tOlal sound il1lell'iil~' eng-ulfs you? (b) Whal is the combined sOllnd lc\"CI? 19. A Irain sounds its horn as it approachc" an imer:-.ection. The h01"11 Cill jusl be heard al ,t le\'cl of 50 dB by ,Ill ohscl"\'er 10 km 'Iway. (a) \\'hat is lhe tring slip~ from it-. normal tension or 6,00 x 101 1': to 5..10 x 10 2 ?,\, "hal heat f1-cf]ucno' is heard when the hammer 'itri .... e~ the t\\'O ~tring .. simulta-
~>-
nl.'ol1sh~
'" FIGURE P14.46
[If]
56. The G strin~ nn a \ iolin has a fundamental frequcnC\' of 196 I-b. It is 30.0 cm long and ha" a Illa ..s of 0.500 g. While lhi" '>trin~ i-; ..ounding, a ncar/)\' \io/inis, effcctiyely :)hortcns Ihe G stl'ing: 011 her idC'llticll \iolin (b\ 'iliding her finger dO\\'1l lhe ~trillg) until a beat freqlleu(\" of 2.00 Hz b heard bCl\\ccn the two ::.trin~.,. \\'hen lhat occurs. \\'hat is the dlectiw length of her strillg~
r\ 60.00-(m guitar sIring Hilder a tt'nsioll 01' 50.000:\1 ha" a mass pn IInit length of 0,100 00 ~/(,1l1. What i.. the highe...t rC<jOll. (h) Opening- hole:-. in the side eneni\l~h !)hOlten~ the.' !eng-Ih of ,hc re'lOIl,tlll colulllll, If Ihe hi,!{huJllc a bod\ tl'mperallln: of37~C.)
ADDITIONAL PROBLEMS 62. Thc intensil\ lewl (Iran orchc.',>Lra is W> dB. A single \ iolin reaches a !t'\-e! of 7.0 X In ' dB. \'\'hal is the ratio of the "/lund ililellSil) of tht' filII orchestra to the illlC'llsit)' 01 a sing-Ie \'iulill:
496
Chapter 14
Sound
63. ASSlIllle a IOlldspeaker broadcast> soulid equ"ll)' ill "II directions and produces ~ollnd with a level of 103 dB at a distance of 1.60 III from its ccmer. (
(ll!=-=
m,
3.6 X 10
9.0 X IO~2
Ill/S 2
17~
9.11 X 10 ," kg
4.0 X 10
17
mis'
.......... Remarks The graVilaLional force between the charged constituents of the atom is negligible compared with the dcclric force between them. The electric force is so SHang, however, that any net charge on an object quickl)' atlraCls nearhy opposite charges, neutral il.ing the ol~jecl. As a rc\uh. gravity plays a greater role in the mechanics of 1l'1Oving ol~ecls ill cvcryday life. QUESTION 15.1 I I" the distance between two charges is doubled, by Wllilt facLor is the magnitude of the clectric force changcd? EXERCISE 15.1
Find I hc magnitude of the electric force bet\',ieen t\\10 protons separated by I fcmLol11eter (10- 1:' 01), approximately the di.."tance hetween two protons in the nucleus of a helium atom. The answer may not appear large, bUl if not for the 51 rang nuclear force, the two prolOns \vottld accelerate in opposite directions at over I X 10~9 Ill/S2 ! Answer
2 X 10'2 ~
The Superposition Principle When a number of" separate ch:.uges act on the charge 01" interest, each exerts an electric force. These electric forces call all be computed separately, one itin~
exerted at a distance in that the force is no\\' exerted by something-the nelclthat is in the sarnc locaLiol1 as the charged object. Figure 15.9 shows 3n objcCl with a small positive charge flo placed near a second o~jcC[ with a lllllch larger positive charge Q.
E
The electric field produced by a charge Q at the location of a sillall "tcst" charg'c flo is defined a~ the electric force exened by Qon flo divided by the lest charge 10:
F
-)E""-F
[15.3]
flo
51 unit: newton per coulomb (N/C)
Concepwally and experimentally, the test charge 10 is required to be very small (arbitrarily small, in fact), so it do~n't cause any significant rearrangement of the charge creating the electric rield E. lvlathematically, however, the size of the test charge makes no difflallce ,-frum radially ml11l/onlJ rrom
q.
(h) 11' q is
negalive, lhe electric fidd at !".6:
£';1 =
0
= }o', cos (90') = 0 I,;" = Ie', si II (900) = 3.93
",.1'1',1
= (8.99 X 10" N'm'/C') (5.00 X 10 ':,C)
ri 1.80
Obtain Lhe x-component of E:!, Ilsing the triangle in Figllre 15.12 to find cos 8:
X 10" N/C
(0500 m)-
w" N/C
X
acU 0.300 cos (} = = - - = 0.600 hyp 0500 Ii" = E, cos 0 = (1.80 X ~
Obtain I he y-componcnt in the same way, hUl a miJius sign has to be provided for sin 8 because this component is directed downwards:
1.08
X
10" l'\/C) (0.600)
lO" I\;/C
. opp 0.400 SII'O = - - = - - = 0.800 h)'p 0.500
E" = E, sill 0
~
(1.80 X 10" N/C)(-O.800)
= -1.44 X JO" N/C
Sum t.he x-components the resultant vector:
1.0
get the x-component of"
lo~ =
E"
+
Ii," ~ 0
+
1.08 X 10" N/C ~ 1.08 X 10" N/C
510
Electric Forces and Electric Fields
Chapter 15
SUIll the )' -components to get the y-component or the result.ant vector: Use the Pythagorean theorem to rll1d the magnitude of the resuhalll vector:
EJ
= E,y + E" = 0 + 3.93
1':, = 2.49 X 10' N/C
E=VE}+Jo.~'= 2.71 X10'N/C
The inverse tangent function yields the direction of the resultant vector:
(b) Find the force placed at P.
all
(all
-I
(2.49 X 10'• N/C) = 66.6 108 X 10" N/C
0
a charge of2.00 X 1O-R C
Calculate the magniLUde of the force (the direclion is the same as that ofE because the charge is POSiliyc): ~
F= Eq = (2.71 X 10' N/C)(2.00 X 10-' C) 5.42 X 10-3 N
.
.
~.-.-
X 10" N/C - 1.44 X 10' N/C
......
Remarks There were numerous steps to this problem, but each \·...as vcry short. ''''hen attacking such problems, it's important La focus on one small step at a time. The solution comes not frolll a leap of genius, bm [rom the assembly of a number of rclativdy easy parts. QUESTION 15.5 Suppose q'!. were moved slowly to the right. vVhat \·.:auld happen to the angle c/J? EXERCISE 15.5 (a) Place a charge of -7.00 j.LC al point Panel find the magnilude and direction of the electric field at the location o[ q'1 due to 'II and the charge at P. (b) Find the ITIZlgnitude and direction or I he force 011 (h.
Answer
(a) 5.84 X
JO' N/C. = 20.2° (b) F ~ 2.921'1, = 200.°
15.5
ELECTRIC FIELD LINES
A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the electric field vector at any point. These lines, introduced by Michael Faraday and called electric field lines, are related 1.0 the electric field in any region of'space in the following way: ~
1. The electric field vector E is tangent to Ihe electric field lines at each point. 2. The number of lines per unit area through a sllrface perpendicular to the lines is proportional Lo the strength of the electric field in a given region. Note that E is large \".. hen the field lines are close together and small when the lines arc far apart. Figure 15.13a shows some reprcsentative electric field lines for a single positive poinl charge. This t\\'o~dimensionaldrawing contains only the field lines that lie in the plane containing the point charge. The lines are actually directed radially outward [rom the charge in all directions, some"...hatlikc the quills of an angry porcupine. Because a positive test charge placed in this field would be repelled by the charge fj~ the lines are directed radially away from the positive charge. The electric fidd lines for a single negative point charge are directed [Oward the charge (Fig. 15.13b) because a positive lest charge is attracted by a negauve charge. fn either case the lines are radial and extend all the \-va)' to infinity. Note that the lines are closer together as they get near the charge, indicating that t.he strength of the flcld is increasing. Equation 15.6 verifies that this is indeed the case. The rules for drawing electric field lines ror any charge distribution loHmv directly from the relationship between electric field lines and electric field vectors:
15.5
Electric field Lines
Sl1
.~
(a)
Ie)
(b)
FIGURE 15.13 The den!"i, lidd lil1('~ rnr a poitllcliargl". (,1) For.l po~ilin:' pOilllch;lrgt..:, lh~ linc~ radiale Oll1\\'ard. (b) For a Ilt'g"cmat ion of I he electric ri.ckt at variolls locations. Excepl in special cases. they do lIol represent 1he pal h of
a charged parlic1e released in are .. mall picles nflhrc'ad su~pended ill oil, I\hich aligllll'illl the dCClric fidel produced by 11110 rharged Ctlliductors.
QUICK QUIZ 15.6
Rank the magnilUdes of the electric field at poims A., B, and Cin figure 15.15, with the largest magnitude lIrsl. (aJ A, /3, c: (b) A, C, IJ (c) C, A, IJ (el) The answer can't be determined by visual inspection.
APPLYING PHYSICS 15.1
MEASURING ATMOSPHERIC ELECTRIC FIELDS
The electric fielel near the surface of' the Earth in fair weather is about lOO N/C downward. Under a thundercloud, the electric field can be very large, on the order 0[20 000 I\jC. How are these electric fields measuITd?
Explanation A device for measuring these flelds is called the Jield mil/.. figure 15.17 sho\vs the fundamental componclHs ofa field mill: two metal plates parallel to the ground. Each pi
~
] -~ ~
(h)
• I Electric fidel paucrn ofa ch;lrgL:d coneiliCLing plalL: llcar all oppo~ild~· charged p,)inteel condliclOL Small pieces nflhread sllspL:ndcd in oil align with the dcnric field lillL:S. 1\olice th;lllhe electric field is IIlost illlense near the poililed pan of llie condllclor. \I'llere the radius of CIII ....alnre is the smallest. ..\lso.lhe lincsar(' perpendicular \0 the C011dllcton,.
FIGURE 15.18 (a) !\t:galivt:cllargc"al lhesurfa,cofacQnduClor.lfllll' t:leclric f-ield \l'l:r(' al ,Ill allgle lO the ~1l1 face, as ~hOWll. an deetde force wOlild ilL: eXCl"l b. (c) Ifan additional charge of -2Qis placed at the center, find the electric field for r> b. (d) ,~rhat is the distribution of charge on t.he sphere in part (c)?
Strategy For each pan, dra\\' a spherical Gaussian surface in the region of illlerest. Add up the charge inside the Gaussian surface, substitute it and the area illLO Gauss's la\'/, and solve for the electric field. To find the distribution 01" charge in pan (c), LIse Gauss's law in reverse: the charge distribution must be such that the electrostatic field is zero inside a condUClOr.
,
Caussian
t I
e",·,·r,,,e
If
,f/
/
b
--'Z:.-:::'V --....
I
GalJssian surface
" '
+
+
+
\ (,)
(hi
(el
FIGURE 15.29 (Exaillplc 15.7) (a) Tht.· electric field in"ide a ulli~ forml)' charg·ed spherical shell i~ zero. It is also zero for tht: conducting material in the regioll a < r < b. Tlie fidd olll"ide is tht: same as LhaL of a POillt charge Ilaving a total charge Q located :ll the center of the shell. (b) '["he cOllstruclioll of a GauI;si:'l11 surface for calculating the electric lield insirlea spheriral shell. (c) The cOlbtruction ora G;mssi;lll surfan: for e chaptl:r Illlraighlforw
UA
1'1" - 0 J-
~
.>
'm ,
l
2 (l .67 X lO
T
'kg-
)(-1.68XIO
17
J)
1.42 X 10' Ill/s
(b) rind lhe electron's inilial speed given lhal. its spccd has fallcn by halfal x = 0.120 Ill. Apply conservation or energy once again, substituting expressions for the initial and final kinetic energies:
1';KH + 1';1'£ = 0
(~m.~v/ - ~m ..v,2)
+
tiPE = 0
Substitutc the condition vI = ~Vi and subtract the change in potential energy from both sicles: -1';1'1£
Combine terms and solve for v" the initial speed, and subsl illite the change in potential energy found in Examplc 16.11,:
81';1'£ Vi=
3m"
8(3.36 X 10
".Il
3(9.11 X lO " kg)
9.92 X 10(; m/s ~
-
_-
.
_
..........
Remarks Although the changes in potential energy £lssociated with the protoll and electron were similar in magnitude, the effect on their speeds differed dramatically. The change in potential energy had a proponionalely much greater effect on the much lighter electron than 011 the proton. QUESTION 16.2 True or False: lfa proton and electron both Illove t.hrollgh the same displacement in an electric field, the change in potential energy associated with the proton must be equal in magnitude and opposite in sign to the change in pOlential energy associated with the electron. EXERCISE 16.2 Refer to Exercise 16.!. Find the electron's speed at. x
Answer
1.35 X 10
7
lll/S
= -0.180111.
The ans\\'cr is 4.5% of" the speed of light.
Electric Potential In Chapter ]5 iL was convenienl to define an electric field E relat.ed to the electric force = (iF.... In this 'tva)' the propert.ies 01" nxed collections or charges could be easily studied, and the force on any particle in the electric field could be obtained simpl}' by multiplying by the panicle's charge 'I. For the same reasons, it's
F
Chapter 16
536
Electrical Energy cnd Capacitance
useful to deAne an eleclrir IJo/en/iat dijfPTf'ure !:iP/': = q!:iV: Potential difference between
two points
-+
~lf reLued
to the potential energy by
The elcClric pOlential difference 8\f between points i\ and 13 is the change in electric potential cnel-gy as a charge 1 moves from A to B divided by the
charge 'I: !:iPE !:iV=V,,-VA = - -
q
[16.2]
51 unit: joule per coulomb, or volt O/C, or V) This definition is completely general, although in Illany cases calculus ,\-'ould be required to compute the change in potential energy of the system. Because electric potential energy is a scalar quantity, electric potential is also a scalar quantity. From Equation 16.2, we see that electric potential difference is a measure of the change in electric potential energy per unit charge. Ahernatcly, the electric potential diff-erence is the work per unit charge that would have [Q be done by sornc force to move a charge from point A to point B in the electric held. The SI unit of electric potential is the joule pCI' coulomb, called the volt (V). From the definition ofthat. unit, 1 J of work must. be done to move a J-C charge bctwecl1two points that are at a potential di f"lerence of 1 V. In the process o[ moving through a potential difference of I V, the l-C charge gains] J of energy. For the special case of a uniform electric Held such as that. between charged pantllel plates, dividing Equation 16.l hy qgives !:iP/,'
- - = ~E.:D.x
q
TIP 16.1 Potential and Potential Energy Electric potential is characteristic
of the fi~kl only. independent or a Lest charge thaI mar be placed in that field. On lhe other hand, potential energy is a characteristic of the charge-field system due to an interaction between the field and a charge placed in the field.
,.
FIGURE 16.3 & ll'iA)
Comparing this equation \·...ith Equation 16.2, we find that !:i\!= -E,tix
[16.3]
Equation 16.3 shows that potential difference also has units of electric lIeld times distance. It then l"ol1o\-v& that the 51 unit of t.he electric field, the newton per coulomb, can also be expressed as VOllS per meter:
J N/C = I Vim Because Equation 16.3 is directly related to Equation 16.J, remember that it's valid only [or the systcm consisting of a unil-orm electric field and a charge moving in one dimension. Released from rest, positive charges accelerate sponL,lIleous!}' from regions of high potential to low potential. If a positive charge is given some initial velocity in the direction of high potential, it can move in that direclion, but will slow and finall>' turn around, just like a ball tossed upwards in a gravity field. ~egative charges clo exactly the opposite: released from rest, they accelerate from regions of [ow potential l.oward regions of high potential. 'Nork must be done on negative charges to makc them go in thc direction of lower electric potential.
QUICK QUIZ 16.2 If a negatively charged panicle is placed at rest in an electric potential field thaI. increases in the positive x-direction, will the panicle (a) accelerate in the positive ;\"-c1irection, (b) accelerate in the negative x-direction, or (c) remain al rest?
(Quick Qui7.:f.es 16..1
QUICK QUIZ 16.3 Figure 16.3 is a graph of an electric potential as a function of" position. If a positively charged particle is placed at point 1\, what will its subsequent motion be? \'Vill it (a) go to the right, (b) go to lhe left, (c) remain at point A, or (d) oscillale around poim B?
16.1
Potential Difference ond Electric Potential
537
QUICK QUIZ 16.4 If a negatively charged particle is placed al poilH B in Figure 16.3 and given a very slllall kick to the right, ,·\'hat will its subsequent Illotion be? \Vill it (a) go to the riglll and nOI rewrn, (b) go to the left, (c) remain at point 13. or (d) oscillate around point B? A n application of potential di ffercnce is the 12-V battery 10Ulld in an automobile. Such a battery maintains a potential difference across its terminals, with the positive terminal 12 V higher in potential than the negative terminal. Tn practice the negative terminal is usually connected to the metal body of the car, ,vhich can be considered to be at a potcntial of zero volts. The battery provides the electrical current necessary to operatc headlights, a radio, pO\\'>er \.. . indows, motors, and so forth. Now consider a charge of +1 C, to be moved around a circuit that contains the battery connectcd to some of these external devices. As the charge is moved inside the bauer)' frolll the negative terminal (at 0 V) to the positive terminal (at 12 V), the work done on the charge by the battery is 12 J. Every coulomb of positive charge that leaves the positive terminal of" the battery carries an energy of 12 J. As the charge moves through the external circuit I.Oward the negative terminal, it gives tip its 12 J of electrical energy to the external devices. \'Vhen the charge reaches the negative terminal, its electrical energy is zero again. At this point, the battery takes over and restores 12.J of energy to the charge as it is moved from the Ilegativc 1.0 the positive terminal, enabling it to makc another transit orthe circuit. The actual amount of charge that lea\'es the bauery each second and trayerses the circuit depends on the properties of the external devices, as seen in the next chapter.
EXAMPLE 16.3 Goal
APPLICATION Automobile Batteries
TV Tubes and Atom Smashers
Relate electric potential to an electric field and conservatiun of energy.
Problem In atom smashers (also known as cyclotrons and linear accelerators) charged particles are accelerated in much the same way they are accelerated in TV tubes: t.hrough potential differences. Suppose a proton is injected at a speed of 1.00 X IOfi m/s between two plates 5.00 cm apart, as shown in Figure 16.4. The proton subsequently accelerates across the gap and exit.s through the opening. (a) What must the electric potential difference be if the exitspeed is LO be 3.00 X 10 6 m/s? (b) \·Vhat is the magnitude of the electric field between the plates, assun1ing it's constant? Strategy Csc conservation of energy, writing the change in potential energy in terms of the change in electric potential, Ln~ and solve for ~E For part (b), solve Equation 16.3 [or the electric fielcl.
fi:)v.+
+
+
,
+ + LOI\' pote1ltial
5.00 cm FIGURE 16.4 (Example 16.3):\ proton enters a ca\'iry and accclenllcs from one charged plate 1,£\I'ard the other in ;HI e1eclric field E .
. . . . . . ...,
" .. Solution (a) Find the electric potential yielding the desired exit speed of the proton. Apply conservation of energy, writing the potential energy in terms of the electric potential:
t!.KE+ t!.PE= t!.K1:'+ qt!.v=
Solve I he energy equation lor the ch "V, =C,QFrom Active Figure ]6.20a,
\,,'C
see lhaL il\'=
dl~
+ LiV~
[16.13)
where LillI and LlV2 are the potential differences across capacitors C1 and C2 (a consequence of the conservation of energy). The potential difference across any number of capacitors (or other circuit ele~ ments) in series equals the sum of the potential differences across the individual capacitors. SubstiLUting these expressions into Equation 16.13 and noting that Li.V:;:: Q/Ceq , \VC have
JL= Q+R Ct'q
Canceling Q, we arrive
aL
C1
C2
the following relationship:
I
1
1
Ceq
C\
C2
-=-+-
series ) ( combination
[16.14]
If this analysis is applied to three or more capacitors connected in series, the equivalent capacitance is found to be
l
1
J
1
CC(l
C1
C2
C:\
-=-+-+-+
series ) ( combination
[16.15)
As \.. 'e \-vill shO\v in Example 16.7, Equation ]6.15 implies that the equivalent capacitance of a series combination is always smaller than any individual capacitance in the combination. QUICK QUIZ 16.8 A capacitor is designed so lhal one plale is large and the other is small. If the plates are connected to a battery, (a) the large plate has a greater charge than the small plate, (b) the large plate has less charge than the small plate, or (c) the plates have equal, but opposite, charge.
EXAMPLE 16.8
Four Capacitors Connected in Series
Goal Find an equivalent capacitance of capacitors in series, and the charge and \'ollage on each capacitor. Problem Four capacilOrs are connected in series ,..· ith a battery, as in Figure J6.2l. (a) Calculate the capacitance or- the equivalent capacitor. (b) Compute t.he charge on lhe 12~,uF capacitor. (c) Find the voltage drop across the 12~,uF capacitor. Strategy Combine all [Jle capacitors into a single, equivalent capacitor using Equa~ Lion J6.15. Find the charge on this equivalent capacitor lIsing C:;:: Q/~V This charge is the same as on the individual capacitors. Usc this same equation again to find the voltage drop across the 12-fLF capacitor.
+ 18 V
FIGURE 16.21 (Example 16.8) Four capacitor~ connccled ill !'cries.
16.8
.
,...
Combinations of Capacitors ..................... -
553
...
~
Solution (a) Calculate the eqllivalent capacitancc of thc series. 1
1
1
1
I
Coq
~.o fLF
6.0 fLF
12 fLF
24 fLF
-=---+--+--+--
Apply EquaLion 16.15:
C~q =
1.6 fLF
(b) Compute thc charge on the 12-ILF capacitor. The desired charge equals the charge on the equivalent capaciwr:
Q= (;,." "'V = (1.6X 10-hF)(18V)
=
29JLC
(c) Find the voltage drop across the 12-p.F capacitor.
Notice that the equivalent capacitance is less than that of any of the individual capacitors. The relationbe used to find the "oltage drops on the other cap~1CiLOrs,jLlstas in pan (c).
QI~V can
QUESTION 16.8 Over which capacitor is the voltage drop thc smallest? The largest? EXERCISE 16.8 The 24-,uF capacitor is removed from the circuit, leaving only three capacitors in series. Find (a) the equivalent capacitance, (b) the charge on the 6-p.F capaciLOr. and (c) t.he voltage drop across the 6-JLF capacitor.
I. Combine capacitors that are in series or in parallel. following' the derived formulas. 2. Redraw the circuit aftcr every' combination. 3. Repeat the first. t.wo steps until there is only a single equivalent capacitor. 4. Find the charge on the single equivalent. capacitor, using C =
(V~V.
5. Work backwards through the diagrams to the original one, finding the charge and voltage drop across each capacitor along the \\fay. To do this, use the follm\'ing collection of facts: Q/~V
A. The capacitor equation: C =
B. Capacitors in parallel: Ce'l = C1
+
C'-!
C. Capacitors in parallel all have the same voltage difference, DoV, as does their equivalent. capacitor.
.
D. CapacItors
.
.
111 SCt"lCS: -
1
= -
1
+
1
~
C"q Cl C'1 E. CapaciLors in series all have Lhe same capacitor.
char~e,
Q, as docs Lheir equivalent
554
Electrical Energy and Capacitance
Chapter 16
EXAMPLE 16.9 Goal
Equivalent Cap.acitance
Solve a complex combination ofscries and parallel capacitors.
Problem (a) Calculate the equivalent capacilance between (f and b lor the combination of capacitors shown in Fig"lIre 16.22a. All capacilances arc in rnicror'.lrads. (b) Ira 12-V battery is conncClcd across the system between points a and b, find the charge on the 4.0-jLf capacitor in the first diagram and the voltage drop across it.
:!.o
{}
---!f---
a 6.0 b
4.0
Strotegy For part (aJ. usc Equations 16.12 Ie) (al Idl Ihl and 16.1.'> to reduce the combination step by FIGURE 16.22 (Exalllple 16.9) To flild the cquivalclIl capacitance orthc circuit ill step, as indicated in the figure. For part (b), (a). usc the series and paralld rult;s described ill the lcx110 sLlccessively rc(lucc the to find the charge on the 4·.0-p,F capacitor, cirCllit as indicated ill (Il), (c), i.lIld (ork n.:quircd lo charge 1he capacitor to a fin"ll ch;:lJ"ge I I I Q is the area IInder tht· slr;ligh1 tine, whidl cfjltals Q A~/2.
Chapter 16
556
Electrical Energy and Capacitance
APPLICATION
II
Defibcillatars
Large capacitors can slore enough electrical energy to cause severe burns or even death iF they are discharged so that the flow o[ charge can pass through the hearL. Under the proper conditions, hmvevcr, they can be used to susLain life by slopping cardiac fibrillation in heart attack victims. \'\Then fibrillation occurs, the heart produces a rapid, ilTcgular pauenl of beats. A fast discharge of eleClrical
energy through the heart can relurn the organ to its normal betH pattern. Emergency medical teams usc port.able ddibriliaLOfs that contain bauerics capable or charging a capacitor to a high voltage. (The circuitry aCluall)' permits the capacitor to be charged to a much higher voltage than the baner)'.) fn this case and others (camera nash units and lasers used lor fusion experiments), capacitors serve as energy reservoirs that Gill be slowly charged and then quickly discharged to provide large amounts of energy in ,l short pulse. The stored electrical energy is released through the hean by conducting electrodes, called paddles, placed on both sides of the victim's chest. The paramedics must wait between applications of electrical energy because of the time it takes the capaciLors to become fully charged. The high voltage on the capacitor can be obtained from a low-voltage battery in a portable machine through lhe phenomenon of electromagnetic induction, l:O be studied in Chapter 20.
EXAMPLE 16.10 Goal
Apply energy and power concepts to a capacitor.
Problem A fully charged defibrillator contains 1.20 kJ o[ energy stored in a 1.10 X ]0- 4 F capacitor. In a discharge through a patient, 6.00 X -10~.1 of electrical energy are delivered in 2.50 ms. (a) Find the voltage needed to store 1.20 ~r in the unit. (b) ',Vhat average power is delivered to the patient? Strategy Because we know the energy stored and the capacitance, we can use Equation 16.17 to I1nd the required voltage in pan (a). For part (b), dividing the energy delivered by the time gives the average pO\..·er.
,..
.
• • • • • • ••• , • • • • • • • • • • • • •""'1
Solution (a) Find the voltage needed to store L.20 kJ in the unit. Solye Equation 16.17 for L\V:
Energy stored = ~C Li. V 2 I'>v=
2 X (energy stored) C
2(1.20 X lO"J) 1.l0X10"F
4.67 X 10' V (b) What average power is delivered to the patient? Divide the energy delivered by the time:
2P av
energy delivered
= --=----
1'>1
6.00X10'J 2.50 X ] 0 3,
2.40 X 10"W
.....
.............. , .. the study of Re circuits in Chap-
~
Remarks The power delivered by a draining capacitor isn't COl1slanL as we'lll1nd in ter 18. For thal reason, we were able to nnd only an average power. Capacitors are necessary in defibrillators because they can deliver energy far more quickly than batteries. Batteries provide current through relatively slow chemical reactions, whereas capacitors can quickly release charge that has already been produced and stored. QUESTION 16.10 If the voltage across the capacitor ,,,'ere doubled, would the energy stored be (a) halved, (b) doubled, or (c) quadrupled?
16.10
Capacitors with Dielectrics
557
EXERCISE 16.10 (a) Find the encrg·ycOIuained in a 2.50 X ]0-:' F parallel-plate capacitor ifit holds 1.75 X 10 3 Cofcharge. (b) \r\'hat's the.: voltage between Ihc plates? (c) \Vhat new voltage will result in a doubling of the sLOred energy? Answers
(a) 6.13 X 10-2 J
(b) 70.0 V
APPLYING PHYSICS 16.1
(cJ 99.0 \'
MAXIMUM ENERGY DESIGN
Hmo\! should three capacitors and two batteries be connected so that the capaciLOrs will sLOre the maxiIllUIll possible energy?
potential difference, so we would like to maximize each of these fJuantities. I f the three capacitors are connected in parallel, their capacitances add, and if the batteries are in series, their potential differences, similarly, also add together.
Explanation The eneq{Y stored in the capacitor is proportiollal to the capacitance and the square uf the
QUICK QUIZ 16.9 A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Do the foIIO\-\'ing quantities increase, decrease, or stay the same? (a) C (b) Q (c) I.ibetween the plates (d) ,,11 (e) energy stored in the capacitor
16.10
CAPACITORS WITH DIELECTRICS
A dielectric is an insulating material, such as rubber, plastic, or \.. . axed paper. \'\Ihcn a dielectric is inserted between the plates of a capaciLOr, the capacitance increases. fr the dielectric completel), fills the space between the plates, the capacitance is multiplied by the factor K, called the dielectric constant. The following experiment illustrat.es 1 he eHcct 01" a dielectric in a capacitor. Consider a parallel-plate capacitor uf charge Qo and capacitance Co in the absence or a dielectric. The potential diHerence across the capacitor plates can be measured, and is given by a\~) = Qo/Cil (Fig. 16.24a). Because the capacitor is not COI1necLed La an external circuit, thcre is no pathway for charge to leave or be added lO the plates. If a dielectric is nuw inserted between the plales as in Figure 16.24b, the voltage across the plates is Teduc('(/' by the !"acLOr K to the value
"\\.
"11= -
K
Dit:l~etri(
/
+
+
6\~)
~
(a)
(J \
,It'
(h)
FIGURE 16.24 (3) With air between the plates, the vollage across the capacitor is .1.1'0' the capacil;lncc is e,), and the charge is Q(I. (b) \Vith a dielectric hetween the plates. lhe {·harl4c n':lllains at '20. bllltlH: vo!laj.\·c ,llld capacit,lI)cc change.
558
Chapter 16
Electrical Energy and Capacitance
Because K > 1, .1Vis less than 6..\10 _ Because the charge Qo on the capacilOrdoesll't change, we conclude that. the capacitance in the presence of the dielectric must change to the value
or [16.18]
According lO this result, the capacitance is mullijJlied by the [aclor K when the dielectric fills the region bcnveen the plates. For a parallel-plate capacilOr, where the capacitance in the absence of a dielectric is Co = EnA/d, we can express (he capacitance in the presence ofa dielectric as C=
/~. I
"" I
··,.,··, , .. W .'
I
T··
...
........
FIGURE 16.25 Dielectric breakdO\\ II in air. Spark!> an: produn:d when a large "It('mating voltagr: is applied ,tCI"U;"S tIll.: \\'irc~ by a hig-h\"ohage induclioll coil power sllpp1r.
t\
[16.19]
KEO-
rl
From this result, it appears that the capacitance could be made very large by decreasing d, the separaLioll bCl\veen the plates. In practice the lowest value of d is limited by the electric discharge that can occur through the dielectric material separating the plates. For any given plate separation, there is a maximum electric field that can be produced in the dielectric before it breaks down and begins La conduct. This maxilllulll electric field is called lhe dielectric strength, and for air it.s value is about 3 X lOll Vjm. Most insulating mat.erials have dielectric strengths greater than that of air, as indicated by the yalues listed in Table 16.1. Figure 16.25 sho\Vs an instance of dielectric breakdown in air. Commercial capacimrs are ortcn n1ade by using mctal foil intcrlaced with thin sheets of paraffin-impregnated paper or Mylar®, which serves as the dielectric material. These alternat.e layers of metal foil and dielectric arc rolled into a small cylinder (Fig_ 16.26a). One type of a high-voltage capacitor consists of a number of interwoven metal plates immersed in silicone oil (fig. 16.26b). Small capacitors are often constructed [rom ceramic materials_ Variable capacitors (typically 10 pF t.o 500 pF) usually consist or two inten\loven sets of metal plates, one fixed and the other movable, with air as the dielectric. An electrolytic capacitor (Fig. 16.26c) is often used to store large amounts of charge at relatively low voltages. 1L consists of a metal foil in contact with an e1ec·
TABLE 16.1 Dielectric Constants and Dielectric Strengths of Various Materials at Room Temperature Dielect.-ic Dielectric Constant K Strength (Vim) Material Air
1.00059
Bakclilc®
24 X 106
Neoprcne rubber
4.9 3.78 6.7
Nylon
3.4
14 X 101;
Paper
3.7 2.;)6
J() X lOr,
24 X 10'
5.6 2.;)
15 X
Fused quartz
Polystyrene Pyrex® glass Silicone oil
Strontium l.itanate Teflon® Vacuum \-V;llcr
233 2.1 1.00000
80
3 X 10'
8 X 10' 12 X 10'
1'1 X lO" IOtl
8X lOt, 60 X IO€>
16,10
""0
•
I
559
FIGURE 16.26 I hret· colllllleiTial tapadtHI fiesig'n~. (a) A tubular c;IP~1( itor whCJ\c plate.. 10 Iravel from the ligh1 switch to the lig-hlbulb lor the liglllbulb lU op..:r:llc. Electrons alreadv in the filament of the lightbulb move in response to the c1ec1rk Held set up by 1he bauer}'. Also, the baltcl)' dClcs not provide electrons to the circllit; it provides e//('rf!J' to the CXiS1 illg electrons.
Drift SReed of Electrons
Calculate a drift speed and compare it with lhe rms speed of an electron gas.
Problem A copper wire of cross-sectional arca 3,00 X 1O-i ; m~ carries a current of 10.0 A. (a) Assuming each copper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use the ideal gas model to compare the drif't speed with the randall) rms speed an electron would have at 20.0°C. The density of copper is 8.92 g/cm 3 . and its atomic mass is 63.5 ll. All the variables in Equation 17.2 are known exccpt [or 17, the number of' frec charge carriers per unit volume. 11 by recalling that one mole of copper contains an Avogadro's number (6.02 X 1023 ) of atoms and each atom contribut.es one charge carrier to the mctal. The volume of one mole can be found from copper's known density and atomic mass. The atomic mass is the same, numerically. as the number of grams in a mole of the substance. . Strategy
\·Vc can find
~........
~
Solution (a) Find t.he drift speed of the elcctrons. Calculate the volume of one mole of copper [rom its density and its atomic
In
63.5 g
p
8.92 gl Cln'
\1= - =
= 7.12 cm~
mass:
Convert the volume from cm 3 to m 3 :
7.12CI11~(~)3 10-
7.12 XlO-Gm:1
(Ill
Divide Avogadro's number (the number of electrons in onc mole) by the volume per mole to obtain the number density: Solve Equation 17.2 for the drift speed and substitute:
11
=
6.02 X lO~n electrons/mole 7.12 X 10 flm~/molc
= 8.46 X 102R eleClrons/m,1
I v" = - IIqll
10.0 Cis (8.46 X 10" eleclrons/m')(1.60 X 10 19 C)(3.00 X 10
v" =
2.46 X 10
~·I
fi
111 2)
mls
(b) rind the rms speed of a gas of electrons at 20.0°C. Apply Equalion 10.18:
Convert the tcmperature to the Keh'in scale and substitut.e valucs:
v
IS( 1.38
= noh
\
X 10 '''J/K)(293 K) 9.11 X 10 " kg
~ 1.15 X 10'
................................ - .....
mls
.
...
574 Remark
Chapter 17
Current and Resistance
The drift speed of an electron in a wire is very small, only about one-billionth of its random thermal
speed.
QUESTION 17.2 True or False: The drift velocity in a \\"ire of a given composition is inversely proponional to the number density of charge carriers. EXERCISE 17.2
,,\That current in a copper wire \vith a cross-sectional area of 7.50 X 10-7 m::! would result in a drift speed equal to 5.00 X 10- 4 m / 5?
Answer
5.08 A
Example 17.2 shows that drift speeds are typically "cry small. In fact, the drift speed is much smaller than the average speed between collisions. Electrons traveling at 2.46 X 10-'1 mis, as in the example, ,,,'ould take about 68 min to travel J Ill! 1n view of this 10''1' speed, ".. hy does a ligillbulh LUrn on almost inst3maneously ,,:hen a switch is thrown? Think of the flow of water through a pipe. If a drop of water is forced into one end of a pipe that. is already filled ""ith ,",,'atcr, a drop must be pushed ouL the other end of the pipe. Although it may take an individual drop a long time to make it through the pipe, a flm'\' initiated at one end produces a similar flmv at the other end very quickly. Another familiar analogy is the motion of a bicycle chain. \'Vhen the sprocket moves one link, the other links all move rnore or less immediately, even though it takes a given link some time [0 make a complete rotation. Tn a conductor, the electric field driving the free electrons travels at a speed close to that of light, so ,vhcn yOll flip a light s,,,·itch, the message for the electrons to start moving through the wire (the electric field) reaches them at a speed on the order of 101'1 m/s!
QUICK QUIZ 17.2 Suppose a current-carrying wire has a cross·sectiollal area that gradually becomes smaller along the wire so that the wire has the shape ora very long, truncated cone. I-Im,,' does the drift speed vary along the wire? (a) It slows down as the cross section becomes smaller. (b) It speeds up as the cross section becomes smaller. (c) It doesn't change. (d) More information is needed.
17.3 CURRENT AND VOLTAGE MEASUREMENTS IN CIRCUITS To study electric current in circuits, we need to understand how to measure currents and volt.ages. The circuit shown in Figure 17.5£1 is a drawing of the actual circuit necessary for measuring the current in Example 17.1. Figure 17.5b shov..s a stylized figure called a circuit diagram that represents the actual circuit of Figure Ii.5a. This circuit consists of only a bauery and a lightbulb. The word circuil.means "a closed loop of some sort around which current circulates." The battery pumps charge through the bulb and around the loop. i\o charge would flow ,..ithout a complete conducting path from the positive terminal of the battery into one side ofthe bulb, out the other side, Clnd through the copper conducting wires back to the negative terminal of the battery. The most important quantities that characterize how the bulb works in different situations are the current / in the bulb and the potential difference IlVacross the bulb. To measure the current in the bulb, we place an ammeter, the device for measuring current, in line '"\lith the bulb so there is no path for the current La bypass the meter; all the charge passing through the bulb lllust also pass through the ammeter. The voltmeter measures the potential difference, or volt-
Remarks From Table 17.1, the resistivity of Nichrome is about 100 times that or copper, a typical good conductor. Thererore, a copper wire of the same radius would have a resistancc per unit length of onl)' 0.052 Dim, and a 1.00-m length of copper \"ire ofthc S(1me radius \,'ould carry the same current (2.2 A) with an applied voltage of only O.115Y.
Becausc of its rcsistance heatcrs.
La
oxidation, Nichrome is often used for heating elemcnts in toasters. irons. and elcctric
QUESTION 17.3
Would replacing the Nichrome with copper result in a higher current or 100,,'er current? EXERCISE 17.3
',Vhat is the resistance of a 6.0-m length or Nichrome \\:ire that has a radius 0.321 mm? calTY \vhen connected to a 120-V source? Answers
I-IO\\'
much current does it
28 fl; 4.3 A
QUICK QUIZ 17.6 Suppose an electrical wire is replaced with one h;'H'ing every linear dimension doubled (i.e., the length and radius have twice their original values). Does the wire now have (a) more resistance than berore, (b) less resistance, or (c) the same resiswncer
17.5
TEMPERATURE VARIATION OF RESISTANCE
The resistivity p, and hence the resistance, of a conductor depends on a number of [actors. One 01" the most impon£lnt is the temperature o[ the metaL For most metals, reSistivity increases \"ith increasing temperature. This correlation can be understood as rollows: as the temperature of the material increases, its constituent atoms vibrate \"\'ith greater amplitudes. As a result, the electrons find it more difficult to gel by those atoms,just as it is more difficult to weave through (I crowded room when the people are in motion th I" ~
I
"
power is delivered to the bulb. This current spike at the bcg'illning of operation is the reason lightbulbs ofLen fail immediately after they are lllrned on. As the filament 'varms, its resistance rises and the current decreases. As a result, the power delivered La the bulb decreases and the bulb is less likely to burn Ollt.
~ J"
> I, = J" > I, = If
I,
b
~I'
(Quick Quiz
17.~)
QUICK QUIZ 17.9 T\vo resistors, A and B, arc connected in a series circuit with a battery. The resistance of A is twice thal ofB. \Vhich resistor dissipates more pm,vcr? (a) Resistor A does. (b) Resistor B does. (c) More informat.ion is needed. QUICK QUIZ 17.10 The diameter of wire A is greater than the diameter of wire B, bUllhcir lengths and resistivities are identical. For a given voltage difrerence across the ends, what is the rel£lLionship between fPA and 2Jl B , the dissipated power ["or wires A and B, respectively? (a) 'J'., ~ (YlB (b) (Yl,\ < 01'B (c) 01'A > 01'"
EXAMPLE 17.5 The Cost of Lighting Uj) Your Life Goal
Apply the electric power concept and calculate the cost. of power usage lIsing kilowatt-hours.
Problem A circuit provides a maximum current of 20.0 A at an operating voltage of 1.20 X lO~ V. (a) How many 75 "V bulbs can operate with this voltage source? (b) At $0.120 per kilov'iau-hour, how much does it cost to operate these bulbs for 8.00 h'
r
Strategy Find the necessary power with QP = .6Vthcn divide by 75.0 '''' per blllb to gCllhe total number of bulbs. To find the cost, converL pm.. . er to kilowatts and multiply by the number of hours, then multiply by the cost per kilowatt-hour. ,...-
~
Solution (a) Finclthe number of bulbs that can be lighled. Substiwte into Equat.ion l7.8 to get the Lota] power:
(Yl"" .., ~ I L'.V = (20.0 A)(L20 X 10' V)
Di\-ide the total power by the pm-vel' per bulb Lhe number of bulbs:
Number of bulbs
1.0
geL
([PIOlal
- --
QPbulb -
= 2.40 X
2.40 X 10,1 \V
75.0 W
=
10' W
32.0
17.6
Electrical Energy and Power
583
(b) Calculatc the cost of this electricity for an 8.00-h day. Enero"\' = '!PI = (2.'10 X 10' W)(
Finclthe energy in kilowatt-hours:
~.
1.00 kW )(8.00 h) 1.00 X 10'W
= 19.2 kWh
Cost = (19.2 kWh)($0.12/kWh)
Multiply the energy by the cost per kilmvatt-hour:
$2.30 • ' , •••• , ' , •...A
Iro.. • • • , • • • • •
Remarks This amount of cnergy might correspond to what a small orflcc uscs in a working day, thow.. thaI lhe current in a circuit depends on lhe resistance of the bauery, so a ballel·y can 'I be considered a source of constant current. Even the terminal "oltage of a bauery givt;n by Equalion 18.1 can't bl.' considered conSlant because I he internal resistance can ch;lIlge (due LO "'anning, for example, during the operation ofthc baIter)'). A banery is, however, a source of constant emr.
596
Chapter 18
Direct-Current Circuits
'" ~I
(h)
(a)
(01
ACTIVE FIGURE 18.2 t\ series COllllcction of 111'0 resi.~lors, HI ,[Ild U:!.. The S
(a)
insulator
(hi
jltlllpCr COllll('Cli'JII.
18,2
the string of" lights, finding them would be a lengthy and annoying task. Christmas lights use special bulbs that have an insuJ:ued loop of ,·.. ire (ajumper) across the conduc[~ lllg supports to the bulb filaments (Fig. 18.3). If the f1Iamcnl breaks and the bulb fails, the bulb's resis~
Resistors in Series
597
tance increases dramatically. As a result, mOst of the applied voltage appears across the loop of wire_ This voltage causes the insulation around the loop of 'v ire to burn, causing the metal wire ro make electrical COil tact with the supports. This produces a conducting path through the bulb, so the other bulbs remain lit.
QUICK QUIZ 18.3 In Figure 18.4 the current is measured with the ammeleI' at the bottom 01" the circuit. ","'hen the switch is opencd, docs the reading on the ammeter (a) increase, (b) decrease, or (c) llot change? QUICK QUIZ 18.4 The circuit in Figure ]8.4 consists of £\'\'0 resistors, a s\vil.ch, an ammeter, and a bauery. \Vhen the switch is closed, power ~
>
1
I
I
I
Ref]
R]
R'l.
R?,
-=-+-+I
I
]
I
II
He',
3.0 fl
6.0 fl
9.0 fl
IS fl
He, =
11
-=--+--+--~--
IS
D
=
1.6 fl
600
Chopter 18
Direct-Current Circuits
(d) Compmc the power dissipated by the equivalent resistance. (if
Remarks There's something imponanl to notice in pan (a): the smallest 3.0 [l resistor carries the largest current, whereas the other, larger resistors of' 6.0!1 and 9.0 [l carry smaller currents. The largest current is ah'\'ays found in the path orIeast resistance. In pan (b) the power could also be [ouncll,.vith r;p = (.~V)2/R. Note that @\ = 108 VIi', but is rounded to 110 W because there are only two significant figures. Finall)', notice that the total power dissipated in the equivalent resistor is the same as the sum of the power dissipated in the individual resistors, as it should be.
TI P 18.2 Don't Forget to Flip It! The mOSI COlllmon llIistake in calculating the equivalent resislance for resistors in parallel is to rorgel to inverl the answer afrer summing the reciprocals. Don't rorgetlo nip it:
QUESTION 18.2 If a founh resiswr were: added in parallel to the other three, haw would the equivalelH resistance change? (a) It would be larger. (b) II would be smaller. (c) More information is required to determine the effect. EXERCISE 18.2 Suppose the resistances in the example are 1.0 fl., 2.0 0, and 3.0.n, respectively, and a new voltage source is provided. If the current measured in the 3.0-0 resistOr is 2.0 A, find (a) the potential difference provided by the ne\v battery and the cllrrems in each of the remaining resistors, (b) the pO\.. .' er delivered to each resistor and the LOlal power, (c) the equivalem resistance, and (d) the total current and the power dissipated by the equivalent resistor. (a) £ = 6.0 V, I, = 6.0 A, J, = 3.0 A (b) '!J'"q = 66 W
Answers
(if,
= 36 W, (if, = 18 W, '!J',
~ 12 W, (ifw,
= 66 W (c) ff f1
(d) 1=11 A,
QUICK QUIZ 18.7 Suppose )'ou have three identical lightbulbs. some wire, and a bauery. YOll connect one Iightbulb to the batter)' and take nOte of its brightness. You add a second lightbulb, connecting it in parallel with the previous lightbulbs, and again take note of the brightness. Repeat the process ,·... ith the thi rei lightbulb, connecting it in parallel ,.. .Iith the other two. As the lightbulbs are added, what happens 1.0 (a) the brightness of the lightbulbs? (b) The individual currents in the lightbulbs? (c) The power delivered b), the bauery' (d) The lifetime of the balter)'? (l\eglect the battery's internal resistance.) QUICK QUIZ 18.8 Trthe lightbulbs in Quick Quiz 18.7 are connected one by one in series instead of in parallel, what happens to (a) the brightness of the lightbulbs? (b) The individual currents in the lightbulbs? (c) The pm...· er delivered by the battery? (d) The lifetime of the battery? (Again, neglect the battery's internal resistance.)
APPLICATION Circuit Breakers
Household circuits are always wired so that the electrical devices are connected in parallel, as in Active Figure 18.6a. In this way each device operates independently of the others so that if one is switched off, the others remain on. For example, if one of the lightbulbs in Active Figure 18.6 '.. ·ere removed from its socket, the other would continue to operate. Equally important is that each device operates at the sallle voltage. If t.he devices 'were connected in series, the voltage across any one device would depend on hm..· lllany devices were in the combination and on their individual resistances. In many household circuits, circuit breakers are used in series with other circuit elements for safety purposes. A circuit breaker is designed to switch off and open the circuit at some maximUll1 value of the current (typically 15 A or 20 A) that depends on the nature of the circuit. IT a circuit breaker were not used, excessive currents caused by operating several devices simultaneously could result in exces· sive ,,,ire temperatures, perhaps causing a fire. In older home construction, fuses
18.3
601
Resistors in Parallel
\vere llsed in place of circuit breakers. "\Then the current in a circuit exceeded some value. the conductor in a fuse melted and opened the circuit. The disadvantage of fuses is that they are destroyed in the process of opening the circuit, whereas circuit breakers can be reset.
APPLYING PHYSICS 18.2
L1GHTBULB COMBINATIONS
Compare the brightness of the four identical lightbulbs shown in Figure 18.9. \·Vhat happens if bulb A fails and so cannot conduct current? \'Vhat irC fails? \'\'hat ifD fails? Explanation Bulbs A and B are connected in series across the emf of t.he bauery, whereas bulb C is cOllneClcd by itself across the battery. This means the voltage drop across C has the same magnitude as the bat.tery emf, whereas this same emf is split between bulbs A and B. As a result, bulb C will glow more brightly than either of bulbs A and B, which will glow equally brightly. Bulb 0 has a wire connected across it-a shon circuit-so the potcntial difference across bulb D is zero and it doesn't glow. )fbulb A fails, B goes out, but C stays lit. lrC fails, there is no effect on the other bulbs. If D fails, the e\"ent is undetectClble because 0 was not gIO\...· ing initially.
APPLYING PHYSICS 18.3
A
~
B
~
~c
::
-f FIGURE 18.9
D
(Applying Physics
18.2)
THREE-WAY LIGHTBULBS
Figure 18.10 illustrates how a three-way lightbulb is constructed to provide three levels of light intensity. The socket or the lamp is equipped with a three~way switch for selecting diffcrentlight intensities. The bulb contains twO filaments. \Nhy arc the f1laments connected in parallel? Explain hm.. . the two filaments are used to provide the three dirferentlight intensities. Explanation If the fllaments were connected in series and one of them were to fail, there would be no current in the bulb anclthe bulb would not glow, regardless of the position orthe switch. \'\Then the fllaments are connected in parallel and one of them (say, the 75-W filament) fails, howc\Ter, the bulb will still operate in one of the s\"itch positions because there is current. in the other (100·\'\') filament. The three light. intensities arc made possible by selecting one of three values offilament resistance, lIsing a single valuc of 120 V for the applied voltage. The 75-\V filament offers Olle value of resistance, the 100-\0\' filament offers a second value, and the third resistance is obtained by combining the two filaments in parallel. \'\'hen switch 51 is closed and s\vitch 52 is opened, only the 75-'W filament carries current. \,Vhen s\.. . itch 5 J is open cd and swit.ch 5:! is closed, only the 100-\'\' filament carries current. \,Vhen both S\'I,jlches are closed, both filaments carry current and a total illumination corresponding to 175 \0\' is obtained.
PROBLEM-SOLVING STRATEGY
SIMPLIFYING CIRCUITS WITH RESISTORS l. Combine all resistors in series by summing the individual resistances and
draw the new, simplified circuit diagram. Usefu I facts:
= R1 + R'J, + R:~ + The current in each resistor is the same.
Rt,Xl
lOO-W filament
FIGURE 18.10 18.3)
(Applying Physics
602
Chapter 18
Direct-Current Circuits
2. Combine all resistors in parallel by summing the reciprocals or the resistances and then laking the reciprocal of the resull. Dnlw t he new, simplified circuit diagram. 1 Useful facts: - - =
Ro,
I
I
I
R,
H,
+- +- +
~
R,
The potential difference across each resisLOr is the same.
3. Repeallhc first two steps as necessary, until no further combinations can be made. If there is only a single battery in the circuit, the result ""ill LlsuaUy be a single equivalent resistor in series 'with the bancry. 4. Usc Ohm's law, 8.\1= JR, to determine the current in rhe cqlliv~dent resistor. Then work back\varcls through the diagrams, applying the lIseful facts listed in step 1 or step 2 to find the currents in the other resisLOrs. (In morc complex circuits, Kirchhoff's rules \",'ill be needed, as described in the next section).
EXAMPLE 18.3
Equivalent Resistance
Goal Solve a problem involving both series and parallel resistors. Problem Four resistors are connected as shown in Figure 18.lla. (a) Find the equivalent resistance between points a and c. (b) \tVhat is the current in each resisLOr if a 42-V battery is connected between a and c? Strategy Reduce the circuit in steps, as shm'\'n in Figures 18Jlb and l8.llc, lIsing the sum rule for resistors in series and the reciprocal-sum rule for resistors in parallel. Finding the currents is a malleI' of applying Ohm's law \"hile ' . .· orking backwards through the di'1grallls.
G.on
-I
8.0 II (a)
"
4.0 II
I
I,t ,
b
1,1 ~.O
II
!
12.0 II
2.0 II
(b)(~
FIGURE 18.11 (Example 18.3) The rOllr resisLOrs shown ill (a) call be reduced in sLeps LO all c(!llivalcllt I'l-fl resisLOI".
14.0 II (0)
~
" ~
.
• • • • • • • • • • • ' ' '111
Solution (a) Find the equivalent resistance of the circuil. The 8.0-0 and 4.0-D resistors are in series, so use the sum rule to Ilnd the equivalent resistance between (f. ancl b: I
1
I
1
1
I
11'" = (6.00)1, = (3.00)1, = 6.0 V and .'>V" = (120)1, = 36 V; therefore, a"f1f = aVab + .lV/x = 42 V, as expected. QUESTION 18.3 'Which of the original resistors dissipates energy at the grealesl rate? EXERCISE 18.3 Suppose the series resistors in Example 18.~ are now 6.00 fl and 3.00 n \vhile the parallel resistors are 8.00 fl (top) and 4,00 n (bottorn), and the battery provides an emf of 27.0 V. Find (a) the equivalent resistance and (b) the currents /, 11, and '2'
Answers
18.4
(a) 11.70 (b) 1= 2.31 A, I, = 0.770 A,
I~
= 1.54 A
KIRCHHOFF'S RULES AND COMPLEX DC CIRCUITS
As demonstrated in the preceding section, we call analyze simple circuits using Ohm's law and (he rules for series and parallel combinations of resistors. There arc, howc\'cr, man)' ways in \"'hich resistors can be connected so that the circuits formed can't be reduced to a single equivalent resistor. The procedure for analyzing mOl-c complex circuits can be facilitated by the use of t\.. .o simple rules called Kirchhoff's rules: (a)
I. The sum of the currents entering an)' junction mllSt equal the Slim of the
currellls leaving that junction. (This rule is often referred to as the junction rule.) 2. The slim of the potential differences across all the elements around an)' closed circuit loop must be zero. (This rule is usually called the loop rule.) Thejulletion rule is a statement of (o1l.'wnJation ofclwrgi'. \\'hate\"cr current enters a gi\'en point in a circuit must leave that poilll because charge can'l build up or disappear at a point. If we apply this rule to the junction in Figure 18.12a, we gel I, =
I~
+
I,
Figure 18.12b represents a mechanical analog or the circuit shown in Figure 18.12£1. In this analogwmer nows through a hranched pipe with no leaks. The now rate inlo the pipe equals the total now rate out of the two branches. The loop rule is equivalent to the principle of C01JSI'f1.H1lion oj nJergy. Any charge thaI. rnoves around any closed loop in a circuit (st:lrting and ending at the same
-
FJowin
(b) FIGURE 18.12 (a) A schematic diagram illuslraling Kirchhofrsjullclion rule, COllserva, ion of charge requires I hal \\ hatc\er current elller'i ajllllc,ioll musllt'a\"c ,haljunctioll. 111 this case. therefore. II = I'!. + 13, (b) A mechanical analog of the jllllcLion rule: lhe 11l;L flow in mUSl equallhc nt:L f1owOlll.
604
Chapter 18
Direct-Current Circuits
-
I
(a)
•
fI
•
'N'I
.:1\'= \'I,-I~= -1/1
-
b
I
(b) fI
~\'= \'.- \'US lime ;tlkr the swilch is closed.
E- Time constant of on RC circuit
608
Chapter 18
Direct-Current Circuits
In other word:), in a lime equallo one time constant, the capacitor loses 63.2% of its initial charge. Because av = q/C. the voltage across the capaciLOr also decreases exponentially with time according to the equation .6. V = Be-I.'m:, where e (,..·hich equals Q/C) is the initial voltage across the fully charged capaciLOr.
APPLYING PHYSICS 18.4
TIMED WINDSHIELD WIPERS
~Ian}' automobiles are equipped with windshield wipers thal can be used intermittently during a liglu rainfall. How does the operation of this feature depend on the charging and discharging of a capacilOr?
ferent values of R through a multi position switch. The brief time thal the wipers remain on and the lime they are off are determined by the value of the time constant of the circuit.
Explanation The wipers are part of an RC circuit ",·ith lime constant 'hal can be varied by selcCling dif-
APPLYING PHYSICS 18.5
BACTERIAL GROWTH
In biological applications concerned with popUlation growth, an equation is used that is simil(1r to the exponential equations encountered in the analysis of RCcircuits. Applied to a number of bacteria, this equation is !\~-=
N1 2"
where Nris the number ofbactcria present aft.er n doubling times, iVj is the number present initial1y, and n is the number of gro\,.lth C}'cles or doubli ng times, Doubling times vary according to the organism. The doubling time for the bacteria responsible for leprosy is abo1ll30 days. and that for the salmonella bacteria
APPLYING PHYSICS 18.6
III responsible for food poisoning is about. 20 minutes. Suppose only 10 salmonella baCleria find their wa}' onto a turkey leg after your Than ksgiving meal. Four hours later you come back for a midnight. snack. How l11 = J ~f~ we can calculate the CUITCIll carried b}' each appliance. The toaster, rated at] 000 \"1, dra'\IS a CUlTent of I 000/120 = 8.33 A. The micrO\-\,ave oven, rated at800 \·V, dra\...,s a current of 6.67 A, and the heater, ratcd at 1 300 \·V, draws a currCll[ of 10.8 A. If thc three appliances are operated simultaneously, they drm.. · a LOtal current of 25.8 A. Therefore, the breakcr should be able to handle at least this much current, or elsc it will be tripped. As an alternative, the toaster and microwave oven could operale on one 20-A circuit and the heater on a separate 20-A circuit. Many heavy-dUly appliances, such as electric ranges and clothes dryers. require 240 V 1.0 operate. The power company supplies this voltage by providing. in addition to a live wire that is 1.20 V above ground potential, another wire, also considered live. that is 120 V below ground potential (fig. 18.22). Therefore, the potential drop across the n.l,'o live wires is 240 V. An appliance operating from a 240-\1 line requires hal r the current of one operating from a 120-V line; conscquently, smaller ,.. ·ires can be used in the higher-voltage circuit withollt becoming overheated.
FIGURE 18.21 A circuit breaker lhal uses a binLt:lallic sLrip for il~ operation.
APPLICATION Fuses and Circuit Breakers
+120 V
-120 V
I
(a) ~
J
(1.67 X 10- 27 kg)" = 2.77 X 10- 12 N
...............
lrr.. • • • . . • •
Remarks The initial acceleration is also in the positive z-direction. Because the direction ofV' changes, however, the subsequent direction orthe magnetic force also changes. In applying right-h£lnd rule number 1 to find the direction, it was imponantLO take into consideration the charge. A negatively charged panicle accelerates in the opposite direction. QUESTION 19.2 Can a constant magnetic Aeld change t.he speed ofa charged particle? Explain. EXERCISE 19.2 Calculate the acceleration oran electron that moves through the same magnetic field as in Example 19.2, at the same velocity as the proton. The mass of an electron is 9.11 X 10- 31 kg.
Answer
3.04 X 10 1M m/s~ in the negalive z-direcrion
19.4
MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR
Ifa magnetic field exerts a (orce on a single charged particle when it moves through a magnetic field, it should be no surprise that magnetic forces are exerted on a current-carrying \\lire as well (see Fig. 19.9). Because the current is a collection of many charged particles in motion, the resultant force on the \'lire is due to the SUIll of the individual forces on the charged particles. The force on the panicles is transmitted to the "bulk" of the 'wire through collisions with the atoms making up the wire. Some explanation is in order concerning notation in manv of the figures. To indicate the direction orB, we lise the following conventions: ~
If B is directed into the page, as in FigurS 19.10, we use a series of green crosses, representing the t.ails of arrows. If B is directed out of the page, we use a series of green dots, representing the tips of arrows. If B lies in the plane of the page, we usc a series of green field lines ·with arrowheads. ~
The force on a current~Glrrying conductor can be demonstrated by hanging a wire between the poles ora magnet, as in Figure ]9.10. In this figure, the magnetic
;;
]
..._-
FIGURE 19.9 Thisapparaws demOnStrales the force on a currentcarrying conductor in an external magnelic field. Why does the bar swing {Jwa_~ from the magnet afler lhe SI\"ltch is closed?
FIGURE 19.10 A segmell1 ofa flexible \'ertical wire partially stretched beth'cen thc poles ofa magnet, with LiH; field (green crosses) direcled imo the page. (a) When there is no current in the wire, it remains vertical. (b) When [he current is upward. the wire deflects to the left. (cl ·When the current is downward, the wire deflects to the right.
(b)
(el
634
Chapter 19
Magnetism
A
field is direcled into lhe page and covers the region within the shaded area. The v.'ire denects to the right or left when it carries a current. \Ve can quantify this discussion by considering a straight segment of wire of _ . 0 Icng,th and cross-sectional area A carrying current I in a uniforJTI external maunetic field B, as in Figure ]9.11. \!Ve assume that the magnetic field is perpendicular (Q the wire and is directed into the page. A force of magnitude 1\11.1), = qvdB is exerted on each charge carrier in the wire, \,,!here v" is the drift velocity of the charge. To find the total force on the wire, \ve Illultiply the force on one charge carrier by the number of carriers in the segment. Because the volume of the segIllent is A[, the number of carriers is nAt', where 11. is the number of carriers per unit volume. Hence, the magnitude of the total magnetic force on the wire of length .f is as follows: Tota I force = force all each charge carrier X total number of carriers ~
FIGURE 19.11 Aseetiollora wire conwining moving ch~Hres in an eXlernalmaglictic field B-
TIP 19.2 The Origin of the Magnetic Force on a Wire When a magnetic field is applied at some angle to a wire carrying a current, of the magnetic field B:
(47T X 1O-'T·m/A)(5.00A) 27T(4.00 X 10 'm)
= -'------'-------'----','---,----"-
2.50 X 10- 4 T
With the right thumb pointing in the direction of the current in Figure 19.27, the fingers curl into the page at ths.location of the proton. The angle 0 between and B is therefore 90°.
v
(b) Compute the magnetic force exerted by the wire on the proton. Substitute into Equation 19.1, which gives the magnilUde of the magnetic force on a charged particle:
F= qvBsin 0 = (1.60 X 10-" C)(1.50 X J03 m/ s) X (2.50 X 1O-'T)(sin900) =
Find the direction of the magnetic force with righthand rule number I: trr... . . • • • • • • .
6.00 X 1O- 211 N
Point your right fingers in the direction orv, curling them into the page toward B. Your thumb points to the left, which is thc direction of the magnetic force .
. . . . . . . . . . . . . . . ..
... . .. . . . . ......
Remarks The location of the proton is important. On the left-hand side, the wire's magnet.ic field points outward and the magnetic force on the proton is 1.0 the right.
19.8
Magnetic Force Between Two Parallel Conductors
645
QUESTION 19.7 How ,.. .' ould the magnetic force differ if it acted on an electron moving with the same velocity? EXERCISE 19.7 Find (a) the magnetic field created by the wire and (b) the magnetic force on a helium~3 nucleus located 7.50 mill to the left of the wire in Figure 19.27, traveling 2.50 X 103 m/s opposite the direction of the current. (See the data table presented in Example 19.5 on page 641). Answers
(a) 1.33 X 10-.1 T
(b) 1.07 X 10 '" N, direcled
lO
the ler, in Figure 19.27
19.8 MAGNETIC FORCE BETWEEN TWO PARALLEL CONDUCTORS As we have seen, a magnetic force acts on a current-carrying conductor when the conductor is placed in an external magnetic field. Because a conductor carrying a current creates a magnetic field around itself, it is easy to understand lhal two current-carrying wires placed close together exert magnetic forces on each other. Consider two long, straight, parallel wires separated by the distance d and carry· ing currents 11 and I'.! in the same direction, as shown in Active Figure 19.28. \\fire I is directly above wire 2. \Vhat's the magnetic force 011 one '...· ire due to a magnetic field set up by the other wire~ In this calculation we are finding the force on wire 1 due to the magnetic field of wire 2. The current''!. sets up magnetic field B'!, at wire 1. The direction ofB:! is perpendicular to the wire, as shown in the figure. Using Equation 19.11, we nnd that the magnitude of this magnetic field is J-Lo
'2
IJ, = 27Td According to Equ~ion 19.5, the magnitude of the magnelic force on wire I in the presence of field U:! due to ' 2. is
F = 13,' I
.! I
e=
(1'-"")' e 27Td
I
=
2
--"'!I'-i-";;:~~ 1 \
ACTIVE FIGURE 19.28 Two parallel wires, oriented \'enically. carry .'>1 cady currents and exerl forces 011 each other. The field B;- al wire I duc to wire 2 produces a force all win: I givcn by 1"1 :: B.,. i1f:. The force i~ .Ill racli,c if the current~ ha\'e lhe !;amc direction, as !;hO\\,ll, alld rt.:plllsivc irthe t\\'o currents have 0PP0l>ilC directions.
l'-oJ,',e 21Td
\Ale can rewrite this relationship in terms of the force per unit length:
FI
e=
}Loll I'},
27Td
[19.14]
The direction of Fl is downward, toward wire 2, as indicated by right-hand rule nurnber 1. This calculation is completely symmetric, which means thal the force on wire 2 is equal to and opposite F\, as expected from Newton's third law of action-reaction. We have shown thal parallel conductors carrying currents in lhe same direction attract each olher. You should use the approach indicated by Figure 19.28 and lhe steps leading to Equation 19.14 to show that parallel conductors carrying currents in Opposile directions 1-e/JP[ each other. The force bel ween two parallel wires carrying a current is used to define the SI unit of currem, lhe ampere (A), as follows:
F\
If two long, parallel wires I m apart carry lhe same current and the magnetic force per unit length on each wire is 2 X 10 7 N/m, the current is defined to be I A.
~
Definition of the ampere
The Siunito( charge, the coulomb (C), can now be defined in lerms of the ampere as follows: If a conductor carries a steady current of 1 A, the quantity of charge that flows through any cross section in I sis 1 C.
f'- Definition of the coulomb
646
Chapter 19
Magnetism
QUICK QUIZ 19.5 Which of the following actions would double the magnitude of the magnetic force per unit length between [".;0 parallel currentcarrying wires? Choose all correct answers. (a) Double one of the currents. (b) Double the distance behveen them. (c) Reduce the distance between them by half. (d) Double both currents. QUICK QUIZ 19.6 [f, in Figure 19.28, f l ~ 2 A and f, = 6 A, which of the following is true? (a) F I = 31'2 (b) F I ~ f'2 (c) F I ~ F2 /3
EXAMPLE 19.8 Goal
Levitating a Wire
Calculate the magnetic force of one current-carrying 'v'ire on a parallel currenl-calTying wire.
Problem Two wires, each having a weight per unit length of 1.00 X 10- 4 N/m, are parallel with one directly above the other. Assume the ,.. .,ires carry currents that are equal in magnitude and opposite in direction. The wires are 0.10 m apart, and the sum of the magnetic force and gravitational force on the upper \.. . ire is zero. Find the current in the wires. (Neglect Earth's magnetic field.) Strategy The upper wire must be in equl1ibrium under the forces of magnetic repulsion and gravity. Set the sum of the forces equal to zero and solve for the unknovvn current, J. , _ ...
~
Solution
~
F gra...
Set the sum of the forces equal to zero and substitute the appropriate expressions. Notice thaL the magnetic force between the wires is repulsive.
The currents are equal, so II = /2 substitutions and solve for 12 :
= I.
~
+ F mag
0
1 2 = -'-(2_,,_d-,-)(,--"-,,,gl,--C.:...)
Make these
/La
,
f- =
Substitute g'iven values, finding 1 2 , then take the square root. Notice that the weight per unit length, mg/ is given.
(2,,·0.100 m)(l.OO X 10- 4 N 1m) _
~
.,
50.0 A-
(4" X 10 'T·m)
e,
~.-
=
1= 7.07 A ,
-_.~
Remark Exercise 19.3 showed that using Earth's magnetic field to levitate a wire required extremely large currents. Currents in wires can create much stronger magnetic fields than Earth's magnetic field in regions near the wire. QUESTION 19.8 Why can't cars be constructed that can magnetically levitate in Earth's magnetic field? EXERCISE 19.8 If the current in each \vire is doubled, hmv far apart should the wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal? Answer
0.400 m
19.9
MAGNETIC FIELDS OF CURRENT LOOPS AND SOLENOIDS
The strength of the magnetic field set up by a piece of wire carrying a current can be en hanced at a specific location if the wire is formed into a loop. You can understand this by considering the eHen of several small segments of the current loop, as in Figure 19.29. The small segment at r.he top of the loop, labeled LlX I , produces a magner.ic field of magnitude Bj at the loop's center, directed out of the page. The
19.9
direction ofB can be \"erified using right~hand rule number 2 for a long, straight wire. Imagine holding- the wire with your right hand, \vith rour thumb pointingi!) the direClion of' the cllrrent. Your tingers then curl around in the direClion ofB. /\. segment 01" lenglll 6.x 2 at the bottom orth/J during the interval 6./, the average ernfinduced in the circuit during time llt is 6 cl)/J 8= - N -
Faraday's Law of Induction
~
Faraday's law
~
Lenz's law
667
[20.2]
6.1
Because cDn = BA cos 0, a change of any of the factors B, A, or f) with time produces an emf. \Ve explore the effect of a change in each of these facIal's in the following seClions. The minus sign in Equation 20.2 is included to indicate the polarity of the induced emf. This polarity determines \vhich of two directions current will flow in a loop, a direction given by Lenz's law: The current caused by the induced emf travels in the direction that creates a magnetic field with flux opposing the change in lhe original flux through the circuit. Lenz's law says that if the magnetic flux through a loop is becoming more positive, say, then lhe induced emf creates a current and associated magnetic field that produces negative magnetic nux. Some misl]kenly think this "counler magnetic field" creat.ed by the induced current, called Bind ('-ind'-jor induced), will always point in a direction opposite the applied magnetic field B, but thaI is only Irue half the time! Figure 20.5a shows a field penetrating a loop. The graph in Figure 20.5b shows thalthe n~.~lgniLUde of the magnetic field B shrinks withJime, which me1 alii of I he shower or Whll might tollch a water pipe, providing : is no/. thesameas that produced by a direct current of the same value, however. Thc rcason is that the alternating current has this maximum value for only an instant of time during a cycle. The important quantity in an AC circuit is a special kind of average value of current, called the rms current: the direct current that dissipates the same alllount of energy in a resistor that is dissipated by the acwal ailernating current. To find the rms current, we first square the current, then find its average value, and finally take the square root of this average value. Hence, the rms current is the square root of the average (mean) of the square of the current.. Because i 2 varies as sin 2 27rft, the avcrage value of i 2 is 4J~1il){ (Fig. 21.3b, page 698),l Thercfore, the rms current 'rills is related to the maximum value of the alternating current JlllilX by [21.2]
This cquation says that all alternating current with a maximulll value of 3 A produces the same heating effect in a resistor as a direct current of (3/V2) A. We can therefore say that the average power dissipatcd in ,,1 resistor that carries alternating current fis
IWe call show lhal (i~)", "" J';",./'t. ;IS follows: Till: currenl in Ihe circuil \ 0 and lav> 0
EXAMPLE 21.1 Goal
What Is the rms Current?
Perform basic AC circuit. calculations lor a purely resistive circuit..
Problem An AC voltage source has an output of llv = (2.00 X 102 V) sin 27TJI. This source is connected 1.00 X 10 2 n resistor as ill Figure 21.1. Find the nns voltage and nns current in the resistor.
LO
a
Strategy Compare the expression for the voltage output just given with the general form, llv = 6.V:l1 dcfined as
~
Xc -
1 27rJC
[21.5)
\,\'hen Cis in farads and lis in hertz, the unit of" Xc is the ohm. Notice that 27iJ= w, the angular rrequency. From Equation 21.5, as the frequency of the voltage source increases, the capacitivc reactance Xc (the impeding effect of the capacitor) decreases, so the current increases. At high frequency, there is less time available to charge the capacitor, so less charge and voltage accumulate on the capacitor, which uanslates into less opposition to the flow or charge and, consequently, a higher current. The analogy between capacitive reactance and resistance means that we can write an equation of the same form as Ohm's law to describe AC circuits containing capacitors. This equation relates the nns voltage and rills current in the circuit to the capacitive reactance:
.r
FIGURE 21.5
Pluts of curn:nt and
voltage ,lcross a rapaCilOl" \'I'rsus lilTW in an .. \C circuit. The milage lags I he currem by 90~.
EXAMPLE 21.2 Goal
[21.6)
A Purel Co ocitive AC Circuit
Perform basic AC circuit calculations for a capacitive circuit.
Problem An 8.00-JLF capacitor is COlll1eCleclto the terminals of an AC generator with an rills voltage of 150 X 10' V and a frequency of60.0 I-Iz. Find the capacitive reaClance and the rms current in the circuit. Strategy Substitute values into Equations 21.5 and 21.6.
21.3
Inductors in an AC Circuit
.
r"' . • . . • . .
Solution
Substitute the values ofJand C into Equation 2 I.5:
Solve Equation 2l.6 for the current and substiwte the values ror X(:and the nns volLage to find the rms current:
I Xc = - . 21'JC
l
= -::--;-::-::-:--cc-c-:-::-:-,--------,o--,21'(60.0 l-lz)(8.00 X 10 "F) 1.50 X 10' V 332 fl
,
332
..
~
n
0.452 A
.
11.. • • • • • • • •
Remark
701
~
Again, notice how similar the technique is to that oranalyzing a DC circuit with a resistor.
QUESTION 21.2 True or False: The larger the capacitance ora capacilOr, the larger the capacitive reactance. EXERCISE 21.2 If the Crequency is doubled, what happens
LO
the capacitive reactance and the rms current?
Answer
Xc is halved, and
21.3
INDUCTORS IN AN AC CIRCUIT
'nils
is doubled.
No'lo\' consider an AC circuit consisting only of an inductor conneCted to the terminals oran AC source, as in Active Figure 21.6. (In £Illy real circuit there is some resistance in the ,,,,'ire Conning the inductive coil, but 'lo\ie ignore this consideration for now.) The changing current output of the generator produces XL Positive if X c < Xl.
Not,.; III each cOlse ,," AC \"Ollagl: (1101 sI1O\\,.) is applied acruss the: combi'Mlion of dClllcnt~ (IIml is, ..crOss the dots).
QUICK QUIZ 21.3 If switch A is closed in Figure 21.12, what happens to the impedance of the circuit? (a) It increases. (b) It decreases. (e) It doesn't change. QUICK QUIZ 21.4 Suppose X,. > Xc. If switch A is closed in Figure 21.12, what happens to the phase angle? (a) It increases. (b) 1L decreases. (e) It doesn'L change. QUICK QUIZ 21.5 Suppose XL> Xc. If switch A is left open and switch B is closed in Figure 21.12, what happens to the phase angle? (a) It increases. (b) It decreases. (c) It doesn't change.
FIGURE 21.12 21.3-21.6)
(Quick Quiucs
QUICK QUIZ 21.6 Suppose XL> Xcin Figure 21.12 and, with both switches opell, a piece of iron is slipped into the inductor. During this process, whal happens to the brighuless of the bulb' (a) It increases. (b) It decreases. (c) It doesn't change.
PROBLEM-SOLVING STRATEGY
RLC CIRCUITS The following procedure is recommended for solving series RLC circuit problems: 1. Calculate the inductive and capacitive reactances, XI. and >4.."
2. Use Xl and Xc together with the resistance I? to calculate the impedance Z of the circuit.
3. Find the maximum current or maximum voltage drop with the equivalent orOhm's law, .11~11aJ( = Ima"Z. 4. Calculate the voltage drops across the individual elements with the appropriate variations of Ohm's law: ~"/l.max = Jm;l>.!?-' .6.Vf .,max = 'nHlxX", and AVe,In,IX = 'l1l:lxXe· 5. Obtain the phase angle using tan q> = (X,. - Xd/R.
NIKOLA TESLA (1856-1943) Teslo was born in Croatia, but spent most of his professional life as on inventor in the United States. He was a key figure in the development of alternating-current electricity, high-voltage transformers, and the transport of electrical power via AC transmission lines. Tesla's viewpoint was at odds with the ideas of Edison, who commiHed himself to the use of direct current in power transmission. Teslo's AC approach won out.
706
Chapter 21
EXAMPLE 21.4 Goal
Alternoting~Current Circuits and Electromagnetic Waves
An RLC Circuit
Analyze a series NLC AC circuit and find the phase angle.
Problem A series RLC AC circuit has resistance R = 2.50 X ]02 n, inductance L = 0.600 H, capacitance C = 3.50 j.LF, frequency I= oU.O Hz, and maximum voltage liV;IMS = 1.50 X IO~ V. Find (a) the impedance of the circuit, (b) the maximum current in the circuit, (c) t.he phase angle, and (d) the maximum voltages across the clements. Strategy Calculate the inducLivc and capacitive reactances, which can be llsed with the resistance impedance and phase angle. The impedance and Ohm's lalA,' yield the maximum current.
.
-
~
...........
La
calculate the
_
~
Solution (a) Find the impedance of the circuit. First, calcuJ;;ue the inductive and capacitive reactances:
X,. = 27TJL = 226 n
SubstilUte these results and the resistance R into Equation 21.13 to obtain the impedance of the Circtlit:
z = VII' + (X,. -
Xc = 1/27TJC = 758 Il
Xc)'
= V(2.50 X 10' 0)'
+ (2260 - 7580)' = 588 Il
(b) Find the maximum current in the circuit.
Usc Equation 21.12, the cqulvalentofOhm's law, to find the maximum current:
r.
,=
rna);
1.50 X 10 2 V 5880
1l'v:11.1:" Z
0.255 A
(c) Find the phase angle. X,. - Xc
Calculate the phase angle between the current and the voltage 'with Equation 21.15:
~ tan-I (2260 - 758 Il) = -64.8" 2.50 X 10'Il
R.
(d) Find the maximum voltages across the elements.
Substitule into the ';Ohm's law" expressions fol' each individual type of' current element: ~VJ."",,, ~ {",,,,X,. = (0.255 A)(2.26 X 10 2 Il) ~ 57.6 V ~Vc.",,, lrrrr... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Remarks Because the circuit is morc capacitive than inductive (Xc> X,), ep is negative. A negative phase angle means that the current leads the applied voltage. Notice also that the sum of the maximum voltages across the elements is 6.VR + .6.V,. + 6.vc = 314 V, which is much greater than the maximum voltage 01' the generator, 150 V. As we saw in Quick Quiz 2].2. the sum or the Ill:aximum voltages is a meaningless quantity because when alternating voltages are added, both (Iuti:,. mnjJl'iludes and theirtJhases must be taken into account. We know that the maximum voltages across the various c1ernents occur at different limes, so it docsn'l make sense to add all the maximum values. The correct way to "add" the voltages is through Equation 21.10.
=
!",,,Xc
=
(0.255 A)(7.58
X ]02
Il) = 193 V
.
~
QUESTION 21.4 True or False: In an RLC circuit, the impedance must always be greater than or equal to the resistance. EXERCISE 21.4 Analyze a series RLC AC circuit for which R = 175 n, L = 0.500 H, C = 22.5t.tF, J= 60.0 Hz, and ~V""" = 325 V. Find (a) the impedance. (b) the maximum current, (c) the phase angle, and (d) the maximum voltages across I he elements.
Answers (a) 1890 (b) 1,72 A (c) 22.0" (d) 301 V, .6. \11,.max = 324 V, .6.'vC.max = 203 V
~VJ which gives, from Equations 2l.5 and 21.8,
27Tfr,L
fu
=
1
=--
27TfoC
1 27TVLC
[21.19]
Figure 2-1.13 is a plot of current as a fUIlCllon of frequency for a circuit containing a fixed value f'or both the capacitance and the inductance. From Equation 2-1.18, it must be concluded that the current would become infinitc at resonancc when N = O. Although Equation 21.18 predicts this result, real circuits always have some resistance, which limits the value of the current. The tuning circuit of a radio is an in"lportant application of a series resonance circuit. The radio is tuned to a particular station (.."hich transmits a specific radiofrequency signal) by varying a capacitor, \I.lhich changes the resonance frequcnc)' of the lUning circuit. 'A'hen this resonance frequency matches that of the incoming radio wave, the current in the tuning circuit increases.
21.6
APPLYING PHYSICS 21.2
Resonance in a Series RLC Circuit
709
METAL DETECTORS AT THE COURTHOUSE
When you walk through the doorway ora courthouse metal detector, as the person in Figure 21.14 is doing, you are really walking through a coil of many turns. How mighllhe metal det.ector \",'ork? Explanation The mctal detect.ur is essentially a resonant circuit. The portal you step through is an inductor (a large loop of conducting wire) that is pan of the circuit. The frequency of the circuit. is lUned to the resonant frequency of the circuit when there is no rnetal in the inductor. \-Vhen you walk through with metal in your POCkCL, you change the effective inductance of the resonance circuit, resulting in a change in the currcnt in the circuit. This change in cllrrenl is detected, and an elect.ronic circuit causcs a sOllnd to bc emitted as an alarm.
EXAMPLE 21.6 Goal
~
FIGURE 21.14
(Apply-
ing Ph)'Sics 21.2) A
COUl"t-
hOlISt:: lTIt::1al dClcclor.
~
"• ...._---~
A Circuit in Resonance
Understand resonance frequcncy and its relat.ion to inductancc, capacitance, and the rms current.
Problem Consider a series RLC circuit for which R = 1.50 X 102 .0, j, = 20.0 mH, 6.Y;'I1lS = 20.0 Y, and J= 796 s I (a) Determine the value of t.he capacitance for which the rms current is a maximum. (b) Find t.he maximum nns cur.. rent in the circuit. Strategy The current is a maximum at the resonance frequency Io, which should be set equal to the driving frequency, 796 S-I. The resulLing equation can be solved for C. For part (b), substitute into Equation 21.18 to get the maximum rms current. . • • • • • • • • • - - • ""'II!
~
Solution (a) Find t.he capacitance giving the maximum current. in the circuit (the resonance condition). Solve the resonance frequency for the capacitance:
I
fc'=9-rpC _1T v LL1 C = -47T-;;-'Ji,:7.,cL
Insertlhe given values, substiwting the source frequency for the resonance frequenc)\L:
I
C = '17T'(796 Hz)'(20.0 X 10
3
H)
2.00 X 10- 6 F
(b) Find the maximum rms current in the circuit. The capacitive and inductive reactances are equal, so Z = Ii = 1.50 X 10' fl. Substitute into Equation 21.18 to find the rms current: lrr.. • • • • • •
20.0 V 1.50 X 10' n
0.133 A
.
~
Remark Because the impedance Z is in the denominator of Equatiun 21 J8, the maximum current will always occur when XI. = Xc because that yields the minimum value of Z.
QUESTION 21.6 True or False: The magnitude of thc current in an RLCcircuit is never larger than the rms cunent.
710
Alternating-Current Circuits and Electromagnetic Waves
Chapter 21
EXERCISE 21.6 Consider a series RLC circuit l'or which R = 1.20 X 10' fl, C = 3,10 X 10-5 F, t;v.,,,,, = 35,0 Y, and I= 60,0,-1, (a) Determine the value of the inductance for which the rms current is a maximum. (b) Find the maximum rInS current in the circuit. Answers
(a) 0,227 H
(b) 0,292 A
s
z\
Primary (inplll)
l=:==:'~
Secondary (ompul)
FIGURE 21.15 An ideal transformer consists oft\l'o coils wound on the same soft iron core. An AC \'ollagc d( 'r is applied 10 the primary coil, and the output voltage ~\l~ is observed across lhc load resistance Rafter Ihe switch is dosed.
21.7
THE TRANSFORMER
IL's often necessary to change a small AC voltage Lo a larger one or vice versa. Such changes are effected with a device called a transformer. In its simplest form the AC transformer consists of two coils of wire wound around a core of soft iron, as shown in Figure 2] .15. The coil on the left, \\'·hich is connected to the input AC voltage source and has N] turns, is called the primary winding, or the pri·"U.L't)'. The coil on the right, which is connected to a resistor R and consists of N 2 turns, is the secoru.ia-ry. The purposes of the common iron core arc to increase the magnetic flux and to provide a medium in \",-hich nearly all the flux through one coil passes through the other. V\'hen an input AC voltage L\VI is applied to the primary, the induced voltage across it is given by t;1l
-N-I
t;t
[21.20]
where (1)8 is the magnetic flux through each lllrll. If we assume that no nux leaks from the iron core, then the flux through each turn of the primary equals the nux through each turn of the secondary_ Hence, the voltage across the secondary coil is [21.21] The term 6.(I>Il/6.t is common ro Equations 21.20 and 21.2.1 and can be algebraically eliminated, g"iving N,
t; If., = - - t; V, N,
[21.22]
vVhcn N'2 is greater than N I • 6.\12 exceeds L~.vl and the transformer is referred to as a sle/J-up transfonna ''''hen N2 is less than N I , making aV2 less than dVI , \'ie have a ste/J-down transfonne):
In on ideol transformer, the input power equals the output power ~
By Faraday's law, a voltage is generated across the secondary only when there is a change in the number of flux lines passing through the secondary. The input current in the primary must therefore change with time, which is what happens when an alternating current is used. "Vhen the input at the primary is a direct current, however, a voltage output occurs at the secondary only at the instant a switch in the primary circuit is opened or closed. Once the current in the primary reaches a steady value, the output voltage at the secondary is zero. rt may seem that a transformer is a device in which it is possible (Q get something [or nothing. For example, a step-up transformer can change an input voltage from, say, 10 V to 100 V. This means that each coulomb of charge leaving the secondary has 100 J of energy, whereas each coulomb of charge entering the primary has only 10 J of energy. That is not the case, hovvever, because the power input to the primary equals the power output at tbe secondary:
[, t;v, = [, t;v,
[21.23]
Although the voltage at the secondary may be, say, ten times greater than the vol[age at the primary, the (,'urre7ll in the secondary will be smaller than the primary's
21.7
current by a f'actor of' ten. Equation 21.23 assumes an ideal transformer in which there are no power losses between the primary and the secondary. Real transformers typically have power efficiencies ranging from 90% to 99%. Power losses occur because of such facLOrs as eddy currents induced in the iron core of the trallSfonner, which dissipate energy in the fon11 of j2R losses. When electric power is transmitted over large distances, it's economical to use a high voltage and a low current. because t.he pm-vcr lost via resistive heating in the transmission lines varies as /2R. ff a utility company can reduce the current b)1 a factor of ten, [or cxamplc, thc power loss is reduced by a factor of one hundred. In practice, the voltage is st.epped up to around 230 000 Vat the generating station, thcn stepped down to around 20000 Vat a distribution station, and finally stepped down to 120 V at the customer's utility pole.
EXAMPLE 21.7 Gool
Distributin Power to
0
The Transformer
711
APPLICATION Long-Distance Electric Power Transmission
Cit
Understand transformers and their role in reducing power loss.
Problem A generator at a utility company produces 1.00 X \0' A of current al 4.00 X 10 3 V. The vollagc is stepped up to 2.40 X 10'"' V by a transformer bcfolT being sent 011 a high-voltage trl\llsmission line across a rural area to a city. Assume the efTective resistance of the power line is 30.0 n and that the transformers are ideal. (a) Determine the percentage 0[' power lost in t.he transmission line. (b) What percentage of" the original power would be lost in the transmission lille if the voltage were llot stepped up? ".
.
Strategy Solving this problem is just a matler of substitution into the equation [or transformers and t he equation for power loss. To obtain the fraction of power losl., it's also necessary to compute the power output of the generator: the current times the potential difference creat.ed by the generator.
""
.
Solution (a) Determine the percentage of power lost in the line. (1.00 X 10' A)(4.00 X LO'V)
Substitutc into Equation 21.23 to find the current in the transmission line:
1.67 A
2.40 X 105 V
\'1',,,,,
= I,'R = (1.67 A)'(30.0 D) = 83.7 W
Now lise Equation 21.]6 to find the power lost in the transmission line:
(1)
Calculate the power output of the generator:
\'I' = I, Jl.V, = (1.00 X 10' A)(4.00 X 10' V) ~ 4.00 X 10" W
Finally. divide Q"Plo.'il by the power outpHt and multiply by 1001.0 rind the percentage or power lost;
% power 10Sl
=
83.7 W ) X 100 ( 4.00 X \0',W
=
0.0209%
(b) ''''hat percentage of the original power would be lost in the transmission line if t.he vohage were not stepped up? = I'R = (1.00 X 10' A)'(30.0 D) = 3.00 X 105 W
Replace the stepped-up current in Equation (1) by the original currentofl.OO X 10 2 A:
\'1',,,,,
Calculate the percentage loss, as before:
% power
\OSl =
X W)
3.00 10" ( 4.00 X 10-',Wx 100 = 75%
712
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
This cylindrical step-down transformer drops the voltage from 4000 V LO 220 V lor dclh'ery La a group of residences.
Remarks This example illustrates the advantage of high-voltage transmission lines. At the city, a transformer at a substation steps the voltage back down La about 4 000 V, and this voltage is maintained across utility lines throughout the city. When the power is to be used at a home or business, a transformer on a utility pole near the establishment reduces the volt-
age
LO
240Vor 120V.
QUESTION 21.7 If the voltage is stepped lip to double the amount in this problem, by what [actor is the power loss
changed? (a) 2 (b) no change (c) ~ (d) ~
-1l
~
i?'....._ ~
EXERCISE 21.7 Suppose the sallle generator has the voltage stepped up to only 7.50 X 10 4 V and the resistance of the Iinc is 85.00. Find the percentage of power lost in this case.
Answer
~c ~
!
Q
0.604%
21.8
MAXWELL'S PREDICTIONS
During the early stages of their study and development, electric and magnetic phenomena were thought to be unrelated. In -1865, however, James Clerk Maxwell (1831-1879) provided a mathematical theory that showed a close relationship between all electric and magnetic phenomena. In addition to unifying the formerly separate fields of electricity and magnetism, his brilliant theory predicted that electric and magnetic fields can move through space as waves. The theory he developed is based on the following [our pieces ofinformation:
1. Electric field lines originate on positive charges and terlllinate on negative charges. 2. Magnetic field lines always form closed loops; they don't begin or end anywhere. 3. A varying magnetic field induces an emf and hence an electric field. This fact is a statement or Faraday's law (Chapter 20). 4. Magnetic fields are generated by moving charges (or currems), as sUlllmarized in Ampere's law (Chapter 19).
JAMES CLERK MAXWELL Scottish Theoretical Physicist (1831-1879)
Maxwell developed the electromagnetic theory of light ond the kinetic theory of gases, ond he explained the nature of Saturn's rings and color vision. MaxwelJ's successful interpretation af the electromagnetic field resulted in the equations that bear his nome. Formidable mathematical ability combined with great insight enabled him to lead the way in the study of electromagnetism and kinetic theory.
The first statement is a consequence of the nature of the electrostatic force bcn.. .een charged particles, given by Coulomb's law. It embodies the fact that free charges (electric monopoles) exist in nat.ure. The second statement-that magnetic fields form continuous loops-is exemplified by the magnetic field lines around a long, straight wire, which are closed circles, and the magnetic field lines or a bar magnet, which form closed loops. It says, in contrast to the first statement, that free magnetic charges (magnetic monopoles) don't exist in nature. The third statement is equivalent to Faraday's law of induClion, and the fourth is equivalent to Ampere's law. In one of the greatest theoretical developments of the 19th century, Maxwell used these four statements within a corresponding mathematical framework to prove that electric and magnetic fields play symmetric roles in nature. It was already known [rom experimenLS that a changing magnetic field produced an electric field according LO Faraday's law. Maxwell believed thalnature was symmetric, and he therefore hypothesized that a changing electric field should produce
21.9
Hertz's Confirmation of Maxwell's Predictions
713
a magnetic field. This hrpmhesis could not be proven experimentally at the time it \vas developed because the magnetic fields generated by changing electric fields are generally very weak and therer-ore difficult 1.0 detect. Tojustify his hypothesis, :\1axwcll searched for other phenomena [hat might be explained by it. He turned his attention to the Illotioll of rapidly oscillating (accel~ erating) charges, such as those in a conducting rod connected to (Ill alternating voltage. Such charges are accelerated and, according to Maxwell's predictions, g"enerate changing electric and magnetic fields. The changing fields cause electromagnetic disturbances that travel through space as waves, similar to the spreading water waves created by a pebble thrown into a pool. The waves sent out by the oscillating charges are Auctuating electric and magnetic fields, so they are called electromagnetic waves. From Faraday's law and from Max\,,·e1l's 0\\'11 generalization of Ampere's law, MaK\,,·e11 calculated the specd of the waves to be equal to the spced of light, c = 3 X ]0 8 m/s. He concluded that visible light and other electromagnetic waves consist of fluctuating electric and magnetic fields traveling through empty space, with each varying field inducing the other~ His was truly one of the greatest discoveries of science, on a par with NeWlon's discovery of the laws of motion. Like Newton's laws, it had a profound innuence on later scientific developments.
21.9
HERTZ'S CONFIRMATION OF MAXWELL'S PREDICTIONS
In 1887, after Maxwell's death, Heinrich Hertz (1857-1894) was the first to generate and detect electromagnctic waves in a laboratory sctting, using I.e circuits. Tn such a circuit a charged capacilor is connected to an inductor, as in Figurc 21.16. 'VVhen the switch is closed, oscillations occur in the current in the circuit and in the charge on the capacilOr. If the resistance of the circuit is neglected, no energy is dissipated and the oscillations continue. In the following analysis, we neglect the resistance in the circuit. Vve assume the capacitor has an initial charge of Qmax and the switch is closed at t = O. W'hen thc capacitor is fully charged, the total energy in the circuit is stOl'cd in the electric field of the capacitor and is equal to Q711:1)/2C. At this time, the current is zero, so no energy is stored in the inductor. As the capacitor begins 10 discharge, the energy stored in its electric Aeld decreases. At the same time, the current increases and energy equal to L1'2/2 is now stored in the magnetic field of the inductor. Thus, energy is transferred from the electric field of the capacitor to the magnetic field of the inductor. \Vhen the capacitor is fully discharged, it stores no energy. Al this time, the current reaches its maximum valuc and all the energy is stored in the inductor. The process then repeats in the reverse direct.ion. The cnergy continues to transfer between the inductor and thc capacitor, corresponding to oscillations in the current and charge. As we saw in Section 21.6, the frequency of oscillation of an LC circuit is called the Tesonancejrequf'1/c)' of the circuit and is given by
j"
=
+ C
FIGURE 21.16 A simpk' LCcircuit. The capacitor has an inilial charge of QrnH' and the switch is closed ;ll I = O.
TnduClion coil
1 27TVLC
The circuit Hertz used in his in\"estigations of electromagnetic waves is similar to that just discussed and is shown schemat ically in Figure 21.17. An induction coil
(a large coil of wire) is connected 1.0 two meta) spheres with a narrow gap between them to form a capacitor. Oscillations are initiated in the circuit by shon \"oltage pulses sent via the coil to the spheres, charging one positive, the other negative. Because L and Care quite small in t.his circuit, the frequency of oscillation is quite high./= 100 MHz. This circuit is called a transmitter because it produces electromagnetic waves. Several meters from the transmiuer circuit, Hertz placed a second circuit, the receiver, which consist.ed ofa single loop of wire connected to two spheres. It. had its own effective inductance, capacitance, and natural frequency 01" oscillation.
FIGURE 21.17 A schemalic diagram of Hertz's app3ratus for gcnenHing :l11d detecti]lg e1cnromagnl:tic wan:s. The transmitter C011sist~ of l\~'O spherical den rodes connected to an induction coil. which prO' ides short \'olla~e sl1rgc.~ to the ~phert's. sc.:ujug up o~cil1adol1s i1l1hc discharg(:. The rtcei,"er is a !lead)\" single loop of wire ('ol1l:linillg a second spark gap.
714
Chapter 21
Alternating-Current Circuits and Electromagnetic Waves
HEINRICH RUDOLF HERTZ German Physicist (1857-!894) Hertz made his most important discovery of rodio waves in 1887. After
finding that the speed of a radio wave was the same as thot of light, Hertz showed that rodio WQves, like light
waves, could be reflected, refracted, and diffracted. Hertz died of blood poisoning at the age of 36. During his short life, he mode many contributions to science. The hertz, equal
Hertz found that energy \vas being sent from the transmit.ter LO the receiver when the resonance frequency of the receiver was adjusted to match that. of the transmitter. The energy translcr was detected when the voltage across t.he spheres in the receiver circuit became high enough to produce ionization in the air, which c, Bm,IX
[21.27]
2JLo
where h:llla " and Billa" an: the 'JI/.axirn'll.fn values or E and B. The quamity I is analogous to the intensity of sound '...·avcs introduced in Chapter 14. From Equation 21.26, \-vc see that Ema " = (Bmax = BIIl Divide both sides by ill. obtaining the force il/J/ill exerted by the Jighl on the sail:
2Mill :1p= -2U =2'!J>ill --=-c
,
ill>
/- = -
~
2M
-
ill =
c
c
2(1 340 W/m')(1.00 X 106 m 2 )
= -'-----'----'-'---;:-------"-
3.00X10"m/s
c
8.93:\
(b) Find the time it takes to change the radial speed b), 1.00
km/s. F
=- =
SubSliwtc the force into NeWlOn's second law and soh'e for the acceleration of the sail:
II
Apply lhe kinematics velocit)' equation:
v = al
Solve for t:
...
111
+
8.93:\ 2.50 X 10' kg
X Hr I m/s 2
DO
1= _V_-_V_" =
a
= 3.57
.,-:.1,,0,,0,-,-X-=..1::.0·.,-;"","1/"s,,, 3.57 X 10 1 mis'
2.80 X 106 s
.
..
720
Chopter 21
Alternating-Current Circuits and Electromagnetic Waves
Remarks The answer LO parL (b) is a little over a 111onth. vVhile the acceleration is very 10\...·• there are no fuel costs, and within a few momhs the velocity can change sufficiently [Q allow the spacecraft to reach any planet in the solar system. Such spacecraft may be useful for certain purposes and are highly economical, but require a considerable amount of patience.
pose you were flaming in space and pointed the laser beam away from you. What v·muld your acceleration be? Assume your tolal mass, including equipment, is 72.0 kg and the force is direCled through your center of mass. !-lin/.: The change in momentum is the same as in the nonrefieClive case. (c) Compare the acceleration found ill pan (b) \\Iith the acceleration of gravity ofa space station of mass 1.00 X lOG kg, if the station's center of mass is 100.0 III away.
QUESTION 21.9 By what facLor will the force exerted by the Sun's light be changed when the spacecraft is twice as far from the Sun' (a) no change (b) ~ (c) t (d) k EXERCISE 21.9 A laser has a power of 22.0 \V and a beam J'adius of 0.500 mm. (a) Find the imensity of the laser. (b) Sup-
21.12
...
Answers (a) 2.80 X 10' Wlm' (b) 1.02 X 10-9 mis' (c) 6.67 X 10-9 m/s2 . I r you were planning 1.0 use your laser welding torch as a thruster [Q get you back La the station, don't bother because the force of gravity is stronger. Better yet, get somebody to toss you a line.
THE SPECTRUM OF ELECTROMAGNETIC WAVES
All electromagnetic waves travel in a vacuum with the speed o[ light, t. These waves transport energy and momenl.Urn [rol11 some source La a receiver. In 1887 Hertz successfully generated and detected the radio-frequency electromagnetic \'laves predicted by Maxwell. Maxwell himself had recognized as electromagnetic g \-,'aves both visible light and the infrared radiation discovered in 1800 by William E ~ Herschel. ft is now known that other forms of electromagnetic waves exist that are ~ disl.ing·uished by their frequencies and wavelengths. Because all electromagnetic waves travel through free space with a speed c, their ~III:"d
fX ~
35"
The angle a is complemenwfy to the angle of incidence, f3il1C' for M 2: Apply the law ofrefleClion a second time, obtaining {3rcf: lrrrr... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
Remark
f3ref = f3inc = 55°
.
~
NOlice the heavy reliance on elementary geometry and trigonometry in these renection problems.
QUESTION 22.1 Tn general, what is the relationship between the incident angle (Jille and the final reflected angle f3refwhen the angle benveen the mirrors is 90.OO? (a) 8 in e: + f3ref = 90.0° (b) 8 ine - f3 ref = 90.0" (c) 8ine + {3rcf = 1800 EXERCISE 22.1 Repeat the problem if the angle of incidence is 550 and the second mirror makes an angle of 100 0 with the first mirror.
Answer
45"
Refraction of Light '''lhen a ray of light traveling through a transparent medium encoumers a boundary leading into another transparent medium, as in Active Figure 22.6a, part of the ray is reflected and part enters the second medium. The ray that cillers the second medium is bent at the boundary and is said to be reJ'ncted. The incident ray. the reflected ray. the refracted ray, and the normal at the point of incidence all lie in the same plane. The angle of refraction, 82 , in Active Figure 22.6a depends on the properties of the 1:\\'0 media and on the angle of incidence, through the relationship sin 8,)
~~-
sin 8 J
V9
=
-=-
= constant
[22.3]
VI
where VI is the speed of light in medium I and v'!. is the speed of light in medium 2. Note that the angle of refraction is also measured v,lith respect to the normal. In Section 22.7 we derive the laws of reflection and refraction llsing Huygens' principle.
22.3
Incident rd.Y
l\'ormal
.\
V,
t\ir Glass
737
ACTIVE FIGURE 22.6 (a) A. rayobliqllel) incidcnI on an air-glass illlcrfacc. The refracted r"y is bClllloward lhe normal because 1.':/< lit, (b) Light incidcnl 011 lhe Lucite block bend.. bOlh \,'hen il enler.. the bloc-I.. and \\hen it leaves the block.
Rdlecled ray
I
The Law of Refraction
I
~. :821
, I
B Rcfr.lCled
ray (a)
(hI
Experiment shows that the path of a light ray through a refracting surface is reversible. For example, the ray in Acti\'c Figure 22.6a travels from point A to point B. If the ray originated at B, it would follow the same path to reach point A, but the reflected ray would be in the glass.
QUICK QUIZ 22.2 rfbearn 1 is the incoming beam in Active Figure 22.6b, \Vh ich of the other four beams are due to refleClion? \Vhich arc due to refraction? When light moves [rom a material in which its speed is high to a material in which its speed is 100'l"er, the angle of refraction 9'1 is less than the angle of incidence. The refractcd ray thercfore bcnds tov./arc! t.he normal, as shown in Acti\'c Figure 22,7£1. If the ray moves from a material in which it travels slov./Iy to a material in which it travels more rapidly, 8 2 is greater t.han 9 1, so the ray bends away from I.he normal, as shown in Active Figure 22.7b.
Normal "I
I I I
(JI >(J'1.
01 :
Air
Glass
'nl
W;1Ve
froll!
A' (a)
(h)
Nr:w \'{aVl: fronL
22.6
2 I
3 ~ ~
A'D
A'
f)
!J
). I' ,I JJ ,I
\..-\
6,,
,,f\
,
8'
8,
C
(hi
(a)
Huygens' Principle Applied to Reflection and Refraction The laws of reflection and refraclion were stated earlier in the chapter \v'ithol.lt proof. vVe nm,,· derive these laws using Hllygens' principle. Figure 22.23£1 illustrates the law of reflection. The line AA' represents a wave front of the incident light. As ray 3 travels from A' to C, ray 1 reflects from A and produces a spherical wavelet of radius AD_ (Recall that the radius of a Huygens W3\-elCt is v til.) Because the twO wavelets having radii A'Cand AD are in the same medium, they have the same speed v, so AD = A'e. \'feamvhilc. the spherical \·vavelct centered at B has spread only halfas far as the one centered at A because ray 2 strikes the surface later than ray 1. From J-luygens' principle, we find that the reflected \va\-e [rant is CD, a line tangent \.0 all the outgoing spherical \vavelets. The remainder or our analysis depends on geometry, as summarized in Figure 22.23b. Note that the righl triangles ADC and AA'Care congruent because they have the same hypotenuse, !le, and because AD = A'e. From the figure, we have
A'C sin 8 1 = AC
and
,
AD AC
smO' = I
The rigl1l-hand sides are equal, so sin 8 = sin 0;, and it follows that 0 1 = 8;, which is the law of reflection. Huygens· principle and Figure 22.24£1 can be llsed to deri\'e Snell's law ofrefraction_ 1n the time interval til, ray 1 moves from A. to B and ray 2 moves from A' to C. The radius of the outgoing spherical wavelet centered at A is equal to v2 dr. The distance A'e is equal to VI 6./. Geon1etric considerations 5hm.. . that angle A~4C equals 8] and angle ACE equals 82 . From triangles AA'Cand ACB, we find that and
2
i\kdiuml, speed of liglH
v,,6.l sin O'J = - - -
-
AC
VI
A'
--r-,Icdiulll 2. spc.;ed or light t'2
\
2
(a)
FIGURE 22.24
747
FIGURE 22.23 (a) J-111ygCI1S· COllstruel ion for prm·ing (IIC law of reflecI ion. (b) Triangle _\DC is congr (ICIlI to triangle AA'C.
~
....""-
Huygens' Principle
(b)
(a) I-Illygens' conslruction for proving lhe la\\' ofrefraclion. (II) Overhead view ora band rolling from concrete onto grass.
Chapter 22
748
Reflection and Refraction of Light
Hwe divide the first equation by the second,
\\-'C
sin 8 1
VI
sin 0':1.
v2
From Equation 22.4, though, we know that
VI
gel.
= r/tl. and v 2 = c/n 2 . Therefore,
sin 8)
clnj
112
sin 8'1
cll1'!.
11)
and it foIl0\"·5 that
which is the law of refraction. A mechanical analog of refraction is shown in Figure 22.24b. When the left end of thc rolling barrel reaches the grass, it slows down, while the right end remains on the concrete and moves at its original speed. This difference in speeds causes the barrel to pivot. changing the direCLion of its motion.
22.7 TOTAL INTERNAL REFLECTION
,
.~
/I' . This pll(,tograpll ,hows Ilonparallellighl rav, ('IHering a glass prism. TIl(' bOl\um
IWO I';\\'$
undcq..(o 10t,11
iJllt'rnal,'efknloll alThe IOllgcsl side of the prism. Thl' top three ra~'S an: refracted al the longest side as tlwy le;l\e the prism.
An interesting effect called lolal internal reflection can occur when light encounters the boundary bClween a medium with ], which is an absurd result because the sine oran angle can never begrealer than J. ACTIVE FIGURE 22.25 (... , Ra\-s from a medium with index of rdractioll I'I tnwd 10 a medium with indc:'l. ofr 112" lh th(' allgk of incidcllce illcre"..cs, the all~lc nf rerranion (11 increases until O~ is ~O (1';1\' I). For ('\'('11 larger angles "f illddcllrc, total internal reflection nccur~ (ra\ :». (b) The ll."e perisctlpc.
(h)
(a)
When mediulll 2 is air, the critical angle is small for substances with large indices of refraction, such as diamond, where n = 2.42 and Or = 24.0", By comparison, for crown glass, n = 1.52 and Or = 41.0°. This propeny, combined with proper faceting, causes a diamond to sparkle brilliantly. A prism and the phenomenon of LOLal internal relleClion can alter the direction oftra\'cl ora light beam. Figure 22.26 illustrates two such possibililies. In one case the light beam is deflected by gO.a" (Fig. 22.26a), and in the second case the path of the beam is reversed (Fig. 22.26b). A common application ofLOtal internal reOeclioll is a submarine periscope. In this device two prisms arc arranged as in Figure 22.2fic so that an incidelll beam of light follows the path shm.. . n and the user call "see around corners."
APPLYING PHYSICS 22.4
f
.
e
(i\ppl}'il1g I'h}'sin. 22.4)
EXAMPLE 22.7
Submarine Periscopes
TOTAL INTERNAL REFLECTION AND DISPERSION
A beam ofv,!hite light is incident on the curved edgc ora semicircular piece of glass, as shown in Figure 22.27. The light enters the curved surface along the normal, so it shows no refraction. It encounters the straight side of" the glass rl 2 (c) Light travels slm.. ·cr in the second medium than in the first. (el) The angle of incidence is less than the critical angle. (c) The angle of incidence lTILlSl equall.he refraction angle. 8. A light ray containing both blue and red wavelengths is incident at an angle on a slab of glass. Which of the sketches in Figure MCQ22.8 represents the most likely oULcome? (al A (b) B (c) C (d) 0 (e) none ofLhese
B
C FIGURE MCQ22.8
Problems 9. 'Which color light is bent the most \",hen entering crown glass frol11 air at some positive angle 0 with respect to Lhe normal?
FIGURE P22.25
26. A cylindrical cistern, conslruCled below ground level, is 3.0 m in diameter and 2.0 III deep and is filled to the brim with a liquid whose index of refraction is 1.5. A small ol~iect rests on the bottom of the cistern at its center. I [ow far from the cdge of the cistern can a girl whose eycs are 1.2 m from the ground stand and still see the object? 27. An opaque cylindrical tank with an open top has a diametef of 3.00 III and is completely filled with W;Hef. Whcn the afternoon Sun reaches an angle of 28.0° above the horizon, sunlight ceases to illuminate the bonom of rhe tank. How deep is the rank?
I I
\
beam is sent from the submarine so that the beam strikes the surface 01" the watcr 2.10 X 10 2 m from the shorc. A building stands on the shore, and the lascr beam hits a target ar the top of the building. The goal is to r-ind the height of the target above sea level. (a) Draw a diagram of the silLlation, identifying the two triangles that are imponant LO finding lhe solution. (b) Find the angle of incidence of the beam srriking the water-air interface. (c) Find the angle of refraction. (e1) \o\That angle does the refracted beam make v"'ith respect to the horizontal? (e) Find the height of the target above sea level.
Tumor
FIGURE P22.22
123.1_ A person looking inlO an empty container is able to sec the far edge of the container's botLOm, as shown in Figure P22.23a. The height of the container is h, and its width is d. When the container is completely filled with a nuid of index of refraction II, the person call see a coin at the middle of the container's bonom, as shown in Figure P2~.23b. (a) Show that rhe ratio hid is given by
~= d
' JB ~ 4 - 11'2
(b) Assuming the container has" w'idth of 8.0 Clll and is filled with W
FIGURE P22.S1
,,
11/
by 11= 2A.
,
,,
'_'_=_I_.'_18
FIGURE P22.S6
....-J }
3.] 0
1111ll
758
Reflection and Refraction of Light
Chapter 22
Determine the number ofinlcrnal reflections of the beam before it emerges from the opposite end of the slab. 57. A light ray enters a rectangular block of plastic at an anglt; 0 1 = 45.0 0 and emerges ,u an angle B,! = 76.00, as shown in Figure P22.57. (a) Determine [he index of refraction of the plastic. (b) If the light ray enters [he plastic at a palm L = 50.0 COl from the botlom edge, how long does illake the light ray to travel through the plastic?
\I
159.IFigure P22.59 shows the p,nh of a beam of light through sevnal laycrs with different indices of'refraction. (a) If 8 1 = 30.0 what is the angle 02 of the emerging beam? (b) 'What must the incident angle 8 1 be to have tOlal imer· !1fll rertection at the surLlce between the mediulll with n = 1.20 and the medium with n = LOO? 0
,
,8,
"
.~,
I.
I
Use the data to verify Snell's law of refraction by plolling thc sine of the angle of incidence versus t.he sine of the angle of refraction. From the resulting plot, deduce the index of refraction of water.
,
11 = 1.60
,
,, e,.
11=1.40
'It,,. ,
FIGURE P22.57
,
58. StuciclllS allo\\' a narrow beam of laser light to strike a waleI' surfilce. They arrange to measure the angle of refraction for selected angles of incidence and record the data shown in the following table: Angle of Incidence (degrees)
Angle of Refraction (degrees)
10.0
7.5
20.0 30.0 40.0 50.0 60.0 70.0 80.0
15.1
'"
,
tl = 1.20 ~
1/=
FIGURE P22.59
60. Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on LOp of shcet 2, and a laser beam is directed onlo the sheets from abovc so that it strikes the interface at an angle of 26.5 with the normal. The refracted beam ill sheet 2 makes an angle of 3l.r with the normal. The experiment is repehind tilt' mil for ilt dislance £I. which is equal ill m'lgniluck to [he ol~jecl distanco.: I).
759
760
Chapter 23
Mirrors and Lenses
ACTIVE FIGURE 23.2 A geometric cOllHrucrion 10 locale tbe image of all ol~j I, and upright because M is posit ive. (Sec Fig. 23.13b.) lrrr... • • • • • • • • • • •
Note the characteristics oran image formed by a concave, spherical mirror. \tVhen the ObjCCl is olltside the focal point, the image is inverted and real; al the focal point, the image is formed at infinity; inside the focal point, the image is upright and virtual.
Remarks
QUESTION 23.2 ''''hat location does the image approach as the object gels arbitraril}' far a\vay from the mirror? (a) infinity (b) the focal point (c) the radius ofcurvature of the mirror (d) the mirror itself EXERCISE 23.2 Tr the object distance is 20.0
Answer
nnd the image distance and the magnification of the mirror.
q = 20.0 em, M = -1.00
EXAMPLE 23,3 Goal
CIll,
Images Formed by a Convex Mirror
Calculate propenies ofa convex mirror.
Problem An objecl 3.00 (Ill high is placed 20.0 em from a convex mirror with a focal length of magnilUde 8.00 em. Find (a) the position oft.he image, (b) the magnification of the mirror, and (c) the height of the image.
Strategy This problem again requires only substitution into the mirror ;lI1d magnification equations. rv{ultiplying the object height by the magnification gives the image heighl. ,
~
~
Solution (a) Find the position of the image. Because the mirror is convex, its focal length is negative. Substitute into the mirror equation:
I
1
1
P
'I
I
-+-~-
--'----- + -I 20.0 em
'I
= ---'---
-8.00 em
q = -5.71 em
Solve [or q: (b) Find the magnification of the mirror.
1\1=
Substitute into Equation 23.2:
q
/)
_(-5.71 em) = 0.286 20.0 em
(c) Find the height o[the image.
Multiply the
ol~ject
height by t.he magnification:
h' ~ hM
= (3.00 el11)(0.286) = 0.858 em
~
~
Remarks The negative value of q indicates the image is virtual, or behind the mirror, as in Figure 23.13c. The image is upright because A1 is positive. QUESTION 23.3 True or False: A convex mirror can produce only virtual images. EXERCISE 23.3 Suppose t.he o~ject is moved so that it is 4.00 em from the same mirror. Repeat parts (a) through (c).
Answers
(a) -2.67 em
EXAMPLE 23.4 Goal
(b) 0.668
(c) 2.00 em; the image is upright and vinual.
The Face in the Mirror
Find a focal length from a magnification and an object distance.
Problem When a \voman stands with her face 40.0 em from her face. \,Vhat is the focal length of the mirror?
Col
cosmetic mirror, the upright image is twice as tall as
Strategy To nndfin this example, we must first find q, the image distance. Because the problem stat.es that the image is upright, the magnificat.ion must be positive (in this case, 1\1 = +2), and because .iVl = -ql!), we can delcr!TIme q.
Images Formed by Refraction
23.4
,..
.
.
Solution
-
769 _
~
'I --=2
Obtain q from the magnification equation:
'I Because q is negative, the image is on the opposite side of the mirror and hence is virtual. Substitute q and jJ into the mirror equation and solve rorJ:
~
P -21' ~ -2(40.0
UII) - -80.0 elll
1
40.0 em
80.0 em
f
f= 80.0 em
..........
.
..
Remarks The positive sign for the focal length tells us that the mirror is concave, a fact \..'e already knew because the mirror magnified the object. (A convex mirror would have produced a smaller image.)
QUESTION 23.4 If she moves the mirror closer to her face, what happens to the image? (a) It becomes inverted and smaller. (b) It
ren1ains upright and becomes smaller. (c) It becomes inverted and larger. (d) It remains upright and becomes larger.
EXERCISE 23.4 Suppose a fun~house spherical mirror makes you appear to be one-third your normal height. lfyou are 1.20 m away from the mirror, find its £-ocal length. Is the mirror concave or convex? Answers
23.4
-0.600 m, convex
IMAGES FORMED BY REFRACTION
In this section we describe hm"-' images are formed by refraction at a spherical surface. Consider t\.. . o transparent media with indices of refraction t1 1 and 11 2 , \vhere the boundary between the two media is a spherical surface of radius R (Fig. 23.l5). \,Ve assumc the medium 1.0 the right has a higher index of refraction than the one to the left: 112 > n l • That would be the case for light entering a curved piece of glass from air or for light entering thc water in a fishbowl from air. The rays originating at the objeCt location Dare refracted at the spherical surface and then converge to the image point I. vVe can begin with Snell's law of refraction and use simple geometric techniques to show that the object distance, image distance, and radius of curvalllre are related by the equation
[23.7] Further, the magnification ofa refracting surface is 11.'
nllj
II
n'l/J
M=-= - -
[23.8]
As with mirrors, certain sign conventions hold, depending on circumstances. First note that real images are [armed by refraction on the side of the surface 0/)/)0sil" the side from which the light comes, in contrast to mirrors, where real images arc formed on the S(J.JJ11' side of lhe renecting surface. This makes sense because light reflects off mirrors, so any real images must form on the same side the light comes from. \'\7ith a transparent medium, the rays pass through and naturally form real images on the opposite side. \Ve define the side of the surface \\,'here light rays originate as the front side. The other side is called the back side. Because of the difference in location of real images, the refraction sign conventions for q and R are the opposite of those for reflection. For example, p, q, and R are all positive in Figure 23.15. The sign convcntions for spherical refracting surfaces are summarized in Table 23.2 (page 770).
FIGURE 23.15 An image formed by refraction at a spherical surface. Ra)'s makillg small angles with the principal axis dh'erge from a pOilU objecl at o and pa~~ lhrough the image point!.
770
Chapter 23
Mirrors and Lenses
TABLE 23.2 Sign Conventions for Refl'acting Surfaces Symbol
In Front
ObjeCllocalioll
/J
+
[mag"c location
q
Radius
R
Image height
I,'
Quantity
APPLYING PHYSICS 23.4
Upright
Inverted
Image
Image
+ + +
UNDERWATER VISION
,I
''''hy does a person with normal vision see blurry image if the eyes are opened underwater h'ith no goggles or diving Illask in use?
Explanation The eye presents a spherical refraction surface. The eye normally functions so that light entering from [Jle air is refractcclto form an image in the retina located at the back of the eyeball. The difference in rhe index of refraction bctween ""ater and the eye is smaller than the difference in the index of refraction ben·...een air and rhe eye. Conse-
quently. light entering the eye from the water doesn't undergo as much refraction as does lighlentcring from the air, and the image is formed behind the retina. A diving mask or swimming goggles hayc no optical action of their own; they eIre simply flat pieces or glass or plastic in a rubber lTlounL. Theydo, however, provide a region or air ;,}cUacclllLO the eyes so that the correel refraction relationship is established and images will be in focus.
Flat Refracting Surfaces If the refracting surf'ace is rIat, then R approaches inAnity and Equation 23.7 reduces to /
o~ ,,
1/,
17'1
/J
q 1/2
'1= --I) II,
!----!'ACTIVE FIGURE 23.16 Thc image form cd bra nat refraCting surface is virtual and 011 the S;Jmc side of the surface as the object. \'ote that iflhe Iighl rays are re\"cr~ed ill direction, we h:we the situation (k~crihcd ill Example 22.7.
[23.9]
Frolll Equation 23.9, \ve see that the sign of q is opposite that of /J. Consequently. the image formed by a flat refracting surface is on the same side of the surface as the object. This statemel1L is iliustr 'l.:!..
QUICK QUIZ 23.2 A person spearfishing from a boat sees a fish localcd ~ m from the boat at an apparent depth of 1 m. To spear the fish, should the person aim (a) at, (b) above, or (c) below [he image or the fish? QUICK QUIZ 23.3 True or False: (a) The image of an object placed III front ora concave mirror is ahvays upright. (b) The height of the image oran object placed in front of a concave mirror must be smaller than or equal to the height of I he object. (c) The image of an object pl J), the ray diagram shows that the image is real and inverted. \Vhen the real object is inside the front focal point (/) < j), as in Active Figure 23.25b, the image is virtual and upright. For the cliverging lens of Active Figure 23.25c, the image is virtual and upright.
QUICK QUIZ 23.4 A clear plastic sandwich bag filled with water can act as a crude converging lens in air. If the bag is filled ,vith air and placed under ',,';'Her, is the effective lens (a) converging or (b) diverging?
QUICK QUIZ 23.5
In Aetive Figure 23.25a the blue object arrow is replaced by one that is much taller than the lens. How many rays from the object will strike the lens?
QUICK QUIZ 23.6 An object is plaeed to the left of a converging lens. ",Vhich of the follm.. . ing statements are true, and which are false? (a) The image is alvvays to the right of the lens. (b) The image can be upright or inverted. (c) The image is always smaller or the same size as the object. Your success in vo.'Orking lens or mirror problems will be determined largely by whether you make sign errors ,.. . hen substituting into the lens or mirror equations. The only way to ensure yOll don't make sign errors is to become adept at using the sign conventions. The best ways to do so are to work a multitucle of problems on your m,m and to construct confirming ray diagrams. ,,,,Tatching an instructor or reading the example problems is no substitute for practice.
APPLYING PHYSICS 23.5
VISION AND DIVING MASKS
Diving masks often have a lens built inLO the glass faceplate for divers who don't have perfect vision. This lens allows the individual to dive without the necessity or glasses because the faceplate perfonns the necessary refraction to produce clear vision. Normal glasses have lenses that are curved on both the front and rear surfaces. The lenses in a diving-mask faceplate often have curved surfaces only on the inside oUhe glass. \rVhy is this design desirable? The main reason for curving only the inner surface of the lens in Lhe diving-mask face-
Solution
plate is to enable the diver to see clearly while underwater and in the air. Il"there were curved surFaces on both the front and the back of the diving lens, there would be two refractions. The lens could be designed so that these twO refractions would give clear vision while the diver is in air. \'\then the diver went underwaLer, however, the refraction between the water and the glass at the first interface would differ because the index of refraction oh..· ater is different from that of air. Consequently, the diver's vision wouldn't be clear underwater.
23.6
EXAMPLE 23.7 Goal
Thin Lenses
Ima es Farmed b a Converging Le"'ngs:...
777
_/
~
Calculate geometric quantities associated with a converging lens.
A converging Jens or local length 10.0 cm forms images of an object siLUated at various distances. (a) If the object is placed 30.0 em from the lens, locCite the image, state whether it's real or virtual, and I-ind its magnitica(jon. (b) Repeat the problem when the object is at 10.0 cm and (c) again when the object is 5.00 Clll from the lens.
Problem
All three problems require only substillltion into the thin-lens equat.ion and the associated magnification equation, Equations 23.10 and 23.11, respectively. The conventions of Table 23.3 must be followed.
Strategy
~
. ... '..,.
.
Solution (a) Find the image distance and describe the image when the ol~ject is placed at 30.0 em. I
l
1
-+-=/) q J
The ray diagram is shown in Figure 23.26a. SubstilUte values into the thin-lens equation to locate the image:
J I +30.0 em q
I
= ---,--,-----
~,-----
10.0 em
Solve for q, the image distance. It's positive, so the image is real ancl on the far side of the lens:
q = +15.0 em
The magnification of the lens is obtained fro111 Equation 23.JO. M is negative and less than 1 in absolute valuc, so the image is inverted and smaller than the object:
M=
q
15.0 cm 30.0 em
/)
-0.500
(b) Repeallhe problem, when lhe objeel is placed al 10.0 em. 1
]
10.0 em
q
Locate the image by substituting into the thin-lens equation:
---,--,----- + -
This equation is satisfied only in the limit as q becomes infinite.
q -->
~
1
---,--,----10.0 em
1 -=0 q
(c) Repeal lhe problem when lhe object is placed 5.00 em from the lens.
1
]
5.00 em
q
See the ray diagram shown in Figure 23.26b. SubstiLUte into the thin~lens equation to locate the image:
----,-,----- + -
Solve for 'I, which is negllt.ivc, meaning the image is on the same side as the object. and is virwal:
q = -10.0 em
1
= ---,--,-----
10.0 em
FIGURE 23.26
a 1--+---->1 10.0 em
15.0
{'Ill
J
10.Oem
:).00 em
~30.0cm
10.0 em (al
(b)
(Example 23.7)
Chapter 23
778
Mirrors and Lenses
q
M=
SubstilUte the values of j) and q into the magnification equation. iV1 is positive and larger than I, so the image is upright and double the object size:
l'
em) em
_(-10.0 5.00
+2.00
~
~
Remark scopes.
The ability of a lens to magnify objects led to the inventions of reading glasses, microscopes, and tele-
QUESTION 23.7 If the lens is lIsed
LO
form an image of the sun on (\ screen, hm...., far [rom the lens should the screen be located?
EXERCISE 23.7 Suppose the image o[an object is upright (lnd magnified 1.75 times when t.he object is placed 15.0 cm from a lens. Find the location of [J1C image and the focal length of the lens.
Answers
o~jeet)
(a) -2fi.3 em (virtual, on the samc side as thc
(b) 34.9 em
EXAMPLE 23.8 The Case of a Diver ing Lens Goal
Calculate geometric quantities associated \',1ith a diverging lens.
Problem
Repeal the problem of Example 23.7 ror a divergi-nglens of focal length 10.0 cm.
Strategy Once again, substitlllion into the thin-lens equation and the associated magnification equation, together with the conventions in Tablc 23.3, solve the various parts. The only difference is the negative rocallength.
.
"-'
.
~
Solution (a) Locate t.he image and ils magnification if the object is at 30.0 cm. I
I
1
-+-=!J (/ I
The ray diagTam is given in Figure 23.27a. Apply the thin-lcns equation with j) = 30.0 cm to locate the image:
1
1
---+-= ~o.o
Solve for 'I, which is ncgative and hence virtual:
q=
Substitute into Equation 23.10 to get the magnification. Because J11 is positive and has absolute value Icss than I, lhe image is upright and smaller than the objeCl:
A1 =
elll
~7.50
'I
10.0 em
cm
'I
!J
em)
-7.50 ( 30.0 elll
+0.250
(b) Locate the image and flnd its magnification irthc o~icet is
Solve for q (once again, the result is ncgatiyc, so the image is virlual):
Thin Lenses
779
q = -5.00 em q
Calculate the mxrlmplp (}flhr. n>sltlrs oh'::lined with small
apertures is the sharp image produced by a pinhole camera, with an aperture size of approx imaLcly 0.1 1111ll. In the case of mirrors used for very distant objects, one can eliminate, or at least minimize, spherical aberrat.ion by employing a parabolic rather than spherical surface. Parabolic surfaces are not used in many applications, however, because they are very expensive to make with high-quality optics. Parallel light rays incident on such a surface focus at a common point. Parabolic reflecting surfaces are used in many astronomical telescopes to enhance the image quality. They are also used in flashlights, in which a nearly parallel light beam is produced from a small lamp placed at the [ocus of the refleCLing surface.
Chromatic Aberration Different wavelengths of light refracted by a lens focus at different poims, which gives rise to chromatic aberration. In Chapter 22 we described how the index of refraction ofa material varies Wi111 ""avclength. vVhen "vhite light passes through a lens, for example, violet light rays are refraCLcd more than red light rays (see Fig. 23.31), so the focal leng·th for red light is greater thall for violet light. Other wavelengths (not shown in the figure) would have intermediate focal poims. Chromatic aberration for a diverging lens is opposite that for a converging lens. Chromatic aberration can be greatly reduced by a combination of converging and diverging lenses.
FIGURE 23.31 ChrolllaLie aberration produced by a comcrging lens. Ry=y m- \-)' III =-(1Il+l) d
AL
AL d
--1n=-
d
(5.63 X 10- 7 Ill)( 1.20 Ill) = 2.25 elll 3.00 X 10 r, III
...
..............
~
Remarks This calculation depends on the angle 8 being small because the small-angle approximation ,vas implicitly used. The measurement of the position of the bright fringes yields the wavelength of light, which in turn is a sigIlature of atomic processes, as is discussed in the chapters on modern physics. This kind of measurement therefore helped open the world of the a 1.0 Ill. QUESTION 24.1 True or False: A larger slit creates a larger separation between interference fringes. EXERCISE 24.1 Suppose the same experiment is run with a different light source. If the first-order maximum is found at 1.85 from the centerline, what is the wavelength of the light?
Answer
463 n m
24.3
CHANGE OF PHASE DUE TO REFLECTION
Young's method of producing two coherent light sources involves illuminating a pair of slits with a single source. Another simple, yet ingenious, arrangement for producing an interference pattern with a single light source is known as Lloyd's mirrm: A point source of light ;s placed at. point S, close to a mirror, as illustrated in Figure 24.5. Light waves can reach the viewing point Peither by the direct path SP or by the path irl\'olving reflection from the mirror. The reflected ray can be treated as a ray originating at the source S' behind the mirror. Source 5', which is the image of S, can be considered a virtual source. At points far from the source, an interf-erence pattern due to 'waves from Sand 5;' is observed, just as for tWO real coherent sources. The positions of the dark and bright fringes. however, are revl'J"sed relative to the pattern obtained from tWO real coherent sources (Young's experiment). This is because the coherent sources Sand S' differ in phase b)' 180°, a phase change produced by reflection. To illustrate the point further, consider pi, the point where the mirror intersects the screen. This point is equidistant from Sand S'. If path difference alone were responsible for the phase difference, a bright fringe would be observed at P' (because the path dirf"erence is zero for this point), corresponding to the central fringe of the t\vo-slit interference pall.ern. Instead, we observe a dark fringe at P, from "ihich we conclude that a 180° phase change must be produced by reflection from the mirror. In general, an electromagnetic wave undergoes a phase change of 180° upon reflection from a medium that has an index of refraction higher than the one in which the wave was traveling. An analogy can be drawn between renected light waves and the reflections of a transverse wave on a stretched string when the wave meets a boundary. as in Figtire 24.6 (page 796). The reflected pulse on a string undergoes a phase change of 180° when it is reflectcd from the boundary of a denser string or from a rigid barrier and undergoes no phase change when it is rerJected frolll the boundary of a less dense string. Similarly, an electromagnetic wavc undergoes a 180° phase change \,,'hen reflected frolll the boundary of a mediulll with index of refraction higher than [he one in \vhich it has been traveling. There is no phase change "..· hen the wave is renected from a boundary leading to a medium of lower index of refraction. The transmitted 'wave that crosses the boundary also undergoes no phase change.
CIll
Viewing screen
Real source
s P' / /
Virtualc,"'''' source S'
MiJTor
FIGURE 24.5 L1oyd's mirror. All imerference pattern is produced Oll a screen at Pas a result ofcbe combinalion of the direcl ray (billel and the reflected ray (brown). The reflected ray undergoes a phase change of 180".
796
Chapter 24
Wave Optics
1800
pha~c
change
No phase change
~ Incident wave
Incident wave
-~
III
III
-
?l.2
Free suppon
1/2
(h) FIGURE 24.6 (a) A ray rellct:lillg from a mediulll of higher refractive index undergoes a 180' phase change. The "ighl side shows the analogy with a renecled pulse on a slring. (b) A ray reflecting from a mediulll of lower rcf.-active index undergoes no phasc change.
180 0 phase
Ch"ng~
f
.Air
No phase
c"""Re
Film with index /I
/
Air
24.4
INTERFERENCE IN THIN FILMS
I.nterferencc effects arc commonly observed in thin Alms, such as the thin surf~lce of a soap bubble or thin layers of oil on water. The varied colors observed \vhen incoherent white light is incident on such films result from the interfercnce of""aves rerJected from the two surfaces of the film. Consider a film of uniform thickness t and index of refraction n, as in Figure 24.7. Assume the light rays traveling in air are nearly normalLO the (\'\'0 surfaces of the film. To determine ,.. . hcther the rcrJcned rays interfere constructively or destructively, we fi rst note the follm\'ing facts: 1. An elcClromagnctic ,..'ave traveling from a medium ofindex of refraction Jl 1 toward a medium of index or refraction n2 undergoes a 180 phase change on reflecLion , = tan- I (1.52) = 56.7°. Because nvaries \-vith wavelength for a given substance, Brewster's angle is also a function ofwavclengl.h. Polarization by refleClion is a common phenomenon. Sunlight reflected from watcr, glass, or sIlO''''' is partially polarized. If the surface is horizol1lal, the electric field vector of the reflected light has a strong horizol1Lal component. Sunglasses made of polarizing material reduce the glare, which is the renected light. The transmission axes of the lenses are oriented vertically to ahsorb the strong horizontal component of the reflected light. Because the reflected light is mostly polarized, lllost of the glare can be eliminated ,...irhout removing most of the normal light.
Polarization by Scattering
Unpolarizcd
light
,
-
•,
~
Air molecule
FIGURE 24.29 The scaucring or unpolarilcd sunlight b~ air ll1ol~ eculcs. The 1i~hl ob~ef\'cd ill right angles is linear!} polarilcd because the vibt AI' what is the separation between bright lines in the two experiments? (a) It's the same for both experiments. (b) It's greater in the first experimenl. (c) It's greater in the second experimcnt. (d) It's dependent on the intensity of the light. (e) ~tore informat.ion is required. 8. W'hat causes t.he brightly colored patterns sometimes seen on wet streets covered with a la)'er of oil? Choose the best answer. (a) diffraction and polarization (b) interference and deflection (c) polarization and reflection (d) refraction and diffraction (e) reflcnion and interference 9. 'What happens to light when it is reflected ofT a shiny surface on a sunny day? (a) h undergoes refraction. (b) It is wtally polarized. (c) It is unpolarized. (d) It tends to be partially polarized. (e) More information is required. 10. Consider a wave passing through a single slit. What happens 1.0 t.he width of the ccntralm3ximuIll of its diffraction pat.tern as the slit is made half as wide? (a) It becomes one-fourth as wide. (b) It becomes one-half as ",..· ide. (c) Its width does not change. (d) It becomes twice as wide. (c) It becomes four times wider.
il).CONCEPTUAL QUESTIONS ~
1. Your automobile has two headlights. What sort of interference pattern do rOll expectLO see from them? 'Nhy? 2. Holding your hand at arm's length, you can readily block direct sunlight from your eyes. Why can you not block sound from your ears this \~'ay?
3. Consider a dark fringe in an interference pattern at which almost no light energ)' is arriving. Light from both slits is arriving at this point, but the wavcs cancel. '",'here docs the energy go? 4. If Young's doublc~s]it experiment were performed under water, how would the observed interference pattern be a frected? 5.ln a laboratory accident, you spill two liquids onto \,·ater, neither of \vhich mixes with the water. They both form t.hin films on the water surface. As the films spread and become very thin, yOll not.ice that one film becomes bright and the other black in reflected light. \Vhy might. that be?
6. If white light is used in Young's double-slit experiment rather than monochromatic light, how docs t.he interference pattern change? 7. A lens with ourer radius of curvature R and index of refraClion 11 I-ests on a flat gla~s plate, and the combination is illuminat.ed from white light [rom above. Is there a dark spot or a light spot at the center of the lens? ""Vhat does it mean if the observed rings are noncircular? 8. Often, fingcrprints left on a piece of gla~s such as a windowpane show colored spectra like that from a diffraction grating. ""hy? 9. In everyday experiencc, why are radio waves polarized, whereas light is not? 10. Suppose reflected \vhite light is used [Q obsen;e a thin, t.ransparent coating on glass as the coating material is gradually deposited by evaporation in a vacuum. Describe some color changes that might occur during the process of bu ilei ing up the th ickness of the coating.
Problems
817
11. Would it be possible t.o place a nonreflecth'e coating 011 an airplane to cancel radar waves orvo.'avelengt.h 3 cm?
12. Cert.ain sunglclsses use a polarizing material to reduce the intensity orlight reflected from shiny surfaces, such as \-vat.er or the hood of a car. \'\That orientat.ion or the transmission axis should thc material ha\'c [Q be most effectivc?
13, \,yhy is it so much easier to perform int.erference experiments with a laser than with an ordinary light source? 14. A soap film is held vertically in air and is viewed in reflected light as in Figure CQ2L1.14. Explain why lhe film appears to be dark at.thc top.
FIGURE CQ24.14
PROBLEMS ~
WebAssign
I.~.:l
m
mil
II = 0=
The Problems for this chapter may be assigned online at \A/ebAssign.
= straightfol"ward,
ilHermccli,ltc.
dl-"f.-- (/
FIGURE 25.1 Cross-sectional \"ic\\ ora ~illlple digital camera. The CCD or C~IOS illlag-c sen sur i" IIll': lig-hlSCllSil ivc compollenl uf lile callH.·ra. 111;1 llondigital camera the JighllrolTl the Il.:ll~ fall~ 011[0 pholographic lilln. III n:alil)' /»> If.
823
824
Chapter 25
Optical Instruments
Most cameras also have an aperture of acljustable diameter [0 further control the intensity of the light reaching the Him. ',Vhen all aperture of small diameter is used, only light from the central portion of the lens reaches the film, so spherical abernllion is reduced. The iIlLen:siL)' J uf the light reaching the film is proportional to the area of the lens. Because this area in turn is proportional LO the square of the lens diameter D, the intensiL)' is also proponional to D?. Light intensity is a measure of the rate at v.. hich energy is received by the film per unit area of the image. Because the area of the image is proporLional La rl in Figure 25.1 and q = ! (when fJ » f, so that p can be approximated as infinite), we conclude that the illlensity is also proportional to 1//'2. Therefore, ]-:x D2/f2. The brightness of the image formed on the film depends on the light intensity, so 'we see that it ultimately depends on both the focallengthfand diameter D of the lens. The ratio flD is called theI-number (or focal ratio) of a lens:
f
f-number == D
[25.1]
The I-number is oftcn given as a dcscription of the lens "speed." A lens \..·ith a low.f-number is a "fa~H" lens. Extremely fast lenses, v\'hich have an I-number as low as approximately 1.2, are expensive because of the difficulty of keeping aberrations acceptably small v.lith light rays passing through a large area of the lens. Camera lenses are often marked with a range of j-numbers, such as 1.4,2,2.8,4, 5.6,8, and 11. Anyone of these scltings can be selected by acljusting the aperture, which changes the value of D. Increasing the sening from one I-number to the next-higher value (for example, from 2.8 to 4) decreases the Clrea of the aperture by a [actor of 2. The 1m-vest I-nu mber scui ng all a camera corresponds to a \vide open aperture and the use of the maximum possible lens area. Simple cameras lIsually have a fixed focal length and fixed aperture size, with an f-number of abollt 11. This high value for the f-number allmvs for a large depth of field and means that objects at a wide range of distances from the lens form reasonably sharp images on the film. In other words, the camera doesn't have to be focused. i\tlost cameras \\lith variable I-numbers acUust them automatically.
25.2 THE EYE HI Like a camera, a normal eye focuses light. and produces a sharp image. The mechanisms by which the eye controls the amount of light admitted and adjusts to produce correctly focused images, hm,\'cver, are far more complex, intricate, and effective than those in even the most sophisticated camera. In all respects the eye is a physiological wonder. Figure 23.2a shows the essential parts of the eye. Light entering the eye passes through a transparent structure called the cornea, behind which are a clear liquid (the aqueous humor), a variable aperture (the jJupil, which is an opening in the in's), and the C1)'stalline lens. Most of the refraction occurs at the outer surface of the eye, where the cornea is covered with a film or tears. Relatively little refraction occurs in the crystalline lens because the aqueous humor in contact with the lens has an average index of refraction close to that of the lens. The iris, which is the colored portion of the eye, is a muscular diaphragm that controls pupil size. The iris regulates the amount of light entering the eye by dilating the pupil in low-light conditions and contracting the pupil under conditions of bright light. The I-number range of the eye is from about 2.8 to 16. The cornea-lens system focuses light onto the back surface of the e>'e-the ret~ i17a-'..vhich consists of millions of sensitive receptors called mds ancl (;Ones. ',Vhen stimulated by light, these structures send impulses to the brain via lhe optic nerve, converting them into our conscious view or the world. The process by which the brain performs this conversion is not \\"ell understood and is the su~ject of much
FIGURE 25.2 (a) F.ssenlial pb 01 all ol~jeCliH' ,md all eyepiece. or ondar Il"ns. (b) A CtlillpOlllld llIiuo.'>cope. The lhree-o]~it'cti\l' turret allow., lht' l11>er 10 switch 10 ,c\"(,;ral difrert..'lIt pnw('rs of m where 80 is the angle subtended by t.he object at t.he objective and 8 is the angle subtended by the final image. From t.he triangles in Active Figure 25.8a, and for small angles, we hm'c h' 8=-
and
f"
Ii
The Hubble Space Telescope enables liS 10 see both funhcr inlO space and further back in lime than ever before.
8=-
"
fo
Therefore, the angular magnification of the telescope can be expressed as
8
h'lf,.
;;,
8,
10f"
f"
11/=-=--=-
[25.8]
This equation says that the angular magnification of a t.elescope equals the ratio of the objective focal length to the eyepiece focal length. Here again, the angular magnification is the ratio of the angular size seen wit.h t.he telescope to the angular size seen with the unaided eye. In some applicat.ions-for instance, the observation of relatively nearby objects such as the SUIl, the Moon, or planets-angular magnification is imponant. Stars, hm·\.'cver, ;ue so far away that they always appear as slllall points of light regardless of how much angular magnificat.ion is used. The large research telescopes used to swell' very distant objects must have great diameters to gather as much light as possible. It's difficult and expensive to manufacture such large lenses for refract· ing telescopes. In addition, the heaviness of large lenses leads to sagging, which is another source of aberration. These problems can be partially overcome by replacing the objective lens with a reflecting, concave mirror, usually havjng a parabolic shape so as to avoid spherical aberration. Figure 25.9 shows the design of a typical reflecting telescope. Incom· ing light rays pass down the barrel of the telescope and (Ire reflected by a parabolic mirror at the base. These rays converge toward point A in the figure, where an image would be formed on a photographic plate or another detector. Before this image is formed, however, a small, flat mirror at 1"1 reflects the lig·ht toward an opening in the side of the tube that passes into an eyepiece. This design is said t.o have a Newtonian focus, after its developer. Note that in the reflecting telescope the
~
Angular magnification
of a telescope
.\
1
.
\
En:pie Solution (a) \Vhat is the limiting angle of resolution at a wavelength of6.00 X 10'nm? Substitute D = 2.40 m and A = 6.00 Equation 25.10:
x 10-; III into
()min
= 1.22
~ -_ 1.22 (6.00 X 10D
7Ill)
2.40 m
3.05 X 10- 7 rad (b) \"'hat's the srnallesl lunar crater the Hubble Space Telescope can resolve? The twO opposite sides of the crater must subtend the minimum angle. Use the arc leng[h formula: ~
,
.
s = .. 8 = (3.84 X 10" 1ll)(3.05 X 10-; rad)
.
117m ~
Remarks The distance is so great and the angle so small that using the arc length of a circle isjustified because the circular arc is very nearly a straight line. The Hubble Space Telescope has produced several gigabytes of data e\'ery day since it first began operation. QUESTION 25.7 Is the resolution of a telescope betLcr at the red end of the visible spectrum or the violet end? EXERCISE 25.7 The Hale telescope on Mount Palomar has a diameter of 5.08 In (200 in.). (a) Find the limiting angle of resolution for a wavelength of6.00 X 10 2 nrn. (b) Calculate the smallest crllter diameter the telescope can resolve on the Moon. (c) The answers appear better than what the Hubble can achieve. \Vhy are the answers misleading?
Answers (a) 1.44 X 1O~7 rad (b) 55.3 m (c) Allhough the numbers are beu:er than Hubble's, the Hale telescope must contend with the effects of mmospheric turbulence, so the smaller space-based telescope actually obtains far bctter results.
IL's intercsting to compare the resolution of thc Hale telescope with that of a large radio telescope, such as the system at Arecibo, Puerto Rico, which has a diameter of 1 000 ft (305 m). This telescope detects radio waves at a wavelength of 0.7.:S m. The corresponding minimum angle of resolution can be calculated as 3.0 X 10-3 racl (10 min 19 s of arc), w'hich is more than 1O 000 times larger than the calculated minirllum angle [or the Hale telescope. W'ith such relatively poor resolution, \vhy is Arecibo considered a valuable astronomical inst.rument? Unlike its optical counterparts, Arecibo can see through clouds of dust. The center of our Milky "Vay galaxy is obscured by such dust clouds, which absorb and scatter visible light. Radio waves easily penetrate the clouds, so radio telescopes allow direct observations of the galactic core.
Resolving Power of the Diffraction Grating The diffraction grating studied in Chapter 24 is mOSt useful for making accurate wavelength measurements. Like the prism, it can be used to disperse a spectrum into its components. or the two dC"ices, the grating is better suited to distinguish-
25.6
Resolution of Single-Slii and Circular Apertures
839
ing between twO closely spaced wavelengths. We say thalthe grating spectrometer has a higher H'.wlution than the prism spectrometer. If AI and A2 26.4
Consequences of Special Relativity
26.5
Relativistic Momentum
26.6
Relativistic Energy and the Equivalence of Mass and Energy
26.7
General Relativity
GALILEAN RELATIVITY
To describe a physical event, it's necessary to choose afm11l(> of refere-no'. For example, when you perfor111 an experimcm in a laboratory, you select a coordinate s)'stem, or frame of reference, that is at rest with respect to 1he Iabo.-II
I I
I I
x'
0'
QUICK QUIZ 26.4 You are packing lor a trip La another star, and on your journey yOll will be traveling at a speed of 0.99(. Can you sleep in a smaller cabin than usual, because you will be shoneI' when you lie down? Explain your answer.
i
I' I I
(a)
o
(b)
.,
ACTIVE FIGURE 26.9 A meterstick moves to the righl with il speed u. (a) The meterstick as viewed by an obsen'er ,H rest wil h respect 10 the meterstick. (b) The meterstick as seen by an observer moving Idlh a speed 11 with respect to it. The moving melerstick is always measured to be short!'r than in its own rest frame by a faclorOrVl - v 2/c 2•
observe a rocket moving away from you. Compared with its length when it was at rest on the ground, ""ill yOll measure its length to be (a) shaner, (b) longer, or (c) the same? Compared to the passage of time measured by the watch on your wrist, is the passage of time on the rocket's clock (d) fast.er, (e) slower, or (f) the same? Answer the same quesLions if (he rocket turns around and cOllleS to\vard >'ou. YOll
EXAMPLE 26.2 Goal
Apply the concept of length contraction to a distance.
Problem (a) An observer on Earth sees a spaceship at an altitude of 4350 km moving dO\\'n""ard toward Earth with a speed ofO.970c. ''''hat is the distance from the spaceship to Earth as rneasured by the spaccship's captain? (b) After firing his engines, the captain Illeasures her ship's ahitude as 267 km, ""hcreas the observer on Earth measures it to be 625 km. What is the speed of the spaceship althis instant? Strategy To the captain, Earth is rushing toward her ship at O.970c; hence, the distance between her ship and Earth is contracted. Substitution into Equation 26.9 yields the answer. In part (b) use the same equation, subsLilllting the distances and solving for the speed. "-'
.
....
Solution (a) Find the distance from the ship to Earth as measured by t.he capt.ain. Substitute into Equation 26.4, getting the altitude as measured by lhe captain in the ship:
L
=
Lp'/1
=
1.06
X
v'/c' 10' km
=
(4350 km)VI ~ (0.970c)'/c'
"'"
858
Chapter 26
Relativity
(b) Whal is the subsequent speed or tile spaceship if the Earth observer Illeasures the distance from the ship to Eanh as 625 kill and the captain measures it as 2fi7 km?
Apply the length-contraction equation: Sq\larC both sides of Illis equation and solve for v:
, "(
")
L = L,,- 1 - v-Ie II
~
(v I -
(LI Ly
->
1 - v'l (2
=
(0L)'
= ("Ill - (267 km/625 km)'
v = 0,904c
.
~.
Remark
~
The proper length is always the length measured by an observer at rest with respect to that length.
QUESTION 26,2 As a spaceship approaches an observer at nearly the speed oflighl, the captain directs a beam of yellow light at [he observer. \-Vhat would tlie observer report upon seeing the light? (a) Its wavelength \vollid be shifted toward the red end of the spectrulll. (b) Its \\'' To star
SUII
865
(actual direction)
Earth
for a black hole with the mass of our Sun. At the black hole's center may lurk a sil1guIm'ity, a point of infinite density and curvalLlre where spacetime comes to an end. There is strong evidence for the existence of a black hole havi ng a mass of millions of Suns at the center of our galaxy.
APPLYING PHYSICS 26.1
FASTER CLOCKS IN A "MILE-HIGH CITY"
At.omic clocks are extremely accurate; in fact, an error of 1 sin 3 million years is typical. This error can be described as aboUl one pan in 10 1.'. On t.he other hand, the awmic clock in Boulder, Colorado, is often 15 ns faster than the atomic clock in 'Vashington, D.C., after only one day. This error is about one pan in 6 X ]0 12 , which is about]7 t.imes larger than the lypical error. Ifatomic clocks are so Clccurate, why docs a clock in Boulder not remain synchronous \\o'ilh one in Washington, D.C.?
Explanation According to the general theory of relativity, the passage of time depends on gravity: clocks rUIl more slowly in strong gravitational fields. "Vashington, D.C., is at an elevation very close to sea level, whereas Boulder is about a mile higher in altitude, so the gravitational field at Boulder is weaker than at vVashington, D.C. As a result, an atomic clock runs more rapidly in Boulder than in vVashinglOn, D.C. (This effect has been verified by experiment.)
26.3 Einstein's Principle of Relativity The two basic postulates of the special theory of relativity a re as follows:
2. The length of an object in motion is con/raeted in the direction of mation. The equation for length contraction is
1. The laws of physics are the same in all inertial framcs of reference. 2. The speed of light is the same for all inertial observers, independently of their motion or of the motion of the source of light.
[26,4]
26.4 Consequences of Special Relativity Some of' the consequences of the special theory of relativity are as follows: 1. Clocks in motion relative 10 an observer slow down, a phenomenon known as time dilation. The relationship between time intervals in the moving and at-rest systems is [26.2] where 6.1 is the time interval measured in the system in relative motion with respect to the clock, 1
y
= -:r=~=;;
VI - v'I.'
[26.3]
and u.t p is the proper time interval measured in the system moving with the clock.
\vhere L is the length measured by an observer in motion relative to t.he object and Lp is the proper length measured by an observer for \\'h0111 the object is at rest. 3. Events that are simultaneous for one observer are not simultaneous for another observer in motion relative to the first.
26.5 Reiativistic Momentum The relativistic expression for the momentum of a panicle moving \.. . ith velocity v is
/J
mv ::::=:
v'1 -
v'J.; ('1
= ymv
[26.51
26.6 Relativistic Energy and the Equivalence of Mass and Energy The relativistic expression (or the kinetic energy of an o~jeCl is KE = ymr'2 - mr'l [26.6) where mc'1 is tbe rest energy of the object, ER
T(b) p~ T(c) P< T(eI) P< Tor p>.r; depending on the direction or motion of the spacecraft (e) None of these 6. A cube measured at rest has ,'olume E An observer then passes the cube pal-allel to one of its sides at 0.981', so)' = 5. What is the volume of the cube as measured by the moving observer' (a) 11/125 (b) W25 (c) W5 (eI) V(e).oV 7. Suppose a photon, proton, and electron all have the sallle lotal energy t.:. Rank their momenta from smalleSlto greatest. (a) photon, elcnron, proton (b) proton. photon, electron (c) cleClroll, photon, proton (d) electron, proton, photon (c) proton, eleClron, photoll 8. \Vhal is the y factor of an electron having a kinetic energy or2mc'? (a) I (b) 2 (c) 3 (eI) 4 (e) 5 9. A mass-spring system moving with simple harmonic mOlion has a period T when measured by a ground observer. rr lhe same systcm is then placed in an iner~ tial frame of reference that moves past the ground observer al a speed of 0.50c, b)' what faclor should 7' be multiplied to give the system's period as measured by the gronnd observer' (a) 0..00 (b) 0.87 (c) 1.00 (eI) 1.15 (e) 1.35
10. The pm',:er Olllput of the Sun is 3.7 X lO~li \V. What is rhe rale at which the Sun's mass is converted ilHo energy' (a) 4.1 X 10' kg/s (b) 6.3 X 10" kg/s (c) 7.4 X 10" kg/s (d) 3.7 X 10' kg/s (e) 2.5 X 10" kg/s
CONCEPTUAL QUESTIONS 1. A spacecraft with the shape of a sphere or diameter D moves past an observcr on LHth with a speed of 0.5c What shape does t.he observer measure for the spacecraft as itlllO\'CS past?
2. What twO speed measurements will two observers in relative motion always agree upon? 3. The speed of light in water is 2.30 X 10 8 m/s. Suppose an elcctron is moving through water at 2.50 X 10/1 rn/s. Does this panicle speed violate the principle of relat ivitv?
4. "Vith regard to reference frames, how docs general,-e1ali"ity differ from special relativity? 5. Some distantslar-like objects. called quasars, are receding from us at half the speed oflight (or greater). What is the speed or the light we receive from these quasars? 6. It is said that Einslein, ill his teenage years, asked the question, "What would T see in a mirror if I carried il illllly IWllds and rail at a speed near that of light?" How would you answer I his question?
Problems
867
7. List some ways our day-la-day lives would change if the speed of light were only 50 m/s.
9. Photons of lighl have zero mass. How is it possible thai
8. Two identically constructed clocks are synchronized. One is put into orbit around Earth, and the other remains on Earth. \rVhich clock runs morc slowl}'? When the moving clock returns to Earth, will the two clocks still be synchronized? Discuss from the standpoints of both special and general relativity.
10. Imagine an astronaut on a trip to Sirius, which lies 8 light-years from Earth. Upon arrival at Sirius, t.he astronaut finds that the lrip lasted 6 years. If the u'ip was made at a constant speed of 0.8r, how can Ihe 8-1ight-year dist.ance be reconciled ",,'ith the 6-year duration?
the)' have moment urn?
PROBLEMS ~
The Problems for Ihis chapter may be WebAssignassigncd online al \'VebAssign. I, :!, :\ = straiglHforward, il\lerlnt'di~lte, cil,llienging
m
= denotes guided problem
II =
denotes ellhanced content problem biomedical application denotes filII solution available in Student Solulions Almwal/
mil = 0=
Stud)' Guide
SECTION 26.4 CONSEQUENCES OF SPECIAL RELATIVITY OJlfastronams could travel at tJ = 0.950c, we on Earth would say it takcs (4.20/0.950) = 4.42 ycars to reach Alph Z. This difference can be panially understood by recognizing thalas the number ofprototls increases, the strength of the Coulomb force increases, which lends to break the nucleus apart. As a result, more ncutl"OIlS arc needed to keep the nucleus stable because neutrons are afl-ected only by the attractive nuclear forces. fn effect, the additional neutrons "diILllC" the nuclear charge density. Eventually, when Z = 83, the repulsive forces between prolOns cannot be cornpens3lcd for by the addition of Ilclltrons. Elements that contain more than 83 protons don't hayc st
I~C
+ iHe
[29.22]
The LOtal mass on the left side of the equation is the sum ofl.he lllassofTH (2.014 102 u) and Lhe mass of IjN (14.003074 u), which equals 16.017 176 u. Similarly, the mass on the right side of tile equation is the sum of the mass of l~C (12.000000 u) plus Lhe mass ofp-Ie (4.002602 u), for a Lotal of 16.002 602 u. Thus, the LOLal mass before the reaction is greater than the total mass after the reaction. The mass difference in Lhe reacLion is equal to l6.017 176 u - 16.002 602 u = 0.014574 u. This "lost" mass is converted to the kinetic energy oCtile nuclei prcselll after t..he reaction. 1n energy units, 0.014574 u is equivalent to 13.576j\·leV of kinetic energy carried away by the carbon and helium nuclei. The energy required to balance the equation is called the Qvalue of the reaction. rn Equation 29.22, the Q value is 13.576 NieV. Nuclear reactions in which there is a release of energy-that is, positive Qvalues-are said to be exothermic reactions. The energy balance sheet isn't complete, however, because we must also consider the kinetic energy of the incident particle before the collision. As an example, assume the demeron in Equation 29.22 has a kinetic energy of5 MeV. Adding this value La our Q value, v·:e find thaL the carbon and helium nuclei have a total kinetic energy of ]8.576 J'vleV [ol}m..· ing the reaction. Now consider the reaction
lAO + IH
[29.23]
Before the re~lCtion, the total Illass is the SUIll of the masses of the alpha particle and the niLrogen nucleus: 4.002 602 u + 14.003074 u = 18.005676 u. After the reaction, the total mass is the sum of the masses of the oxygen nucleus and the proton: 16.999133 u + l.007 825 u = 18.006 %8 u.ln this case the LOral mass afLer Lhe reaction is greater than the total Illass before the reaction. The mass deficit is 0.001 282 ll, equivalent to an energy deficit of L194 MeV. This deficit is expressed b}' the negative Q value of the reaction, -l.J94 ~1eV. Reactions with negative Qvalues are called endothermic reactions. Such reactions won't take place unless the incoming panicle has at least enough kinetic energy to overcome the energy deficit. At first it might appear that the reaction in Equation 29.23 can take place if the incoming alpha particle has a kinelic energy of 1.194 MeV. In practice, however, the alpha panicle must have more energy than that. H it has an energy of only 1.194 MeV, energy is conserved; careful analysis, though, shows that momentum isn't, \',1hich can be understood by recognizing that the incoming alpha particle has some momenUl1ll before the reaction. If its kinetic energy is only L.194 MeV, however, lhe products (oxygen and a proton) \volild be created \'lith zero kinetic energy and thus zero momentum. [t can be shown that to conserve both energy and momentulll, lhe incoming particle must have a min imum kinetic energy given by
KE",;" =
(1+ Mm) IQI
[29.24]
29.7
Medical Applications of Radiation
where 111 is the mass of the incident particle, /'vI is the mass of the target, and the absolute value of the Q value is used. For the reaction gi\·en by Equation 29.23, we find that
KE m ;" ~ ( 1
4.002602 )
+ 14.003074 1- J.I94 Me"l - 1.535 MeV
This minimum value of the kinetic energy 01" the incoming particle is called the threshold energy. The lluclear reaction shown in Equation 29.23 won't occur if the incoming alpha panicle has a kinetic energy of less than 1.535 MeV, but. can occur if its kinetic energy is equallO or greater 1hall 1.535 MeV. QUICK QUIZ 29.5 Ifthe Qvalue of an endothermic reaction is -2.17 "leV, the minimum kinetic energy needed ill the reactantllllclei for the reaction to occur must be (a) equal to 2.17 MeV, (bl greater than 2.17 MeV, (c) less than 2.17 MeV, or (d) cxactly halfof2.17 McV.
29.7
MEDICAL APPLICATIONS OF RADIATION III
Radiation Damage in Matter Radiation absorbed by matter can cause severe damage. The degree and kind of damage depend on several factors, including the type and energy of the radiation and the properties of the absorbing material. Radiation damage in biological organisms is due primarily to ionization effects in cells. The normal function ofa cell may be disrupted when highly reactive ions or radicals are formed as the result of' ionizing radiation. For example, hydrogen and hydroxyl radicals produced from water molecules can induce chemical reactions that may break bonds in proteins and other vital molecules. Large acute doses of radiation are especially dangerous because damage to a great number of molecules in a cell may cause the cellLO die. Also, cells thaL do survive the radiation may become defectivc, which can leadlO cancer. In biological systems it is common 1.0 separate radiation damage into two categorics: somatic damage and genetic damagc. Somatic damage is radiation damage to any cells except the reproductive cells. Such damage can lead to cancer at high radiation levels or seriously alter the characteristics of specific organisms. Genetic damage affects only reproductive cells. Damage 10 the genes in reproductive cells can lead to defective offspri ng. Clearly. we must be concerned abollt the effect of diagnostic treatments, such as x-rays and other forms of exposurc to radiation. Several units are used to quantify radiation exposure and dose. The roentgen (R) is defined as the amount of ionizing radiation that will produce 2.08 X 10 9 ion pairs in 1 cm 3 of air under standard conditions. Equivalently, the roentgen is the amount of radiation that deposits 8.76 x 10-3 J of energy into] kg of air. For most applications, the roentgen has been replaced by the rad (an acronym for 1ildial.ion absorbed dose), defined as follows: One rad is the amount of radiation that deposits 10-2 J of energy into 1 kg of absorbing material. Although the rad is a pcrfectly good physicalullit, it's not the beSlunit for measuring the degree of biological damage produced by radiatioll because the degree of darnagc depends nol only all thc dose, but also on the fJJle of radiation. For example, a given dose of alpha panicles causes about ten times more biological d8111age than an equal dose of x-rays. The RBE (1-e1ative biological effectivcness) facLOt" is defined as the number of rads of x-radiation or gamma radiation that produces the same biological damage as 1 rad of the radiation being used. The RBE factors for different types of radiation are given in Table 29.3 (page 930). NOle that the values are only approximatc because they vary with panicle energy and t.he form of damage.
929
930
Chapter 29
Nuclear Physics
TABLE 29.3 RBE Factors for Several Types of Radiation Radiation
RBE Factor
1.0 1.0-1.7 10-20 4-5 LO 20
X-rays :Oll1d gam ma 1""yS
Beta particles
AlphU, splits, or fissions, into IWO smaller nuclei. In such a reaction the total mass of the products is less than the original mass of the heavy nucleus. The fission of 235C by slow (low·energy) neutrons can be represented by the sequence of events -----?
X
+ Y + neutrons
Problems ond Perspectives
[30.1]
where 236U' is an intermediate state that lasts only for about 10- 12 s before splitting into nuclei X and Y, called fission fragments, Man)' combinations of X and Y satisfy the requirements or conservation of energy and charge. In the fission of uraIlium, about 90 different daughter nuclei can be formed. The process also results in the production of se\'eral (typicall)' two or th ree) neutrons per fission event. On the average, 2,47 neutrons are released per event. A lypical reaction of th is type is [30.2]
937
938
Chapter 30
FIGURE 30.1
A !ludear
Nuclear Energy and Elementary Particles
r.5~ioll
y
l..'\crll:l
iHe
+ c+ + "
or ~He
+
~He
~
~He
+ 2CH)
Thc energy liberated is carried primarily by gamma rays, positrons, and neutrinos, as can bc seen from the reactions. The gamma rays are soon absorbed by the dense gas, raising its terr"lperature. The positrons combine \vith electrons to prodllce gamma rays, which in turn are also absorbed by the gas wit.hin a fe\\' cenI imeters. The neutrinos, howevcr, a1mos1 neyer int.eracl with mat.tcr; hence, they escape from the star, carrying about 2% of the encrgy generated \\lith thclll. These energy-liberating fusion reactions are callcd thermonuclear fusion reactions. The hydrogen (fusion) bomb, first. exploded in 1952, is an example of an uncontrolled thermonuclear fusion reactioll.
Fusion Reactors A great deal of effort is under way 1.0 develop a sustained and controllable fusion power reactor. Cont.rolled fusion is often called the ultimate energy source becallse of the availability of water, its fuel source. For example, if" deuteriulll, the isotope of hydrogen consisting ofa proton and a neutron, \,'ere used as the fuel, 0.06 g of it could be extraClcd rrom I gal or\,v·ater at a cost. OrabOUl four cents. Hence, the fuel
APPLICATION Fusion Reactors
941
942
Chopl.,JO
Nuclear Energy and Elementary Particles
costs of even an inefflcient reaCLQr would be almost illsignificam. An additional advant.age of Fusion reactors is Lhal comparatively few radioactive by-products are formed. As noted in Equation 30.3, the end product of the fusion of hydrogen nuclei is safe, nonradioactive heliulll. Unfortunately, a thermonuclear reactor that can delivc,' a net power output over a reasonable time interval is not yet a ,'cality, and rnany problems fnust be solved before a successful device is constructed. The fusion reactions that appear most promising in the construction of a fusion power reactor involve deuterium (0) and tritium (T), which areisoLOpes of hydrogen. These reactions are
TO + ~D ;0 + TD
--> -->
~He
+ 611
?T +
11-1
Q ~ 3.27 MeV Q ~ 4.03 MeV
[30.41
and
lD
Lawson's criterion ~
Vacuum
/
Plasma FIGURE 30.4
Diagram ora tokamak
used in the magnetic confinement scheme. Tile plasma is trapped within
the spiralillg magnetic fidd lines as shown.
+
iT
-->
lHe + &11
Q = 17.59 MeV
".'here the Qvalues refer to the amount of energy released per reaction. As noted earlier, deuterium is available in almost unlimited quantities from Ollr lakes and oceans and is very inexpensive to extract. Tritiulll, however, is radioaClh'e (Tv:.! = l2.3 yr) and undergoes beta decay [Q 3He. For this reason, tritium doesn't occur naturally to any great extent and must be artificially produced. The fundamental challenge of Iluclear fusion power is to give the nuclei enough kinetic energy to overcome the repulsive Coulomb force benveen them at close proximity. This step can be accomplished by heating the fuel to extremely high temperawres (about .10:) K, far greater than the interior temperawre of the Sun). Such high temperatures are not easy to obtain in a laboratory or a power plant. At these high temperatures, the atoms are ionized and the system then consists of a collection of electrons and nuclei, common ly reI-erred to as a !Jlasma. In addition to the high temperature requirements, two other critical factors determine whether or not a thermonuclear reactor will function: the plasma ion density 11 and the plasma confinement time T, the time the interacting ions are maintained at a temperature equal to or greater than that required for the reaction to proceed. The density and confinerncnl time must both be large enough to ensure that more fusion energy will be released than is required to heat the plasma. Lawson's criterion states that a net power output in a fusion reactor is possible under the follmving conditions: 1"17
2:::
101'1 s/cm 3
Deuteri u m-tritiu min Leraction
117
2:::
IO IG s/cm:\
Deuteriu m-deuteriulll interaction
[30.5]
The problem of plasma confinement time has yet to be solved. How can a plasma be confined at a temperature of 10 8 K for tirnes on the order of 1 s? Most fusion experiments use magnetic field confinement Lo contain a plasma. One device, called a tokamak, has a doughnut-shaped geometry (a toroid), as shmvn in Figure 30.4. This device uses a combination of t\·,;o magnetic fields to confine the plasma inside the doughnut. A strong magnetic field is produced by the current in the windings, and a weaker magnetic field is produced by the current in the toroid. The resulting magnetic field lines are helical, as shown in the figure. In this configuration the field lines spiral around the plasma and pre\'ent it from touching the walls of the vaCUUlll chamber. There are a number of other methods of creating fusion events. In inertial laser confinement fusion, the fuel is put into the form of a small pellct and then collapsed by liltrahigh-pm'ler lasers. Fusion can also take place in a device the size of a TV set and in fact was invented by Philo Farnsworth, one of the fathers of' electronic television. In this method, called inertial electrostatic confinemcnt, positively charged particles are rapidly attracted to\vard a negativcly charged grid. Some of the positivc particles then collide and fuse.
30.3
EXAMPLE 30.2 Goal
Elementary Particles and the Fundamental Forces
Astrofuel on the Moon
Calculate the energy released in a fusion reaction.
Problem
Find lhe energy released in the reaClion of" helium-3 with deuterium: ~I-Ie
Strategy
+
~D
~
JHe
+ :II
The energy released is the difference between the mass energy of the reactants and the products. . ....
,.. ... - ....
Solution Add the masses on the left-hand side and subtract the masses on the right. obtaining llln in atomic mass units:
Convert the mass difference
~
3.016029 u
"
+ 2.014 102 u - 4.002603 u - 1.007825 u
= 0.019703 u 10
an equivalent amount
I~=
(0.019703u) (
of energy in MeV:
931.5 MeV) = 18.35 MeV 1 II
........ - ......
lrro... • • • • • •
Remarks The result is a large amount of energy per reaction. Helium-3 is rarc 011 Earl h but. plentiful on the Moon, where it has become trapped in the nile dust of the lunar soil. Helium p.-
e-
+ V,
+ v"
\'\Thich of the following reactions cannot occur?
+ p --> 2y
rr'l +
(b) n --> p
(b) y
+ :£-
+ P --> n + "0 + P --> K+ + ~+
(eI) "+
30.7
The Eightfold Woy
949
Conservotion of Strangeness The K, 1\, and 2: panicles exhibilullusual properties in their production and decay and hence are called stmnge particles. One unusual property of strange panicles is that they are alwrl)'s produced in pairs. For example. when a pion collides v'lith a proton, two neutral strange particles are produced \\dth high probability (Fig. 30.6) following the reaction:
e-\
~ ~
On the other hand, the reaction 7r- + p+ ~ KU + n has never occurred, even though it violates no known conservation laws and the energy of the pion is sllf~ ricicllllO initiate the reaction. The second peculiar feature o[ strange panicles is that although the}' are proe1uced by the strong interaction at a high rme, 1 hey don', decay into particles that interact. via the strong force at a very high rate. Instead, they decay very slowly, which is characteristic of the weak interaction. Their half-lives are in the range of 10- 10 S [010-8 s; most other panicles that interact via the strong force have much shorter lifetimes on the order of 10-2:i s. To explain these unusual properties of strange particles, a law called conseroali.on oj strangeness \vas imroduced, lOgcther with a new quantulll number S called strangeness. The strangeness numbers for some particles are given in Table 30.2. The production of strange panicles in pairs is explained by assigning S = +1 to one of the particles and S = -1 to the other. All nonstrange panicles are assigned strangeness S = O. The law of conservation of strangeness states thai whenever a nuclear reaction or decay occurs, the sum of the strangeness numbers before the process must equal the sum of the strangeness Ilumbers after the process. The slow decay of strange panicles can be explained by assuming the strong and eleetromagnetic interactions obey the law of conservation of strangeness, 'whereas the weak interaction does no\.. Because the decay reaction involves the loss of one strange panicle, it violates strangeness conservation and hence proceeds slowly via the "veak interaction. 1n checking reactions for proper strangeness conservation, the same procedure as with baryon number conservation and lepton number conservation is followed. Using Table 30.2, counL the strangeness all each side. rf the two results are equal, the reaction conserves strangeness.
APPLYING PHYSICS 30.2
~
~
I ,;.
p\
I a:;
~ ~
~
FIGURE 30.6 This drawing rcpre· SCIII.S tracks of many events obtained
by analyzing a bubble-chamber photograph. The strange particles AO and KO are formed (at the bottom) as the ...- imcracls with a prawn according to the interaction 7T + P -+ Nl + KO. (7'\ole that the lIeutral particle,s leave no Iracks, as indicated by the dashed lines.) The AO and K(I then decay according to lhc interactions All -+ 7r +- P and fell -+ 7r + JL + /)'1'
BREAKING CONSERVATION LAWS
A student claims Lo have observed a decay o[an electron into two neutrinos traveling in opposite directions. \,Vhat conservation h1\\"s would be violatcd by lhis decay? Explanation Several conservation laws would be violated. Conservmion of electric charge would be violated because the negative charge of the eleclroll has disappeared. Conservation of electron lepton number \.. · ould also be violated because there is one lepton before the decay and two afterward. If bOlh neutrinos were electron neutrinos, electron lepton number conservation v.:culd be violated in the final
30.7
"0
state. If one or the product neutrinos were other than an electron neutrino, however, another lepton conservation law would be violated because there were no other leptons in the initial state. Other conservation la\\ls would be obeyed by this decay. Energy can be conscrved; the rest energy of the electron appears as thc kinetic energy (and possibly some small rest energy) of lhe neutrinos. The opposite directions of the two neutrinos' ,"clodties allow for the conservation of momentum. Conservation of baryon number and conservation of other lepton numbers v,,'ould also be upheld in this decay.
THE EIGHTFOLD WAY
Quantities such a5 spin, baryon number, lepton number, and strangeness are labels we associate with particles. Many classification schemes that group panicles into
950
Chapter 30
Nuclear Energy and Elementary Particles
FIGURE 30.7 (a) The hexagonal eightfold.way paltcrll for the eight spin-~ baryolls. This strangeness \er"US charge plm uses a horizolllal axi~ for the str ",- + c(b) ro + p --7 P + ro+ (c)p+p --7 p+ro l (d) p + P --> P + P + n (e) "y + p --> n + n"
22. Determine the type of neutrino or antinelll.rino involved ill each of the following- processes. (a) rot- --7 roO + e+ + ? (b) ? + P --7 J.L- + P + rot(c) IV l --7 P + J.L +? (d) T t --7 11-+ + ? + ? 23. Identify the unknown particle on the left side of the reaction ?+P
3(iJ-le)
This reaction is an attractive possibilit)' because boron is easily obtained from Earth's crust. A disadvantage is th,lt the protons and boron nuclei must have larg"e kinetic energies for the reaction to take placc. This requirement contrasts to the initi'ltion ofuraniulll fission by slow \leuI rollS. (a) How much energy is releascd in each reaction? (b) Why must the reaCl.ant particles have high kinetic cnergies?
SECTION 30.4 POSITRONS AND OTHER ANTIPARTICLES 17. A photon produces a proton-antiproton pair according to the reaction y --7 P + p. What is the minimum possible I"reqllellcy of the photon? What is its wavcleng-Ih?
[ffiJA photon with an energy of 2.09 CeV creates a protonantiproton pair in '.. . hich the proton has a kinetic energy of95.0 MeV. What is the kinetic energy of the antiproton? 19. A neutral pion at. rest deca)'s into two photons according10
Find the energy. momentum, and frequency of each pholon.
--7
n
+ J.L+
24. mil (a) Show that baryon number and charge are conserved ;n the following reactions of a pion with a proton:
+ 17.6 .\tleV
Neglecting relativistic corrections, dctcnnine the kinetic energy acquired by Ihe IlClltrOll.
959
(I)
rr'
+
p
--.
K-
+
(2)
rr" + p
--7
ro+
+ 1+
I+
(b) The first reaction is observed, blll. the second never occurs. Explain these observations. (c) Could the second reaction happen ifit created a third particle? Usa, which panicles in Table ::S0.2 might make it possible? Would the reaction require less energy or morc energy than the reaction of Equatioll (J)? Why?
25. _
Identify the consen'cd quantities in the following pn)cesses. (a) 2- --> (b) 1(0 (c) K
-7
+P
--7
2.0 + n
(d) In --> 1\.0 + "y (e)e-+e- --7 11-+
(f)
p+
n
-->
+ 11--
Afl + 1-
SECTION 30.8 QUARKS AND COLOR 26. The quark composilion of the proton is uud, whereas that of the ncULron is mid. Show that the charge, baryon number, and strangeness of these particles equal the sums of these !lumbers fOI" their quark constituents. 127.lrind the number of dectrons, and of each species of quark, in I L of water. 28. The quark compositions of the KO and AU particles arc (15 and mis, respectively. Show that the charge, baryon number. and strangeness of these panicles equal the sums of these numbers for their fJuark constituems.
29. Identify the particles corresponding (a)
SUll,
(b) tid, (c) sd, and (d) ssd.
to
the quark st.ates
960
Chapter 30
Nuclear Energy and Elementary Particles
30. \"'hat is the elcCLrical charge or the baryons with the quark compositions (,,) ulid and (b) ueld? What are these baryons called?
ADDITIONAL PROBLEMS
[]I):\.
~o panicle tr;weling through malleI' strikes a prown and a k+, and a gamma rny, as \\'ell as a third parlicle, emerges. Use the quark model of each 10 determine the identity of the third panicle.
32. :\ame at least one conservation law that prevents each or the following rC;:lcrions from occurring.
+ P --. s- + "" ~ ",- + v~ p ~ 7[+ + IT! + ",-
(a) ",-
(b) (c)
J.L~
133.Wind the: energy released in the Cusioll reaction I-I
34.
35.
+ ~H
~
~Hc
+ e- + "
Occ.:asjonaJl~',
high-energy muons collide with electrons and produce two neutrinos according to the rcaClion j.L+ + e- ----7 2/1. \'\'"hat kind of neutrinos are they?
am
Each or the following decays is forbidden. For each process, determine a cansen'atian law that is violalCd. (a) JL(b) n
--. .....-7
(c) AO .....-7 (eI) P ~ (c)
Sil
~
+~ p T e- + v.. P + 1Jo e+ + 11(1 n + ...H e-
36. Two protons approach each other with 70.4 MeV of kinetic energy and engage in a reactjon in which a proton and a positive pion cmerge at rest. \lVhat third parlicle, obviously uncharged and therefore difficult to detect, must have been created? 13i.IA 2.0-MeV neUlron is emitted in a fission reactor. If it loses one-half its kinetic energy in each collision with a modeL1LOr atolll. how lll 0, the straight line has a positive slope, as in Figure A.I. If "Ill < 0, the straight line has a negative slope. In Figure A.l, both II/. and b arc positive. Three other possible situations a re shown in Figu re A. 2.
EXAMPLE Suppose the electrical resistance of a metal wire is 5.00.n. at a temperature of 20.0°C and 6.-14 n at 80.0°C. Assuming Lhe resistance changes linearly, what is t.he resistance of the wire aL 60.0°C? Solution Find the equation of the line describing the resistance R and then substitute the new temperature into it. T"..' o points on the graph of resistance versus temperalllre, (20.0°C, 5.00 D) and (So.ooe, 6.14 0), allow computation of the slope: !J.R 6.140 - 5.000 ,> = = 1.90 X 10--fl/"C (1) '/1l = !J. T SO.O°C - 20.0"C Now lise the point-slope formulation ofa line....vith this slope and (20.0°C, .5.00 .0):
R - Ro =m(T -
(2) (3)
R - 5.00 0
~
7~)
(1.90 X 10-2 OJ"C)(T - 20.0°C)
Finally, substitute T = 60.00 into Equation (3) and solve for R, getting R = 5.76 n.
EXERCISES
1. Dra.. " graphs or the follO\ving straight lines: (a)y=5x+3 (b)y=-2x+4 (e)y~-3x-6 2. Find rhe slopes or the straight lines described in Exercise 1. Answers: (a) 5 (b) -2 (e)-3 3. Find the slopes of the straight lines that pass through the following sets of points: (a) (0, -4) and (4.2) (b) (0,0) and (2, -5) (e) (-5, 2) and (4, -2) Answers: (a) 3/2 (b) -5/2 (e) -4/9
A.3
Algebra
4. Suppose an experiment measures the following clisplaccmclllS (in meters) from equilibriulll of a vt:rtical spring due 1.0 allClching weights (in NCWLOIlS): (0.0250 fIl, 22.0 :"J), (0.0750,66.0 N). Find the spring constant, which is the slope oCthe line ill lhe graph ofvY'cight versus displacement Answer: 880 N/m
F. Solving Simultaneous Linear Equations Consider the equation 3x + 5y = 15, \vhich }l;-lS l\"O unknO\,.rns, x and ). Such an equation doesn't have a unique solution. For c:xanlple, note that (x = 0, )' = 3), (x = 5, Y = 0), and (x = 2, y = 9/5) arc all solutions to this equation. rra problem has t\\/O unknowns, a unique SOllllion is possible only ifw'c have L\'/O equations. Tn general, jf a problem has n unknowns, its solution requires 1/ equations. To solve t\vo simultaneous equations involying twO unknowns, x and y. we solve one of the equations for x in terms ofy and SubSI illite this expressiun into the other equation,
EXAMPLE Solve the follmving t\vo simultaneous equations: (1) Solution
!;x
+ y = -8
2x - 2)' = 4
(2)
From Equation (2), we find that x = J + 2. Substitution of this value inl.O Equation (1) giycs
"(y + 2) + Y = -8 6)'=-18 )' =
-3
x=y+2= -I Alternate Solution
Multiply each term in Equation (1) by the [actor 2 and add the result to Equation (2): lOx+2y=~16
2x-2)'=4 J2x=-12 x = -1
Y= x - 2
-3
Two linear equations containing two unknowns can also be solved by a graphical method, If the straight lines corresponding to the t\\'o equations arc plotted in a conventional coordinatc systcm, thc intcrsection of the I.WO lincs represents the solution. For example, consider the two equations
y 5
,,--),=2 x - 2)' = -I
These equations are ploned in Figure A.3. The intcrseCLion of the two lines has the coordinates x = 5, Y = 3, which rcprcsents the sulution to the equations. YOll should check this solution by the analyticaltechniquc discussed abm'c.
FIGURE A.3
A.9
A.10
Appendix A
Mathematics Review
EXAMPLE A block of Illass 1/1 = 2.00 kg travels in the positive x-direction at v, = 5.00 rn/s, while a second block, of Illass iv[ = 4.. 00 kg and leading the first block, travels in the positive x-dircClion at 2.00 m/s. The surface is flicLionless. VVhal. are the blocks' velocities after collision, if that collision is perrenly elastic? Solution As c
' 1 back int.o Equation (4) yields vf =
1.00 Ill/S.
EXERCISES
Solve the following pairs of simultaneous equations involving two unknowns: ANSWERS
J.x+)'=ll x-)'=2 2.98 - T = T - 49 = 3. 6x + 2)' = 8x - 'I" =
x
= 5, J' =
3
lOa 5a
T = 65.3, a = 3.27
6
x=
~,y=-3
28
G. Logarithms and Exponentials Suppose a quantity x is expressed as a power of some quantity a: x= ([.)'
[A.Il]
The number a is called the base number. The logarithm of x with respect to the base a is equal to the exponent to which the base must be raised so as to satisfy the expression x = a J: [A.12] y = loga x Conversely, the antilogarithm of)' is the number x:
x = antilog,,)'
[A.13]
The antilog expression is in fan identical to the exponemial expression in Equation A.11, which is preferable for practical purposes. In pracLice, the two bases most orten used are base 10, called the common logarithm base, and base e = 2.718 . _. , called the nal"UTallogarilhm base. When COlllmon logarithms are lIsed, [A.14]
A.3
Algebm
A.ll
\t\'hcn nalUral logarithms are used, )' ~ In x (or, = e')
[A.I5]
For example. log lO 52 = 1.716, so alltilogllJ 1.716 = IOl.illi = 52. Likewise, 111 .. 52 = 3.951, '\0 antilnl' 3.951 = e~·951 = 52. In general. note lhm yOll can cOllvert bCl\\'cen base 10 and base e with the eqllalil} In x = (2.302 585)log,,,.\
[A.16]
Fin::lll). some useful properties oflogarilhms are log (lib) = log
II
+
log b
In
p
= I
log(alb) = log a - log b log(lI') =
11
I"G) -In
log a
=
a
Logarithms in college physics are used most nOlabl) in the definition of decibel le\'el. Sound imcnsi(y \'aries across several orders of magnitude, making it awkward to compare different intensities. Decibellc\'c1 (onH:'rts these intensities to a morc manageable logarithmic scale.
EXAMPLE (LOGS) Suppose ajcl testing its engines produces a sound intensity of' J = 0.750 'IN at a given location in an airplane hangar, \,Vhat decibel level corresponds 1.0 this sound intensity? Solution
Decibelle\'el {3 is defined by (3 =
where III = I X
10-1~ \'\'/m~
10IOg(.!...) III
is the standard reference il1l.cnsit}', Subsliwte the given information: (3
= 10 log (
0.750 W/m') I" " = 1 19 dB 10 - W/m-
EXAMPLE (ANTILOGS) A collection of four identicalm3chilles create... a decibel lc"c1 of f3 = 87.0 dB in a machine shop, \\'hat sound intensity ,.. .ould be created by only one such machine? Solution \\Te usc the equation of decibel Ic\'cl to find the lOtal sound intensity of the four machines. and then ,,:e divide by 4. From Equation (I):
87.0 dB = 1010g(
,., I ,) 10 - \l'/m-
Di\'idc both sides by 10 and take the antilog of both sides, which means, equivalcl1lly.lO exponentiate: lOX., = IO~f1.:fJ 1 = 10 "·10" = 10
3.:\
iii
'J =
_1_
10- 12 = 5.01 X 10-" W/m'
There are four machines, so this result must be divided b) 4 to get the intensity of" aile machine: 1 = 1.25 X 10
1
W/m'
A.12
Appendix A
Mathematics Review
EXAMPLE EXPONENTIALS) The half-life of tritium is 12.33 rears. (Tritium is the heaviest isoLOpe of hydrogen, with a nucleus consisting of a prown and two neutrons.) If a sample contains 3.0 g of tritium in ilially. how much rell"lains after 20.0 years? Solution
The equaliull giving the 1I11111UCI uf Ilucil:i of a n.tdiuacti,-e substance as a [unction of time is
N=
Ala
(2"I)"
where Nis the number of nuclei remaining. ''\'u is the initial number of nuclei, and 11 is the IlUmbCI- of half-lives. i\'ole lhat this equation is an exponential expression ,.. . ith a base of~. The number of half-lives is gi\'en b}' n = 1/0."2 = 20.0 yr/ 12.33 yr = 1.62. The fractional alTIollnl of tritium that remains after 20.0 )'1" is therefore
(,)'62 = 0.325
-N = No
2
Hence, of the original 3.00 g of tritium, 0.325 X 3.00 g = 0.975 g remains.
A.4
GEOMETRY
Table A.2 gives the areas and volumes [or several geometric shapes used [hrollgh~ oul this texl. These areas and volumes are important in numerous physics applications. A good example is the concept of pressure P, which is the force per unit area. As an equation, it is wriucn P = F/ A. Areas must also be calculated in problems involving the volume rate of fluid flow through a pipe using the equation of continuity, the lensile ~lrcss exerted all a cable by a \vcight, the nne of thermal energy transfer through a barrier, and the densil)' of current through a wire. There arc numerous other applications. Volumes are imponanl in computing the buoyant force excncd by waler on a submerged object, in calculaling densit.ies, and in determining the bulk Slress of" Ouid or gas on an object, which affects its volume. Again, there are numerous other applications.
TABLE A.2 Useful Information for Geometry
Surface area Area =
{w
=41l"T 2
Volume = ~.II"T'
Sphere
Rectangle
Surface area = 2.11"r~ ... 21frt
c
Volume
Are-d = 1ft :1
=.11" rtf
Circumference = 2;rr
Circle
~ b Triangle
./r?
11
i
~
til
l
t Recl:lnglllar box
Surface area = 2(iht (wt /'w) Volume =(wit
A.S
A.5
Trigonometry
TRIGONOMETRY
Some of the most basic facts concerning trigonometry arc presented in Chapter 1, and we encourage yOll to sLUdy me material prcscmed there if you afe having trouble with this branch of mathematics_ The most imponam trigonometric concepts include the P}/thagorean theorem:
[A. I?]
This equation states thal the square distance along the hypotenuse of a right triangle equals the slim of the squares of the legs. It can also be lIsed LO find distances between points in Canesian coordinates and the length of a ,'ceLOr, where .6.x is replaced by the x-compone~ of the veelor and .:lJ is replaced by the )'-compOnenl of the ,'eelor. If the vec[Qr A has components Ax and A~, the magniLUde A of the vector satisfies [A.18]
which has a form completely analogous LO the form of the Pythagorean theorem. Also highly useful are the cosine and sine functions because they relate the length of a veCLOr LO its x- and J-components: [A.19] [A.20j
Ax = A cosO 1\ = AsinO
The direction 8 of a vector in a plane can be dCLcnnined by usc of the Langent fUllction:
e=
tan
A,
~
[A.21]
A.
A relative of the Pythagorean theorem is also frcquclllly uscful:
sin'
e + cos' e =
[A.22]
Details on thc above concepts can be found in the extensive discussions in Chapters 1 and 3. The following are some othcr trigonometric identiLies that can sometimes be useful:
4. Force is a vector quamit), measured in units of ncwtons, N. What must be the angle bel\.. .een lWO concurrentl;' aClillg forces of 5 :\1 and 3 "\1, respectively, if the resultant veOor has;[ magniwdc of8 N? (a) 0° (b) Tbecause
In
+ Nl>
1ll,
(m+NJ)a
-
F
T
'1ll{(
-->
(111.
A.27
+ M)
F -
111.
T
and again the answer is (b).
EXAMPLE 3 A block \'\'ith mass JfI is started at the top oran incline with coefficient or kinetic friction ILk and allowed to slide to the bott.om. A second block \-"ith mass iVf = 2m is also allowed to slide down the incline. If J.Lk is the same for both blocks, how does the acceleration aM of the second block compare with the acceleration fim of the first block? (a) niH = 2a m (b) fl,\J = am (c) aM = ~am
Conceptual Solution The accelereltions are the same because I he force of grm'ity, the force of friction, and the ma side of:-\ewton's second law when applied in this physical context are all proportional 1.0 the mass. Hence, a different mass should not affect the acceleration of the body, and the answer is (b).
EXAMPLE 4 Planet A has twice the mass and L\vice the radius of Planet B. On \vhich planet would an astronaut have a greater weight? (a) Planet A (b) Planet B (c) The astronaut's weight would be unaffected. Conceptuol Solution ''''eight as measured on a given planet is the magnitude of the gravitational force at the surface of that planet. In 0Jewton's law of gravitation, the gravitational force is directly proportional to the mass of each of two bodies but inversely proportional to the square of the distance between them. l-lence, in considering only the masses, planet A exerts twice the gravitational force of planet B. Planet A also has twice the radius of" planet B, howe\"cr, and the inverse square of 2 is~. Overall, therefore, Planet A has a weaker gravilational acceleration at its surface than Planet B by a facLOr of one-half, and the weight of an astronaut will be greater on planet B [answer (b)]. Quantitative Solution \Vrite Newton's law of" gravitation for the weight of an astronaut of mass m on pl,lnet A: \r\'rite the same law for the weight or the astronaLlt on Planel. 13:
Divide Equalion (2) by Equalion (J):
(I)
(2)
II/,Nl nG
W/J=--,-
'"8
1\1J1TJI:!
lVI..j /",/"
A.28
Appendix E
MeAT Skill Builder Study Guide
MB (2'"n)' (2M,,),} Solve [or wlJ:
W II
=
2
2w, [answer (b)]
MULTIPLE·CHOICE PROBLEMS 1. Which of the following will result from the application of a nonzero net force on an object? (a) The velocity of I he object v·lill remain consl.anL (b) The velocity of the object ' . . . il1 remain conSlam, bUllhe direction in which U1C objecllllo....es will change. (c) The velocity of the object will change. (e1) None of the above.
2. Body A has a mass that is twice as great as that of body B. Ir a force acting on body A is half the value ora force acting on body 13, which statement is true? (a) The acceleralion of A will be twice that of 13. (b) The acceleration of A \.. . ill be half that of B. (c) The acceleration of A will be equal 10 thal of B. (eI) The acceleration of A will be one-fourth that of B.
3. \Nhich of the following is a statement of New·ton's second law of motion? (a) For every action there is an equal and opposite reaction. (b) Force and the acceleration it produces are dircctly proporl ion
mv 'Tn
I 1~,1 I 1111 1
+ lW
1-':1 I171. mv I + i\!1
11/ + 1\1 =--->1
M
[11111 [answer (a)].
MULTIPLE-CHOICE PROBLEMS 1. A nonuniform bar 8.00 III long is placed on a pivot 2.00 In [rom the right end of the bar, which is also the lighter end. The center of gravit), of the bar is located 2.00 m from the heavier left encl. [f a weight of H/ = 5.00 X 10 2 N on the right end balances Ihe bar, what must be the weigh! war the bar? (a) 1.25 X 10:? N (b) 2.50 X 1O:? N (c) 5.00 X 10' N (d) 1.00 X 10";\1
2. A rod ornegligiblc mass is 10 m in length.lfa 3D-kg object is suspended from the left end of the rod and a 20-kg object rrom the right end, where must the pivot poim be placed to ensure equilibrium? (a) 4 m from the 30-kg object (b) ~I III from the 20-kg object (c) 8 m from the 3D-kg object (d) 5 m from the 20-kg object 3. A car with a mass or8.00 X IOC' kg is stalled all a road. A tnlck with a Jll d (c) D < rI
Conceptual Solution Kinetic energy is directly proportional to mass, and because the truck is morc massive than the car, its kinetic energy is greater and hence it might be thought thiu a greater braking distance \vould be required. The friction force, however, is also directly proportional LO mass, so the friction force is proportionately greater lor rhe truck than the car. All other things being equal, the truck and car stop in the same distance, D = d, '...· hich is ans,\'er (a). Quantitative Solution \'\Trite the expression for the force of rriction, where is I he normal force: Now Lise Equation (4) from Example 1:
SubsLitute the expressions for the kinetic friction force from Equation (1):
11
(I)
(2)
F, = iL' n = iL,mg
Fk"rlu:k
D
1~I,r.lrd
1 ., ~A1v~ 1 ? 2mv~
(iL,Mg)J)
~j\1v2
MIJ
111
IJ
(iLl,mg) rI
1 ? 2ml)~
md
m
rI
The answer is (a), as expected:
EXAMPLE 3 A car undergoes uniform acceleration from rest. \·\'hal. can be said about the instantaneous power delivered by the engine? (a) The instantaneous power incn.:ases with increasing speed, (b) The instantancous power remains the samc. (c) The instantaneous power decreases with increasing speed. Conceptual Solution The correct answer is (a). Because the acceleration is uniform, the force is constanl.. The installlaneous power is givcn by qp = Fv, so with increasing velocity the delivered pmver must also increase.
EXAMPLE 4 A circus stuntman is blasted straight up out or a cannon. Irail' resistance is negligible, \.. . hich of the follmving is true abollt his lOta) mechanical energy as he rises? (3) It increases. (b) ft remains the same, (c) It decreases. Conceptual Solution [11 the absence of non conservative forces, mechanical energy is conserved, so it. will remain the same, and ans\",cr (b) is correcL Application Irthe stunl.man has mass 65.0 kg and his initial vclocity is 8.00 m/s, find the maximum height reached. Solution Apply conscrvation of mechanical energy:
flKE+ fll'E= 0
1
.,
The final velocity is zero, and initial height may' be taken as zero:
0-
Solve for II/and substitute values:
hr = - ' =
'I1nVI-
+ mghJ - 0
= 0
v·':!.
(8.00111/s)'
2g
2(9.80111/s')
3.27 111
A.34
Appendix E
MeAT Sk;1I Bu;lder Study Gu;de
MULTIPLE·CHOICE PROBLEMS 1. If the speed at which ')/('1.64 s)'
= 9.78 I11/S'
2. (a). Wavelength is veloci!y divided by frequency. The formula does nOl depend upon the type 01" wave involved.
,.\ = v/f=
(15 m/s)/5.0 s
1=
3.0 m
3. (e). Two waves are completel)' out of phase when t.heir anti nodes coincide so lhat each crest on one wave coincides with a trough on the other. This sitnation occurs when the W'. (h) a-decay. (c) y-ckeav. (d) posilron em ission. 12. Wh;H is the hall:lire of a radiOllllclide if 1/1601" its initial mas.., is present ;\fter (a) Jr) min (b) ~O lllin (c) 4r:. mill (d) 60 min
~
h?
13. The hall~lifc oQiNa is '2.6 y. If Xgrams of this sodilllll isolOpe are iniLially presellt. how ",uch;, len afler I~ yr? (a) X/32 (b) X/13 (e) X/8 (d) X/f, Answers I. (c). ~Olice that the mass llumber is unchanged. whel'L'as Llw atOmic Humber has heen reduced by I, \.. . hich implies that a proton has changed into a nClllron. The emiLling panicle must ha"e;l charge eC(ualto a proton hut ;111 atomic mass number of lero. The positron is lhe only choice that has both these attributes. 2. (a). The ;l1omic number fore, Z = 14.
i~
Z. Conservat.ion of charge llleans lhal I.)
=Z+
3. (d). First. tinct the lHllllbcr or hall~li\'es and Ihell multiply by t.he "duc or the g-eL t.he elapsed time:
~~:=i=G)' U/
1/
=
I, There-
half~lire to
3
= JIll·:! = 3(lO min) = 30 min
4. (b). An a-panicle is a ~I-Ie helium nuclem, III a-decay, two protons arc efTcni\'ely rCIllO\'ed. 5, (b). J3-dcc3Y emit.s a high-energy electron. jle. In dlt' p1"Oce:-.~. a neulron decays into a proton plus Ihe emiucd electron and all alllint"lIl.rino, The number of nuclcons rcmains Ilnchallged, with the proton replacing the neutron ill the sum ornUCleOlIS. 6. (d). The llla..;s number is the
SlIlll
OfllclltrollS and protons: 6 + 7
= 13 lIucleon..;.
7. (c). !,,;oLOpes of all clement. have the same atomic number but dilTcl-ent number:-. of" neutrons. so their mass numbers are different. The atomic number or X is 63. 8, (a). The Illas..; numbers must be the same on hOlh ..,ides 01' the reCloioll. II" A is the Illass number of X, then 14 + 4 = 17 + .-I., so il = L. As for alomic number, i + 2 = 8 + Z. Therefore, Z = I, which describes a proton, 1H. 9. (b). The number of nelll.rons is the mass number minlls llle atomic number: i\ - Z 141l - 54
~
=
86.
10. (d). The hall:lifc is a const.alll that depends on the identity of the nuclide, not. on the . Each a-particle lhen acquires I\m electrons to form a neutral helium alom. 12. (b). In every haIr-life, the mass decreases to halfils previousyaluc: 1/16:::: 1/2-t, It takco; fOllr hall~livcs to decay down to 1/16 the original mass. Each must be 30 min long
hecause the entire process takes two Itouro;. 13. (a). In 13 yr, there ,.,..ill be 5 hall:livcsoI"2.(i yr cacll (5 X 2.6 = 13). The isotope dccreascs
to 1/2:'
= 1/32 orits original amount.
A.S1
ANSWERS TO QUICK QUIZZES, EXAMPLE QUESTIONS, ODD-NUMBERED MULTIPLE CHOICE QUESTIONS, CONCEPTUAL QUESTIONS, AND PROBLEMS CHAPTER 1 EXAMPLE QUESTIONS 1. F'1Isc.." 2. True 3. U.9Sm:!
4. 2X.O rn/ .. =
(2S.0J~)(2.21I1li
.h) = 62.7 lIIi/h
s 1.00 In S The amwer i~ slighlh diffen:nl because l!l(' dilfert'nl ('"ol1\(.'r..ion faClor~ WCIT rounded. leading 10 !>l1lall, tlllp' ('dic. l;lbk c1iffcren{cs inlhe finill.It1.'i\\(:-r~.
5. (60.0 "';")' 1.00 h 6. An ;In""{"f of 101:! Ct'll" is witllin an order 01 mOign illldc: of Ill(' gin'>n answer, COl re ..ponrtillJ.:" 10 sli~htlr dilTcrc!H choicc~ in Ihe HlllllllC ('stimaliol1s. COllsequcntl). !OI:! cells is al~o" rCil = I.~ X 1O-~ "-g 55. The \'alue of k. a dimemionle£:o> constant. cannUl be found b,' dimensional anal\·sis. 57. - 10 '1 m 59. -10' tuners (assnmes one Iliner pel 10 OO(J residents and a pnplIlillion of /.5 million) 61. (a) 3.16 x 10~ s (b) Between lOlU \'1 and lOll VI'
3, The \"e1ocit\ ,'s. timc gnlph (a) ha~ a cnn . . t;\Ilt ~Iop(', indicating a cOIlSUIll ;:Iccelel-ation, which is rl·prc· . . cnted b\ the accd~ l'ralion \'s. lime graph (e). Graph (b) r('preselll~ an ohject with inue.. :.ing speed, but a:. Lime pl·ogl"es~es. Iht'lilll'~ (11;1\\11 langcm to the cun'e ha"e illCl"e:Jsing ~Iope:-,. Sinn' thl' ;Kcell'lation i, cqllallo Ille -",lope 01 Lhe lallgellllillc. ,he i!e' 180 s, 320 S, and 330 s.. 8. The upwalTljump would :>.Iightl), increase the ball's initial \'clocit>', slighLly increasing [lie maximulll height. 9. I) 10. The engine should be Ilrt:-([ agi-lill;1I
CONCEPTUAL QUESTIONS 1. Yes. If the velocil)' oflhe panicle is 1I0n/era, tile panicle is in
5.
7. 9.
3. (a) 52.9 klll/h (h) 90.0 km 5. (a) Boal A Will~ by liO km (b) U 7. (a) 180 knl (b) 6::H klll/h 9. (a) 4.0 m/s (h) -4.0 Il1ls (c; 0 (d) 2.0 m/s II. I .:~2 h 13. 2.80 11,218 kill 17. (a) 5.00 Ill;:" (b) -2..')(1 m/s (c) 0 (d)
I. :'\lo. The objecllllay not he lra\'dill~ in (l sLraighlline. IfLhe initial and fillal positions an; in [he same place. for example.
3,
PROBLEMS I. = 0.02 "i
15. 2i4 kl11/h
EXAMPLE QUESTIONS
5. The graphical ~Ollitioll and a parabola.
A.S3
mOlion. If the acceleration is leI"C), [he velocil> of the panicle is unchanging or is COllst'llll. Yes. Iflllisoc("lIr~, Ille .nllle lime. Wl-iting expressions for the position versus Lilla: for each vchicle and equaling the 1.'\\0 gives a qua(h-at ic equat i(lIl in I whose solution is eiLhel' 1 I A s or 13.6 s. The 'irst solution. IIA s. is the time of the collision. The collision occurs wIlen the van is 212 m from the ol'iginal po~ition of"Suc·:>. car. 37. 200 III 39. (: = 3.00 X 10'!·mi/h. (v.f.J) = 0
A
B
and arc in Ihe same dircclion. The reSUltalll will he 1.('1"0 when the two vectors are equal in magnitude lind opposite in direction. 3. (a) Atlhe top orthe pn~ieClilc's nighl. ils vclocil}' is horizolltal alld its acceleration is downward. This is Ihe only poi lit :H which Lhe velocit)' and ;lcce\eration vectors are perpcndicubu. (b) If the projcClilc is thrown straiglll IIp 01' down, thell the velocity and acceleration will be paralld thmughout the motion. For any Olher kind of projectile Illotion. the vclocit} and acceleration vectors arc never parallel. 5. (a) Thc acceleration is zero. since bOlh the tn::lgniwde and direct ion of the velocily n;ma in consl l'cgards the pr(~jcClilc. gravilY pm\'ides an acceleration thai
v
G6.2~
45, 18.0 s 4,7. (a) ·1O.!l m/s (b) Rathl,;r l.han falling like;1 rock, the skier glides through the air like a bird. proionging-Ihejlllllp. 49. 68.6 km/h 51. (a) 1.52 X 10:1 111 (b) 36.1 s (c) 4.05 X 10: 1 m 53. H"H"'" = 18 Ill. H\l,'" = 7.9 III 55. (a) '12 ntis (b) 3.8 s (c) r.,. = :l4 m/s. Vv = -13 IH/s;
lJ=:17 ll1/s 57. 61. 63. 65.
7.5 mls in the dil'ecuoll the ball was thrown 11/2 7.5 mill 10.8 III g . 1 . 67. (b))' = Ax'! with ;\ ~ - - , where lJ· IS the Illlll.l.lc \'e OCltv 2v,~
1
(c) 1'1.5 mls 69. (a) 3!l.1" 01" 54.9 (b) 42.2 III 01' 85.4 m. respecti\·t:!y 71. (a) 2U.0 above lhe hOI"ilOlHaf (b) 3.05 s 73. (a) 2.1 nt/~ (b) 360 m horiLOlltall~' from the base of cliff
•
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
7. (b) 8. (h) By exerting an up\\'ard Force conlponent
reduce tlte nOI"mal force ()ITe of k.i net ic friction.
011
Oil the sled. you the gmund and so reduce the
7.
EXAMPLE QUESTIONS 1. Othel' than the forces rnentioned in (he pmblem, the force or gr{l\,ity pulls dowll\\';Hds all the boal. Because the boat doesn't sink. a force exened by the waler all the boat must oppose the gl';wity rorce. (Tn Chapter 9this force will be identified as the buoyancy force.) 2. False. The angle at which lhe forces are applied is also imporI.Jlll in detennining the direclioll of the acceleration n:Cl()]". 3.0.2 N
4.2g 5. The tensions would double. 6. The magnitude of the tension force would be greater. and the magniLllde 01" the normal force would be smaller. 7. Doubling the weight doubles the mass. which halves both the acceleration and displacement. 8. A gentler slope ll1e(lns a smaller angle and hence a smaller acceleration dowll the slope. Consequently, the car would Lake longer to reach the bottom of tile hill. 9. The scale reading is grCatel" than the \...· eight of the I'ish dm· ing the first accdel·,ltioll phase. \'\!hen the velocity becomes constant, the scale reading is equal 10 the weight. When tbe cle,·ator slows dowll, the scale reading is less than the weight. 10. A1tach one end oCthe cable to the object to be lil"ted and the other end Lo a platform. Place liglnel' weights on t.he platform until the total mass of the weights and platform exceeds the mass of the heavy object. II. A larger stalic friction coeHicient would increase lilt:" maximum angle. 12, The coefficiellt ofkinctic friction ,\'ould be larger than in the example. 13, Both the acceleration and the tension increase when In'.! is incI·eased. 14. The top block would slide ofT the back end of the lower block.
MULTIPLE-CHOICE QUESTIONS I. (a)
3. (eI) 5. 7. 9. II. 13. 15.
(b) (b) (a)
(d) (e)
(a)
CONCEPTUAL QUESTIONS I. (a) Two externJI fOI"Ces aCI on the ball. (i) One is a downward gravitational force exerted by Earth. (ii) The second force on the ball is an upward Ilormallorce exerted by the hand. The reactions to these forces are (i) all upward gr'lvitational force exerted by the ball on Earth and (ii) a downward force exerted by the ball on the ham!. (b) Afler the ball leaves the ham!. the only external force act'ing on the ball is the gravi-
9.
11.
13.
A.S5
tational force cxertcd by Earth. The reflction is all upward grm'itational force exerted by the ball on Eanh. The coefficient of static friction is larger than that of k.inctic friction. To start the box moving. you must cOUlllerb,·Jlance the maximum static friction force. This fOITe exceeds the kinetic frict ion force that you must counlerbabnce 10 maint.ain the constant velocity 01" the bo:\ once it start!; mO\·ing. The inertia 01" the su itcase IvOtJid keep it moving forward as the bus stops. There would be no tendellcy fOI' the suitcase to be thrown backward 10ward the passenger. Tile case should be dismissed. The force causing an automobile to move is the friction between the tires and the roadway as the automobile attempts to push the roadway backward. The fOITe driving a propeller ail"j)lanc forwal-d is the reaction force exened by the air on the propeller as the rotating pmpeller pLlslles the air backward (the action). In a rowboat, the rower pushes the water backward with the oars (the action). The water pushes fOr\\-arc! on the oars and hence the boat (the reaction). \Vhcn rhe bus starts moving. Claudette's mass is accelerated by the force exerted by the back of the seat on her body. Clark is standing. however, and the only force acting 011 him is the friction between his shoes and the noor of the bus. Thus, when thc bus starts moying, his !"eet accelel'ate fOI'w'lrd, but the l'est of his body experiences almost no acceleral.ing force (only t.hat due to his being fll.tached to his accelerating feCI!). As a consequence, his body tends 10 stay almost at rest, according to Newton's first law, I'elalive to the ground. Relalive 10 Cbudeue, howe\'er, he is moving towal"d her and falls into her lap. Both performers won Academy Awards. The tension in the rope is the maximum fOITe that OCCllrs in fmlhdircnions. In this case, then. since both are pulling \\'ith a force of magnitildc 200 N, the tcnsion is 200 N. If the rope does nor move, then the force on each athlete mus1 equal zero. Tberefore, each athlete exerts 200 J\ against the grolllld. (a) As the lllan takes lhe step, the anion is the force his fOOl exerts on Earth; the reacrion is tile force exerted by Earth on his foot. (b) I-Jere, the anion is the fOITe exerted by lhe snowball on the girl's back; the reaction is the lorce exel"(ed br the girl's back on the snowball. (c) This action is the fOI'ce exerted by the glove nn the ball; the reanion is the force exerted by the ball on the glove. (d) This action is the force exerted by the air molecules on the window: the reaction is rhe [olTe exerted by the window all the air molecules. In each case, we could equall), well inter'change the terms Maction" and "reaction."
PROBLEMS 1.2XI0 4 r-.; 3. (a) 121\" (b) 3,0 m/s 2 5. 3.71 ~,58.7 N, 2.27 kg 7.9.6N 9. lAX 10J:'\I
II. (a) 0.200 m/s 2 (b) 10.0 m (c) 2.00 m/s 13. (a) 0.2.1] kg/m (b) 7.35 m/s 2 15. l.J X 10-1 N
17. (a) 600 N in \'cnical cable, 997 N in inclined cable. 796 N in hol"izolllal cable (b) Irthe poilll.ofau at ~23° (c) 7BEl .J. The lost kinetic cncl'gy is tran.'ifornwd into mher form.'i of eller6')' .'iuch as lhennal cnerh'J' and sound. 49. 5.59 m/s llonh 5\. (a) 2.50 m/s ill -60" (b) elaslic collision 53. 1.78 X \0:\ N on truck driver. 8.89 X 1O~;\ on car dri\'er 55. (a) 8/3 Ill/s (incident panicle), 32/3 m/s (target p,Hlicle) (Il) -16/::1 m/s (incidem p:lrlicle). 8/31ll/s (targ-t'l pilnicle) (c) 7.1 X 10 '2.1 in case (a). :tnd 2.8 X 1O-~.1 ill C:I~e (b). The incident panicle loses mOl'C kim·tic energy in ca~(' (:1). in which the larget lll:l~S is 1.0 g. 57. 1.\ X \0:1 N (upward) 59. 71.9 N 61. (a) 3 (b) ~ 63. (a) -2.33 tn/s. '1.67 m/s (b) 0.277 m (c) 2.98111 (e1) IA91ll 65. ( T"""n' = "'R"""u,' = Tn ",."H,,,,,,,.,./R,.'k"."1'
(2)
Thi~ torque will be panly absorhed hy frirtion in tlte feed heads (which we assume to be slTlall): sOllle will be ahsorhed by friction in the source reel. Another sm:lll amount of the torqlle will be absorbed by thc increasing alll;ulidc a rubher balloon is greater th,ul ;UlllOspheric pressure became it lllust also excr! a fOITC ,lgainsl the elastic force of the rubber. 5. Al higher aitillldt:. the column orair above a gi\clI al"ea is progrl'ssively shoneI' anrllcss dense, so Ille weighl or the air column is reduced. Prcssurt..: is caused by 1 he weight of the air column, so the presslll-e is also reduced. 6. As fluid pours Oilithroligh a single opening. the air" inside the can above Lilt, fluid t'xpallcls imo a largcr \"ollllnC', reducing the pressure to below ;ltillospheric pressure. Air must then ('mer the same opening going the opposite dircction, resulting in disrupted fluid flow. A separatc openin~ for ait" intake Ill;tint,aills air pressure inside the call without disnlpling tlte flow of lhe fluid. 7. True 8. False:' 9. (a) 10. Tlw aluminum cuhe would float free orthe hOUlllll. 11. The speed Ofl)ll' hlond in tile.: nalTo\\'ed region illcre;lses. 12. A factor of2 13. The speed slo\\l~\\"ilh time. 14. Substitllting AI = rI~ into Eqll. so the effcel orair drag is less in Dt.:nvcr than it would be ill a cilY such as l\"ew York, The reduced air draR" means a well-hit ball will go farther. benefit· ing home-run hillcrs. On the.: other hand, curve ball pitchers prefer to throll' at lower al1ittldes where tiLe highn dcnsity ait" produces greater defleCling forces on a spinning ball. 3. She eXel"lS enough pressure IlI1 the flOOl' to denl or plll1C1I1fC the (Joor covcring. Tlu.: large pressllre is CiIIIS('d by the fact lhat her II'cight is distribuled O\'f'r Ihe "ery sillall crossseClional area of her high heels. Ifyoll are the homeowner, you Illighl want to suggest that she remove her hig-h Ireels and pill all some slippers. 5. If you think orllle grain stored in the silo as a fluid, the pressure the grain exerts 011 the walls of the si 10 incn:ase1> with incrca'iing depth,just as waler pressure ill ,1 lake illcreases with illcreasing depth, Thus, the spacing uet\\'een hplacemclll is de~cribed by x(1) = A cos wi. where A is lhe distance from the cenler of Ihe wheel (0 the cn"mkpin. 25. 0.63 s 27. (a) 1.0 s (hj 0.28m/' (cj 0.25 m/s 29. (a) 5.98 m/s (b) 206 N/rn (c) O.2~H m 31. (a) 11.0 N toward the left (b) 0.881 oscillaliolls 33. 1/ = ± wA sin Wi. II = - (tPA cos wi 35. (a) 1.:)05 (1)) 0.5.59 III 37. (a) slow (b) 9:47 39. (a) LLmh = 25 cm, L M... , = 9.4 em. (b) "'1..U"lh = 111\\;".. = 0.25 kg 41. {a} 4.13 cm (b) lOA cm (c) 5.56 X IQ-:! s (d) 187 cm '5 43. ( I'),,, ~ 6.0 V (d) ?f',,, ~ 6.0 w. '!Po ~ 0.60 IV, '!Pm ~ 0.54 W. 71'/, ~ 0.36 W, 'lP"t = 1.5 \0\', CJ1'dh = 6.0 VI 49. (a) 12.4 V (b) 9.65 V
Answers to Quick Quizzes, Example Questions, Odd-Numbered Multiple-Choice Questions, Conceptual Questions, and Problems
A.72
51. (a) R~rn = 3R,R~~''''d
=
e'
e'
2R(b)1J"lloCn = 31?'~d""'d = 21?
(c) Lamps A and B increase in brigillness.lamp C goes oui.
53. 112 V. 0.200 n 55. (a) I? t = R~ - tR[ (b) R" = 2.8 n (inadcqU ligllll}' bound as the lritiullliluclell~.
13.
fcrenl Tlul1lben ofnt::utrons. This h'ill'Tsull in a v,u·iety of differenl physical properties for lhe nllclei, including the obdoll.~ olle of mass. The chemical behavior. howc\·el·, is gm'emed b) til(" elemen"s electrons. All isolopes ofa given element 11;1\'c the .~:lmc nu mbcr or electrons and, thcrefi,re. the ~amc chcmical behtLlhlung.876 Brew'Her. David, 812 Brewsler's angle, 812 Brcwslcr's law, 812, S 15 Bright fringes conditions for. 792-793. 792f 815 in thin films. 796-797. 815 local ion of. 793. 81.:c, Brili~h t hermaillnit (BIll). 35~ Broll1illc (Br), ground-stale configuration. 9041 Brown dwarf s(ars. 21 ~ Bulk lllodulus (B). 27t. ~7It. 311 Bulk stres~. See\'olullIc ~ll'C'na III ics L.-.w of l"cf'l"aCljon (Snell), 737~7'12,
