THÔNG TIN TÀI LIỆU
This page intentionally left blank MATHEMATICAL FORMULAS* Quadratic Formula If ax2 ϩ bx ϩ c ϭ 0, then x ϭ Derivatives and Integrals Ϫb Ϯ 1b2 Ϫ 4ac 2a nx n(n Ϫ 1)x2 ϩ ϩ 1! 2! (x2 Ͻ 1) : ͵ ͵ ͵ : Let u be the smaller of the two angles between a and b Then : : a и b ϭ b и a ϭ axbx ϩ ayby ϩ azbz ϭ ab cos u : ͉ ˆi : : : : a ϫ b ϭ Ϫb ϫ a ϭ ax bx ϭ ˆi ͉ ay by ˆj ay by ͉ ͉ az a Ϫ ˆj x bz bx ͉ kˆ az bz dx 2x2 ϩ a2 ex dx ϭ ex ϭ ln(x ϩ 2x2 ϩ a2) x dx ϭϪ 2 3/2 (x ϩ a ) (x ϩ a2)1/2 dx x ϭ 2 (x2 ϩ a2)3/2 a (x ϩ a2)1/2 Cramer’s Rule ͉ ͉ az a ϩ kˆ x bz bx Two simultaneous equations in unknowns x and y, ͉ ay by a1x ϩ b1 y ϭ c1 and xϭ ͉ ͉ yϭ ͉ ͉ : |a ϫ b | ϭ ab sin u Trigonometric Identities sin a Ϯ sin b ϭ sin (a cos a ϩ cos b ϭ cos Ϯ b) cos (a (a ϩ b) cos ͉ ͉ ϭ c1b2 Ϫ c2b1 a1b2 Ϫ a2b1 ͉ ͉ ϭ a1c2 Ϫ a2c1 a1b2 Ϫ a2b1 c1 c2 b1 b2 a1 a2 b1 b2 a1 a2 c1 c2 a1 a2 b1 b2 and ϯ b) (a Ϫ b) *See Appendix E for a more complete list SI PREFIXES* Factor 24 10 1021 1018 1015 1012 109 106 103 102 101 a2x ϩ b2 y ϭ c2, have the solutions ϭ (aybz Ϫ by az)iˆ ϩ (azbx Ϫ bzax)jˆ ϩ (axby Ϫ bxay)kˆ : cos x dx ϭ sin x d x e ϭ ex dx Products of Vectors : sin x dx ϭ Ϫcos x d cos x ϭ Ϫsin x dx Binomial Theorem (1 ϩ x)n ϭ ϩ ͵ ͵ ͵ d sin x ϭ cos x dx Prefix yotta zetta exa peta tera giga mega kilo hecto deka Symbol Y Z E P T G M k h da Factor –1 10 10–2 10–3 10–6 10–9 10–12 10–15 10–18 10–21 10–24 Prefix Symbol deci centi milli micro nano pico femto atto zepto yocto d c m m n p f a z y *In all cases, the first syllable is accented, as in ná-no-mé-ter www.freebookslides.com This page intentionally left blank halliday_FM_extended_i-xxiv_v2.0.1.qxd 12-03-2014 12:29 Page i www.freebookslides.com PRINCIPLES OF PHYSICS T E N T H E D I T I O N www.freebookslides.com This page intentionally left blank halliday_FM_extended_i-xxiv_v2.0.1.qxd 12-03-2014 12:29 Page iii www.freebookslides.com Halliday & Resnick PRINCIPLES OF PHYSICS T E N T H E D I T I O N International Student Version J EAR L WALK E R CLEVELAND STATE UNIVERSITY halliday_FM_extended_i-xxiv_v2.0.2.qxd 3/25/14 9:53 AM Page iv www.freebookslides.com Copyright ©2014, 2011, 2008, 2005, 2003 John Wiley & Sons Singapore Pte Ltd Cover image from ©Samot/Shutterstock Founded in 1807, John Wiley & Sons, Inc has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support For more information, please visit our website: www.wiley.com/go/citizenship All rights reserved This book is authorized for sale in Europe, Asia, Africa and the Middle East only and may not be exported The content is materially different than products for other markets including the authorized U.S counterpart of this title Exportation of this book to another region without the Publisher’s authorization may be illegal and a violation of the Publisher’s rights The Publisher may take legal action to enforce its rights No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, website http://www.wiley.com/go/permissions ISBN: 978-1-118-23074-9 Printed in Asia 10 halliday_FM_extended_i-xxiv_v2.0.1.qxd 12-03-2014 12:29 Page v www.freebookslides.com B R I E F C O N T E N T S Measurement 24 Electric Potential Motion Along a Straight Line 25 Capacitance Vectors 26 Current and Resistance Motion in Two and Three Dimensions 27 Circuits Force and Motion—I 28 Magnetic Fields Force and Motion—II 29 Magnetic Fields Due to Currents Kinetic Energy and Work 30 Induction and Inductance Potential Energy and Conservation of Energy 31 Electromagnetic Oscillations and Alternating Center of Mass and Linear Momentum Current 10 Rotation 32 Maxwell’s Equations; Magnetism of Matter 11 Rolling, Torque, and Angular Momentum 33 Electromagnetic Waves 12 Equilibrium and Elasticity 34 Images 13 Gravitation 35 Interference 14 Fluids 36 Diffraction 15 Oscillations 37 Relativity 16 Waves—I 38 Photons and Matter Waves 17 Waves—II 39 More About Matter Waves 18 Temperature, Heat, and the First Law of 40 All About Atoms Thermodynamics 41 Conduction of Electricity in Solids 19 The Kinetic Theory of Gases 42 Nuclear Physics 20 Entropy and the Second Law of Thermodynamics 43 Energy from the Nucleus 21 Coulomb’s Law 44 Quarks, Leptons, and the Big Bang 22 Electric Fields Appendices/Answers to Checkpoints and Odd-Numbered Problems/Index 23 Gauss’ Law v halliday_FM_extended_i-xxiv_v2.0.1.qxd 12-03-2014 12:29 Page vi www.freebookslides.com C O N T E Measurement 1-1 MEASURING THINGS, INCLUDING LENGTHS What Is Physics? Measuring Things The International System of Units Changing Units Length Significant Figures and Decimal Places 1-2 TIME Time T S Adding Vectors by Components 40 Vectors and the Laws of Physics 41 3-3 MULTIPLYING VECTORS Multiplying Vectors 44 44 REVIEW & SUMMARY PROBLEMS 49 What Is Physics? 53 Position and Displacement 54 4-2 AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY REVIEW & SUMMARY PROBLEMS Average Velocity and Instantaneous Velocity Motion Along a Straight Line 11 2-1 POSITION, DISPLACEMENT, AND AVERAGE VELOCITY What Is Physics? 11 Motion 12 Position and Displacement 12 Average Velocity and Average Speed Instantaneous Velocity and Speed 11 Projectile Motion 61 61 4-5 UNIFORM CIRCULAR MOTION 13 16 4-6 RELATIVE MOTION IN ONE DIMENSION 16 Relative Motion in Two Dimensions 21 Constant Acceleration: A Special Case Another Look at Constant Acceleration 2-5 FREE-FALL ACCELERATION REVIEW & SUMMARY 73 27 29 34 5-2 SOME PARTICULAR FORCES Some Particular Forces 35 3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS 80 What Is Physics? 80 Newtonian Mechanics 81 Newton’s First Law 81 Force 82 Mass 83 Newton’s Second Law 84 27 PROBLEMS Vectors 34 3-1 VECTORS AND THEIR COMPONENTS vi PROBLEMS Force and Motion—I 80 5-1 NEWTON’S FIRST AND SECOND LAWS 25 Graphical Integration in Motion Analysis What Is Physics? 34 Vectors and Scalars 34 Adding Vectors Geometrically Components of Vectors 36 71 24 25 28 72 71 21 2-6 GRAPHICAL INTEGRATION IN MOTION ANALYSIS 40 69 4-7 RELATIVE MOTION IN TWO DIMENSIONS 2-4 CONSTANT ACCELERATION REVIEW & SUMMARY 69 18 18 Free-Fall Acceleration 67 67 Relative Motion in One Dimension 2-3 ACCELERATION 56 Average Acceleration and Instantaneous Acceleration 4-4 PROJECTILE MOTION 2-2 INSTANTANEOUS VELOCITY AND SPEED 55 4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION Uniform Circular Motion Unit Vectors 53 6 Acceleration 50 Motion in Two and Three Dimensions 4-1 POSITION AND DISPLACEMENT 53 1-3 MASS Mass N 5-3 APPLYING NEWTON’S LAWS 40 Newton’s Third Law 92 Applying Newton’s Laws REVIEW & SUMMARY 88 88 92 94 100 PROBLEMS 100 59 58 halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 58 www.freebookslides.com 58 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS Sample Problem 4.02 Two-dimensional velocity, rabbit run v ϭ 2v2x ϩ v2y ϭ 2(Ϫ2.1 m /s)2 ϩ (Ϫ2.5 m /s)2 For the rabbit in the preceding sample problem, find the velocity : v at time t ϭ 15 s KEY IDEA ϭ 3.3 m /s and We can find : v by taking derivatives of the components of the rabbit’s position vector (Answer) vy ϭ tanϪ1 vx ϭ tanϪ1 m /s Ϫ2.5 Ϫ2.1 m /s ϭ tanϪ1 1.19 ϭ Ϫ130Њ (Answer) Check: Is the angle Ϫ130° or Ϫ130° ϩ 180° ϭ 50°? Calculations: Applying the vx part of Eq 4-12 to Eq 4-5, we find the x component of : v to be dx d ϭ (Ϫ0.31t ϩ 7.2t ϩ 28) dt dt ϭ Ϫ0.62t ϩ 7.2 y (m) 40 vx ϭ 20 (4-13) At t ϭ 15 s, this gives vx ϭ Ϫ2.1 m/s Similarly, applying the vy part of Eq 4-12 to Eq 4-6, we find vy ϭ At t ϭ 15 s, this gives vy ϭ Ϫ2.5 m/s Equation 4-11 then yields v ϭ (Ϫ2.1 m /s)iˆ ϩ (Ϫ2.5 m /s)jˆ , 60 80 x (m) –40 (4-14) : 40 –20 dy d ϭ (0.22t2 Ϫ 9.1t ϩ 30) dt dt ϭ 0.44t Ϫ 9.1 20 (Answer) which is shown in Fig 4-5, tangent to the rabbit’s path and in the direction the rabbit is running at t ϭ 15 s To get the magnitude and angle of : v , either we use a vector-capable calculator or we follow Eq 3-6 to write x –60 These are the x and y components of the vector at this instant v –130° v at t ϭ 15 s Figure 4-5 The rabbit’s velocity : Additional examples, video, and practice available at WileyPLUS 4-3 AVERAGE ACCELERATION AND INSTANTANEOUS ACCELERATION Learning Objectives After reading this module, you should be able to 4.08 Identify that acceleration is a vector quantity and thus has both magnitude and direction and also has components 4.09 Draw two-dimensional and three-dimensional acceleration vectors for a particle, indicating the components 4.10 Given the initial and final velocity vectors of a particle and the time interval between those velocities, determine the average acceleration vector in magnitude-angle and unit-vector notations 4.11 Given a particle’s velocity vector as a function of time, determine its (instantaneous) acceleration vector 4.12 For each dimension of motion, apply the constantacceleration equations (Chapter 2) to relate acceleration, velocity, position, and time Key Ideas ● If a particle’s velocity changes from : v1 ⌬t, its average acceleration during ⌬t is : a avg ϭ ● As : v2 to : v in time interval : ⌬v Ϫ: v1 ϭ ⌬t ⌬t ⌬t is shrunk to 0, : aavg reaches a limiting value called either the acceleration or the instantaneous acceleration : a: : dv : aϭ dt ● In unit-vector notation, : a ϭ ax iˆ ϩ ay jˆ ϩ azkˆ , where ax ϭ dvx /dt, ay ϭ dvy /dt, and az ϭ dvz /dt halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 59 www.freebookslides.com 4-3 AVE RAG E ACCE LE RATION AN D I NSTANTAN EOUS ACCE LE RATION Average Acceleration and Instantaneous Acceleration When a particle’s velocity changes from : v to : v in a time interval ⌬t, its average : acceleration a avg during ⌬t is average change in velocity ϭ , acceleration time interval : : a avg ϭ or : v2 Ϫ : v1 ⌬v ϭ ⌬t ⌬t (4-15) If we shrink ⌬t to zero about some instant, then in the limit : a avg approaches the instantaneous acceleration (or acceleration) : a at that instant; that is, : aϭ : dv dt (4-16) If the velocity changes in either magnitude or direction (or both), the particle must have an acceleration We can write Eq 4-16 in unit-vector form by substituting Eq 4-11 for : v to obtain : aϭ ϭ d (v iˆ ϩ vy jˆ ϩ vz kˆ ) dt x dvx ˆ dvy ˆ dvz ˆ i ϩ j ϩ k dt dt dt We can rewrite this as : a ϭ ax iˆ ϩ ay jˆ ϩ az kˆ, (4-17) where the scalar components of : a are ax ϭ dvx , dt ay ϭ dvy , dt dvz dt and az ϭ (4-18) To find the scalar components of : a , we differentiate the scalar components of : v Figure 4-6 shows an acceleration vector : a and its scalar components for a particle moving in two dimensions Caution: When an acceleration vector is drawn, as in Fig 4-6, it does not extend from one position to another Rather, it shows the direction of acceleration for a particle located at its tail, and its length (representing the acceleration magnitude) can be drawn to any scale y These are the x and y components of the vector at this instant ax ay a of a particle and the Figure 4-6 The acceleration : a scalar components of : a Path O x 59 halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 60 www.freebookslides.com 60 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS Checkpoint Here are four descriptions of the position (in meters) of a puck as it moves in an xy plane: r ϭ 2t iˆ Ϫ (4t ϩ 3)jˆ (1) x ϭ Ϫ3t ϩ 4t Ϫ and y ϭ 6t Ϫ 4t (3) : (2) x ϭ Ϫ3t Ϫ 4t and (4) : r ϭ (4t Ϫ 2t)iˆ ϩ 3jˆ y ϭ Ϫ5t ϩ Are the x and y acceleration components constant? Is acceleration : a constant? Sample Problem 4.03 Two-dimensional acceleration, rabbit run For the rabbit in the preceding two sample problems, find the acceleration : a at time t ϭ 15 s has the same tangent as Ϫ35° but is not displayed on a calculator, we add 180°: Ϫ35° ϩ 180° ϭ 145° (Answer) KEY IDEA a because it gives This is consistent with the components of : a vector that is to the left and upward Note that : a has the same magnitude and direction throughout the rabbit’s run because the acceleration is constant That means that we could draw the very same vector at any other point along the rabbit’s path (just shift the vector to put its tail at some other point on the path without changing the length or orientation) This has been the second sample problem in which we needed to take the derivative of a vector that is written in unit-vector notation One common error is to neglect the unit vectors themselves, with a result of only a set of numbers and symbols Keep in mind that a derivative of a vector is always another vector We can find : a by taking derivatives of the rabbit’s velocity components Calculations: Applying the ax part of Eq 4-18 to Eq 4-13, we find the x component of : a to be dvx d ϭ (Ϫ0.62t ϩ 7.2) ϭ Ϫ0.62 m /s2 dt dt Similarly, applying the ay part of Eq 4-18 to Eq 4-14 yields the y component as ax ϭ ay ϭ dvy d ϭ (0.44t Ϫ 9.1) ϭ 0.44 m /s2 dt dt We see that the acceleration does not vary with time (it is a constant) because the time variable t does not appear in the expression for either acceleration component Equation 4-17 then yields 40 a ϭ (Ϫ0.62 m /s2)iˆ ϩ (0.44 m /s2)jˆ , 20 : (Answer) which is superimposed on the rabbit’s path in Fig 4-7 To get the magnitude and angle of : a , either we use a vector-capable calculator or we follow Eq 3-6 For the magnitude we have a ϭ 2a 2x ϩ a 2y ϭ 2(Ϫ0.62 m /s2)2 ϩ (0.44 m /s2)2 ϭ 0.76 m/s2 (Answer) For the angle we have ϭ tanϪ1 ay ax y (m) 20 40 60 80 –20 –40 a 145° x –60 ϭ tanϪ1 0.44 m/s2 ϭ Ϫ35Њ Ϫ0.62 m/s2 However, this angle, which is the one displayed on a calculator, indicates that : a is directed to the right and downward in Fig 4-7 Yet, we know from the components that : a must be directed to the left and upward To find the other angle that x (m) These are the x and y components of the vector at this instant a of the rabbit at t ϭ 15 s The rabbit Figure 4-7 The acceleration : happens to have this same acceleration at all points on its path Additional examples, video, and practice available at WileyPLUS halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 61 www.freebookslides.com 4-4 PROJ ECTI LE M OTION 61 4-4 PROJECTILE MOTION Learning Objectives After reading this module, you should be able to 4.13 On a sketch of the path taken in projectile motion, explain the magnitudes and directions of the velocity and acceleration components during the flight 4.14 Given the launch velocity in either magnitude-angle or unit-vector notation, calculate the particle’s position, displacement, and velocity at a given instant during the flight 4.15 Given data for an instant during the flight, calculate the launch velocity Key Ideas ● In projectile motion, a particle is launched into the air with a speed v0 and at an angle u0 (as measured from a horizontal x axis) During flight, its horizontal acceleration is zero and its vertical acceleration is Ϫg (downward on a vertical y axis) ● The equations of motion for the particle (while in flight) can be written as x Ϫ x0 ϭ (v0 cos 0)t, y Ϫ y0 ϭ (v0 sin 0)t Ϫ 12 gt 2, vy ϭ v0 sin 0 Ϫ gt, v2y ϭ (v0 sin 0 )2 Ϫ 2g(y Ϫ y0) ● The trajectory (path) of a particle in projectile motion is parabolic and is given by y ϭ (tan 0)x Ϫ gx , 2(v0 cos 0)2 if x0 and y0 are zero ● The particle’s horizontal range R, which is the horizontal distance from the launch point to the point at which the particle returns to the launch height, is v2 R ϭ sin 20 g Projectile Motion We next consider a special case of two-dimensional motion: A particle moves in a v but its acceleration is always the freevertical plane with some initial velocity : g , which is downward Such a particle is called a projectile (meanfall acceleration : ing that it is projected or launched), and its motion is called projectile motion A projectile might be a tennis ball (Fig 4-8) or baseball in flight, but it is not a duck in flight Many sports involve the study of the projectile motion of a ball For example, the racquetball player who discovered the Z-shot in the 1970s easily won his games because of the ball’s perplexing flight to the rear of the court Our goal here is to analyze projectile motion using the tools for twodimensional motion described in Modules 4-1 through 4-3 and making the assumption that air has no effect on the projectile Figure 4-9, which we shall analyze soon, shows the path followed by a projectile when the air has no effect The v that can be written as projectile is launched with an initial velocity : : v ϭ v0x iˆ ϩ v0y jˆ (4-19) v0 The components v0x and v0y can then be found if we know the angle u0 between : and the positive x direction: v0x ϭ v0 cos u0 and v0y ϭ v0 sin u0 (4-20) r and velocity During its two-dimensional motion, the projectile’s position vector : v change continuously, but its acceleration vector : vector : a is constant and always directed vertically downward.The projectile has no horizontal acceleration Projectile motion, like that in Figs 4-8 and 4-9, looks complicated, but we have the following simplifying feature (known from experiment): In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other Richard Megna/Fundamental Photographs Figure 4-8 A stroboscopic photograph of a yellow tennis ball bouncing off a hard surface Between impacts, the ball has projectile motion halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 62 www.freebookslides.com 62 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS A y O + y Vertical motion ➡ Horizontal motion y This vertical motion plus this horizontal motion produces this projectile motion v0y O O v0 v0y Vertical velocity x v0x Projectile motion θ0 Launch velocity Launch angle x v0x O Launch Launch y y vy vy Speed decreasing O O v vx x vx x O Constant velocity y y vy = Stopped at maximum height vx x O O v vy = x O Constant velocity y y vx Speed increasing vy vy vx O x O v x O Constant velocity y y vx vy O Constant velocity x vx O θ vy v Figure 4-9 The projectile motion of an object launched into the air at the origin of a coordinate system and with launch velocity : v at angle u0 The motion is a combination of vertical motion (constant acceleration) and horizontal motion (constant velocity), as shown by the velocity components x halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 63 www.freebookslides.com 63 4-4 PROJ ECTI LE M OTION This feature allows us to break up a problem involving two-dimensional motion into two separate and easier one-dimensional problems, one for the horizontal motion (with zero acceleration) and one for the vertical motion (with constant downward acceleration) Here are two experiments that show that the horizontal motion and the vertical motion are independent Two Golf Balls Figure 4-10 is a stroboscopic photograph of two golf balls, one simply released and the other shot horizontally by a spring.The golf balls have the same vertical motion, both falling through the same vertical distance in the same interval of time The fact that one ball is moving horizontally while it is falling has no effect on its vertical motion; that is, the horizontal and vertical motions are independent of each other A Great Student Rouser In Fig 4-11, a blowgun G using a ball as a projectile is aimed directly at a can suspended from a magnet M Just as the ball leaves the blowgun, the can is released If g (the magnitude of the free-fall acceleration) were zero, the ball would follow the straight-line path shown in Fig 4-11 and the can would float in place after the magnet released it The ball would certainly hit the can However, g is not zero, but the ball still hits the can! As Fig 4-11 shows, during the time of flight of the ball, both ball and can fall the same distance h from their zero-g locations The harder the demonstrator blows, the greater is the ball’s initial speed, the shorter the flight time, and the smaller the value of h Checkpoint At a certain instant, a fly ball has velocity v ϭ 25iˆ Ϫ 4.9jˆ (the x axis is horizontal, the y axis is upward, and : v is in meters per second) Has the ball passed its highest point? : Richard Megna/Fundamental Photographs Figure 4-10 One ball is released from rest at the same instant that another ball is shot horizontally to the right Their vertical motions are identical The ball and the can fall the same distance h M th -g ro The Horizontal Motion Can h Ze Now we are ready to analyze projectile motion, horizontally and vertically We start with the horizontal motion Because there is no acceleration in the horizontal direction, the horizontal component vx of the projectile’s velocity remains unchanged from its initial value v0x throughout the motion, as demonstrated in Fig 4-12 At any time t, the projectile’s horizontal displacement x Ϫ x0 from an initial position x0 is given by Eq 2-15 with a ϭ 0, which we write as x Ϫ x0 ϭ v0x t Because v0x ϭ v0 cos u0, this becomes x Ϫ x0 ϭ (v0 cos u0)t (4-21) The Vertical Motion The vertical motion is the motion we discussed in Module 2-5 for a particle in free fall Most important is that the acceleration is constant Thus, the equations of Table 2-1 apply, provided we substitute Ϫg for a and switch to y notation Then, for example, Eq 2-15 becomes y Ϫ y0 ϭ v0yt Ϫ 12gt ϭ (v0 sin 0)t Ϫ 12gt 2, (4-22) where the initial vertical velocity component v0y is replaced with the equivalent v0 sin u0 Similarly, Eqs 2-11 and 2-16 become vy ϭ v0 sin u0 Ϫ gt (4-23) and pa v2y ϭ (v0 sin 0)2 Ϫ 2g(y Ϫ y0) (4-24) G Figure 4-11 The projectile ball always hits the falling can Each falls a distance h from where it would be were there no free-fall acceleration halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 64 www.freebookslides.com 64 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS As is illustrated in Fig 4-9 and Eq 4-23, the vertical velocity component behaves just as for a ball thrown vertically upward It is directed upward initially, and its magnitude steadily decreases to zero, which marks the maximum height of the path The vertical velocity component then reverses direction, and its magnitude becomes larger with time The Equation of the Path We can find the equation of the projectile’s path (its trajectory) by eliminating time t between Eqs 4-21 and 4-22 Solving Eq 4-21 for t and substituting into Eq 4-22, we obtain, after a little rearrangement, y ϭ (tan 0)x Ϫ Jamie Budge Figure 4-12 The vertical component of this skateboarder’s velocity is changing but not the horizontal component, which matches the skateboard’s velocity As a result, the skateboard stays underneath him, allowing him to land on it The Horizontal Range The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial height (the height at which it is launched) To find range R, let us put x Ϫ x0 ϭ R in Eq 4-21 and y Ϫ y0 ϭ in Eq 4-22, obtaining R ϭ (v0 cos u0)t ϭ (v0 sin 0)t Ϫ 12gt y Rϭ II Figure 4-13 (I) The path of a fly ball calculated by taking air resistance into account (II) The path the ball would follow in a vacuum, calculated by the methods of this chapter See Table 4-1 for corresponding data (Based on “The Trajectory of a Fly Ball,” by Peter J Brancazio, The Physics Teacher, January 1985.) 98.5 m v20 sin 20 g (4-26) This equation does not give the horizontal distance traveled by a projectile when the final height is not the launch height Note that R in Eq 4-26 has its maximum value when sin 2u0 ϭ 1, which corresponds to 2u0 ϭ 90° or u0 ϭ 45° The horizontal range R is maximum for a launch angle of 45° However, when the launch and landing heights differ, as in many sports, a launch angle of 45° does not yield the maximum horizontal distance a Path I (Air) Range Maximum height Time of flight Rϭ x Table 4-1 Two Fly Balls 2v20 sin 0 cos 0 g Using the identity sin 2u0 ϭ sin u0 cos u0 (see Appendix E), we obtain I 60° (4-25) Eliminating t between these two equations yields and range v0 (trajectory) This is the equation of the path shown in Fig 4-9 In deriving it, for simplicity we let x0 ϭ and y0 ϭ in Eqs 4-21 and 4-22, respectively Because g, u0, and v0 are constants, Eq 4-25 is of the form y ϭ ax ϩ bx2, in which a and b are constants This is the equation of a parabola, so the path is parabolic and Air reduces height gx 2(v0 cos 0)2 Path II (Vacuum) 177 m 53.0 m 76.8 m 6.6 s 7.9 s a See Fig 4-13 The launch angle is 60° and the launch speed is 44.7 m/s The Effects of the Air We have assumed that the air through which the projectile moves has no effect on its motion However, in many situations, the disagreement between our calculations and the actual motion of the projectile can be large because the air resists (opposes) the motion Figure 4-13, for example, shows two paths for a fly ball that leaves the bat at an angle of 60° with the horizontal and an initial speed of 44.7 m/s Path I (the baseball player’s fly ball) is a calculated path that approximates normal conditions of play, in air Path II (the physics professor’s fly ball) is the path the ball would follow in a vacuum halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 65 www.freebookslides.com 65 4-4 PROJ ECTI LE M OTION Checkpoint A fly ball is hit to the outfield During its flight (ignore the effects of the air), what happens to its (a) horizontal and (b) vertical components of velocity? What are the (c) horizontal and (d) vertical components of its acceleration during ascent, during descent, and at the topmost point of its flight? Sample Problem 4.04 Projectile dropped from airplane In Fig 4-14, a rescue plane flies at 198 km/h (ϭ 55.0 m/s) and constant height h ϭ 500 m toward a point directly over a victim, where a rescue capsule is to land (a) What should be the angle f of the pilot’s line of sight to the victim when the capsule release is made? y v0 O φ Tr aje Calculations: In Fig 4-14, we see that f is given by ϭ tanϪ1 x , h Lin eo (4-28) Here we know that x0 ϭ because the origin is placed at the point of release Because the capsule is released and v is equal to not shot from the plane, its initial velocity : the plane’s velocity Thus, we know also that the initial velocity has magnitude v0 ϭ 55.0 m/s and angle u0 ϭ 0° (measured relative to the positive direction of the x axis) However, we not know the time t the capsule takes to move from the plane to the victim To find t, we next consider the vertical motion and specifically Eq 4-22: y Ϫ y0 ϭ (v0 sin 0)t Ϫ 12gt (4-29) Here the vertical displacement y Ϫ y0 of the capsule is Ϫ500 m (the negative value indicates that the capsule moves downward) So, Ϫ500 m ϭ (55.0 m/s)(sin 0Њ)t Ϫ 12 (9.8 m/s2)t (4-30) Solving for t, we find t ϭ 10.1 s Using that value in Eq 4-28 yields x Ϫ ϭ (55.0 m/s)(cos 0°)(10.1 s), (4-31) or x ϭ 555.5 m f si gh t θ v Figure 4-14 A plane drops a rescue capsule while moving at constant velocity in level flight While falling, the capsule remains under the plane (4-27) where x is the horizontal coordinate of the victim (and of the capsule when it hits the water) and h ϭ 500 m We should be able to find x with Eq 4-21: x Ϫ x0 ϭ (v0 cos u0)t cto ry h KEY IDEAS Once released, the capsule is a projectile, so its horizontal and vertical motions can be considered separately (we need not consider the actual curved path of the capsule) x Then Eq 4-27 gives us ϭ tanϪ1 555.5 m ϭ 48.0Њ 500 m (Answer) v? (b) As the capsule reaches the water, what is its velocity : KEY IDEAS (1) The horizontal and vertical components of the capsule’s velocity are independent (2) Component vx does not change from its initial value v0x ϭ v0 cos u0 because there is no horizontal acceleration (3) Component vy changes from its initial value v0y ϭ v0 sin u0 because there is a vertical acceleration Calculations: When the capsule reaches the water, vx ϭ v0 cos u0 ϭ (55.0 m/s)(cos 0°) ϭ 55.0 m/s Using Eq 4-23 and the capsule’s time of fall t ϭ 10.1 s, we also find that when the capsule reaches the water, vy ϭ v0 sin u0 Ϫ gt ϭ (55.0 m/s)(sin 0°) Ϫ (9.8 m/s2)(10.1 s) ϭ Ϫ99.0 m/s Thus, at the water v ϭ (55.0 m /s)iˆ Ϫ (99.0 m /s)jˆ : (Answer) v are From Eq 3-6, the magnitude and the angle of : v ϭ 113 m/s and u ϭ Ϫ60.9° Additional examples, video, and practice available at WileyPLUS (Answer) halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 66 www.freebookslides.com 66 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS Sample Problem 4.05 Launched into the air from a water slide One of the most dramatic videos on the web (but entirely fictitious) supposedly shows a man sliding along a long water slide and then being launched into the air to land in a water pool Let’s attach some reasonable numbers to such a flight to calculate the velocity with which the man would have hit the water Figure 4-15a indicates the launch and landing sites and includes a superimposed coordinate system with its origin conveniently located at the launch site From the video we take the horizontal flight distance as D ϭ 20.0 m, the flight time as t ϭ 2.50 s, and the launch angle as 0 ϭ 40.0° Find the magnitude of the velocity at launch and at landing we know that the horizontal velocity component vx is constant during the flight and thus is always equal to the horizontal component v0x at launch We can relate that component, the displacement x Ϫ x0, and the flight time t ϭ 2.50 s with Eq 2-15: KEY IDEAS That is a component of the launch velocity, but we need the magnitude of the full vector, as shown in Fig 4-15b, where the components form the legs of a right triangle and the full vector forms the hypotenuse We can then apply a trig definition to find the magnitude of the full velocity at launch: v cos0 ϭ 0x , v0 and so v0x 8.00 m/s v0 ϭ ϭ cos u0 cos 40؇ (1) For projectile motion, we can apply the equations for constant acceleration along the horizontal and vertical axes separately (2) Throughout the flight, the vertical acceleration is ay ϭ Ϫg ϭ Ϫ9.8 m/s2 and the horizontal acceleration is ax ϭ Calculations: In most projectile problems, the initial challenge is to figure out where to start There is nothing wrong with trying out various equations, to see if we can somehow get to the velocities But here is a clue Because we are going to apply the constant-acceleration equations separately to the x and y motions, we should find the horizontal and vertical components of the velocities at launch and at landing For each site, we can then combine the velocity components to get the velocity Because we know the horizontal displacement D ϭ 20.0 m, let’s start with the horizontal motion Since ax ϭ 0, x Ϫ x0 ϭ v0xt ϩ 12axt (4-32) Substituting ax ϭ 0, this becomes Eq 4-21 With x Ϫ x0 ϭ D, we then write 20 m ϭ v0x(2.50 s) ϩ 12 (0)(2.50 s)2 v0x ϭ 8.00 m/s ϭ 10.44 m/s Ϸ 10.4 m/s (Answer) Now let’s go after the magnitude v of the landing velocity We already know the horizontal component, which does not change from its initial value of 8.00 m/s To find the vertical component vy and because we know the elapsed time t ϭ 2.50 s and the vertical acceleration ay ϭ Ϫ9.8 m/s2, let’s rewrite Eq 2-11 as y vy ϭ v0y ϩ ayt v0 θ0 x Launch and then (from Fig 4-15b) as Water pool vy ϭ v0 sin 0 ϩ ayt (4-33) Substituting ay ϭ Ϫg, this becomes Eq 4-23.We can then write vy ϭ (10.44 m/s) sin (40.0؇) Ϫ (9.8 m/s2)(2.50 s) D (a) v0 Launch velocity θ0 v0x (b) v0y ϭ Ϫ17.78 m/s v0x θ0 Landing velocity v Now that we know both components of the landing velocity, we use Eq 3-6 to find the velocity magnitude: vy (c) Figure 4-15 (a) Launch from a water slide, to land in a water pool The velocity at (b) launch and (c) landing v ϭ 2v2x ϩ v2y ϭ 2(8.00 m/s)2 ϩ (Ϫ17.78 m/s)2 ϭ 19.49 m/s Ϸ 19.5 m/s Additional examples, video, and practice available at WileyPLUS (Answer) halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 67 www.freebookslides.com 67 4-5 U N I FOR M CI RCU L AR M OTION 4-5 UNIFORM CIRCULAR MOTION Learning Objectives After reading this module, you should be able to 4.16 Sketch the path taken in uniform circular motion and explain the velocity and acceleration vectors (magnitude and direction) during the motion 4.17 Apply the relationships between the radius of the circular path, the period, the particle’s speed, and the particle’s acceleration magnitude Key Ideas ● If a particle travels along a circle or circular arc of radius r at constant speed v, it is said to be in uniform circular motion a of constant magnitude and has an acceleration : v2 aϭ r The direction of : a is toward the center of the circle or circular arc, and : a is said to be centripetal The time for the particle to complete a circle is 2r Tϭ v T is called the period of revolution, or simply the period, of the motion Uniform Circular Motion A particle is in uniform circular motion if it travels around a circle or a circular arc at constant (uniform) speed Although the speed does not vary, the particle is accelerating because the velocity changes in direction Figure 4-16 shows the relationship between the velocity and acceleration vectors at various stages during uniform circular motion Both vectors have constant magnitude, but their directions change continuously The velocity is always directed tangent to the circle in the direction of motion The acceleration is always directed radially inward Because of this, the acceleration associated with uniform circular motion is called a centripetal (meaning “center seeking”) acceleration As we prove next, the magnitude of this acceleration : a is aϭ v2 r (centripetal acceleration), (4-34) where r is the radius of the circle and v is the speed of the particle In addition, during this acceleration at constant speed, the particle travels the circumference of the circle (a distance of 2pr) in time Tϭ 2r v (period) (4-35) T is called the period of revolution, or simply the period, of the motion It is, in general, the time for a particle to go around a closed path exactly once Proof of Eq 4-34 To find the magnitude and direction of the acceleration for uniform circular motion, we consider Fig 4-17 In Fig 4-17a, particle p moves at constant speed v around a circle of radius r At the instant shown, p has coordinates xp and yp Recall from Module 4-2 that the velocity : v of a moving particle is always tangent to the particle’s path at the particle’s position In Fig 4-17a, that means : v is perpendicular to a radius r drawn to the particle’s position Then the angle u that : v makes with a vertical at p equals the angle u that radius r makes with the x axis The acceleration vector always points toward the center v v a a The velocity vector is always tangent to the path a v Figure 4-16 Velocity and acceleration vectors for uniform circular motion halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 68 www.freebookslides.com 68 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS y v The scalar components of : v are shown in Fig 4-17b With them, we can write the velocity : v as θ p r θ xp v ϭ vx iˆ ϩ vy jˆ ϭ (Ϫv sin )iˆ ϩ (v cos )jˆ : yp x (4-36) Now, using the right triangle in Fig 4-17a, we can replace sin u with yp /r and cos u with xp /r to write vxp ˆ vyp ˆ : j vϭ Ϫ iϩ (4-37) r r a of particle p, we must take the time derivative of this To find the acceleration : equation Noting that speed v and radius r not change with time, we obtain (a) y d: v v dyp ˆ aϭ iϩ ϭ Ϫ dt r dt : v θ vy vx x vr dxp ˆ j dt (4-38) Now note that the rate dyp /dt at which yp changes is equal to the velocity component vy Similarly, dxp /dt ϭ vx, and, again from Fig 4-17b, we see that vx ϭ Ϫv sin u and vy ϭ v cos u Making these substitutions in Eq 4-38, we find : aϭ Ϫ v2 v2 cos iˆ ϩ Ϫ sin jˆ r r (4-39) This vector and its components are shown in Fig 4-17c Following Eq 3-6, we find (b) a ϭ 2a 2x ϩ a 2y ϭ y a , we find the angle f shown in Fig 4-17c: as we wanted to prove To orient : ax a φ v2 v2 v2 2(cos )2 ϩ (sin )2 ϭ 11 ϭ , r r r tan ϭ ay x (c) Figure 4-17 Particle p moves in counterclockwise uniform circular motion (a) Its v at a certain position and velocity : a v (c) Acceleration : instant (b) Velocity : ay Ϫ(v2/r) sin ϭ ϭ tan ax Ϫ(v2/r) cos : Thus, f ϭ u, which means that a is directed along the radius r of Fig 4-17a, toward the circle’s center, as we wanted to prove Checkpoint An object moves at constant speed along a circular path in a horizontal xy plane, with the center at the origin When the object is at x ϭ Ϫ2 m, its velocity is Ϫ(4 m/s) ˆj Give the object’s (a) velocity and (b) acceleration at y ϭ m Sample Problem 4.06 Top gun pilots in turns “Top gun” pilots have long worried about taking a turn too tightly As a pilot’s body undergoes centripetal acceleration, with the head toward the center of curvature, the blood pressure in the brain decreases, leading to loss of brain function There are several warning signs When the centripetal acceleration is 2g or 3g, the pilot feels heavy At about 4g, the pilot’s vision switches to black and white and narrows to “tunnel vision.” If that acceleration is sustained or increased, vision ceases and, soon after, the pilot is unconscious — a condition known as g-LOC for “g-induced loss of consciousness.” What is the magnitude of the acceleration, in g units, of a pilot whose aircraft enters a horizontal circular turn with a vi ϭ (400ˆi ϩ 500ˆj) m/s and 24.0 s later leaves the velocity of : turn with a velocity of : v f ϭ (Ϫ400ˆi Ϫ 500 ˆj) m/s? KEY IDEAS We assume the turn is made with uniform circular motion Then the pilot’s acceleration is centripetal and has magnitude a given by Eq 4-34 (a ϭ v2/R), where R is the circle’s radius Also, the time required to complete a full circle is the period given by Eq 4-35 (T ϭ 2pR/v) Calculations: Because we not know radius R, let’s solve Eq 4-35 for R and substitute into Eq 4-34 We find 2v aϭ T To get the constant speed v, let’s substitute the components of the initial velocity into Eq 3-6: v ϭ 2(400 m/s)2 ϩ (500 m/s)2 ϭ 640.31 m/s halliday_c04_053-079v3.0.1.qxd 2/27/14 4:59 PM Page 69 www.freebookslides.com 69 4-6 R E L ATIVE M OTION I N ON E DI M E NSION To find the period T of the motion, first note that the final velocity is the reverse of the initial velocity This means the aircraft leaves on the opposite side of the circle from the initial point and must have completed half a circle in the given 24.0 s Thus a full circle would have taken T ϭ 48.0 s Substituting these values into our equation for a, we find aϭ 2(640.31 m/s) ϭ 83.81 m/s2 Ϸ 8.6g 48.0 s (Answer) Additional examples, video, and practice available at WileyPLUS 4-6 RELATIVE MOTION IN ONE DIMENSION Learning Objective After reading this module, you should be able to 4.18 Apply the relationship between a particle’s position, velocity, and acceleration as measured from two reference frames that move relative to each other at constant velocity and along a single axis Key Idea : v PA ● When two frames of reference A and B are moving relative to each other at constant velocity, the velocity of a particle P as measured by an observer in frame A usually differs from that measured from frame B The two measured velocities are related by ϭ: v PB ϩ : v BA, where : v BA is the velocity of B with respect to A Both observers measure the same acceleration for the particle: : a PA ϭ : a PB Relative Motion in One Dimension Suppose you see a duck flying north at 30 km/h To another duck flying alongside, the first duck seems to be stationary In other words, the velocity of a particle depends on the reference frame of whoever is observing or measuring the velocity For our purposes, a reference frame is the physical object to which we attach our coordinate system In everyday life, that object is the ground For example, the speed listed on a speeding ticket is always measured relative to the ground The speed relative to the police officer would be different if the officer were moving while making the speed measurement Suppose that Alex (at the origin of frame A in Fig 4-18) is parked by the side of a highway, watching car P (the “particle”) speed past Barbara (at the origin of frame B) is driving along the highway at constant speed and is also watching car P Suppose that they both measure the position of the car at a given moment From Fig 4-18 we see that xPA ϭ xPB ϩ xBA (4-40) The equation is read: “The coordinate xPA of P as measured by A is equal to the coordinate xPB of P as measured by B plus the coordinate xBA of B as measured by A.” Note how this reading is supported by the sequence of the subscripts Taking the time derivative of Eq 4-40, we obtain d d d (xPA) ϭ (xPB) ϩ (x ) dt dt dt BA Thus, the velocity components are related by vPA ϭ vPB ϩ vBA (4-41) This equation is read: “The velocity vPA of P as measured by A is equal to the Frame B moves past frame A while both observe P y y Frame A Frame B P vBA xBA xPB x x xPA = xPB + xBA Figure 4-18 Alex (frame A) and Barbara (frame B) watch car P, as both B and P move at different velocities along the common x axis of the two frames At the instant shown, xBA is the coordinate of B in the A frame Also, P is at coordinate xPB in the B frame and coordinate xPA ϭ xPB ϩ xBA in the A frame halliday_c04_053-079v3.0.1.qxd 2/27/14 5:00 PM Page 70 www.freebookslides.com 70 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS velocity vPB of P as measured by B plus the velocity vBA of B as measured by A.” The term vBA is the velocity of frame B relative to frame A Here we consider only frames that move at constant velocity relative to each other In our example, this means that Barbara (frame B) drives always at constant velocity vBA relative to Alex (frame A) Car P (the moving particle), however, can change speed and direction (that is, it can accelerate) To relate an acceleration of P as measured by Barbara and by Alex, we take the time derivative of Eq 4-41: d d d (v ) ϭ (v ) ϩ (v ) dt PA dt PB dt BA Because vBA is constant, the last term is zero and we have aPA ϭ aPB (4-42) In other words, Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle Sample Problem 4.07 Relative motion, one dimensional, Alex and Barbara In Fig 4-18, suppose that Barbara’s velocity relative to Alex is a constant vBA ϭ 52 km/h and car P is moving in the negative direction of the x axis (a) If Alex measures a constant vPA ϭ Ϫ78 km/h for car P, what velocity vPB will Barbara measure? KEY IDEAS Calculation: The initial velocity of P relative to Alex is vPA ϭ Ϫ78 km/h and the final velocity is Thus, the acceleration relative to Alex is v Ϫ v0 Ϫ (Ϫ78 km/h) m/s ϭ t 10 s 3.6 km/h ϭ 2.2 m/s (Answer) aPA ϭ We can attach a frame of reference A to Alex and a frame of reference B to Barbara Because the frames move at constant velocity relative to each other along one axis, we can use Eq 4-41 (vPA ϭ vPB ϩ vBA) to relate vPB to vPA and vBA Calculation: We find vPB ϭ Ϫ130 km/h (c) What is the acceleration aPB of car P relative to Barbara during the braking? KEY IDEA Ϫ78 km/h ϭ vPB ϩ 52 km/h Thus, to relate the acceleration to the initial and final velocities of P (Answer) Comment: If car P were connected to Barbara’s car by a cord wound on a spool, the cord would be unwinding at a speed of 130 km/h as the two cars separated (b) If car P brakes to a stop relative to Alex (and thus relative to the ground) in time t ϭ 10 s at constant acceleration, what is its acceleration aPA relative to Alex? KEY IDEAS To calculate the acceleration of car P relative to Alex, we must use the car’s velocities relative to Alex Because the acceleration is constant, we can use Eq 2-11 (v ϭ v0 ϩ at) To calculate the acceleration of car P relative to Barbara, we must use the car’s velocities relative to Barbara Calculation: We know the initial velocity of P relative to Barbara from part (a) (vPB ϭ Ϫ130 km/h) The final velocity of P relative to Barbara is Ϫ52 km/h (because this is the velocity of the stopped car relative to the moving Barbara) Thus, v Ϫ v0 Ϫ52 km/h Ϫ (Ϫ130 km/h) m/s ϭ t 10 s 3.6 km/h ϭ 2.2 m/s (Answer) aPB ϭ Comment: We should have foreseen this result: Because Alex and Barbara have a constant relative velocity, they must measure the same acceleration for the car Additional examples, video, and practice available at WileyPLUS halliday_c04_053-079v3.0.1.qxd 2/27/14 5:00 PM Page 71 www.freebookslides.com 71 4-7 R E L ATIVE M OTION I N TWO DI M E NSIONS 4-7 RELATIVE MOTION IN TWO DIMENSIONS Learning Objective After reading this module, you should be able to 4.19 Apply the relationship between a particle’s position, velocity, and acceleration as measured from two reference frames that move relative to each other at constant velocity and in two dimensions Key Idea : v PA ● When two frames of reference A and B are moving relative to each other at constant velocity, the velocity of a particle P as measured by an observer in frame A usually differs from that measured from frame B The two measured velocities are related by ϭ: v PB ϩ : v BA, : where v BA is the velocity of B with respect to A Both observers measure the same acceleration for the particle: : : a PA ϭ a PB Relative Motion in Two Dimensions y Our two observers are again watching a moving particle P from the origins of reference frames A and B, while B moves at a constant velocity : v BA relative to A (The corresponding axes of these two frames remain parallel.) Figure 4-19 shows a certain instant during the motion At that instant, the position vector of the origin of B relative to the origin of A is : r BA.Also, the position vectors of particle P are : r PA rela: tive to the origin of A and r PB relative to the origin of B From the arrangement of heads and tails of those three position vectors, we can relate the vectors with : : : r PA ϭ r PB ϩ r BA (4-43) : By taking the time derivative of this equation, we can relate the velocities v PA and : v PB of particle P relative to our observers: : v PA ϭ : v PB ϩ : v BA (4-44) P y rPB rPA vBA rBA Frame A x Frame B x Figure 4-19 Frame B has the constant two-dimensional velocity : v BA relative to frame A The position vector of B relative to A is : r BA The position vectors of particle P are : r PB r PA relative to A and : relative to B By taking the time derivative of this relation, we can relate the accelerations : a PA and : a PB of the particle P relative to our observers However, note that because : v BA is constant, its time derivative is zero Thus, we get : a PA ϭ : a PB (4-45) As for one-dimensional motion, we have the following rule: Observers on different frames of reference that move at constant velocity relative to each other will measure the same acceleration for a moving particle Sample Problem 4.08 Relative motion, two dimensional, airplanes In Fig 4-20a, a plane moves due east while the pilot points the plane somewhat south of east, toward a steady wind that blows to the northeast The plane has velocity : v PW relative to the wind, with an airspeed (speed relative to the wind) of 215 km/h, directed at angle u south of east The wind has velocity : v WG relative to the ground with speed 65.0 km/h, directed 20.0° east of north What is the magnitude of the velocity : v PG of the plane relative to the ground, and what is ? KEY IDEAS The situation is like the one in Fig 4-19 Here the moving particle P is the plane, frame A is attached to the ground (call it G), and frame B is “attached” to the wind (call it W) We need a vector diagram like Fig 4-19 but with three velocity vectors Calculations: First we construct a sentence that relates the three vectors shown in Fig 4-20b: halliday_c04_053-079v3.0.1.qxd 2/27/14 5:00 PM Page 72 www.freebookslides.com 72 CHAPTE R M OTION I N TWO AN D TH R E E DI M E NSIONS velocity of plane velocity of plane velocity of wind ϭ ϩ relative to ground relative to wind relative to ground (PG) (PW) (WG) This is the plane's actual direction of travel N vPG This relation is written in vector notation as v PW ϩ : v WG v PG ϭ : : We need to resolve the vectors into components on the coordinate system of Fig 4-20b and then solve Eq 4-46 axis by axis For the y components, we find E θ (4-46) This is the plane's orientation N (a) ϭ Ϫ(215 km/h) sin u ϩ (65.0 km/h)(cos 20.0°) Solving for u gives us ϭ sinϪ1 vPG y (65.0 km/h)(cos 20.0Њ) ϭ 16.5Њ 215 km/h vWG This is the wind direction vPG,y ϭ vPW,y ϩ vWG,y or 20° vPW θ (Answer) vWG vPW x Similarly, for the x components we find The actual direction is the vector sum of the other two vectors (head-to-tail arrangement) vPG,x ϭ vPW,x ϩ vWG,x v PG is parallel to the x axis, the component Here, because : vPG,x is equal to the magnitude vPG Substituting this notation and the value u ϭ 16.5°, we find vPG ϭ (215 km/h)(cos 16.5°) ϩ (65.0 km/h)(sin 20.0°) ϭ 228 km/h (Answer) (b) Figure 4-20 A plane flying in a wind Additional examples, video, and practice available at WileyPLUS Review & Summary Position Vector The location of a particle relative to the ori: gin of a coordinate system is given by a position vector r , which in unit-vector notation is As ⌬t in Eq 4-8 is shrunk to 0, : v avg reaches a limit called either the velocity or the instantaneous velocity : v: : r ϭ xiˆ ϩ yjˆ ϩ zkˆ : (4-1) Here x ˆi , y ˆj , and z kˆ are the vector components of position vector : r, and x, y, and z are its scalar components (as well as the coordinates of the particle) A position vector is described either by a magnitude and one or two angles for orientation, or by its vector or scalar components Displacement If a particle moves so that its position vector : changes from : r to r 2, the particle’s displacement ⌬: r is ⌬r ϭ : : r2 Ϫ : r (4-2) The displacement can also be written as ⌬: r ϭ (x2 Ϫ x1)iˆ ϩ ( y2 Ϫ y1)jˆ ϩ (z2 Ϫ z1)kˆ ϭ ⌬xˆi ϩ ⌬yˆj ϩ ⌬zkˆ vϭ v ϭ vx ˆi ϩ vy ˆj ϩ vzkˆ , : Average Acceleration and Instantaneous Acceleration v in time interval ⌬t, its If a particle’s velocity changes from : v to : average acceleration during ⌬t is (4-8) : Ϫ: v1 ⌬v ϭ ⌬t ⌬t (4-15) As ⌬t in Eq 4-15 is shrunk to 0, : a avg reaches a limiting value called a: either the acceleration or the instantaneous acceleration : : aϭ : ⌬: r ϭ ⌬t : v2 (4-3) Average Velocity and Instantaneous Velocity If a parti- : v avg (4-11) where vx ϭ dx /dt, vy ϭ dy /dt, and vz ϭ dz /dt The instantaneous velocity : v of a particle is always directed along the tangent to the particle’s path at the particle’s position a avg ϭ cle undergoes a displacement ⌬ r in time interval ⌬t, its average velocity : v avg for that time interval is (4-10) which can be rewritten in unit-vector notation as : (4-4) d: r , dt In unit-vector notation, d: v dt a ϭ ax iˆ ϩ ay jˆ ϩ azkˆ , : where ax ϭ dvx /dt, ay ϭ dvy /dt, and az ϭ dvz /dt (4-16) (4-17) ... 15 :50:45 15 : 51: 43 15 :52: 41 15:53:39 15 :54:37 15 :55:35 15 :56:33 D 12 :03:59 12 :02:52 12 : 01: 45 12 :00:38 11 :59: 31 11: 58:24 11 :57 :17 E 12 :03:59 12 :02:49 12 : 01: 54 12 : 01: 52 12 : 01: 32 12 : 01: 22 12 : 01: 12 18 Because... What Is Physics? 11 58 Discovering the Nucleus 11 58 11 78 11 78 11 20 11 24 11 53 11 77 11 77 11 18 X Rays and the Ordering of the Elements The p-n Junction 11 48 The Junction Rectifier 11 49 The Light-Emitting... Speck of dust Penicillin molecule Uranium atom Proton Electron ϫ 10 53 ϫ 10 41 ϫ 10 30 ϫ 10 22 ϫ 10 15 ϫ 10 12 ϫ 10 7 ϫ 10 3 ϫ 10 Ϫ3 ϫ 10 ? ?10 ϫ 10 ? ?17 ϫ 10 Ϫ25 ϫ 10 Ϫ27 ϫ 10 Ϫ 31 (1- 7) with an uncertainty of ? ?10
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