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iv CONTENTS 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2 Acceleration 31 2.1 The law of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.1.1 Inertial reference frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.2 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.2.1 Average and instantaneous acceleration . . . . . . . . . . . . . . . . . . . . . 35 2.2.2 Motion with constant acceleration . . . . . . . . . . . . . . . . . . . . . . . . 39 2.2.3 Acceleration as a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 2.2.4 Acceleration in different reference frames . . . . . . . . . . . . . . . . . . . . 41 2.3 Free fall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 2.4 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 2.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.5.1 Motion with piecewise constant acceleration . . . . . . . . . . . . . . . . . . . 46 2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 3 Momentum and Inertia 51 3.1 Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.1.1 Relative inertia and collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 3.1.2 Inertial mass: definition and properties . . . . . . . . . . . . . . . . . . . . . 55 3.2 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.2.1 Conservation of momentum; isolated systems . . . . . . . . . . . . . . . . . . 56 | University Physics I Classical Mechanics_Page_7_Chunk4201 |
CONTENTS v 3.3 Extended systems and center of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.3.1 Center of mass motion for an isolated system . . . . . . . . . . . . . . . . . . 59 3.3.2 Recoil and rocket propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.4 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.5.1 Reading a collision graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.5.2 Collision in different reference frames, center of mass, and recoil . . . . . . . 65 3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4 Kinetic Energy 69 4.1 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.1.1 Kinetic energy in collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 4.1.2 Relative velocity and coefficient of restitution . . . . . . . . . . . . . . . . . . 74 4.2 “Convertible” and “translational” kinetic energy . . . . . . . . . . . . . . . . . . . . 76 4.2.1 Kinetic energy and momentum in different reference frames . . . . . . . . . . 79 4.3 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 4.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 4.4.1 Collision graph revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 4.4.2 Inelastic collision and explosive separation . . . . . . . . . . . . . . . . . . . . 83 4.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5 Interactions and energy 89 | University Physics I Classical Mechanics_Page_8_Chunk4202 |
vi CONTENTS 5.1 Conservative interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.1.1 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 5.1.2 Potential energy functions and “energy landscapes” . . . . . . . . . . . . . . 94 5.2 Dissipation of energy and thermal energy . . . . . . . . . . . . . . . . . . . . . . . . 96 5.3 Fundamental interactions, and other forms of energy . . . . . . . . . . . . . . . . . . 98 5.4 Conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 5.5 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 5.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5.6.1 Inelastic collision in the middle of a swing . . . . . . . . . . . . . . . . . . . . 103 5.6.2 Kinetic energy to spring potential energy: block collides with spring . . . . . 104 5.7 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 5.7.1 Two carts colliding and compressing a spring . . . . . . . . . . . . . . . . . . 106 5.7.2 Getting the potential energy function from collision data . . . . . . . . . . . . 107 5.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 6 Interactions, part 2: Forces 113 6.1 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 6.1.1 Forces and systems of particles . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.2 Forces and potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.3 Forces not derived from a potential energy . . . . . . . . . . . . . . . . . . . . . . . . 120 6.3.1 Tensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 | University Physics I Classical Mechanics_Page_9_Chunk4203 |
CONTENTS vii 6.3.2 Normal forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 6.3.3 Static and kinetic friction forces . . . . . . . . . . . . . . . . . . . . . . . . . 123 6.3.4 Air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 6.4 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 6.5 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 6.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 6.6.1 Dropping an object on a weighing scale . . . . . . . . . . . . . . . . . . . . . 129 6.6.2 Speeding up and slowing down . . . . . . . . . . . . . . . . . . . . . . . . . . 130 6.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 7 Impulse, Work and Power 137 7.1 Introduction: work and impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 7.2 Work on a single particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 7.2.1 Work done by the net force, and the Work-Energy Theorem . . . . . . . . . . 140 7.3 The “center of mass work” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 7.4 Work done on a system by all the external forces . . . . . . . . . . . . . . . . . . . . 142 7.4.1 The no-dissipation case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 7.4.2 The general case: systems with dissipation . . . . . . . . . . . . . . . . . . . 147 7.4.3 Energy dissipated by kinetic friction . . . . . . . . . . . . . . . . . . . . . . . 150 7.5 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 7.6 In summary . . . . . . . . . . . . . | University Physics I Classical Mechanics_Page_10_Chunk4204 |
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 | University Physics I Classical Mechanics_Page_10_Chunk4205 |
viii CONTENTS 7.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 7.7.1 Braking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 7.7.2 Work, energy and the choice of system: dissipative case . . . . . . . . . . . . 154 7.7.3 Work, energy and the choice of system: non-dissipative case . . . . . . . . . . 155 7.7.4 Jumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 7.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 8 Motion in two dimensions 161 8.1 Dealing with forces in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 161 8.2 Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 8.3 Inclined planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 8.4 Motion on a circle (or part of a circle) . . . . . . . . . . . . . . . . . . . . . . . . . . 169 8.4.1 Centripetal acceleration and centripetal force . . . . . . . . . . . . . . . . . . 169 8.4.2 Kinematic angular variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 8.5 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 8.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 8.6.1 The penny on the turntable . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 8.7 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 8.7.1 Staying on track . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 8.7.2 Going around a banked curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 8.7.3 Rotating frames of | University Physics I Classical Mechanics_Page_11_Chunk4206 |
reference: Centrifugal force and Coriolis force . . . . . . . 184 | University Physics I Classical Mechanics_Page_11_Chunk4207 |
CONTENTS ix 8.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 9 Rotational dynamics 191 9.1 Rotational kinetic energy, and moment of inertia . . . . . . . . . . . . . . . . . . . . 191 9.2 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 9.3 The cross product and rotational quantities . . . . . . . . . . . . . . . . . . . . . . . 197 9.4 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 9.5 Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 9.6 Rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 9.7 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 9.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 9.8.1 Torques and forces on the wheels of an accelerating bicycle . . . . . . . . . . 213 9.8.2 Blocks connected by rope over a pulley with non-zero mass . . . . . . . . . . 214 9.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 10 Gravity 221 10.1 The inverse-square law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 10.1.1 Gravitational potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . 224 10.1.2 Types of orbits under an inverse-square force . . . . . . . . . . . . . . . . . . 227 10.1.3 Kepler’s laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 10.2 Weight, acceleration, and the equivalence principle . . . . . . . . . . . . . . . . . . . 236 10.3 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 | University Physics I Classical Mechanics_Page_12_Chunk4208 |
x CONTENTS 10.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 10.4.1 Orbital dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 10.4.2 Orbital data from observations: Halley’s comet . . . . . . . . . . . . . . . . . 245 10.5 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 10.5.1 Tidal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 10.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 11 Simple harmonic motion 253 11.1 Introduction: the physics of oscillations . . . . . . . . . . . . . . . . . . . . . . . . . 253 11.2 Simple harmonic motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 11.2.1 Energy in simple harmonic motion . . . . . . . . . . . . . . . . . . . . . . . . 259 11.2.2 Harmonic oscillator subject to an external, constant force . . . . . . . . . . . 260 11.3 Pendulums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 11.3.1 The simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 11.3.2 The “physical pendulum” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 11.4 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 11.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 11.5.1 Oscillator in a box (a basic accelerometer!) . . . . . . . . . . . . . . . . . . . 268 11.5.2 Meter stick as a physical pendulum . . . . . . . . . . . . . . . . . . . . . . . . 270 11.6 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 11.6.1 Mass on a spring damped by friction with a surface | University Physics I Classical Mechanics_Page_13_Chunk4209 |
. . . . . . . . . . . . . . 272 | University Physics I Classical Mechanics_Page_13_Chunk4210 |
CONTENTS xi 11.6.2 The Cavendish experiment: how to measure G with a torsion balance . . . . 273 11.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 12 Waves in one dimension 279 12.1 Traveling waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 12.1.1 The “wave shape” function: displacement and velocity of the medium. . . . . 281 12.1.2 Harmonic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 12.1.3 The wave velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 12.1.4 Reflection and transmission of waves at a medium boundary . . . . . . . . . 286 12.2 Standing waves and resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 12.3 Conclusion, and further resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 12.4 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 12.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 12.5.1 Displacement and density/pressure in a longitudinal wave . . . . . . . . . . . 295 12.5.2 Violin sounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 12.6 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 12.6.1 Chain of masses coupled with springs: dispersion, and long-wavelength limit. 299 12.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 13 Thermodynamics 303 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 13.2 Introducing temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 | University Physics I Classical Mechanics_Page_14_Chunk4211 |
xii CONTENTS 13.2.1 Temperature and heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . 304 13.2.2 The gas thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 13.2.3 The zero-th law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 13.3 Heat and the first law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 13.3.1 “Direct exchange of thermal energy” and early theories of heat . . . . . . . . 309 13.3.2 The first law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 311 13.4 The second law and entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 13.4.1 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 13.4.2 The efficiency of heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 13.4.3 But what IS entropy, anyway? . . . . . . . . . . . . . . . . . . . . . . . . . . 317 13.5 In summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319 13.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 13.6.1 Calorimetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 13.6.2 Equipartition of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 13.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 | University Physics I Classical Mechanics_Page_15_Chunk4212 |
Preface Students: if this is too long, at the very least read the last four paragraphs. Thank you! For many years Eric Mazur’s Principles and Practice of Physics was the required textbook for University Physics I at the University of Arkansas. In writing this open-source replacement I have tried to preserve some of its best features, while at the same time condensing much of the presentation, and reworking several sections that did not quite fit the needs of our curriculum: primarily, the chapters on Thermodynamics, Waves, and Work. I have also skipped entirely the chapter on the “Principle of Relativity,” and instead distributed its contents among other chapters: in particular, the Galilean reference frame transformations are now introduced at the very beginning of the book, as are the law of inertia and the concept of inertial reference frames. Over the past few decades, there has been a trend to increase the size of introductory physics text- books, by including more and more visual aids (pictures, diagrams, boxes. . .), as well as lengthier and more detailed explanations, perhaps in an attempt to reach as many students as possible, and maybe even to take the place of the instructor altogether. It seems to me that the result is rather the opposite: a massive (and expensive) tome that no student could reasonably be expected to read all the way through, at a time when “TL;DR” has become a popular acronym, and visual learning aids (videotaped lectures, demonstrations, and computer simulations) are freely available everywhere. Our approach at the University of Arkansas, developed as a result of the work of Physics Education Research experts John and Gay Stewart, is based instead on two essential facts. First, that different students learn differently: some will learn best from a textbook, others will learn best from a lecture, and most will only really learn from a hands-on approach, by working out the answers to questions themselves. Second, just about everybody will benefit from repeated presentations of the material to be learned, in different environments and even from slightly different points of view. In keeping with this, we start by requiring the students to read the textbook material before coming to lecture, and also take an online “reading quiz” where they can check their understanding of what xiii | University Physics I Classical Mechanics_Page_16_Chunk4213 |
xiv CONTENTS they have read. Then, in the lecture, they will have an opportunity to see the material presented again, as a sort of executive summary delivered by, typically, a different instructor, who will also be able to answer any questions they might have about the book’s presentation. Additionally, the instructor will directly test the students’ understanding by means of conceptual questions asked of the whole class, which are to be answered using clickers. This prompts the students to think harder about the material, and encourages them to discuss it on the spot with their classmates. Immediately after each lecture, the students will have a lab activity where they will be able to verify experimentally the concepts and principles to which they have just been introduced. Finally, every week they will have an “open response” homework assignment where they get to apply the principles, mathematically, to concrete problems. For both the labs and the homework, additional assistance is provided by a group of dedicated teaching assistants, who are often able to explain the material to the students in a way that better relates to their own experience. In all this, the textbook is expected to play an important role, but certainly not to be the students’ only (nor even, necessarily, the primary) source of understanding or knowledge. Its job is to start the learning process, and to stand by to provide a reference (among possibly several others) afterwards. To fulfill this role, perhaps the most essential requirement is that it should be readable, and hence concise enough for every reading assignment to be of manageable size. This (as well as a sensible organization, clear explanations, and a minimal assortment of worked-out exercises and end-of chapter problems) is what I have primarily tried to provide here. A student who wants more information than provided in this textbook, or alternative explanations, or more worked-out examples, can certainly get these from many other sources: first, of course, the instructor and the teaching assistants, whose essential role as learning facilitators has to be recognized from the start. Then, there is a variety of alternative textbooks available: the best choice, probably, would still be Mazur’s Principles and Practice of Physics, since it uses the same terminology and notation, introduces the material in almost the same sequence, and has tons of worked-out examples and self-quiz conceptual questions. That book is available on reserve in the Physics library, where it can be consulted by anybody. If a student feels the need for an alternative textbook that they can actually take home, one option is, of course, to actually buy Mazur’s (which is what everybody had to do before); another option is to explore other open-source textbooks available online, which have been around longer than this one and benefit from more worked-out exercises and a more conventional presentation. One such book is University Physics I at openstax.org (https://openstax.org/details/books/university-physics-volume-1). Finally, there are also a large number of other online resources, although I would advise the students to approach them with caution, since not all of them may be totally rigorous, and some may end up being more confusing than helpful. Some of my students have found the Khan Academy lectures and/or the “Flipping Physics” lectures helpful; I personally would recommend the lectures of Walter Lewin at MIT, if only for the wide array of cool demonstrations you can see there. | University Physics I Classical Mechanics_Page_17_Chunk4214 |
CONTENTS xv One last word, for the students who may have read this far, concerning the use of equations and “proofs” in this book. It is essential to the nature of physics to be able to cast its results in mathematical terms, and to use math to explain and predict new results; hence, equations and mathematical derivations are integral parts of any physics textbook. I have, however, tried to keep the math as simple as possible throughout, and I would not want a lengthy mathematical derivation to get in the way of your reading assignment. If you are reading the text and come upon several lines of math, skim them at first to see if they make sense, but if you get stuck do not spend too much time on them: move on to the bottom line, and keep reading from there. You can always ask your instructor or TA for help with the math later. I would, however, encourage you to return, eventually, to any bit of math that you found challenging the first time around. Do try to go through all the algebra yourself! I have occasionally skipped intermediate steps, just to keep the math from overwhelming the text: but these are typically straightforward manipulations (multiplying or dividing both sides by something, moving something from one side of the equation to the other, multiplying out a parenthesis or, conversely, pulling out a common factor or denominator. . .). If you actually work out, on your own, all the missing steps, you will find it’s a great way to improve your algebra skills. This is something that will make it much easier for you to deal with the homework and the exams later on this semester—and for the rest of your career as well. P.S. A few possibly useful features. The pdf version of this book is, of course searchable: you can use command-F on the Mac or control-F in Windows to search the text for any term (hint: try searching the textbook before hitting Google!!). Hopefully, this will make up a little for the lack of an actual index, which I just haven’t found the time to compile yet. The text is also pretty thoroughly hyperlinked: every equation number and figure number quoted is a link. Clicking on the equation number (for instance, the (6.30), in this sentence) will take you to where the equation was first displayed (in this case, in Chapter 6). Then you can use command-left arrow (on the Mac) or alt-left arrow (on Windows) to go back to the page you were reading before you clicked on the link. Command- (or alt-) left and right arrows will let you go back and forth between the two pages. (Note: these keyboard shortcuts only work in Adobe Acrobat, as far as I know.) The same thing works if you click on the reference to a figure number (actually that takes you to the caption to the figure, so you may need to scroll up a bit to see the figure). Finally, the entries in the table of contents are links to the respective sections as well (except for the preface, which for some reason does not work; oh, well. . . if you are reading this, at least you made it here!). | University Physics I Classical Mechanics_Page_18_Chunk4215 |
Chapter 1 Reference frames, displacement, and velocity 1.1 Introduction Classical mechanics is the branch of physics that deals with the study of the motion of anything (roughly speaking) larger than an atom or a molecule. That is a lot of territory, and the methods and concepts of classical mechanics are at the foundation of any branch of science or engineering that is concerned with the motion of anything from a star to an amoeba—fluids, rocks, animals, planets, and any and all kinds of machines. Moreover, even though the accurate description of processes at the atomic level requires the (formally very different) methods of quantum mechanics, at least three of the basic concepts of classical mechanics that we are going to study this semester, namely, momentum, energy, and angular momentum, carry over into quantum mechanics as well, with the last two playing, in fact, an essential role. 1.1.1 Particles in classical mechanics In the study of motion, the most basic starting point is the concept of the position of an object. Clearly, if we want to describe accurately the position of a macroscopic object such as a car, we may need a lot of information, including the precise shape of the car, whether it is turned this way or that way, and so on; however, if all we want to know is how far the car is from Fort Smith or Fayetteville, we do not need any of that: we can just treat the car as a dot, or mathematical point, on the map—which is the way your GPS screen will show it, anyway. When we do this, we say that are describing the car (or whatever the macroscopic object may be) as a particle. 1 | University Physics I Classical Mechanics_Page_19_Chunk4216 |
2 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY In classical mechanics, an “ideal” particle is an object with no appreciable size—a mathematical point. In one dimension (that is to say, along a straight line), its position can be specified just by giving a single number, the distance from some reference point, as we shall see in a moment (in three dimensions, of course, three numbers are required). In terms of energy (which is perhaps the most important concept in all of physics, and which we will introduce properly in due course), an ideal particle has only one kind of energy, what we will later call translational kinetic energy; it cannot have, for instance, rotational kinetic energy (since it has “no shape” for practical purposes), or any form of internal energy (elastic, thermal, etc.), since we assume it is too small to have any internal structure in the first place. The reason this is a useful concept is not just that we can often treat extended objects as particles in an approximate way (like the car in the example above), but also, and most importantly, that if we want to be more precise in our calculations, we can always treat an extended object (mathematically) as a collection of “particles.” The physical properties of the object, such as its energy, momentum, rotational inertia, and so forth, can then be obtained by adding up the corresponding quantities for all the particles making up the object. Not only that, but the interactions between two extended objects can also be calculated by adding up the interactions between all the particles making up the two objects. This is how, once we know the form of the gravitational force between two particles (which is fairly simple, as we will see in Chapter 10), we can use that to calculate the force of gravity between a planet and its satellites, which can be fairly complicated in detail, depending, for instance, on the relative orientation of the planet and the satellite. The mathematical tool we use to calculate these “sums” is calculus—specifically, integration—and you will see many examples of this. . . in your calculus courses. Calculus I is only a corequisite for this course, so we will not make a lot of use of it here, and in any case you would need multidimensional integrals, which are an even more advanced subject, to do these kinds of calculations. But it may be good for you to keep these ideas on the back of your mind. Calculus was, in fact, invented by Sir Isaac Newton precisely for this purpose, and the developments of physics and mathematics have been closely linked together ever since. Anyway, back to particles, the plan for this semester is as follows: we will start our description of motion by treating every object (even fairly large ones, such as cars) as a “particle,” because we will only be concerned at first with its translational motion and the corresponding energy. Then we will progressively make things more complex: by considering systems of two or more particles, we will start to deal with the internal energy of a system. Then we will move to the study of rigid bodies, which are another important idealization: extended objects whose parts all move together as the object undergoes a translation or a rotation. This will allow us to introduce the concept of rotational kinetic energy. Eventually we will consider wave motion, where different parts of an extended object (or “medium”) move relative to each other. So, you see, there is a logical progression here, with most parts of the course building on top of the previous ones, and energy as one of the main connecting themes. | University Physics I Classical Mechanics_Page_20_Chunk4217 |
1.2. POSITION, DISPLACEMENT, VELOCITY 3 1.1.2 Aside: the atomic perspective As an aside, it should perhaps be mentioned that the building up of classical mechanics around this concept of ideal particles had nothing to do, initially, with any belief in “atoms,” or an atomic theory of matter. Indeed, for most 18th and 19th century physicists, matter was supposed to be a continuous medium, and its (mental) division into particles was just a mathematical convenience. The atomic hypothesis became increasingly more plausible as the 19th century wore on, and by the 1920’s, when quantum mechanics came along, physicists had to face a surprising development: matter, it turned out, was indeed made up of “elementary particles,” but these particles could not, in fact, be themselves described by the laws of classical mechanics. One could not, for instance, attribute to them simultaneously well-defined positions and velocities. Yet, in spite of this, most of the conclusions of classical mechanics remain valid for macroscopic objects, because, most of the time, it is OK to (formally) “break up” extended objects into chunks that are small enough to be treated as particles, but large enough that one does not need quantum mechanics to describe their behavior. Quantum properties were first found to manifest themselves at the macroscopic level when dealing with thermal energy, because at one point it really became necessary to figure out where and how the energy was stored at the truly microscopic (atomic) level. Thus, after centuries of successes, classical mechanics met its first failure with the so-called problem of the specific heats, and a completely new physical theory—quantum mechanics—had to be developed in order to deal with the newly-discovered atomic world. But all this, as they say, is another story, and for our very brief dealings with thermal physics—the last chapter in this book—we will just take specific heats as given, that is to say, something you measure (or look up in a table), rather than something you try to calculate from theory. 1.2 Position, displacement, velocity Kinematics is the part of mechanics that deals with the mathematical description of motion, leaving aside the question of what causes an object to move in a certain way. Kinematics, therefore, does not include such things as forces or energy, which fall instead under the heading of dynamics. It may be said, then, that kinematics by itself is not true physics, but only applied mathematics; yet it is still an essential part of classical mechanics, and its most natural starting point. This chapter (and parts of the next one) will introduce the basic concepts and methods of kinematics in one dimension. | University Physics I Classical Mechanics_Page_21_Chunk4218 |
4 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY 1.2.1 Position As stated in the previous section, we are initially interested only in describing the motion of a “particle,” which can be thought of as a mathematical point in space. (Later on we will see that, even for an extended object or system, it is often useful to consider the motion of a specific point that we call the system’s center of mass.) A point in three dimensions can be located by giving three numbers, known as its Cartesian coordinates (or, more simply, its coordinates). In two dimensions, this works as shown in Figure 1.1 below. As you can see, the coordinates of a point just tell us how to find it by first moving a certain distance x, from a previously-agreed origin, along a horizontal (or x) axis, and then a certain distance y along a vertical (or y) axis. (Or, of course, you could equally well first move vertically and then horizontally.) x axis y axis r x y θ Figure 1.1: The position vector, ⃗r, of a point, and its x and y components (the point’s coordinates). The quantities x and y are taken to be positive or negative depending on what side of the origin the point is on. Typically, we will always start by choosing a positive direction for each axis, as the direction along which the algebraic value of the corresponding coordinate increases. This is often chosen to be to the right for the horizontal axis, and upwards for the vertical axis, but there is nothing that says we cannot choose a different convention if it turns out to be more convenient. In Figure 1.1, the arrows on the axes denote the positive direction for each. Going by the grid, the coordinates of the point shown are x = 4 units, y = 3 units. In two or three dimensions (and even, in a sense, in one dimension), the coordinates of a point can | University Physics I Classical Mechanics_Page_22_Chunk4219 |
1.2. POSITION, DISPLACEMENT, VELOCITY 5 be interpreted as the components of a vector that we call the point’s position vector, and denote by ⃗r (sometimes boldface letters are used for vectors, instead of an arrow on top; in that case, the position vector would be denoted by r). A vector is a mathematical object, with specific geometric and algebraic properties, that physicists use to represent a quantity that has both a magnitude and a direction. The magnitude of the position vector in Fig. 1.1 is just the length of the arrow, which is to say, 5 length units (by the Pythagorean theorem, the length of ⃗r, which we will often write using absolute value bars as |⃗r|, is equal to ! x2 + y2); this is just the straight-line distance of the point to the origin. The direction of ⃗r, on the other hand, can be specified in a number of ways; a common convention is to give the value of the angle that it makes with the positive x axis, which I have denoted in the figure as θ (in this case, you can verify that θ = tan−1(y/x) = 36.9◦). In three dimensions, two angles would be needed to completely specify the direction of ⃗r. As you can see, giving the magnitude and direction of ⃗r is a way to locate the point that is completely equivalent to giving its coordinates x and y. By the same token, the coordinates x and y are a way to specify the vector ⃗r that is completely equivalent to giving its magnitude and direction. As I stated above, we call x and y the components (or sometimes, to be more specific, the Cartesian components) of the vector ⃗r. In a sense all the vectors that will be introduced later on this semester will derive their geometric and algebraic properties from the position vector ⃗r, so once you know how to deal with one vector, you can deal with them all. The geometric properties (by which I mean, how to relate a vector’s components to its magnitude and direction) I have just covered, and will come back to later on in this chapter, and again in Chapter 8; the algebraic properties (how to add vectors and multiply them by ordinary numbers, which are called scalars in this context) I will introduce along the way. For the first few chapters this semester, we are going to be primarily concerned with motion in one dimension (that is to say, along a straight line, backwards or forwards), in which case all we need to locate a point is one number, its x (or y, or z) coordinate; we do not then need to worry particularly about vector algebra. Alternatively, we can simply say that a vector in one dimension is essentially the same as its only component, which is just a positive or negative number (the magnitude of the number being the magnitude of the vector, and its sign indicating its direction), and has the algebraic properties that follow naturally from that. The description of the motion that we are aiming for is to find a function of time, which we denote by x(t), that gives us the point’s position (that is to say, the value of x) for any value of the time parameter, t. (See Eq. (1.10), below, for an example.) Remember that x stands for a number that can be positive or negative (depending on the side of the origin the point is on), and has dimensions of length, so when giving a numerical value for it you must always include the appropriate units (meters, centimeters, miles. . .). Similarly, t stands for the time elapsed since some more or less arbitrary “origin of time,” or time zero. Normally t should always be positive, but in special cases it may make sense to consider negative times (think of how we count years: “AD” would correspond to “positive” and “BC” would correspond to negative—the difference being that there is actually no year zero!). Anyway, t also is a number with dimensions, and must be reported with | University Physics I Classical Mechanics_Page_23_Chunk4220 |
its appropriate | University Physics I Classical Mechanics_Page_23_Chunk4221 |
6 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY units: seconds, minutes, hours, etc. x (m) t (s) Figure 1.2: A possible position vs. time graph for an object moving in one dimension. We will be often interested in plotting the position of an object as a function of time—that is to say, the graph of the function x(t). This may, in principle, have any shape, as you can see in Figure 1.2 above. In the lab, you will have a chance to use a position sensor that will automatically generate graphs like that for you on the computer, for any moving object that you aim the position sensor at. It is, therefore, important that you learn how to “read” such graphs. For example, Figure 1.2 shows an object that starts, at the time t = 0, a distance 0.2 m away and to the right of the origin (so x(0) = 0.2 m), then moves in the negative direction to x = −0.15 m, which it reaches at t = 0.5 s; then turns back and moves in the opposite direction until it reaches the point x = 0.1 m, turns again, and so on. Physically, this could be tracking the damped oscillations of a system such as an object attached to a spring and sliding over a surface that exerts a friction force on it (see Example 11.5.1). 1.2.2 Displacement In one dimension, the displacement of an object over a given time interval is a quantity that we denote as ∆x, and equals the difference between the object’s initial and final positions (in one dimension, we will often call the “position coordinate” simply the “position,” for short): ∆x = xf −xi (1.1) Here the subscript i denotes the object’s position at the beginning of the time interval considered, and the subscript f its position at the end of the interval. The symbol ∆will consistently be used throughout this book to denote a change in the quantity following the symbol, meaning the | University Physics I Classical Mechanics_Page_24_Chunk4222 |
1.2. POSITION, DISPLACEMENT, VELOCITY 7 difference between its initial value and its final value. The time interval itself will be written as ∆t and can be expressed as ∆t = tf −ti (1.2) where again ti and tf are the initial and final values of the time parameter (imagine, for instance, that you are reading time in seconds on a digital clock, and you are interested in the change in the object’s position between second 130 and second 132: then ti = 130 s, t2 = 132 s, and ∆t = 2 s). You can practice reading offdisplacements from Figure 1.2. The displacement between ti = 0.5 s and tf = 1 s, for instance, is 0.25 m (xi = −0.15 m, xf = 0.1 m). On the other hand, between ti = 1 s and tf = 1.3 s, the displacement is ∆x = 0 −0.1 = −0.1 m. Notice two important things about the displacement. First, it can be positive or negative. Positive means the object moved, overall, in the positive direction; negative means it moved, overall, in the negative direction. Second, even when it is positive, the displacement does not always equal the distance traveled by the object (distance, of course, is always defined as a positive quantity), because if the object “doubles back” on its tracks for some distance, that distance does not count towards the overall displacement. For instance, looking again at Figure 1.2 , in between the times ti = 0.5 s and tf = 1.5 s the object moved first 0.25 m in the positive direction, and then 0.15 m in the negative direction, for a total distance traveled of 0.4 m; however, the total displacement was just 0.1 m. In spite of these quirks, the total displacement is, mathematically, a useful quantity, because often we will have a way (that is to say, an equation) to calculate ∆x for a given interval, and then we can rewrite Eq. (1.1) so that it reads xf = xi + ∆x (1.3) That is to say, if we know where the object started, and we have a way to calculate ∆x, we can easily figure out where it ended up. You will see examples of this sort of calculation in the homework later on. Extension to two dimensions In two dimensions, we write the displacement as the vector ∆⃗r = ⃗rf −⃗ri (1.4) The components of this vector are just the differences in the position coordinates of the two points involved; that is, (∆⃗r)x (a subscript x, y, etc., is a standard way to represent the x, y . . . component of a vector) is equal to xf −xi, and similarly (∆⃗r)y = yf −yi. | University Physics I Classical Mechanics_Page_25_Chunk4223 |
8 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY x (m) y (m) ri xi yi rf Δr xf yf Figure 1.3: The displacement vector for a particle that was initially at a point with position vector ⃗ri and ended up at a point with position vector ⃗rf is the difference of the position vectors. Figure 1.3 shows how this makes sense. The x component of ∆⃗r in the figure is ∆x = 3−7 = −4 m; the y component is ∆y = 8−4 = 4 m. This basically shows you how to subtract (and, by extension, add, since ⃗rf = ⃗ri + ∆⃗r) vectors: you just subtract (or add) the corresponding components. Note how, by the Pythagorean theorem, the length (or magnitude) of the displacement vector, |∆⃗r| = ! (xf −xi)2 + (yf −yi)2, equals the straight-line distance between the initial point and the final point, just as in one dimension; of course, the particle could have actually followed a very different path from the initial to the final point, and therefore traveled a different distance. 1.2.3 Velocity Average velocity If you drive from Fayetteville to Fort Smith in 50 minutes, your average speed for the trip is calculated by dividing the distance of 59.2 mi by the time interval: average speed = distance ∆t = 59.2 mi 50 min = 59.2 mi 50 min × 60 min 1 hr = 71.0 mph (1.5) (this equation, incidentally, also shows you how to convert units, and how you will be expected to | University Physics I Classical Mechanics_Page_26_Chunk4224 |
1.2. POSITION, DISPLACEMENT, VELOCITY 9 work with significant figures this semester: the rule of thumb is, keep four significant figures in all intermediate calculations, and report three in the final result). The way we define average velocity is similar to average speed, but with one important difference: we use the displacement, instead of the distance. So, the average velocity vav of an object, moving along a straight line, over a time interval ∆t is vav = ∆x ∆t (1.6) This definition has all the advantages and the quirks of the displacement itself. On the one hand, it automatically comes with a sign (the same sign as the displacement, since ∆t will always be positive), which tells us in what direction we have been traveling. On the other hand, it may not be an accurate estimate of our average speed, if we doubled back at all. In the most extreme case, for a roundtrip (leave Fayetteville and return to Fayetteville), the average velocity would be zero, since xf = xi and therefore ∆x = 0. It is clear that this concept is not going to be very useful in general, if the object we are tracking has a chance to double back in the time interval ∆t. A way to prevent this from happening, and also getting a more meaningful estimate of the object’s speed at any instant, is to make the time interval very small. This leads to a new concept, that of instantaneous velocity. Instantaneous velocity We define the instantaneous velocity of an object (a “particle”), at the time t = ti, as the mathe- matical limit v = lim ∆t→0 ∆x ∆t (1.7) The meaning of this is the following. Suppose we compute the ratio ∆x/∆t over successively smaller time intervals ∆t (all of them starting at the same time ti). For instance, we can start by making tf = ti + 1 s, then try tf = ti + 0.5 s, then tf = ti + 0.1 s, and so on. Naturally, as the time interval becomes smaller, the corresponding displacement will also become smaller—the particle has less and less time to move away from its initial position, xi. The hope is that the successive ratios ∆x/∆t will converge to a definite value: that is to say, that at some point we will start getting very similar values, and that beyond a certain point making ∆t any smaller will not change any of the significant digits of the result that we care about. This limit value is the instantaneous velocity of the object at the time ti. When you think about it, there is something almost a bit self-contradictory about the concept of instantaneous velocity. You cannot (in practice) determine the velocity of an object if all you are given is a literal instant. You cannot even tell if the object is moving, if all you have is one instant! Motion requires more than one instant, the passage of time. In fact, all the “instantaneous” velocities that we can measure, with any instrument, are always really average velocities, only the | University Physics I Classical Mechanics_Page_27_Chunk4225 |
10 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY average is taken over very short time intervals. Nevertheless, the fact is that for any reasonably well-behaved position function x(t), the limit in Eq. (1.7) is mathematically well-defined, and it equals what we call, in calculus, the derivative of the function x(t): v = lim ∆t→0 ∆x ∆t = dx dt (1.8) x (m) t (s) 0 5 5 10 (tf , xf) Δx Δt (ti , xi) Figure 1.4: The slope of the green segment is the average velocity for the time interval ∆t shown. As ∆t becomes smaller, this approaches the slope of the tangent at the point (ti, xi) In fact, there is a nice geometric interpretation for this quantity: namely, it is the slope of a line tangent to the x-vs-t curve at the point (ti, xi). As Figure 1.4 shows, the average velocity ∆x/∆t is the slope (rise over run) of a line segment drawn from the point (ti, xi) to the point (tf, xf) (the green line in the figure). As we make the time interval smaller, by bringing tf closer to ti (and hence, also, xf closer to xi), the slope of this segment will approach the slope of the tangent line at (ti, xi) (the blue line), and this will be, by the definition (1.7), the instantaneous velocity at that point. This geometric interpretation makes it easy to get a qualitative feeling, from the position-vs-time graph, for when the particle is moving more or less fast. A large slope means a steep rise or fall, and that is when the velocity will be largest—in magnitude. A steep rise means a large positive velocity, whereas a steep drop means a large negative velocity, by which I mean a velocity that is given by a negative number which is large in absolute value. In the future, to simplify sentences like this one, I will just use the word “speed” to refer to the magnitude (that is to say, the absolute value) of the instantaneous velocity. Thus, speed (like distance) is always a positive number, by definition, whereas velocity can be positive or negative; and a steep slope (positive or negative) means the speed is large there. | University Physics I Classical Mechanics_Page_28_Chunk4226 |
1.2. POSITION, DISPLACEMENT, VELOCITY 11 Conversely, looking at the sample x-vs-t graphs in this chapter, you may notice that there are times when the tangent is horizontal, meaning it has zero slope, and so the instantaneous velocity at those times is zero (for instance, at the time t = 1.0 s in Figure 1.2). This makes sense when you think of what the particle is actually doing at those special times: it is just changing direction, so its velocity is going, for instance, from positive to negative. The way this happens is, it slows down, down. . . the velocity gets smaller and smaller, and then, for just an instant (literally, a mathematical point in time), it becomes zero before, the next instant, going negative. We will be coming back to this “reading of graphs” in the lab and the homework, as well as in the next chapter, when we introduce the concept of acceleration. Motion with constant velocity If the instantaneous velocity of an object never changes, it means that it is always moving in the same direction with the same speed. In that case, the instantaneous velocity and the average velocity coincide, and that means we can write v = ∆x/∆t (where the size of the interval ∆t could now be anything), and rewrite this equation in the form ∆x = v ∆t (1.9) which is the same as xf −xi = v (tf −ti) Now suppose we keep ti constant (that is, we fix the initial instant) but allow the time tf to change, so we will just write t for an arbitrary value of tf, and x for the corresponding value of xf. We end up with the equation x −xi = v (t −ti) which we can also write as x(t) = xi + v (t −ti) (1.10) after some rearranging, and where the notation x(t) has been introduced to emphasize that we want to think of x as a function of t. This is, not surprisingly, the equation of a straight line—a “curve” which is its own tangent and always has the same slope. (Please make sure that you are not confused by the notation in Eq. (1.10). The parentheses around the t on the left-hand side mean that we are considering the position x as a function of t. On the other hand, the parentheses around the quantity t −ti on the right-hand side mean that we are multiplying this quantity by v, which is a constant here. This distinction will be particularly important when we introduce the function v(t) next.) Either one of equations (1.9) or (1.11) can be used to solve problems involving motion with constant velocity, and again you will see examples of this in the homework. Motion with changing velocity | University Physics I Classical Mechanics_Page_29_Chunk4227 |
12 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY If the velocity changes with time, obtaining an expression for the position of the object as a function of time may be a nontrivial task. In the next chapter we will study an important special case, namely, when the velocity changes at a constant rate (constant acceleration). For the most general case, a graphical method that is sometimes useful is the following. Suppose that we know the function v(t), and we graph it, as in Figure 1.5 below. Then the area under the curve in between any two instants, say ti and tf, is equal to the total displacement of the object over that time interval. The idea involved is known in calculus as integration, and it goes as follows. Suppose that I break v (m/s) t (s) t1 t2 t3 ... tf v1 v2 v(t) Figure 1.5: How to get the displacement from the area under the v-vs-t curve. down the interval from ti to tf into equally spaced subintervals, beginning at the time ti (which I am, equivalently, going to call t1, that is, t1 ≡ti, so I have now t1, t2, t3, . . . tf). Now suppose I treat the object’s motion over each subinterval as if it were motion with constant velocity, the velocity being that at the beginning of the subinterval. This, of course, is only an approximation, since the velocity is constantly changing; but, if you look at Figure 1.5, you can convince yourself that it will become a better and better approximation as I increase the number of intermediate points and the rectangles shown in the figure become narrower and narrower. In this approximation, the displacement during the first subinterval would be ∆x1 = v1(t2 −t1) (1.11) where v1 = v(t1); similarly, ∆x2 = v2(t3 −t2), and so on. | University Physics I Classical Mechanics_Page_30_Chunk4228 |
1.2. POSITION, DISPLACEMENT, VELOCITY 13 However, Eq. (1.11) is just the area of the first rectangle shown under the curve in Figure 1.5 (the base of the rectangle has “length” t2 −t1, and its height is v1). Similarly for the second rectangle, and so on. So the sum ∆x1 +∆x2+. . . is both an approximation to the area under the v-vs-t curve, and an approximation to the total displacement ∆t. As the subdivision becomes finer and finer, and the rectangles narrower and narrower (and more numerous), both approximations become more and more accurate. In the limit of “infinitely many,” infinitely narrow rectangles, you get both the total displacement and the area under the curve exactly, and they are both equal to each other. Mathematically, we would write ∆x = " tf ti v(t) dt (1.12) where the stylized “S” (for “sum”) on the right-hand side is the symbol of the operation known as integration in calculus. This is essentially the inverse of the process know as differentiation, by which we got the velocity function from the position function, back in Eq. (1.8). This graphical method to obtain the displacement from the velocity function is sometimes useful, if you can estimate the area under the v-vs-t graph reliably. An important point to keep in mind is that rectangles under the horizontal axis (corresponding to negative velocities) have to be added as having negative area (since the corresponding displacement is negative); see example 1.5.1 at the end of this chapter. Extension to two dimensions In two (or more) dimensions, you define the average velocity vector as a vector ⃗vav whose com- ponents are vav,x = ∆x/∆t, vav,y = ∆y/∆t, and so on (where ∆x, ∆y, . . . are the corresponding components of the displacement vector ∆⃗r). This can be written equivalently as the single vector equation ⃗vav = ∆⃗r ∆t (1.13) This tells you how to multiply (or divide) a vector by an ordinary number: you just multiply (or divide) each component by that number. Note that, if the number in question is positive, this operation does not change the direction of the vector at all, it just scales it up or down (which is why ordinary numbers, in this context, are called scalars). If the scalar is negative, the vector’s direction is flipped as a result of the multiplication. Since ∆t in the definition of velocity is always positive, it follows that the average velocity vector always points in the same direction as the displacement, which makes sense. To get the instantaneous velocity, you just take the limit of the expression (1.13) as ∆t →0, for each component separately. The resulting vector ⃗v has components vx = lim∆t→0 ∆x/∆t, etc., which can also be written as vx = dx/dt, vy = dy/dt, . . .. All the results derived above hold for each spatial dimension and its corresponding velocity com- ponent. For instance, the graphical method shown in Figure 1.5 can always be used to get ∆x if | University Physics I Classical Mechanics_Page_31_Chunk4229 |
14 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY the function vx(t) is known, or equivalently to get ∆y if you know vy(t), and so on. Introducing the velocity vector at this point does cause a little bit of a notational difficulty. For quantities like x and ∆x, it is pretty obvious that they are the x components of the vectors ⃗r and ∆⃗r, respectively; however, the quantity that we have so far been calling simply v should more properly be denoted as vx (or vy if the motion is along the y axis). In fact, there is a convention that if you use the symbol for a vector without the arrow on top or any x, y, . . . subscripts, you must mean the magnitude of the vector. In this book, however, I have decided not to follow that convention, at least not until we get to Chapter 8 (and even then I will use it only for forces). This is because we will spend most of our time dealing with motion in only one dimension, and it makes the notation unnecessarily cumbersome to keep having to write the x or y subscripts on every component of every vector, when you really only have one dimension to worry about in the first place. So v will, throughout, refer to the relevant component of the velocity vector, to be inferred from the context, until we get to Chapter 8 and actually need to deal with both a vx and a vy explicitly. Finally, notice that the magnitude of the velocity vector, |⃗v| = # v2x + v2y + v2z, is equal to the instantaneous speed, since, as ∆t →0, the magnitude of the displacement vector, |∆⃗r|, becomes the actual distance traveled by the object in the time interval ∆t. 1.3 Reference frame changes and relative motion Everything up to this point assumes that we are using a fixed, previously agreed upon reference frame. Basically, this is just an origin and a set of axes along which to measure our coordinates, as shown in Figure 1. There are, however, a number of situations in physics that call for the use of different reference frames, and, more importantly, that require us to convert various physical quantities from one reference frame to another. For instance, imagine you are on a boat on a river, rowing downstream. You are moving with a certain velocity relative to the water around you, but the water itself is flowing with a different velocity relative to the shore, and your actual velocity relative to the shore is the sum of those two quantities. Ships generally have to do this kind of calculation all the time, as do airplanes: the “airspeed” is the speed of a plane relative to the air around it, but that air is usually moving at a substantial speed relative to the earth. The way we deal with all these situations is by introducing two reference frames, which here I am going to call A and B. One of them, say A, is “at rest” relative to the earth, and the other one is “at rest” relative to something else—which means, really, moving along with that something else. (For instance, a reference frame at rest “relative to the river” would be a frame that’s moving along | University Physics I Classical Mechanics_Page_32_Chunk4230 |
1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 15 with the river water, like a piece of driftwood that you could measure your progress relative to.) In any case, graphically, this will look as in Figure 1.6, which I have drawn for the two-dimensional case because I think it makes it easier to visualize what’s going on: A B xAB xBP xAP yBP yAP yAB P rBP rAP rAB Figure 1.6: Position vectors and coordinates of a point P in two different reference frames, A and B. In the reference frame A, the point P has position coordinates (xAP , yAP ). Likewise, in the reference frame B, its coordinates are (xBP , yBP ). As you can see, the notation chosen is such that every coordinate in A will have an “A” as a first subscript, while the second subscript indicates the object to which it refers, and similarly for coordinates in B. The coordinates (xAB, yAB) are special: they are the coordinates, in the reference frame A, of the origin of reference frame B. This is enough to fully locate the frame B in A, as long as the frames are not rotated relative to each other. The thin colored lines I have drawn along the axes in Figure 1.6 are intended to make it clear that the following equations hold: xAP = xAB + xBP yAP = yAB + yBP (1.14) Although the figure is drawn for the easy case where all these quantities are positive, you should be able to convince yourself that Eqs. (1.14) hold also when one or more of the coordinates have negative values. | University Physics I Classical Mechanics_Page_33_Chunk4231 |
16 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY All these coordinates are also the components of the respective position vectors, shown in the figure and color-coded by reference frame (so, for instance, ⃗rAP is the position vector of P in the frame A), so the equations (1.14) can be written more compactly as the single vector equation ⃗rAP = ⃗rAB + ⃗rBP (1.15) From all this you can see how to add vectors: algebraically, you just add their components sepa- rately, as in Eqs. (1.14); graphically, you draw them so the tip of one vector coincides with the tail of the other (we call this “tip-to-tail”), and then draw the sum vector from the tail of the first one to the tip of the other one. (In general, to get two arbitrary vectors tip-to-tail you may need to displace one of them; this is OK provided you do not change its orientation, that is, provided you only displace it, not rotate it. We’ll see how this works in a moment with velocities, and later on with forces.) Of course, I showed you already how to subtract vectors with Fig. 1.3: again, algebraically, you just subtract the corresponding coordinates, whereas graphically you draw them with a common origin, and then draw the vector from the tip of the vector you are subtracting to the tip of the other one. If you read the previous paragraph again, you can see that Fig. 1.3 can equally well be used to show that ∆⃗r = ⃗rf −⃗ri, as to show that ⃗rf = ⃗ri + ∆⃗r. In a similar way, you can see graphically from Fig. 1.6 (or algebraically from Eq. (1.15)) that the position vector of P in the frame B is given by ⃗rBP = ⃗rAP −⃗rAB. The last term in this expression can be written in a different way, as follows. If I follow the convention I have introduced above, the quantity xBA (with the order of the subscripts reversed) would be the x coordinate of the origin of frame A in frame B, and algebraically that would be equal to −xAB, and similarly yBA = −yAB. Hence the vector equality ⃗rAB = −⃗rBA holds. Then, ⃗rBP = ⃗rAP −⃗rAB = ⃗rAP + ⃗rBA (1.16) This is, in a way, the “inverse” of Eq. (1.15): it tells us how to get the position of P in the frame B if we know its position in the frame A. Let me show next you how all this extends to displacements and velocities. Suppose the point P indicates the position of a particle at the time t. Over a time interval ∆t, both the position of the particle and the relative position of the two reference frames may change. We can add yet another subscript, i or f, (for initial and final) to the coordinates, and write, for example, xAP,i = xAB,i + xBP,i xAP,f = xAB,f + xBP,f (1.17) Subtracting these equations gives us the corresponding displacements: ∆xAP = ∆xAB + ∆xBP (1.18) | University Physics I Classical Mechanics_Page_34_Chunk4232 |
1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 17 Dividing Eq. (1.18) by ∆t we get the average velocities1, and then taking the limit ∆t →0 we get the instantaneous velocities. This applies in the same way to the y coordinates, and the result is the vector equation ⃗vAP = ⃗vBP + ⃗vAB (1.19) I have rearranged the terms on the right-hand side to (hopefully) make it easier to visualize what’s going on. In words: the velocity of the particle P relative to (or measured in) frame A is equal to the (vector) sum of the velocity of the particle as measured in frame B, plus the velocity of frame B relative to frame A. The result (1.19) is just what we would have expected from the examples I mentioned at the beginning of this section, like rowing in a river or an airplane flying in the wind. For instance, for the airplane ⃗vBP could be its “airspeed” (only it has to be a vector, so it would be more like its “airvelocity”: that is, its velocity relative to the air around it), and ⃗vAB would be the velocity of the air relative to the earth (the wind velocity, at the location of the airplane). In other words, A represents the earth frame of reference and B the air, or wind, frame of reference. Then, ⃗vAP would be the “true” velocity of the airplane relative to the earth. You can see how it would be important to add these quantities as vectors, in general, by considering what happens when you fly in a cross wind, or try to row across a river, as in Figure 1.7 below. vEb vRb vER Figure 1.7: Rowing across a river. If you head “straight across” the river (with velocity vector ⃗vRb in the moving frame of the river, which is flowing with velocity ⃗vER in the frame of the earth), your actual velocity relative to the shore will be the vector ⃗vEb. This is an instance of Eq. (1.19), with frame A being E (the earth), frame B being R (the river), and “b” (for “boat”) standing for the point P we are tracking. As you can see from this couple of examples, Equation (1.19) is often useful as it is written, but 1We have made a very natural assumption, that the time interval ∆t is the same for observers tracking the particle’s motion in frames A and B, respectively (where each observer is understood to be moving along with his or her frame). This, however, turns out to be not true when any of the velocities involved is close to the speed of light, and so the simple addition of velocities formula (1.19) does not hold in Einstein’s relativity theory. (This is actually the first bit of real physics I have told you about in this book, so far; unfortunately, you will have no use for it this semester!) | University Physics I Classical Mechanics_Page_35_Chunk4233 |
18 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY sometimes the information we have is given to us in a different way: for instance, we could be given the velocity of the object in frame A (⃗vAP ), and the velocity of frame B as seen in frame A (⃗vAB), and told to calculate the velocity of the object as seen in frame B. This can be easily accomplished if we note that the vector ⃗vAB is equal to −⃗vBA; that is to say, the velocity of frame B as seen from frame A is just the opposite of the velocity of frame A as seen from frame B. Hence, Eq. (1.19) can be rewritten as ⃗vAP = ⃗vBP −⃗vBA (1.20) For most of the next few chapters we are going to be considering only motion in one dimension, and so we will write Eq. (1.19) (or (1.20)) without the vector symbols, and it will be understood that v refers to the component of the vector ⃗v along the coordinate axis of interest. A quantity that will be particularly important later on is the relative velocity of two objects, which we could label 1 and 2. The velocity of object 2 relative to object 1 is, by definition, the velocity which an observer moving along with 1 would measure for object 2. So it is just a simple frame change: let the earth frame be frame E and the frame moving with object 1 be frame 1, then the velocity we want is v12 (“velocity of object 2 in frame 1”). If we make the change A→1, B→E, and P→2 in Eq. (1.20), we get v12 = vE2 −vE1 (1.21) In other words, the velocity of 2 relative to 1 is just the velocity of 2 minus the velocity of 1. This is again a familiar effect: if you are driving down the highway at 50 miles per hour, and the car in front of you is driving at 55, then its velocity relative to you is 5 mph, which is the rate at which it is moving away from you (in the forward direction, assumed to be the positive one). It is important to realize that all these velocities are real velocities, each in its own reference frame. Something may be said to be truly moving at some velocity in one reference frame, and just as truly moving with a different velocity in a different reference frame. I will have a lot more to say about this in the next chapter, but in the meantime you can reflect on the fact that, if a car moving at 55 mph collides with another one moving at 50 mph in the same direction, the damage will be basically the same as if the first car had been moving at 5 mph and the second one had been at rest. For practical purposes, where you are concerned, another car’s velocity relative to yours is that car’s “real” velocity. Resources A good app for practicing how to add vectors (and how to break them up into components, mag- nitude and direction, etc.) may be found here: https://phet.colorado.edu/en/simulation/vector-addition. Perhaps the most dramatic demonstration of how Eq. (1.19) works in the real world is in this episode of Mythbusters: https://www.youtube.com/watch?v=BLuI118nhzc. (If this link does not work, do a search for “Mythbusters cancel momentum.”) They shoot a ball from the bed of a truck, | University Physics I Classical Mechanics_Page_36_Chunk4234 |
1.4. IN SUMMARY 19 with a velocity (relative to the truck) of 60 mph backwards, while the truck is moving forward at 60 mph. I think the result is worth watching. (Do not be distracted by their talk about momentum. We will get there, in time.) A very old, but also very good, educational video about different frames of reference is this one: https://www.youtube.com/watch?v=sS17fCom0Ns. You should try to watch at least part of it. Many things will be relevant to later parts of the course, including projectile motion, and the whole discussion of relative motion coming up next, in Chapter 2. 1.4 In summary 1. To describe the motion of an object in one dimension we treat it as a mathematical point, and consider its position coordinate, x (often shortened to just the position), as a function of time: x(t). 2. Numerically, the position coordinate is the distance to a chosen origin, with a positive or negative sign depending on which side of the origin the point is. For every problem, when we introduce a coordinate axis we need to specify a positive direction. Starting from the origin in that direction, the position coordinate is positive and increasing, whereas going from the origin in the opposite direction (negative direction) it becomes increasingly negative. 3. The displacement of an object over a time interval from an initial time ti to a final time tf is the quantity ∆x = xf −xi, where xf is the position of the object at the final time (or, the final position), and xi the position at the initial time (or initial position). 4. The average velocity of an object over the time interval from ti to tf is defined as vav = ∆x/∆t, where ∆t = tf −ti. 5. The instantaneous velocity (often just called the velocity) of an object at the time t is the limit value of the quantity ∆x/∆t, calculated for successively shorter time intervals ∆t, all with the same initial time ti = t. This is, mathematically, the definition of the derivative of the function x(t) at the time t, which we express as v = dx/dt. 6. Graphically, the instantaneous velocity of the object at the time t is the slope of the tangent line to the x-vs-t graph at the time t. 7. The instantaneous velocity of an object is a positive or negative quantity depending on whether the object, at that instant, is moving in the positive or the negative direction. 8. For an object moving with constant velocity v, the position function is given by [Eq. (1.10)]: x(t) = xi + v (t −ti) | University Physics I Classical Mechanics_Page_37_Chunk4235 |
20 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY where ti is an arbitrarily chosen initial time and xi the position at that time. This can also be written in the form given by Eq. (1.9). The argument (t) on the left-hand side of (1.10) is optional, and ti is often set equal to zero, giving just x = xi + vt. This, however, is not quite as generally applicable as the result (1.9) or (1.10). 9. For an object moving with changing velocity, the total displacement in between times ti and tf is equal to the total area under the v-vs-t curve in between those times; areas below the horizontal (t) axis must be treated as negative. 10. In two or more dimensions one introduces, for every point in space, a position vector whose components are just the Cartesian coordinates of that point; then the displacement vector is defined as ∆⃗r = ⃗rf −⃗ri, the average velocity vector is ⃗vav = ∆⃗r/∆t, and the instantaneous velocity vector is the limit of this as ∆t goes to zero. Vectors are added by adding their components separately; to multiply a vector by an ordinary number, or scalar, we just multiply each component by that number. 11. When tracking the motion of an object, “P”, in two different reference frames, A and B, the position vectors are related by ⃗rAP = ⃗rAB + ⃗rBP , and likewise the velocity vectors: ⃗vAP = ⃗vAB + ⃗vBP . Here, the first subscript tells you in which reference frame you are measuring, and the second subscript what it is that you are looking at; ⃗rAB is the position vector of the origin of frame B as seen in frame A, and ⃗vAB its velocity. | University Physics I Classical Mechanics_Page_38_Chunk4236 |
1.5. EXAMPLES 21 1.5 Examples 1.5.1 Motion with (piecewise) constant velocity You leave your house on your bicycle to go visit a friend. At your normal speed of 9 mph, you know it takes you 6 minutes to get there. This time, though, when you have traveled half the distance you realize you forgot a book at home that you were going to return to your friend, so you turn around and pedal at twice your normal speed, get back home, grab the book, and start offagain for your friend’s house at 18 mph (imagine you are really fit to pull this off!) (a) How far away from you does your friend live? (b) What is the total distance you travel on this trip? (c) How long did the whole trip take? (d) Draw a position versus time and a velocity versus time graph for the whole trip. Use SI units for both graphs. Neglect the time it takes you to stop and turn around, and also the time it takes you to run into your house and grab the book (in other words, assume those changes in your direction of motion happen instantly). (e) Show explicitly, using your v-vs-t graph, that the graphical method of Figure 1.5 gives you the total displacement for your trip. Solution I am going to work out this problem using both miles and SI units, the first because it seems most natural, and the second because we are asked to use SI units for part (d), so we might as well use them from the start. In general, you should use SI units whenever you can. If you are unsure of what to do in a specific problem, ask your instructor! (a) We are told that at 9 miles per hour it would take 6 minutes to get there, so let us use ∆x = v∆t (1.22) with v = 9 mph and ∆t = 6 min. We have to either convert the hours to minutes, or vice-versa. Again, in this case it seems easiest to realize that 6 min equals 1/10 of an hour, so: ∆x = $ 9 miles hr % × 0.1 hr = 0.9 miles. (1.23) In SI units, 9 mph = 4.023 m/s, and 6 min = 360 s, so ∆x = 1448 m. (b) This is just a matter of keeping track of the distance traveled in the various parts of the trip. You start by riding half the distance to your friend’s house, which is to say, 0.45 miles, and then you ride that again back home, so that’s 0.9 miles, and then you’re back where you started, so you still have to go the 0.9 miles to your friend’s house. So overall, you ride for 1.8 miles, or 2897 m. | University Physics I Classical Mechanics_Page_39_Chunk4237 |
22 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY (c) The whole trip consists, as detailed above, of 0.45 miles at 9 mph, and the rest, which is 1.35 miles, at 18 mph. Applying ∆t = ∆x/v to each of these intervals, we get a total time of ∆t = 0.45 miles 9 mph + 1.35 miles 18 mph = 0.125 hours = 0.125 × 60 min = 7.5 min = 7.5 × 60 s = 450 s (1.24) (d) The graphs are shown below. Details on how to get them follow. x (m) v (m/s) t (s) t (s) • First interval: from t = 0 to t = 180 s (3 min, which is what it would take to cover half the distance to your friend’s house at 9 mph). The velocity is a constant v = 4.023 m/s. For the position graph, use Eq. (1.10) with xi = 0, ti = 0 and v = 4.023 m/s. • Second interval: from t = 180 s to t = 270 s (it takes you half of 3 min, which is to say 90 s, to cover the same distance as above at twice the speed). The velocity is a constant v = −8.046 m/s | University Physics I Classical Mechanics_Page_40_Chunk4238 |
1.5. EXAMPLES 23 (twice what it was earlier, but in the opposite direction). For the position graph, use Eq. (1.10) with xi = 724 m (this is half of the distance to your friend’s house, and the starting position for this interval), ti = 180 s and v = −8.046 m/s. • Third interval: from t = 270 s to t = 450 s. The velocity is a constant v = 8.046 m/s (same speed as just before, but in the opposite direction). For the position graph, use Eq. (1.10) with xi = 0 m (you start back at your house), ti = 270 s and v = 8.046 m/s. If you are familiar with the software package Mathematica, the position graph was produced using the command Plot[If[t<180, 4.023 t, If[t<270, 4.023*180-8.046 (t-180), 8.046 (t-270)]], {t, 0, 450}] and the velocity graph was produced using Plot[If[t<180, 4.023 , If[t<270, -8.046, 8.046]], {t, 0, 450}] (and then connecting the horizontal lines by hand, which is not necessary, but helps to visualize what’s going on). The graphs could also have been produced using the free plotting software package Gnuplot (avail- able here: http://www.gnuplot.info/download.html) with the following commands: gnuplot> set dummy t gnuplot> f(t) = t<180 ? 4.023*t : t<270 ? 4.023*180-8.046*(t-180) : 8.046*(t-270) gnuplot> plot [0:450] f(t) The first line sets the default independent variable to t (instead of x, which is what Gnuplot ex- pects). The second line defines the piecewise function using the ternary operator (? :) borrowed from the C programming language. The third line plots the function over the range indicated. (e) For this we need to find the area under the v-vs-t graph we just plotted. Basically, we have three rectangles: the first one has base 180 units (s) and height 4 units (m/s), so its area is 4×180 = 720 (m). The second rectangle has base 90 units and height −8 (negative, because it is below the horizontal axis!), so its area is −720. The last one has base 180 units again (from 270 to 450) and height 8, so its area is 8 × 180 = 1440. So the total area “under” the v-vs-t curve is 720 −720 + 1440 = 1440 meters which is (approximately) your total displacement, that is, the 9 miles to your friend’s house. (Of course, we would have obtained a more accurate result if we had used the more accurate values for the “heights” of 4.023, −8.046, and 8.046, but if all we have to go by is the graph, such accuracy is pretty much impossible.) | University Physics I Classical Mechanics_Page_41_Chunk4239 |
24 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY 1.5.2 Addition of velocities, relative motion This example was inspired by the “race on a moving sidewalk” demo at http://physics.bu.edu/~duffy/classroom.html. Please go take a look at it! Two girls, Ann and Becky (yes, A and B) decide to have a race while they wait for a plane at a nearly-deserted airport. Ann will run on the moving walkway, to the end of it (which is 30 m away) and back, whereas Becky will run alongside her on the (non-moving) floor, also 30 m out and back. The walkway moves at 1 m/s, and the girls both run at the same constant speed of 5 m/s relative to the surface they are standing on. (a) Relative to the (non-moving) floor, what is Ann’s velocity for the first leg of her race, when she is moving in the same direction as the walkway (take that to be the positive direction)? What is her velocity for the return leg? (b) How long does it take each of the girls to complete their race? (c) When both girls are running in the positive direction, what is Becky’s velocity relative to Ann? (That is, how fast does Ann see Becky move, and in what direction?) (d) When Ann turns around and starts running in the negative direction, but Becky is still running in the positive direction, what is Becky’s velocity relative to Ann? (e) What is the total distance Ann runs in the moving walkway’s frame of reference? Solution I am going to solve this in the format that you will be required to use this semester for most of the homework and exam problems. I will not be able to do this for every single example, but you should! Please follow this carefully. To begin with, you must draw a sketch of the situation described in the problem, detailed enough to include all the relevant information you are given. Here is mine: vWA= 5 m/s vFW= 1 m/s vFB= 5 m/s 30 m going out Alice Becky vWA= –5 m/s vFW= 1 m/s vFB= –5 m/s coming back | University Physics I Classical Mechanics_Page_42_Chunk4240 |
1.5. EXAMPLES 25 Note that I have drawn one picture for each half of the race, and that all the information given in the text of the problem is there. The figure makes it clear also the notation I will be using for each of the girls’ velocities, and to see at a glance what is happening. You should next state what kind of problem this is and what basic result (theorem, principle, or equation(s)) you are going to use to solve it. For this problem, you could say: “This is a relative motion/reference frame transformation problem. I will use Eq. (1.19) ⃗vAP = ⃗vBP + ⃗vAB as well as the basic equation for motion with constant velocity:” ∆x = v∆t After that, solve each part in turn, and make sure to show all your work! Part (a): Let F stand for the floor frame of reference, and W the walkway frame. In the notation of Section 1.3, we have vF W = 1 m/s. For the first leg of her race, we are told that Ann’s velocity relative to the walkway is 5 m/s, so vW A = 5 m/s. Then, by Eq. (1.19) (with the following change of indices: A →F, B →W, and P →A), vF A = vF W + vW A = 1 m s + 5 m s = 6 m s (1.25) (when you see an equation like this, full of subscripts, it is a good practice to read it out, mentally, to yourself: “Ann’s velocity relative to the floor equals the velocity of the walkway relative to the floor plus Ann’s velocity relative to the walkway.” Then take a moment to see if it makes sense! Here is a place where the picture can be really helpful.) For the return leg, use the same formula, but note that now her velocity relative to the walkway is negative, vW A = −5 m/s, since she is moving in the opposite direction: vF A = vF W + vW A = 1 m s −5 m s = −4 m s (1.26) Part (b): Relative to the floor reference frame, we have just seen that Ann first covers 30 m at a speed of 6 m/s, and then the same 30 m at a speed of 4 m/s, so her total time is ∆tA = 30m 6m/s + 30m 4m/s = 5 s + 7.5 s = 12.5 s (1.27) whereas Becky just runs 30 m at 5 m/s both ways, so it takes her 6 s either way, for a total of 12 s, which means she wins the race. | University Physics I Classical Mechanics_Page_43_Chunk4241 |
26 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY Part (c): The quantity we want is written, in the notation of Section 1.3, vAB (“velocity of Becky relative to Ann”). To calculate this, we just need to know the velocities of both girls in some frame of reference (the same for both!), then subtract Ann’s velocity from Becky’s (this is what Eq. (1.21) is saying). In this case, if we just choose the floor’s reference frame, we have vF A = 6 m/s and vF B = 5 m/s, so vAB = vF B −vF A = 5 m s −6 m s = −1 m s (1.28) The negative sign makes sense: Ann sees Becky falling behind her, so relative to her Becky is moving backwards, which is to say, in the direction we have identified as negative. Part (d): Again we use the same equation, and Becky’s velocity is still the same, but now Ann’s velocity is vF A = −4 m/s (note the negative sign!), so vAB = vF B −vF A = 5 m s − & −4 m s ' = 9 m s (1.29) Part (e): You may find this a bit surprising, but if you think about it the explanation for why Ann lost the race, despite her running at the same speed as Becky relative to the surface she was standing on, has to be that she actually ran a longer distance on that surface! Since she was running for a total of 12.5 s at a constant speed (not velocity!) of 5 m/s in the walkway frame, then in that frame she ran a distance d = |v|∆t = 5 × 12.5 = 62.5 m. That is the total length of walkway that she actually stepped on. | University Physics I Classical Mechanics_Page_44_Chunk4242 |
1.6. PROBLEMS 27 1.6 Problems Problem 1 x (m) t (s) The above figure is the position (in meters) versus time (in seconds) graph of an object in motion. Only the segments between t = 1 s and t = 2 s, and between t = 4 s and t = 5 s, are straight lines. The peak of the curve is at t = 3 s, x = 4 m. Answer the following questions, and provide a brief justification for your answer in every case. (a) At what time(s) is the object’s velocity equal to zero? (b) For what range(s) of times is the object moving with constant velocity? (c) What is the object’s position coordinate at t = 1 s? (d) What is the displacement of the object between t = 1 s and t = 4 s? (e) What is the distance traveled between t = 1 s and t = 4 s? (f) What is the instantaneous velocity of the object at t = 1.5 s? (g) What is its average velocity between t = 1 s and t = 3 s? Problem 2 A particle is initially at xi = 3 m, yi = −5 m, and after a while it is found at the coordinates xf = −4 m, yf = 2 m. (a) On the grid below (next page), draw the initial and final position vectors, and the displacement vector. (b) What are the components of the displacement vector? (c) What are the magnitude and direction of the displacement vector? (You can specify the direction by the angle it makes with either the positive x or the positive y axis.) | University Physics I Classical Mechanics_Page_45_Chunk4243 |
28 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY x y Problem 3 Marshall Dillon is riding at 30 mph after the robber of the Dodge City bank, who has a head start of 15 minutes, but whose horse can only make 25 mph on a good day. How long does it take for Dillon to catch up with the bad guy, and how far from Dodge City are they when this happens? (Assume the road is straight, for simplicity.) Problem 4 The picture below shows the velocity versus time graph of the first 21 seconds of a race between two friends, “Red” and “Green.” (a) Who is ahead at t = 10 s, and by how much? (b) Who passes the 100 m marker first? 5 10 15 20 0 0 5 t (s) v (m/s) | University Physics I Classical Mechanics_Page_46_Chunk4244 |
1.6. PROBLEMS 29 Problem 5 You are trying to pass a truck on the highway. The truck is driving at 55 mph, so you speed up to 60 mph and move over to the left lane. If the truck is 17 m long, and your car is 3 m long (a) how long does it take you to pass the truck completely? (b) How far (along the highway) have you traveled in that time? Note: to answer part (a) look at the problem from the perspective of the truck driver. How far are you going relative to him, and how far would it take you to cover 20 m at that speed? Problem 6 Suppose the position function of a particle moving in one dimension is given by x(t) = 5 + 3t + 2t2 −0.5t3 (1.30) where the coefficients are such that the result will be in meters if you enter the time in seconds. What is the particle’s velocity at t = 2 s? There are two ways you can do this: • If you know calculus, calculate the derivative of (1.30) and evaluate it at t = 2 s. • If you do not yet know how to take derivatives, calculate the limit in the definition (1.8). That is to say, calculate ∆x/∆t with ti = 2 s and ∆t equal, first, to 0.1 s, then to 0.01 s, and then to 0.001 s. You will need to keep more than the usual 4 decimals in the intermediate calculations if you want an accurate result, but you should still report only 3 significant digits in the final result. Problem 7 Suppose you are rowing across a river, as in Figure 1.7. Your speed is 2 miles per hour relative to the current, which is moving at a leisurely 1 mile per hour. If the river is 10 m wide, (a) How far downstream do you end up? (b) To row straight across you would need to have an upstream velocity component (relative to the current). How large would that be? (c) If your rowing speed is still only 2 miles per hour, how long does it take you to row across the river now? | University Physics I Classical Mechanics_Page_47_Chunk4245 |
30 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY | University Physics I Classical Mechanics_Page_48_Chunk4246 |
Chapter 2 Acceleration 2.1 The law of inertia There is something funny about motion with constant velocity: it is indistinguishable from rest. Of course, you can usually tell whether you are moving relative to something else. But if you are enjoying a smooth airplane ride, without looking out the window, you have no idea how fast you are moving, or even, indeed (if the flight is exceptionally smooth) whether you are moving at all. I am actually writing this on an airplane. The flight screen informs me that I am moving at 480 mph relative to the ground, but I do not feel anything like that: just a gentle rocking up and down and sideways that gives me no clue as to what my forward velocity is. If I were to drop something, I know from experience that it would fall on a straight line—relative to me, that is. If it falls from my hand it will land at my feet, just as if we were all at rest. But we are not at rest. In the half second or so it takes for the object to fall, the airplane has moved forward 111 meters relative to the ground. Yet the (hypothetical) object I drop does not land 300 feet behind me! It moves forward with me as it falls, even though I am not touching it. It keeps its initial forward velocity, even though it is no longer in contact with me or anything connected to the airplane. (At this point you might think that the object is still in contact with the air inside the plane, which is moving with the plane, and conjecture that maybe it is the air inside the plane that “pushes forward” on the object as it falls and keeps it from moving backwards. This is not necessarily a dumb idea, but a moment’s reflection will convince you that it is impossible. We are all familiar with the way air pushes on things moving through it, and we know that the force an object experiences depends on its mass and its shape, so if that was what was happening, dropping objects of different masses and shapes I would see them falling in all kind of different ways—as I would, in fact, if I 31 | University Physics I Classical Mechanics_Page_49_Chunk4247 |
32 CHAPTER 2. ACCELERATION were dropping things from rest outdoors in a strong wind. But that is not what we experience on an airplane at all. The air, in fact, has no effect on the forward motion of the falling object. It does not push it in any way, because it is moving at the same velocity. This, in fact, reinforces our previous conclusion: the object keeps its forward velocity while it is falling, in the absence of any external influence.) This remarkable observation is one of the most fundamental principles of physics (yes, we have started to learn physics now!), which we call the law of inertia. It can be stated as follows: in the absence of any external influence (or force) acting on it, an object at rest will stay at rest, while an object that is already moving with some velocity will keep that same velocity (speed and direction of motion)—at least until it is, in fact, acted upon by some force. Please let that sink in for a moment, before we start backtracking, which we have to do now on several accounts. First, I have used repeatedly the term “force,” but I have not defined it properly. Or have I? What if I just said that forces are precisely any “external influences” that may cause a change in the velocity of an object? That will work, I think, until it is time to explore the concept in more detail, a few chapters from now. Next, I need to draw your attention to the fact that the object I (hypothetically) dropped did not actually keep its total initial velocity: it only kept its initial forward velocity. In the downward direction, it was speeding up from the moment it left my hand, as would any other falling object (and as we shall see later in this chapter). But this actually makes sense in a certain way: there was no forward force, so the forward velocity remained constant; there was, however, a vertical force acting all along (the force of gravity), and so the object did speed up in that direction. This observation is, in fact, telling us something profound about the world’s geometry: namely, that forces and velocities are vectors, and laws such as the law of inertia will typically apply to the vector as a whole, as well as to each component separately (that is to say, each dimension of space). This anticipates, in fact, the way we will deal, later on, with motion in two or more dimensions; but we do not need to worry about that for a few chapters still. Finally, it is worth spending a moment reflecting on how radically the law of inertia seems to contradict our intuition about the way the world works. What it seems to be telling us is that, if we throw or push an object, it should continue to move forever with the same speed and in the same direction with which it set out—something that we know is certainly not true. But what’s happening in “real life” is that, just because we have left something alone, it doesn’t mean the world has left it alone. After we lose contact with the object, all sorts of other forces will continue to act on it. A ball we throw, for instance, will experience air resistance or drag (the same effect I was worrying about in that paragraph in parenthesis in the previous page), and that will slow it down. An object sliding on a surface will experience friction, and that will slow it down too. Perhaps the closest thing to the law of inertia in action that you may get to see is a hockey puck sliding on the ice: it is remarkable (perhaps even a bit frightening) to see how little it slows down, but even so the ice does a exert a (very small) frictional force that would bring the puck to a stop eventually. | University Physics I Classical Mechanics_Page_50_Chunk4248 |
2.1. THE LAW OF INERTIA 33 This is why, historically, the law of inertia was not discovered until people started developing an appreciation for frictional forces, and the way they are constantly acting all around us to oppose the relative motion of any objects trying to slide past each other. This mention of relative motion, in a way, brings us full circle. Yes, relative motion is certainly detectable, and for objects in contact it actually results in the occurrence of forces of the frictional, or drag, variety. But absolute (that is, without reference to anything external) motion with constant velocity is fundamentally undetectable. And in view of the law of inertia, it makes sense: if no force is required to keep me moving with constant velocity, it follows that as long as I am moving with constant velocity I should not be feeling any net force acting on me; nor would any other detection apparatus I might be carrying with me. What we do feel in our bodies, and what we can detect with our inertial navigation systems (now you may start to guess why they are called “inertial”), is a change in our velocity, which is to say, our acceleration (to be defined properly in a moment). We rely, ultimately, on the law of inertia to detect accelerations: if my plane is shaking up and down, because of turbulence (as, in fact, it is right now!), the water in my cup may not stay put. Or, rather, the water may try to stay put (really, to keep moving, at any moment, with whatever velocity it has at that moment), but if the cup, which is connected to my hand which is connected, ultimately, to this bouncy plane, moves suddenly out from under it, not all of the water’s parts will be able to adjust their velocities to the new velocity of the cup in time to prevent a spill. This is the next very interesting fact about the physical world that we are about to discover: forces cause accelerations, or changes in velocity, but they do so in different degrees for different objects; and, moreover, the ultimate change in velocity takes time. The first part of this statement has to do with the concept of inertial mass, to be introduced in the next chapter; the second part we are going to explore right now, after a brief detour to define inertial reference frames. 2.1.1 Inertial reference frames The example I just gave you of what happens when a plane in flight experiences turbulence points to an important phenomenon, namely, that there may be times where the law of inertia may not seem to apply in a certain reference frame. By this I mean that an object that I left at rest, like the water in my cup, may suddenly start to move—relative to the reference frame coordinates—even though nothing and nobody is acting on it. More dramatically still, if a car comes to a sudden stop, the passengers may be “projected forward”—they were initially at rest relative to the car frame, but now they find themselves moving forward (always in the car reference frame), to the point that, if they are not wearing seat belts, they may end up hitting the dashboard, or the seat in front of them. | University Physics I Classical Mechanics_Page_51_Chunk4249 |
34 CHAPTER 2. ACCELERATION Again, nobody has pushed on them, and in fact what we can see in this case, from outside the car, is nothing but the law of inertia at work: the passengers were just keeping their initial velocity, when the car suddenly slowed down under and around them. So there is nothing wrong with the law of inertia, but there is a problem with the reference frame: if I want to describe the motion of objects in a reference frame like a plane being shaken up or a car that is speeding up or slowing down, I need to allow for the fact that objects may move—always relative to that frame—in an apparent violation of the law of inertia. The way we deal with this in physics is by introducing the very important concept of an inertial reference frame, by which we mean a reference frame in which all objects will, at all times, be observed to move (or not move) in a way fully consistent with the law of inertia. In other words, the law of inertia has to hold when we use that frame’s own coordinates to calculate the objects’ velocities. This, of course, is what we always do instinctively: when I am on a plane I locate the various objects around me relative to the plane frame itself, not relative to the distant ground. To ascertain whether a frame is inertial or not, we start by checking to see if the description of motion using that frame’s coordinates obeys the law of inertia: does an object left at rest on the counter in the laboratory stay at rest? If set in motion, does it move with constant velocity on a straight line? The Earth’s surface, as it turns out, is not quite a perfect inertial reference frame, but it is good enough that it made it possible for us to discover the law of inertia in the first place! What spoils the inertial-ness of an Earth-bound reference frame is the Earth’s rotation, which, as we shall see later, is an example of accelerated motion. In fact, if you think about the grossly non-inertial frames I have introduced above—the bouncy plane, the braking car—they all have this in common: that their velocities are changing; they are not moving with constant speed on a straight line. So, once you have found an inertial reference frame, to decide whether another one is inertial or not is simple: if it is moving with constant velocity (relative to the first, inertial frame), then it is itself inertial; if not, it is not. I will show you how this works, formally, in a little bit (section 2.2.4, below), after I (finally!) get around to properly introducing the concept of acceleration. It is a fundamental principle of physics that the laws of physics take the same form in all inertial reference frames. The law of inertia is, of course, an example of such a law. Since all inertial frames are moving with constant velocity relative to each other, this is another way to say that absolute motion is undetectable, and all motion is ultimately relative. Accordingly, this principle is known as the principle of relativity. | University Physics I Classical Mechanics_Page_52_Chunk4250 |
2.2. ACCELERATION 35 2.2 Acceleration 2.2.1 Average and instantaneous acceleration Just as we defined average velocity in the previous chapter, using the concept of displacement (or change in position) over a time interval ∆t, we define average acceleration over the time ∆t using the change in velocity: aav = ∆v ∆t = vf −vi tf −ti (2.1) Here, vi and vf are the initial and final velocities, respectively, that is to say, the velocities at the beginning and the end of the time interval ∆t. As was the case with the average velocity, though, the average acceleration is a concept of somewhat limited usefulness, so we might as well proceed straight away to the definition of the instantaneous acceleration (or just “the” acceleration, without modifiers), through the same sort of limiting process by which we defined the instantaneous velocity: a = lim ∆t→0 ∆v ∆t (2.2) Everything that we said in the previous chapter about the relationship between velocity and position can now be said about the relationship between acceleration and velocity. For instance (if you know calculus), the acceleration as a function of time is the derivative of the velocity as a function of time, which makes it the second derivative of the position function: a = dv dt = d2x dt2 (2.3) (and if you do not know calculus yet, do not worry about the superscripts “2” on that last expres- sion! It is just a weird notation that you will learn someday.) Similarly, we can “read off” the instantaneous acceleration from a velocity versus time graph, by looking at the slope of the line tangent to the curve at any point. However, if what we are given is a position versus time graph, the connection to the acceleration is more indirect. Figure 2.1 (next page) provides you with such an example. See if you can guess at what points along this curve the acceleration is positive, negative, or zero. The way to do this “from scratch,” as it were, is to try to figure out what the velocity is doing, first, and infer the acceleration from that. Here is how that would go: Starting at t = 0, and keeping an eye on the slope of the x-vs-t curve, we can see that the velocity starts at zero or near zero and increases steadily for a while, until t is a little bit more than 2 s (let us say, t = 2.2 s for definiteness). That would correspond to a period of positive acceleration, since ∆v would be positive for every ∆t in that range. | University Physics I Classical Mechanics_Page_53_Chunk4251 |
36 CHAPTER 2. ACCELERATION x (m) t (s) 0 5 5 10 Figure 2.1: A possible position vs. time graph for an object whose acceleration changes with time. Between t = 2.2 s and t = 2.5 s, as the object moves from x = 2 m to x = 4 m, the velocity does not appear to change very much, and the acceleration would correspondingly be zero or near zero. Then, around t = 2.5 s, the velocity starts to decrease noticeably, becoming (instantaneously) zero at t = 3 s (x = 6 m). That would correspond to a negative acceleration. Note, however, that the velocity afterwards continues to decrease, becoming more and more negative until around t = 4 s. This also corresponds to a negative acceleration: even though the object is speeding up, it is speeding up in the negative direction, so ∆v, and hence a, is negative for every time interval there. We conclude that a < 0 for all times between t = 2.5 s and t = 4 s. Next, as we just look past t = 4 s, something else interesting happens: the object is still going in the negative direction (negative velocity), but now it is slowing down. Mathematically, that corresponds to a positive acceleration, since the algebraic value of the velocity is in fact increasing (a number like −3 is larger than a number like −5). Another way to think about it is that, if we have less and less of a negative thing, our overall trend is positive. So the acceleration is positive all the way from t = 4 s through t = 5 s (where the velocity is instantaneously zero as the object’s direction of motion reverses), and beyond, until about t = 6 s, since between t = 5 s and t = 6 s the velocity is positive and growing. You can probably figure out on your own now what happens after t = 6 s, reasoning as I did above, but you may also have noticed a pattern that makes this kind of analysis a lot easier. The acceleration (as those with a knowledge of calculus may have understood already), being proportional to the second derivative of the function x(t) with respect to t, is directly related to the curvature of the x-vs-t graph. As figure 2.2 below shows, if the graph is concave (sometimes | University Physics I Classical Mechanics_Page_54_Chunk4252 |
2.2. ACCELERATION 37 called “concave upwards”), the acceleration is positive, whereas it is negative whenever the graph is convex (or “concave downwards”). It is (instantly) zero at those points where the curvature changes (which you may know as inflection points), as well as over stretches of time when the x-vs-t graph is a straight line (motion with constant velocity). x x x t t t a > 0 a < 0 a = 0 Figure 2.2: What the x-vs-t curves look like for the different possible signs of the acceleration. Figure 2.3 (in the next page) shows position, velocity, and acceleration versus time for a hypothetical motion case. Please study it carefully until every feature of every graph makes sense, relative to the other two! You will see many other examples of this in the homework and the lab. Notice that, in all these figures, the sign of x or v at any given time has nothing to do with the sign of a at that same time. It is true that, for instance, a negative a, if sustained for a sufficiently long time, will eventually result in a negative v (as happens, for instance, in Fig. 2.3 over the interval from t = 1 to t = 4 s) but this may take a long time, depending on the size of a and the initial value of v. The graphical clues to follow, instead, are: the acceleration is given by the slope of the tangent to the v-vs-t curve, or the curvature of the x-vs-t curve, as explained in Fig. 2.2; and the velocity is given by the slope of the tangent to the x-vs-t curve. (Note: To make the interpretation of Figure 2.3 simpler, I have chosen the acceleration to be “piecewise constant,” that is to say, constant over extended time intervals and changing in value discontinuously from one interval to the next. This is physically unrealistic: in any real-life situa- tion, the acceleration would be expected to change more or less smoothly from instant to instant. We will see examples of that later on, when we start looking at realistic models of collisions.) | University Physics I Classical Mechanics_Page_55_Chunk4253 |
38 CHAPTER 2. ACCELERATION Position (meters) Acceleration (m/s2) Velocity (m/s) Time (seconds) 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 25 30 35 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 -5 -10 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 -5 -10 Time (seconds) Time (seconds) Figure 2.3: Sample position, velocity and acceleration vs. time graphs for motion with piecewise-constant acceleration. | University Physics I Classical Mechanics_Page_56_Chunk4254 |
2.2. ACCELERATION 39 2.2.2 Motion with constant acceleration A particular kind of motion that is both relatively simple and very important in practice is motion with constant acceleration (see Figure 2.3 again for examples). If a is constant, it means that the velocity changes with time at a constant rate, by a fixed number of m/s each second. (These are, incidentally, the units of acceleration: meters per second per second, or m/s2.) The change in velocity over a time interval ∆t is then given by ∆v = a ∆t (2.4) which can also be written v = vi + a (t −ti) (2.5) Equation (2.5) is the form of the velocity function (v as a function of t) for motion with constant acceleration. This, in turn, has to be the derivative with respect to time of the corresponding position function. If you know simple derivatives, then, you can verify that the appropriate form of the position function must be x = xi + vi(t −ti) + 1 2 a (t −ti)2 (2.6) or in terms of intervals, ∆x = vi∆t + 1 2 a (∆t)2 (2.7) Most often Eq. (2.6) is written with the implicit assumption that the initial value of t is zero: x = xi + vit + 1 2 at2 (2.8) This is simpler, but not as general as Eq. (2.6). Always make sure that you know what conditions apply for any equation you decide to use! As you can see from Eq. (2.5), for intervals during which the acceleration is constant, the velocity vs. time curve should be a straight line. Figure 2.3 (previous page) illustrates this. Equation (2.6), on the other hand, shows that for those same intervals the position vs. time curve should be a (portion of a) parabola, and again this can be seen in Figure 2.3 (sometimes, if the acceleration is small, the curvature of the graph may be hard to see; this happens in Figure 2.3 for the interval between t = 4 s and t = 5 s). The observation that v-vs-t is a straight line when the acceleration is constant provides us with a simple way to derive Eq. (2.7), when combined with the result (from the end of the previous chapter) that the displacement over a time interval ∆t equals the area under the v-vs-t curve for that time interval. Indeed, consider the situation shown in Figure 2.4. The total area under the segment shown is equal to the area of a rectangle of base ∆t and height vi, plus the area of a | University Physics I Classical Mechanics_Page_57_Chunk4255 |
40 CHAPTER 2. ACCELERATION triangle of base ∆t and height vf −vi. Since vf −vi = a∆t, simple geometry immediately yields Eq. (2.7), or its equivalent (2.6). v (m/s) t (s) ti tf vi vf v(t) Figure 2.4: Graphical way to find the displacement for motion with constant acceleration. Lastly, consider what happens if we solve Eq. (2.4) for ∆t and substitute the result in (2.7). We get ∆x = vi∆v a + (∆v)2 2a (2.9) Letting ∆v = vf −vi, a little algebra yields v2 f −v2 i = 2a∆x (2.10) This is a handy little result that can also be seen to follow, more directly, from the work-energy theorems to be introduced in Chapter 71. 1In fact, equation (2.10) turns out to be so handy that you will probably find yourself using it over and over this semester, and you may even be tempted to use it for problems involving motion in two dimensions. However, unless you really know what you are doing, you should resist the temptation, since it is very easy to use (2.10) incorrectly when the acceleration and the displacement do not lie along the same line. You should use the appropriate form of a work-energy theorem instead. | University Physics I Classical Mechanics_Page_58_Chunk4256 |
2.2. ACCELERATION 41 2.2.3 Acceleration as a vector In two (or more) dimensions we introduce the average acceleration vector ⃗aav = ∆⃗v ∆t = 1 ∆t (⃗vf −⃗vi) (2.11) whose components are aav,x = ∆vx/∆t, etc.. The instantaneous acceleration is then the vector given by the limit of Eq. (2.11) as ∆t →0, and its components are, therefore, ax = dvx/dt, ay = dvy/dt, . . . Note that, since ⃗vi and ⃗vf in Eq. (2.11) are vectors, and have to be subtracted as such, the acceler- ation vector will be nonzero whenever ⃗vi and ⃗vf are different, even if, for instance, their magnitudes (which are equal to the object’s speed) are the same. In other words, you have accelerated motion whenever the direction of motion changes, even if the speed does not. As long as we are working in one dimension, I will follow the same convention for the acceleration as the one I introduced for the velocity in Chapter 1: namely, I will use the symbol a, without a subscript, to refer to the relevant component of the acceleration (ax, ay, . . .), and not to the magnitude of the vector ⃗a. 2.2.4 Acceleration in different reference frames In Chapter 1 you saw that the following relation (Eq. (1.19)) holds between the velocities of a particle P measured in two different reference frames, A and B: ⃗vAP = ⃗vAB + ⃗vBP (2.12) What about the acceleration? An equation like (2.12) will hold for the initial and final velocities, and subtracting them we will get ∆⃗vAP = ∆⃗vAB + ∆⃗vBP (2.13) Now suppose that reference frame B moves with constant velocity relative to frame A. In that case, ⃗vAB,f = ⃗vAB,i, so ∆⃗vAB = 0, and then, dividing Eq. (2.13) by ∆t, and taking the limit ∆t →0, we get ⃗aAP = ⃗aBP (for constant ⃗vAB) (2.14) So, if two reference frames are moving at constant velocity relative to each other, observers in both frames measure the same acceleration for any object they might both be tracking. The result (2.14) means, in particular, that if we have an inertial frame then any frame moving at constant velocity relative to it will be inertial too, since the respective observers’ measurements | University Physics I Classical Mechanics_Page_59_Chunk4257 |
42 CHAPTER 2. ACCELERATION will agree that an object’s velocity does not change (otherwise put, its acceleration is zero) when no forces act on it. Conversely, an accelerated frame will not be an inertial frame, because Eq. (2.14) will not hold. This is consistent with the examples I mentioned in Section 2.1 (the bouncing plane, the car coming to a stop). Another example of a non-inertial frame would be a car going around a curve, even if it is going at constant speed, since, as I just pointed out above, this is also an accelerated system. This is confirmed by the fact that objects in such a car tend to move—relative to the car—towards the outside of the curve, even though no actual force is acting on them. 2.3 Free fall An important example of motion with (approximately) constant acceleration is provided by free fall near the surface of the Earth. We say that an object is in “free fall” when the only force acting on it is the force of gravity (the word “fall” here may be a bit misleading, since the object could actually be moving upwards some of the time, if it has been thrown straight up, for instance). The space station is in free fall, but because it is nowhere near the surface of the earth its direction of motion (and hence its acceleration, regarded as a two-dimensional vector) is constantly changing. Right next to the surface of the earth, on the other hand, the planet’s curvature is pretty much negligible and gravity provides an approximately constant, vertical acceleration, which, in the absence of other forces, turns out to be the same for every object, regardless of its size, shape, or weight. The above result—that, in the absence of other forces, all objects should fall to the earth at the same rate, regardless of how big or heavy they are—is so contrary to our common experience that it took many centuries to discover it. The key, of course, as with the law of inertia, is to realize that, under normal circumstances, frictional forces are, in fact, acting all the time, so an object falling through the atmosphere is never really in “free” fall: there is always, at a minimum, and in addition to the force of gravity, an air drag force that opposes its motion. The magnitude of this force does depend on the object’s size and shape (basically, on how “aerodynamic” the object is); and thus a golf ball, for instance, falls much faster than a flat sheet of paper. Yet, if you crumple up the sheet of paper till it has the same size and shape as the golf ball, you can see for yourself that they do fall at approximately the same rate! The equality can never be exact, however, unless you get rid completely of air drag, either by doing the experiment in an evacuated tube, or (in a somewhat extreme way), by doing it on the surface of the moon, as the Apollo 15 astronauts did with a hammer and a feather back in 19712. This still leaves us with something of a mystery, however: the force of gravity is the only force known to have the property that it imparts all objects the same acceleration, regardless of their mass or constitution. A way to put this technically is that the force of gravity on an object is 2The video of this is available online: https://www.youtube.com/watch?v=oYEgdZ3iEKA. It is, however, pretty low resolution and hard to see. A very impressive modern-day demonstration involving feathers and a bowling ball in a completely evacuated (airless) room is available here: https://www.youtube.com/watch?v=E43-CfukEgs. | University Physics I Classical Mechanics_Page_60_Chunk4258 |
2.3. FREE FALL 43 proportional to that object’s inertial mass, a quantity that we will introduce properly in the next chapter. For the time being, we will simply record here that this acceleration, near the surface of the earth, has a magnitude of approximately 9.8 m/s2, a quantity that is denoted by the symbol g. Thus, if we take the upwards direction as positive (as is usually done), we get for the acceleration of an object in free fall a = −g, and the equations of motion become ∆v = −g ∆t (2.15) ∆y = vi∆t −1 2 g (∆t)2 (2.16) where I have used y instead of x for the position coordinate, since that is a more common choice for a vertical axis. Note that we could as well have chosen the downward direction as positive, and that may be a more natural choice in some problems. Regardless, the quantity g is always defined to be positive: g = 9.8 m/s2. The acceleration, then, is g or −g, depending on which direction we take to be positive. In practice, the value of g changes a little from place to place around the earth, for various reasons (it is somewhat sensitive to the density of the ground below you, and it decreases as you climb higher away from the center of the earth). In a later chapter we will see how to calculate the value of g from the mass and radius of the earth, and also how to calculate the equivalent quantity for other planets. In the meantime, we can use equations like (2.15) and (2.16) (as well as (2.10), with the appropriate substitutions) to answer a number of interesting questions about objects thrown or dropped straight up or down (always, of course, assuming that air drag is negligible). For instance, back at the beginning of this chapter I mentioned that if I dropped an object it might take about half a second to hit the ground. If you use Eq. (2.16) with vi = 0 (since I am dropping the object, not throwing it down, its initial velocity is zero), and substitute ∆t = 0.5 s, you get ∆y = 1.23 m (about 4 feet). This is a reasonable height from which to drop something. On the other hand, you may note that half a second is not a very long time in which to make accurate observations (especially if you do not have modern electronic equipment), and as a result of that there was considerable confusion for many centuries as to the precise way in which objects fell. Some people believed that the speed did increase in some way as the object fell, while others appear to have believed that an object dropped would “instantaneously” (that is, at soon as it left your hand) acquire some speed and keep that unchanged all the way down. In reality, in the presence of air drag, what happens is a combination of both: initially the speed increases at an approximately constant rate (free, or nearly free fall), but the drag force increases with the speed as well, until eventually it balances out the force of gravity, and from that point on the speed does not increase anymore: we say that the object has reached “terminal velocity.” Some objects reach terminal velocity almost instantly, whereas others (the more “aerodynamic” ones) may take a long time to do so. This accounts for the confusion that prevailed before Galileo’s experiments in the early 1600’s. | University Physics I Classical Mechanics_Page_61_Chunk4259 |
44 CHAPTER 2. ACCELERATION Galileo’s main insight, on the theoretical side, was the realization that it was necessary to separate clearly the effect of gravity and the effect of the drag force. Experimentally, his big idea was to use an inclined plane to slow down the “fall” of an object, so as to make accurate measurements possible (and also, incidentally, reduce the air drag force!). These “inclined planes” were just basically ramps down which he sent small balls (like marbles) rolling. By changing the steepness of the ramp he could control how slowly the balls moved. He reasoned that, ultimately, the force that made the balls go down was essentially the same force of gravity, only not the whole force, but just a fraction of it. Today we know that, in fact, an object sliding (not rolling!) up or down on a frictionless incline will experience an acceleration directed downwards along the incline and with a magnitude equal to g sin θ, where θ is the angle that the slope makes with the horizontal: a = g sin θ (inclined plane, taking downwards to be positive) (2.17) (for some reason, it seems more natural, when dealing with inclined planes, to take the downward direction as positive!). Equation (2.17) makes sense in the two extreme cases in which the plane is completely vertical (θ = 90◦, a = g) and completely horizontal (θ = 0◦, a = 0). For intermediate values, you will carry out experiments in the lab to verify this result. We will show, in a later chapter, how Eq. (2.17) comes about from a careful consideration of all the forces acting on the object; we will also see, later on, how it needs to be modified for the case of a rolling, rather than a sliding, object. This modification does not affect Galileo’s main conclusion, which was, basically, that the natural falling motion in the absence of friction or drag forces is motion with constant acceleration (at least, near the surface of the earth, where g is constant to a very good approximation). 2.4 In summary 1. The law of inertia states that, if no external influences (forces) are acting on an object, then, if the object is initially at rest it will stay at rest, and if it is initially moving it will continue to move with constant velocity (unchanging speed and direction). 2. Reference frames in which the law of inertia is seen to hold (when the velocities of objects are calculated from their coordinates in that frame) are called inertial. A reference frame that is moving at constant velocity relative to an inertial frame is also an inertial frame. Conversely, accelerated reference frames are non-inertial. 3. Motion with constant velocity is fundamentally indistinguishable from no motion at all (i.e., rest). As long as the velocity (of the objects involved) does not change, only relative motion can be detected. This is known as the principle of relativity. Another way to state it is that the laws of physics must take the same form in all inertial reference frames (so you cannot single out one as being in “absolute rest” or “absolute motion”). | University Physics I Classical Mechanics_Page_62_Chunk4260 |
2.4. IN SUMMARY 45 4. Changes in velocity are detectable, and, by (1) above, are evidence of unbalanced forces acting on an object. 5. The rate of change of an object’s velocity is the object’s acceleration: the average acceleration over a time interval ∆t is aav = ∆v/∆t, and the instantaneous acceleration at a time t is the limit of the average acceleration calculated for successively shorter time intervals ∆t, all with the same initial time ti = t. Mathematically, this means the acceleration is the derivative of the velocity function, a = dv/dt. 6. In a velocity versus time graph, the acceleration can be read from the slope of the line tangent to the curve (just like the velocity in a position versus time graph). 7. In a position versus time graph, the regions with positive acceleration correspond to a concave curvature (like a parabola opening up), and those with negative acceleration correspond to a convex curvature (like a parabola opening down). Points of inflection (where the curvature changes) and straight lines correspond to points where the acceleration is zero. 8. The basic equations used to describe motion with constant acceleration are (2.4), (2.7) and (2.10) above. Alternative forms of these are also provided in the text. 9. In more than one dimension, a change in the direction of the velocity vector results in a nonzero acceleration, even if the object’s speed does not change. 10. An object is said to be in free fall when the only force acting on it is gravity. All objects in free fall experience the same acceleration at the same point in their motion, regardless of their mass or composition. Near the surface of the earth, this acceleration is approximately constant and has a magnitude g = 9.8 m/s2. 11. An object sliding on a frictionless inclined plane experiences (if air drag is negligible) an acceleration directed downward along the incline and with a magnitude g sin θ, where θ is the angle the incline makes with the horizontal. | University Physics I Classical Mechanics_Page_63_Chunk4261 |
46 CHAPTER 2. ACCELERATION 2.5 Examples 2.5.1 Motion with piecewise constant acceleration Construct the position vs. time, velocity vs. time, and acceleration vs. time graphs for the motion described below. For each of the intervals (a)–(d) you’ll need to figure out the position (height) and velocity of the rocket at the beginning and the end of the interval, and the acceleration for the interval. In addition, for interval (b) you need to figure out the maximum height reached by the rocket and the time at which it occurs. For interval (d) you need to figure out its duration, that is to say, the time at which the rocket hits the ground. (a) A rocket is shot upwards, accelerating from rest to a final velocity of 20 m/s in 1 s as it burns its fuel. (Treat the acceleration as constant during this interval.) (b) From t = 1 s to t = 4 s, with the fuel exhausted, the rocket flies under the influence of gravity alone. At some point during this time interval (you need to figure out when!) it stops climbing and starts falling. (c) At t = 4 s a parachute opens, suddenly causing an upwards acceleration (again, treat it as constant) lasting 1 s; at the end of this interval, the rocket’s velocity is 5 m/s downwards. (d) The last part of the motion, with the parachute deployed, is with constant velocity of 5 m/s downwards until the rocket hits the ground. Solution: (a) For this first interval (for which I will use a subscript “1” throughout) we have ∆y1 = 1 2a1(∆t1)2 (2.18) using Eq. (2.6) for motion with constant acceleration with zero initial velocity (I am using the variable y, instead of x, for the vertical coordinate; this is more or less customary, but, of course, I could have used x just as well). Since the acceleration is constant, it is equal to its average value: a1 = ∆v ∆t = 20m s2 . Substituting this into (2.18) we get the height at t = 1 s is 10 m. The velocity at that time, of course, is vf1 = 20 m/s, as we were told in the statement of the problem. (b) This part is free fall with initial velocity vi2 = 20 m/s. To find how high the rocket climbs, use Eq. (2.15) in the form vtop −vi2 = −g(ttop −ti2), with vtop = 0 (as the rocket climbs, its velocity decreases, and it stops climbing when its velocity is zero). This gives us ttop = 3.04 s as the time at | University Physics I Classical Mechanics_Page_64_Chunk4262 |
2.5. EXAMPLES 47 which the rocket reaches the top of its trajectory, and then starts coming down. The corresponding displacement is, by Eq. (2.16), ∆ytop = vi2(ttop −ti2) −1 2g(ttop −ti2)2 = 20.4 m so the maximum height it reaches is 30.4 m. At the end of the full 3-second interval, the rocket’s displacement is ∆y2 = vi2∆t2 −1 2g(∆t2)2 = 15.9 m (so its height is 25.9 m above the ground), and the final velocity is vf2 = vi2 −g∆t2 = −9.43 m s . (c) The acceleration for this part is (vf3 −vi3)/∆t3 = (−5+9.43)/1 = 4.43 m/s2. Note the positive sign. The displacement is ∆y3 = −9.43 × 1 + 1 2 × 4.43 × 12 = −7.22 m so the final height is 25.9 −7.21 = 18.7 m. (d) This is just motion with constant speed to cover 18.7 m at 5 m/s. The time it takes is 3.74 s. The graphs for this motion are shown earlier in the chapter, in Figure 2.3. | University Physics I Classical Mechanics_Page_65_Chunk4263 |
48 CHAPTER 2. ACCELERATION 2.6 Problems Problem 1 You get on your bicycle and ride it with a constant acceleration of 0.5 m/s2 for 20 s. After that, you continue riding at a constant velocity for a distance of 200 m. Finally, you slow to a stop, with a constant acceleration, over a distance of 20 m. v (m/s) t (s) 0 0 x (m) t (s) 0 0 a (m/s2) t (s) 0 | University Physics I Classical Mechanics_Page_66_Chunk4264 |
2.6. PROBLEMS 49 (a) How far did you travel while you were accelerating at 0.5 m/s2, and what was your velocity at the end of that interval? (b) After that, how long did it take you to cover the next 200 m? (c) What was your acceleration while you were slowing down to a stop, and how long did it take you to come to a stop? (d) Considering the whole trip, what was your average velocity? (e) Plot the position versus time, velocity versus time, and acceleration versus time graphs for the whole trip, in the grids provided above. Values at the beginning and end of each interval must be exact. Slopes and curvatures must be represented accurately. Do not draw any of the curves beyond the time the rider stops (or, if you do, make sure what you draw makes sense!). Problem 2 You throw an object straight upwards and catch it again, when it comes down to the same initial height, 2 s later. (a) How high did it rise above its initial height? (b) With what initial speed did you throw it? (Note: for this problem you should use the fact that, if air drag is negligible, the object will return to its initial height with the same speed it had initially.) Problem 3 You are trying to catch up with a car that is in front of you on the highway. Initially you are both moving at 25 m/s, and the distance between you is 100 m. You step on the gas and sustain a constant acceleration for a time ∆t = 30 s, at which point you have pulled even with the other car. (a) What is 25 m/s, in miles per hour? (b) What was your acceleration over the 30 s time interval? (c) How fast were you going when you caught up with the other car? Problem 4 Go back to Problem 4 of Chapter 1, and use the information in the figure to draw an accurate position vs. time graph for both runners. Problem 5 A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0 s with a constant negative acceleration of −1.5 m/s2 with respect to her direction of motion. She then slides up another icy slope that makes a 20◦angle with the horizontal. (a) How fast was the child going when she reached the bottom of the first slope? How long did it take her to get there? (b) How long was the flat stretch at the bottom? (c) How fast was the child going as she started up the second slope? (d) How far up the second slope did she slide? | University Physics I Classical Mechanics_Page_67_Chunk4265 |
50 CHAPTER 2. ACCELERATION | University Physics I Classical Mechanics_Page_68_Chunk4266 |
Chapter 3 Momentum and Inertia 3.1 Inertia In everyday language, we speak of something or someone “having a large inertia” to mean, essen- tially, that they are very difficult to set in motion. This usage of the word “inertia” is consistent with the “law of inertia” we introduced in the previous chapter (which states, among other things, that an object at rest, if left to itself, will just remain at rest), but it goes a bit beyond that by trying to quantify just how hard it may be to get the object to move. We do know, from experience, that lighter objects are easier to set in motion than heavier objects, but most of us probably have an intuition that gravity (the force that pulls an object towards the earth and hence determines its weight) is not involved in an essential way here. Imagine, for instance, the difference between slapping a volleyball and a bowling ball. It is not hard to believe that the latter would hurt as much if we did it while floating in free fall in the space station (in a state of effective “weightlessness”) as if we did it right here on the surface of the earth. In other words, it is not (necessarily) how heavy something feels, but just how massive it is. But just what is this “massiveness” quality that we associate intuitively with a large inertia? Is there a way (other than resorting to the weight again) to assign to it a numerical value? 3.1.1 Relative inertia and collisions One possible way to determine the relative inertias of two objects, conceptually, at least, is to try to use one of them to set the other one in motion. Most of us are familiar with what happens 51 | University Physics I Classical Mechanics_Page_69_Chunk4267 |
52 CHAPTER 3. MOMENTUM AND INERTIA when two identical objects (presumably, therefore, having the same inertia) collide: if the collision is head-on (so the motion, before and after, is confined to a straight line), they basically exchange velocities. For instance, a billiard ball hitting another one will stop dead and the second one will set offwith the same speed as the first one. The toy sometimes called “Newton’s balls” or “Newton’s cradle” also shows this effect. Intuitively, we understand that what it takes to stop the first ball is exactly the same as it would take to set the second one in motion with the same velocity. But what if the objects colliding have different inertias? We expect that the change in their velocities as a result of the collision will be different: the velocity of the object with the largest inertia will not change very much, and conversely, the change in the velocity of the object with the smallest inertia will be comparatively larger. A velocity vs. time graph for the two objects might look somewhat like the one sketched in Fig. 3.1. -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0 2 4 6 8 10 1 1 2 2 1 m/s v (m/s) t (ms) Figure 3.1: An example of a velocity vs. time graph for a collision of two objects with different inertias. In this picture, object 1, initially moving with velocity v1i = 1 m/s, collides with object 2, initially at rest. After the collision, which here is assumed to take a millisecond or so, object 1 actually bounces back, so its final velocity is v1f = −1/3 m/s, whereas object 2 ends up moving to the right with velocity v2f = 2/3 m/s. So the change in the velocity of object 1 is ∆v1 = v1f −v1i = −4/3 m/s, whereas for object 2 we have ∆v2 = v2f −v2i = 2/3 m/s. It is tempting to use this ratio, ∆v1/∆v2, as a measure of the relative inertia of the two objects, only we’d want to use it upside down and with the opposite sign: that is, since ∆v2/∆v1 = −1/2 we would say that object 2 has twice the inertia of object 1. But then we have to ask: is this a | University Physics I Classical Mechanics_Page_70_Chunk4268 |
3.1. INERTIA 53 reliable, repeatable measure? Will it work for any kind of collision (within reason, of course: we clearly need to stay in one dimension, and eliminate external influences such as friction), and for any initial velocity? To begin with, we have reason to expect that it does not matter whether we shoot object 1 towards object 2 or object 2 towards object 1, because we learned in the previous chapter that only relative motion is detectable, and the relative motion is the same in both cases. Consider, for instance, what the collision in Figure 3.1 appears like to a hypothetical observer moving along with object 1, at 1 m/s. To him, object 1 appears to be at rest, and it is object 2 that is coming towards him, with a velocity of −1 m/s. To see what the outcome of the collision looks like to him, just add the same −1 m/s to the final velocities we obtained before: object 1 will end up moving at v1f = −4/3 m/s, and object 2 would move at v2f = −1/3 m/s, and we would have a situation like the one shown in Figure 3.2, where both curves have simply been shifted down by 1 m/s: -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0 2 4 6 8 10 1 2 v (m/s) t (ms) 1 2 −1 m/s Figure 3.2: Another example (really the same collision as in Figure 1, only as seen by an observer initially moving to the right at 1 m/s). But then, this is exactly what we should expect to find also in our laboratory if we actually did send the second object at 1 m/s towards the first one sitting at rest. All the individual velocities have changed relative to Figure 3.1, but the velocity changes, ∆v1 and ∆v2, are clearly still the same, and therefore so is our (tentative) measure of the objects’ relative inertia. Clearly, the same argument can be used to conclude that the same result will be obtained when both objects are initially moving towards each other, as long as their relative velocity is the same as | University Physics I Classical Mechanics_Page_71_Chunk4269 |
54 CHAPTER 3. MOMENTUM AND INERTIA in these examples, namely, 1 m/s. However, unless we do the experiments we cannot really predict what will happen if we increase (or decrease) their relative velocity. In fact, we could imagine smashing the two objects at very high speed, so they might even become seriously mangled in the process. Yet, experimentally (and this is not at all an obvious result!), we would still find the same value of −1/2 for the ratio ∆v2/∆v1, at least as long as the collision is not so violent that the objects actually break up into pieces. Perhaps the most surprising result of our experiments would be the following: imagine that the objects have a “sticky” side (for instance, the small black rectangles shown in the pictures could be strips of Velcro), and we turn them around so that when they collide they will end up stuck to each other. In this case (which, as we shall see later, is termed a completely inelastic collision), the v-vs-t graph might look like Figure 3.3 below. Now the two objects end up moving together to the right, fairly slowly: v1f = v2f = 1/3 m/s. The velocity changes are ∆v1 = −2/3 m/s and ∆v2 = 1/3 m/s, both of which are different from what they were before, in Figs. 3.1 and 3.2: yet, the ratio ∆v2/∆v1 is still equal to −1/2, just as in all the previous cases. 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 1 2 1 m/s 1 2 v (m/s) t (ms) Figure 3.3: What would happen if the objects in Figure 1 became stuck together when they collided. | University Physics I Classical Mechanics_Page_72_Chunk4270 |
3.1. INERTIA 55 3.1.2 Inertial mass: definition and properties At this point, it would seem reasonable to assume that this ratio, ∆v2/∆v1, is, in fact, telling us something about an intrinsic property of the two objects, what we have called above their “relative inertia.” It is easy, then, to see how one could assign a value to the inertia of any object (at least, conceptually): choose a “standard” object, and decide, arbitrarily, that its inertia will have the numerical value of 1, in whichever units you choose for it (these units will turn out, in fact, to be kilograms, as you will see in a minute). Then, to determine the inertia of another object, which we will label with the subscript 1, just arrange a one-dimensional collision between object 1 and the standard, under the right conditions (basically, no net external forces), measure the velocity changes ∆v1 and ∆vs, and take the quantity −∆vs/∆v1 as the numerical value of the ratio of the inertia of object 1 to the inertia of the standard object. In symbols, using the letter m to represent an object’s inertia, m1 ms = −∆vs ∆v1 (3.1) But, since ms = 1 by definition, this gives us directly the numerical value of m1. The reason we use the letter m is, as you must have guessed, because, in fact, the inertia defined in this way turns out to be identical to what we have traditionally called “mass.” More precisely, the quantity defined this way is an object’s inertial mass. The remarkable fact, mentioned earlier, that the force of gravity between two objects turns out to be proportional to their inertial masses, allows us to determine the inertial mass of an object by the more traditional procedure of simply weighing it, rather than elaborately staging a collision between it and the standard kilogram on an ice-hockey rink. But, in principle, we could conceive of the existence of two different quantities that should be called “inertial mass” and “gravitational mass,” and the identity (or more precisely, the—so far as we know—exact proportionality) of the two is a rather mysterious experimental fact1. In any case, by the way we have constructed it, the inertial mass, defined as in Eq. (3.1), does capture, in a quantitative way, the concept that we were trying to express at the beginning of the chapter: namely, how difficult it may be to set an object in motion. In principle, however, other experiments would need to be conducted to make sure that it does have the properties we have traditionally associated with the concept of mass. For instance, suppose we join together two objects of mass m. Is the mass of the resulting object 2m? Collision experiments would, indeed, show this to be the case with great accuracy in the macroscopic world (with which we are concerned this semester), but this is a good example of how you cannot take anything for granted: at the microscopic level, it is again a fact that the inertial mass of an atomic nucleus is a little less than the sum of the masses of all its constituent protons and neutrons2. Probably the last thing that would need to be checked is that the ratio of inertias is independent 1This fact, elevated to the category of a principle by Einstein (the equivalence principle) is the starting point of the general theory of relativity. 2And this is not just an unimportant bit of trivia: all of nuclear power depends on this small difference. | University Physics I Classical Mechanics_Page_73_Chunk4271 |
56 CHAPTER 3. MOMENTUM AND INERTIA of the standard. Suppose that we have two objects, to which we have assigned masses m1 and m2 by arranging for each to collide with the “standard object” independently. If we now arrange for a collision between objects 1 and 2 directly, will we actually find that the ratio of their velocity changes is given by the ratio of the separately determined masses m1 and m2? We certainly would need that to be the case, in order for the concept of inertia to be truly useful; but again, we should not assume anything until we have tested it! Fortunately, the tests would indeed reveal that, in every case, the expected relationship holds3 −∆v2 ∆v1 = m1 m2 (3.2) At this point, we are not just in possession of a useful definition of inertia, but also of a veritable law of nature, as I will explain next. 3.2 Momentum For an object of (inertial) mass m moving, in one dimension, with velocity v, we define its momen- tum as p = mv (3.3) (the choice of the letter p for momentum is apparently related to the Latin word “impetus”). We can think of momentum as a sort of extension of the concept of inertia, from an object at rest to an object in motion. When we speak of an object’s inertia, we typically think about what it may take to get it moving; when we speak of its momentum, we typically think of that it may take to stop it (or perhaps deflect it). So, both the inertial mass m and the velocity v are involved in the definition. We may also observe that what looks like inertia in some reference frame may look like momentum in another. For instance, if you are driving in a car towing a trailer behind you, the trailer has only a large amount of inertia, but no momentum, relative to you, because its velocity relative to you is zero; however, the trailer definitely has a large amount of momentum (by virtue of both its inertial mass and its velocity) relative to somebody standing by the side of the road. 3.2.1 Conservation of momentum; isolated systems For a system of objects, we treat the momentum as an additive quantity. So, if two colliding objects, of masses m1 and m2, have initial velocities v1i and v2i, we say that the total initial momentum of 3Equation 3.2 actually is found to hold also at the microscopic (or quantum) level, although there we prefer to state the result by saying that conservation of momentum holds (see the following section). | University Physics I Classical Mechanics_Page_74_Chunk4272 |
3.2. MOMENTUM 57 the system is pi = m1v1i + m2v2i, and similarly if the final velocities are v1f and v2f, the total final momentum will be pf = m1v1f + m2v2f. We then assert that the total momentum of the system is not changed by the collision. Mathemat- ically, this means pi = pf (3.4) or m1v1i + m2v2i = m1v1f + m2v2f (3.5) But this last equation, in fact, follows directly from Eq. (3.2): to see this, move all the quantities in Eq. (3.5) having to do with object 1 to one side of the equal sign, and those having to do with object 2 to the other side. You then get m1 (v1i −v1f) = m2 (v2f −v2i) −m1∆v1 = m2∆v2 (3.6) which is just another way to write Equation (3.2). Hence, the result (3.2) ensures the conservation of the total momentum of a system of any two interacting objects (“particles”), regardless of the form the interaction takes, as long as there are no external forces acting on them. Momentum conservation is one of the most important principles in all of physics, so let me take a little time to explain how we got here and elaborate on this result. First, as I just mentioned, we have been more or less implicitly assuming that the two interacting objects form an isolated system, by which we mean that, throughout, they interact with nothing other than each other. (Equivalently, there are no external forces acting on them.) It is pretty much impossible to set up a system so that it is really isolated in this strict sense; instead, in practice, we settle for making sure that the external forces on the two objects cancel out. This is what happens on the air tracks with which you will be doing experiments this semester: gravity is acting on the carts, but that force is balanced out by the upwards push of the air from the track. A system on which there is no net external force is as good as isolated for practical purposes, and we will refer to it as such. (It is harder, of course, to completely eliminate friction and drag forces, so we just have to settle for approximately isolated systems in practice.) Secondly, we have assumed so far that the motion of the two objects is restricted to a straight line—one dimension. In fact, momentum is a vector quantity (just like velocity is), so in general we should write ⃗p = m⃗v and conservation of momentum, in general, holds as a vector equation for any isolated system in three dimensions: ⃗pi = ⃗pf (3.7) | University Physics I Classical Mechanics_Page_75_Chunk4273 |
58 CHAPTER 3. MOMENTUM AND INERTIA What this means, in turn, is that each separate component (x, y and z) of the momentum will be separately conserved (so Eq. (3.7) is equivalent to three scalar equations, in three dimensions). When we get to study the vector nature of forces, we will see an interesting implication of this, namely, that it is possible for one component of the momentum vector to be conserved, but not another—depending on whether there is or there isn’t a net external force in that direction or not. For example, anticipating things a bit, when you throw an object horizontally, as long as you can ignore air drag, there is no horizontal force acting on it, and so that component of the momentum vector is conserved, but the vertical component is changing all the time because of the (vertical) force of gravity. Thirdly, although this may not be immediately obvious, for an isolated system of two colliding objects the momentum is truly conserved throughout the whole collision process. It is not just a matter of comparing the initial and final velocities: at any of the times shown in Figures 1 through 3, if we were to measure v1 and v2 and compute m1v1 + m2v2, we would obtain the same result. In other words, the total momentum of an isolated system is constant: it has the same value at all times. Finally, all these examples have involved interactions between only two particles. Can we really generalize this to conclude that the total momentum of an isolated system of any number of particles is constant, even when all the particles may be interacting with each other simultaneously? Here, again, the experimental evidence is overwhelmingly in favor of this hypothesis4, but much of our confidence on its validity comes in fact from a consideration of the nature of the internal interactions themselves. It is a mathematical fact that all of the interactions so far known to physics have the property of conserving momentum, whether acting individually or simultaneously. No experiments have ever suggested the existence of an interaction that does not have this property. 3.3 Extended systems and center of mass Consider a collection of particles with masses m1, m2, . . ., and located, at some given instant, at positions x1, x2 . . .. (We are still, for the time being, considering only motion in one dimension, but all these results generalize easily to three dimensions.) The center of mass of such a system is a mathematical point whose position coordinate is given by xcm = m1x1 + m2x2 + . . . m1 + m2 + . . . (3.8) Clearly, the denominator of (3.8) is just the total mass of the system, which we may just denote by M. If all the particles have the same mass, the center of mass will be somehow “in the middle” 4For an important piece of indirect evidence, just consider that any extended object is in reality a collection of interacting particles, and the experiments establishing conservation of momentum almost always involve such extended objects. See the following section for further thoughts on this matter. | University Physics I Classical Mechanics_Page_76_Chunk4274 |
3.3. EXTENDED SYSTEMS AND CENTER OF MASS 59 of all of them; otherwise, it will tend to be closer to the more massive particle(s). The “particles” in question could be spread apart, or they could literally be the “parts” into which we choose to subdivide, for computational purposes, a single extended object. If the particles are in motion, the position of the center of mass, xcm, will in general change with time. For a solid object, where all the parts are moving together, the displacement of the center of mass will just be the same as the displacement of any part of the object. In the most general case, we will have (by subtracting xcmi from xcmf) ∆xcm = 1 M (m1∆x1 + m2∆x2 + . . .) (3.9) Dividing Eq. (3.9) by ∆t and taking the limit as ∆t →0, we get the instantaneous velocity of the center of mass: vcm = 1 M (m1v1 + m2v2 + . . .) (3.10) But this is just vcm = psys M (3.11) where psys = m1v1 + m2v2 + . . . is the total momentum of the system. 3.3.1 Center of mass motion for an isolated system Equation (3.11) is a very interesting result. Since the total momentum of an isolated system is constant, it tells us that the center of mass of an isolated system of particles moves at constant velocity, regardless of the relative motion of the particles themselves or their possible interactions with each other. As indicated above, this generalizes straightforwardly to more than one dimension, so we can write ⃗vcm = ⃗psys/M. Thus, we can say that, for an isolated system in space, not only the speed, but also the direction of motion of its center of mass does not change with time. Clearly this result is a sort of generalization of the law of inertia. For a solid, extended object, it does, in fact, provide us with the precise form that the law of inertia must take: in the absence of external forces, the center of mass will just move on a straight line with constant velocity, whereas the object itself may be moving in any way that does not affect the center of mass trajectory. Specifically, the most general motion of an isolated rigid body is a straight line motion of its center of mass at constant speed, combined with a possible rotation of the object as a whole around the center of mass. For a system that consists of separate parts, on the other hand, the center of mass is generally just a point in space, which may or may not coincide at any time with the position of any of the parts, but which will nonetheless move at constant velocity as long as the system is isolated. This is illustrated by Figure 3.4, where the position vs. time curves have been drawn for the colliding | University Physics I Classical Mechanics_Page_77_Chunk4275 |
60 CHAPTER 3. MOMENTUM AND INERTIA objects of Figure 3.1. I have assumed that object 1 starts out at x1i = −5 mm at t = 0, and object 2 starts at x2i = 0 at t = 0. Because object 2 has twice the inertia of object 1, the position of the center of mass, as given by Eq. (3.8), will always be xcm = x1/3 + 2x2/3 that is to say, the center of mass will always be in between objects 1 and 2, and its distance from object 2 will always be half its distance to object 1: |xcm −x1| = 2 3|x1 −x2| |xcm −x2| = 1 3|x1 −x2| Figure 4 shows that this simple prescription does result in motion with constant velocity for the center of mass (the green straight line), even though the x-vs-t curves of the two objects themselves look relatively complicated. (Please check it out! Take a ruler to Fig. 3.4 and verify that the center of mass is, at every instant, where it is supposed to be.) -5 -4 -3 -2 -1 0 1 2 3 4 0 2 4 6 8 10 x1(t) x2(t) xcm x (mm) t (ms) Figure 3.4: Position vs. time graph for the objects colliding in Figure 1. The green line shows the position of the center of mass as a function of time. The concept of center of mass gives us an important way to simplify the description of the motion of potentially complicated systems. We will make good use of it in forthcoming chapters. A very nice demonstration of the motion of the center of mass in two-body one-dimensional collisions can be found at https://phet.colorado.edu/sims/collision-lab/collision-lab_en.html (you need to check the “center of mass” box to see it). | University Physics I Classical Mechanics_Page_78_Chunk4276 |
3.4. IN SUMMARY 61 3.3.2 Recoil and rocket propulsion As we have just seen, you cannot alter the motion of your center of mass without relying on some external force—which is to say, some kind of external support. This is actually something you may have experienced when you are resting on a very slippery surface and you just cannot “get a grip” on it. There is, however, one way to circumvent this problem which, in fact, relies on conservation of momentum itself: if you are carrying something with you, and can throw it away from you at high speed, you will recoil as a result of that. If you can keep throwing things, you (with your store of as yet unthrown things) will speed up a little more every time. This is, in essence, the principle behind rocket propulsion. Mathematically, consider two objects, of masses m1 and m2, initially at rest, so their total mo- mentum is zero. If mass 1 is thrown away from mass 2 with a speed v1f, then, by conservation of momentum (always assuming the system is isolated) we must have 0 = m1v1f + m2v2f (3.12) and therefore v2f = −m1v1f/m2. This is how a rocket moves forward, by constantly expelling mass (the hot exhaust gas) backwards at a high velocity. Note that, even though both objects move, the center of mass of the whole system does not (in the absence of any external force), as discussed above. 3.4 In summary 1. The inertia of an object is a measure of its tendency to resist changes in its motion. It is quantified by the inertial mass (measured in kilograms). 2. A system of objects is called isolated (for practical purposes) when there are no net (or unbalanced) external forces acting on any of the objects (the objects may still interact with each other). 3. When two objects forming an isolated system collide in one dimension, the changes in their velocities are inversely proportional to their inertial masses: ∆v1 ∆v2 = −m2 m1 This may be used, in principle, as a way to define the inertial mass operationally. 4. The inertial mass thus defined turns out to be exactly (as far as we know) proportional to the object’s gravitational mass, which determines the gravitational force of attraction between it and any other object. For this reason, most often we measure an object’s inertial mass simply by weighing it. | University Physics I Classical Mechanics_Page_79_Chunk4277 |
62 CHAPTER 3. MOMENTUM AND INERTIA 5. The momentum of an object of inertial mass m moving with a velocity ⃗v is defined as ⃗p = m⃗v. The total momentum of a system of objects is defined as the (vector) sum of all the individual momenta. 6. (Conservation of momentum) The momentum of an isolated system remains always con- stant, regardless of how the parts that make up the system may interact with one another. 7. In one dimension, the center of mass of a system of particles is a mathematical point whose x coordinate is given by Equation (3.8) above. (In more dimensions, just change the x’s in Eq. (3.8) to y and z to get ycm and zcm.) 8. The center of mass of a system always moves with a velocity ⃗vcm = ⃗psys M where ⃗psys is the total momentum of the system, and M its total mass. 9. It follows from 8 and 6 above that for an isolated system, the center of mass must always be at rest or moving with constant velocity. This result generalizes the law of inertia to extended objects, or systems of particles. | University Physics I Classical Mechanics_Page_80_Chunk4278 |
3.5. EXAMPLES 63 3.5 Examples 3.5.1 Reading a collision graph The graph shows a collision between two carts (possibly equipped with magnets so that they repel each other before they actually touch) on an air track. The inertia (mass) of cart 1 is 1 kg. Note: this is a position vs. time graph! (a) What are the initial velocities of the carts? (b) What are the final velocities of the carts? (c) What is the mass of the second cart? (d) Does the air track appear to be level? Why? (Hint: does the graph show any evidence of acceleration, for either cart, outside of the collision region?) (e) At the collision time, is the acceleration of the first cart positive or negative? How about the second cart? (Justify your answers.) (f) For the system consisting of the two carts, what is its initial (total) momentum? What is its final momentum? (g) Imagine now that one of the magnets is reversed, so when the carts collide they stick to each other. What would then be the final momentum of the system? What would be its final velocity? x (m) t (s) cart 2 cart 1 Figure 3.5: A collision between two carts. Solution (a) All the velocities are to be calculated by picking an easy straight part of each curve and calculating v = ∆x ∆t | University Physics I Classical Mechanics_Page_81_Chunk4279 |
64 CHAPTER 3. MOMENTUM AND INERTIA for suitable intervals. In this way one gets v1i = −1 m s v2i = 0.5 m s (b) Similarly, one gets v1f = 1 m s v2f = −0.5 m s (c) Use this equation, or equivalent (conservation of momentum is OK) m2 m1 = −∆v1 ∆v2 m2 m1 = −1 −(−1) −0.5 −0.5 = 2 so the mass of the second cart is 2 kg. (d) Yes, the track appears to be level because the carts do not show any evidence of acceleration outside of the collision region (the position vs. time curves are straight lines outside of the region approximately given by 4.5 s < t < 5.5 s). (e) The acceleration of the first cart is positive. You can see this either graphically (the curve is like a parabola that opens upwards, i.e., concave), or algebraically (the cart’s velocity increases, going from −1 m/s to 1 m/s) Similarly, the acceleration of the second cart is negative. The curve is like a parabola that opens downwards, i.e., convex; or, algebraically, the cart’s velocity decreases, going from 0.5 m/s to −0.5 m/s. (f) The initial momentum of the system is pi = m1v1i + m2v2i = (1 kg) × & −1 m s ' + (2 kg) × & 0.5 m s ' = 0 The final momentum is pf = m1v1f + m2v2f = (1 kg) × & 1 m s ' + (2 kg) × & −0.5 m s ' = 0 You could also just say that the final momentum should be the same as the initial momentum, since the system appears to be isolated. (g) The momentum should be conserved in this case as well, so pf = 0. The velocity would be vf = pf m1 + m2 = 0 | University Physics I Classical Mechanics_Page_82_Chunk4280 |
3.5. EXAMPLES 65 3.5.2 Collision in different reference frames, center of mass, and recoil An 80-kg hockey player (call him player 1), moving at 3 m/s to the right, collides with a 90-kg player (player 2) who was moving at 2 m/s to the left. For a brief moment, they are stuck sliding together as they grab at each other. (a) What is their joint velocity as they slide together? (b) What was the velocity of their center of mass before and after the collision? (c) What does the collision look like to another player that was skating initially at 1.5 m/s to the right? Give all the initial and final velocities as seen by this player, and show explicitly that momentum is also conserved in this player’s frame of reference. (d) Eventually, the 90-kg player manages to push the other one back, in such a way that player 1 (the 80-kg player) ends up moving at 1 m/s to the left relative to player 2. What are their final velocities in the earth frame of reference? Solution (a) Call the initial velocities v1i and v2i, the joint final velocity vf, and assume the two players are an isolated system for practical purposes. Then conservation of momentum reads m1v1i + m2v2i = (m1 + m2)vf (3.13) Solving for the final velocity, we get vf = m1v1i + m2v2i m1 + m2 (3.14) Substituting the values given, we get vf = 80 × 3 −90 × 2 170 = 0.353 m s (3.15) (b) According to Eq. (3.10), the velocity of the center of mass, vcm, is just the same as what we just calculated (Eq. (3.14) above). This makes sense: after the collision, if the players are moving together, their system’s center of mass has to be moving with them. Also, if the system is isolated, the center of mass velocity should be the same before and after the collision. So the answer is vcm = vf = 0.353 m/s (c) Let me refer to this third player as “player 3,” and introduce a subscript “3” to refer to the quantities as seen in his frame of reference. Let also the subscript “E” denote the original, “Earth” reference frame. From Eq. (1.19), we have then (for player 1, for instance) v31 = v3E + vE1 = vE1 −vE3 (3.16) because v3E, the “velocity of the Earth in player 3’s reference frame,” is clearly equal to −vE3, the negative of the velocity of player 3 relative to the Earth. Basically, what Eq. (3.16) is saying is | University Physics I Classical Mechanics_Page_83_Chunk4281 |
66 CHAPTER 3. MOMENTUM AND INERTIA that to convert all the Earth-frame velocities to the reference frame of player 3, we just need to subtract 1.5 m/s from them. This gives us v31,i = 3 m s −1.5 m s = 1.5 m s v32,i = −2 m s −1.5 m s = −3.5 m s v31,f = v32,f = 0.353 m s −1.5 m s = −1.147 m s (3.17) The total initial momentum in player’s 3 reference frame is then psys,i = m1v31,i + m2v32,i = 80 × 1.5 + 90 × (−3.5) = −195 kg·m s (3.18) and the final momentum is psys,f = (m1 + m2)v31,f = 170 × (−1.147) = −195 kg·m s (3.19) So the total momentum is conserved in player 3’s reference frame. The reason for this is that this is an inertial reference frame, because the velocity of player 3 does not change. (d) For this part of the problem, we are back to the original reference frame (the Earth reference frame), and we can drop the “E” subscript. For this new process, the final velocities from part (a) become the initial velocities, so we have v1i = v2i = 0.353 m/s. [Note: alternatively, since the system is isolated throughout, it would be OK in this case to use the velocities before the collision to calculate its total momentum, which also needs to be conserved in this process.] We are also told that the final velocity of player 1 relative to player 2 is v21,f = v1f −v2f = −1 m/s. So we have two equations to solve: (m1 + m2) × & 0.353 m s ' = m1v1f + m2v2f (conservation of momentum) v1f −v2f = −1 m s (final relative velocity) (3.20) Leaving aside the units for the moment, to make the equations more readable (the final units will work out, if we make sure to use SI units all along), we have: (80 + 90) × 0.353 = 80v1f + 90v2f v1f = v2f −1 (3.21) Now substitute the second equation, which I have “solved” already for v1f, in the first equation and solve for v2f. The result is v2f = 0.824 m/s, which, when substituted back in the relative velocity equation, gives v1f = −0.176 m/s. | University Physics I Classical Mechanics_Page_84_Chunk4282 |
3.6. PROBLEMS 67 3.6 Problems Problem 1 This figure shows the position vs. time graph for two objects before and after they collide. Assume that they form an isolated system. (a) What are the velocities of the two objects before and after the collision? (Hint: you will get a more accurate result if you choose the initial and final times where the lines go exactly through a point on the grid shown.) (b) Given the result in (a), what is the ratio of the inertias of the two objects? 1 m 3 m 2 m 1 s 2 s 3 s object 1 object 1 object 2 object 2 Collision region 1.5 s 1.75 s x (m) t (s) Problem 2 A car and a truck collide on a very slippery highway. The car, with a mass of 1600 kg, was initially moving at 50 mph. The truck, with a mass of 3000 kg, hit the car from behind at 65 mph. Assume the two vehicles form an isolated system in what follows. (a) If, immediately after the collision, the vehicles separate and the truck’s velocity is found to be 55 mph in the same direction it was going, how fast (in miles per hour) is the car moving? (b) If instead the vehicles end up stuck together, what will be their common velocity immediately after the collision? Problem 3 A 4-kg gun fires a 0.012-kg bullet at a 3-kg block of wood that is initially at rest. The bullet is embedded in the block, and they move together, immediately after the impact, with a velocity of 3.5 m/s. | University Physics I Classical Mechanics_Page_85_Chunk4283 |
68 CHAPTER 3. MOMENTUM AND INERTIA (a) What was the velocity of the bullet just before impact? (b) In order to shoot a bullet at this speed, what must have been the recoil speed of the gun? Problem 4 A 2-kg object, moving at 1 m/s, collides with a 1-kg object that is initially at rest. After the collision, the two objects are found to move away from each other at 1 m/s. Assume they form an isolated system. (a) What are their actual final velocities in the Earth reference frame? (b) What is the velocity of the center of mass of this system? Does it change as a result of the collision? Problem 5 Imagine you are stranded on a frozen lake (that means no friction—no traction!), with just a bow and a quiver of arrows. Each arrow has a mass of 0.02 kg, and with your bow can shoot them at a speed of 90 m/s (relative to you—but you might as well assume that this is the arrow’s velocity relative to the earth, since, as you will see, your recoil velocity will end up being pretty small anyway). So you decide to use them to propel yourself back to shore. (a) Suppose your mass (plus the bow and arrows) is 70 kg. When you shoot an arrow, starting from rest, with what speed do you recoil? (b) Suppose you try to be really clever, and tie a string to the arrow, with the other end of the string tied around your waist. The idea is to get the arrow to pull you forward. Will this work? (Hint: remember part (a). What will happen when the string becomes taut?) Problem 6 An object’s position function is given by x1(t) = 5 + 10t (with x1 in meters if t is in seconds). A second object’s position function is x2(t) = 5 −6t. (a) If the first object’s mass is 1/3 the mass of the second one, what is the position of the system’s center of mass as a function of time? (b) Under the same assumption, what is the velocity of the system’s center of mass? | University Physics I Classical Mechanics_Page_86_Chunk4284 |
Chapter 4 Kinetic Energy 4.1 Kinetic Energy For a long time in the development of classical mechanics, physicists were aware of the existence of two different quantities that one could define for an object of inertia m and velocity v. One was the momentum, mv, and the other was something proportional to mv2. Despite their obvious similarities, these two quantities exhibited different properties and seemed to be capturing different aspects of motion. When things got finally sorted out, in the second half of the 19th century, the quantity 1 2mv2 came to be recognized as a form of energy—itself perhaps the most important concept in all of physics. Kinetic energy, as this quantity is called, may be the most obvious and intuitively understandable kind of energy, and so it is a good place to start our study of the subject. We will use the letter K to denote kinetic energy, and, since it is a form of energy, we will express it in the units especially named for this purpose, which is to say joules (J). 1 joule is 1 kg·m2/s2. In the definition K = 1 2mv2 (4.1) the letter v is meant to represent the magnitude of the velocity vector, that is to say, the speed of the particle. Hence, unlike momentum, kinetic energy is not a vector, but a scalar: there is no sense of direction associated with it. In three dimensions, one could write K = 1 2m ( v2 x + v2 y + v2 z ) (4.2) There is, therefore, some amount of kinetic energy associated with each component of the velocity vector, but in the end they are all added together in a lump sum. 69 | University Physics I Classical Mechanics_Page_87_Chunk4285 |
70 CHAPTER 4. KINETIC ENERGY For a system of particles, we will treat kinetic energy as an additive quantity, just like we did for momentum, so the total kinetic energy of a system will just be the sum of the kinetic energies of all the particles making up the system. Note that, unlike momentum, this is a scalar (not a vector) sum, and most importantly, that kinetic energy is, by definition, always positive, so there can be no question of a “cancellation” of one particle’s kinetic energy by another, again unlike what happened with momentum. Two objects of equal mass moving with equal speeds in opposite directions have a total momentum of zero, but their total kinetic energy is definitely nonzero. Basically, the kinetic energy of a system can never be zero as long as there is any kind of motion going on in the system. 4.1.1 Kinetic energy in collisions To gain some further insights into the concept of kinetic energy, and the ways in which it is different from momentum, it is useful to look at it in the same setting in which we “discovered” momentum, namely, one-dimensional collisions in an isolated system. If we look again at the collision represented in Figure 1 of Chapter 3, reproduced below, -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0 2 4 6 8 10 1 1 2 2 1 m/s v (m/s) t (ms) Figure 4.1: Elastic collision in an isolated system. (Figure 3.1.) we can use the definition (4.1) to calculate the initial and final values of K for each object, and for the system as a whole. Remember we found that, for this particular system, m2 = 2m1, so we can just set m1 = 1 kg and m2 = 2 kg, for simplicity. The initial and final velocities are v1i = 1 m/s, | University Physics I Classical Mechanics_Page_88_Chunk4286 |
4.1. KINETIC ENERGY 71 v2i = 0, v1f = −1/3 m/s, v2f = 2/3 m/s, and so the kinetic energies are K1i = 1 2 J, K2i = 0; K1f = 1 18 J, K2f = 4 9 J Note that 1/18 + 4/9 = 9/18 = 1/2, and so Ksys,i = K1i + K2i = 1 2 J = K1f + K2f = Ksys,f In words, we find that, in this collision, the final value of the total kinetic energy is the same as its initial value, and so it looks like we have “discovered” another conserved quantity (besides momentum) for this system. This belief may be reinforced if we look next at the collision depicted in Figure 2 of Chapter 3, again reproduced below. Recall I pointed out back then that we can think of this as being really the same collision as depicted in Figure 3.1, only looked at from another frame of reference (one moving initially to the right at 1 m/s). We will have more to say about how to transform quantities from a frame of reference to another by the end of the chapter. -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0 2 4 6 8 10 1 2 v (m/s) t (ms) 1 2 −1 m/s Figure 4.2: Another elastic collision, equivalent to the one in Figure 1 as seen from another reference frame. (Figure 3.2.) In any case, as observed there, all we need to do is add −1 m/s to all the velocities in the previous problem, so we have v1i = 0, v2i = −1 m/s, v1f = −4/3 m/s, v2f = −1/3 m/s. The corresponding kinetic energies are, accordingly, K1i = 0, K2i = 1 J, K1f = 8 9J, K2f = 1 9J. These are all different | University Physics I Classical Mechanics_Page_89_Chunk4287 |
72 CHAPTER 4. KINETIC ENERGY from the values we had in the previous example, but note that once again the total kinetic energy after the collision equals the total kinetic energy before—namely, 1 J in this case1 . Things are, however, very different when we consider the third collision example shown in Chapter 3, namely, the one where the two objects are stuck together after the collision. 0 0.2 0.4 0.6 0.8 1 0 2 4 6 8 10 1 2 1 m/s 1 2 v (m/s) t (ms) Figure 4.3: A totally inelastic collision. (Figure 3.3.) Their joint final velocity, consistent with conservation of momentum, is v1f = v2f = 1/3 m/s. Since the system starts as in Figure 4.1, its kinetic energy is initially Ksys,i = 1 2J, but after the collision we have only Ksys,f = 1 2(3 kg) $1 3 m s %2 = 1 6 J So kinetic energy is not conserved in this case at all. What this shows, however, is that unlike the total momentum of a system, which is completely unaffected by internal interactions, the total kinetic energy does depend on the details of the interaction, and thus conveys some information about its nature. We can then refine our study of collisions to distinguish two kinds: the ones where the initial kinetic energy is recovered after the collision, which we will call elastic, and the ones where it is not, which we call inelastic. A 1This is, of course, consistent with the principle of relativity I told you about in Chapter 2: if the process in Fig. 4.2 is really the same as the one in Fig. 4.1, only viewed in a different inertial reference frame, then, if energy is seen to be conserved in one frame, it should also be seen to be conserved in the other. More on this below, in Section 4.2.1. | University Physics I Classical Mechanics_Page_90_Chunk4288 |
4.1. KINETIC ENERGY 73 special case of inelastic collision is the one called totally inelastic, where the two objects end up stuck together, as in Figure 4.3. As we shall see later, the kinetic energy “deficit” is largest in that case. I have said above that in an elastic collision the kinetic energy is “recovered,” and I prefer this terminology to “conserved,” because, in fact, unlike the total momentum, the total kinetic energy of a system does not remain constant throughout the interaction, not even during an elastic collision. The simplest example to show this would be an elastic, head-on collision between two objects of equal mass, moving at the same speed towards each other. In the course of the collision, both objects are brought momentarily to a halt before they reverse direction and bounce back, and at that instant, the total kinetic energy is zero. You can also examine Figures 4.1 and 4.2 above, and calculate, from the graphs, the value of the total kinetic energy during the collision. You will see that it dips to a minimum, and then comes back to its initial value (see also Figure 4.5, later in this chapter). Conventionally, we may talk of kinetic energy as being “conserved” in elastic collisions, but it is important to realize that we are looking at a different kind of “conservation” than what we had with the total momentum, which was constant before, during, and after the interaction, as long as the system remained isolated. Elastic collisions do suggest that, whatever the ultimate nature of this thing we call “energy” might be, it may be possible to store it in some form (in this case, during the course of the collision), and then recover it, as kinetic energy, eventually. This paves the way for the introduction of other kinds of “energy” besides kinetic energy, as we shall see in a later chapter, and the possibility of interconversion to take place among these kinds. For the moment, we shall simply say that in an elastic collision some amount of kinetic energy is temporarily stored as some kind of “internal energy,” and after the collision this is converted back into kinetic energy; whereas, in an inelastic collision, some amount of kinetic energy gets irrevocably converted into some “internal energy,” and we never get it back. Since whatever ultimately happens depends on the details and the nature of the interaction, we will be led to distinguish between “conservative” interactions, where kinetic energy is reversibly stored as some other form of energy somewhere, and “dissipative” interactions, where the energy conversion is, at least in part, irreversible. Clearly, elastic collisions are associated with conservative interactions and inelastic collisions are associated with dissipative interactions. This preliminary classification of interactions will have to be reviewed a little more carefully, however, in the next chapter. | University Physics I Classical Mechanics_Page_91_Chunk4289 |
74 CHAPTER 4. KINETIC ENERGY 4.1.2 Relative velocity and coefficient of restitution An interesting property of elastic collisions can be disclosed from a careful study of figures 4.1 and 4.2. In both cases, as you can see, the relative velocity of the two objects colliding has the same magnitude (but opposite sign) before and after the collision. In other words: in an elastic collision, the objects end up moving apart at the same rate as they originally came together. Recall that, in Chapter 1, we defined the velocity of object 2 relative to object 1 as the quantity v12 = v2 −v1 (4.3) (compare Eq. (1.21); and similarly the velocity of object 1 relative to object 2 is v21 = v1 −v2. With this definition you can check that, indeed, the collisions shown in Figs. 4.1 and 4.2 satisfy the equality v12,i = −v12,f (4.4) (note that we could equally well have used v21 instead of v12). For instance, in Fig. 4.1, v12,i = v2i −v1i = −1 m/s, whereas v12,f = 2/3 −(−1/3) = 1 m/s. So the objects are initially moving towards each other at a rate of 1 m per second, and they end up moving apart just as fast, at 1 m per second. Visually, you should notice that the distance between the red and blue curves is the same before and after (but not during) the collision; the fact that they cross accounts for the difference in sign of the relative velocity, which in turns means simply that before the collision they were coming together, and afterwards they are moving apart. It takes only a little algebra to show that Eq. (4.4) follows from the joint conditions of conservation of momentum and conservation of kinetic energy. The first one (pi = pf) clearly has the form m1v1i + m2v2i = m1v1f + m2v2f (4.5) whereas the second one (Ki = Kf) can be written as 1 2m1v2 1i + 1 2m2v2 2i = 1 2m1v2 1f + 1 2m2v2 2f (4.6) We can cancel out all the factors of 1/2 in Eq. (4.6)2, then rearrange it so that quantities belonging to object 1 are on one side, and quantities belonging to object 2 are on the other. We get m1 ( v2 1i −v2 1f ) = −m2 ( v2 2i −v2 2f ) m1 (v1i −v1f) (v1i + v1f) = −m2 (v2i −v2f) (v2i + v2f) (4.7) (using the fact that a2 −b2 = (a + b)(a −b)). Note, however, that Eq. (4.5) can also be rewritten as m1 (v1i −v1f) = −m2 (v2i −v2f) 2You may be wondering, just why do we define kinetic energy with a factor 1/2 in front, anyway? There is no good answer at this point. Let’s just say it will make the definition of “potential energy” simpler later, particularly as regards its relationship to force. | University Physics I Classical Mechanics_Page_92_Chunk4290 |
4.1. KINETIC ENERGY 75 This immediately allows us to cancel out the corresponding factors in Eq (4.7), so we are left with v1i + v1f = v2i + v2f, which can be rewritten as v1f −v2f = v2i −v1i (4.8) and this is equivalent to (4.4). So, in an elastic collision the speed at which the two objects move apart is the same as the speed at which they came together, whereas, in what is clearly the opposite extreme, in a totally inelastic collision the final relative speed is zero—the objects do not move apart at all after they collide. This suggests that we can quantify how inelastic a collision is by the ratio of the final to the initial magnitude of the relative velocity. This ratio is denoted by e and is called the coefficient of restitution. Formally, e = −v12,f v12,i = −v2f −v1f v2i −v1i (4.9) For an elastic collision, e = 1, as required by Eq. (4.4). For a totally inelastic collision, like the one depicted in Fig. 3, e = 0. For a collision that is inelastic, but not totally inelastic, e will have some value in between these two extremes. This knowledge can be used to “design” inelastic collisions (for homework problems, for instance!): just pick a value for e, between 0 and 1, in Eq. (4.9), and combine this equation with the conservation of momentum requirement (4.5). The two equations then allow you to calculate the final velocities for any values of m1, m2, and the initial velocities. Figure 4.4 below, for example, shows what the collision in Figure 4.1 would have been like, if the coefficient of restitution had been 0.6 instead of 1. You can check, by solving (4.5) and (4.9) together, and using the initial velocities, that v1f = −1/15 m/s = −0.0667 m/s, and v2f = 8/15 m/s = 0.533 m/s. -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0 2 4 6 8 10 v (m/s) t (ms) 1 2 Figure 4.4: An e = 0.6 collision between objects with the same inertias and initial velocities as in Figure 1. | University Physics I Classical Mechanics_Page_93_Chunk4291 |
76 CHAPTER 4. KINETIC ENERGY Although, as I just mentioned, for most “normal” collisions the coefficient of restitution will be a positive number between 1 and 0, there can be exceptions to this. If one of the objects passes through the other (like a bullet through a target, for instance), the value of e will be negative (although still between 0 and 1 in magnitude). And e can be greater than 1 for so-called “explosive collisions,” where some amount of extra energy is released, and converted into kinetic energy, as the objects collide. (For instance, two hockey players colliding on the rink and pushing each other away.) In this case, the objects may well fly apart faster than they came together. An extreme example of a situation with e > 0 is an explosive separation, which is when the two objects are initially moving together and then fly apart. In that case, the denominator of Eq. (4.9) is zero, and so e is formally infinite. This suggests, what is in fact the case, namely, that although explosive processes are certainly important, describing them through the coefficient of restitution is rare, even when it would be formally possible. In practice, use of the coefficient of restitution is mostly limited to the elastic-to-completely inelastic range, that is, 0 ≤e ≤1. 4.2 “Convertible” and “translational” kinetic energy Figure 4.5 shows how the total kinetic energy varies with time, for the two objects shown colliding in Figure 4.1 , depending on the details of the collision, namely, on the value of e. The three curves shown cover the elastic case, e = 1 (Figure 4.1), the totally inelastic case, e = 0 (Figure 4.3), and the inelastic case with e = 0.6 of Figure 4.4. Recall that the total momentum is conserved in all three cases. 0 0.1 0.2 0.3 0.4 0.5 0 2 4 6 8 10 e = 1 e = 0.6 e = 0 Ksys (J) t (ms) Figure 4.5: The total kinetic energy as a function of time for the collisions shown in Figures 1, 3 and 4, respectively. | University Physics I Classical Mechanics_Page_94_Chunk4292 |
4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 77 Figure 4.5 shows that the greatest loss of kinetic energy happens for the totally inelastic collision, which, as we will see in a moment, is, in fact, a general result. That being the case, the figure also shows that it may not be always be possible to bring the total kinetic energy down to zero, even temporarily. The reason for this is that, if momentum is conserved, the velocity of the center of mass cannot change, so if the center of mass was moving before the collision, it must still be moving afterwards; and, as mentioned in this chapter’s introduction, as long as there is motion in a system, its total kinetic energy cannot be zero. All of this suggests that it should be possible to break up a system’s total kinetic energy into two parts: one part associated with the motion of the center of mass, which cannot change in any momentum-conserving collision, and one part associated with the relative motion of the parts that make up the system. This second part would vanish irreversibly in a totally inelastic collision, whereas it would recover its original value in an elastic collision. The way to see this mathematically, for a system of two objects with masses m1 and m2, is to introduce the center of mass velocity vcm [Eq. (3.10)] vcm = m1v1 + m2v2 m1 + m2 and the relative velocity v12 = v2 −v1 (Eq. (4.3) above), and observe that the velocities v1 and v2 can be written, respectively, as v1 = vcm − m2 m1 + m2 v12 v2 = vcm + m1 m1 + m2 v12 (4.10) Substituting the equations (4.10) into the expression Ksys = 1 2m1v2 1 + 1 2m2v2 2, one finds that the cross-terms vanish, and all that is left is Ksys = 1 2(m1 + m2)v2 cm + 1 2 m1m2 2 + m2m2 1 (m1 + m2)2 v2 12 A factor of (m1 + m2) may be canceled in the last term, and the final expression takes the form Ksys = Kcm + Kconv (4.11) where the center of mass kinetic energy (or translational energy) is just what one would have if the whole system was a single particle of mass M = m1 + m2 moving at the center of mass speed: Kcm = 1 2Mv2 cm (4.12) and the “convertible energy” Kconv is the part associated with the relative motion, which can be | University Physics I Classical Mechanics_Page_95_Chunk4293 |
78 CHAPTER 4. KINETIC ENERGY made to vanish entirely in an inelastic collision3: Kconv = 1 2 m1m2 m1 + m2 v2 12 = 1 2µv2 12 (4.13) The last equation implicitly defines a useful quantity that we call the reduced mass of a system of two particles, and denote by µ: µ = m1m2 m1 + m2 (4.14) Equation (4.11), with the definitions (4.12) and (4.13), pretty much explains everything that we see going on in Figure 4.5. The total kinetic energy is the sum of two terms, the first of which, Kcm, can never change: it is, in fact, as constant as the total momentum itself, since it involves the center of mass velocity, vcm, which is proportional to the total momentum of the system (recall equation (3.11)). The term that can, and does change, is the second one, the convertible energy. In fact, in an ordinary collision in which the objects do not pass through each other, there must be at least an instant in time when Kconv = 0. This is because it involves the relative velocity, and since the relative velocity must change sign at some point (the objects are initially coming together, but end up moving apart), it must be zero at that time. This explains why all the curves in Fig. 4.5 have the same minimum value (even though they may reach it at different times): that value is clearly Kcm for the system (since Kconv is zero at that time). It is the same for all the curves because all the systems considered have the same total mass and momentum (as determined by the initial velocities)—we just chose them that way. Since Kcm cannot change for an isolated system, the maximum kinetic energy that can be lost in a collision in such a system is the initial value of Kconv, which we would denote as Kconv,i. This is, in fact, completely lost in a totally inelastic collision, since in that case v12,f = 0, and Eq. (4.13) then gives Kconv,f = 0. In fact, using Eq. (4.9), we can relate the final value of the convertible energy to its initial value via the coefficient of restitution: Kconv,f = 1 2µv2 12,f = 1 2µe2v2 12,i = e2Kconv,i (4.15) Thus, for example, in a collision with e = 0.6, the final value of the convertible energy would be only 0.36 times its initial value: 64% of it would have been “lost.” (This is not, however, the same as 64% of the total initial energy, since the latter still includes Kcm, which does not change.) We can also write Eq. (4.15) as ∆Ksys = ( e2 −1 ) Kconv,i = ( e2 −1 ) 1 2µv2 12,i (4.16) since the only possible change in Ksys must come from the convertible energy. 3Although the name “convertible energy” makes sense in this context, it is not, as far as I can tell, in general usage. I have borrowed it from Mazur’s The Principles and Practice of Physics, but you should probably not expect to find it in other textbooks. | University Physics I Classical Mechanics_Page_96_Chunk4294 |
4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 79 Although we have derived the decomposition (4.11) for the very restricted situation of two objects moving in one dimension, the basic result is quite general: first, everything in the derivation works if v1 and v2 are replaced by vectors ⃗v1 and ⃗v2, so the results holds in three dimensions as well. Second, for a system of any number of particles, one still can write Ksys as Kcm+ another term that depends only on the relative motion of all the pairs of particles. This “generalized convertible energy,” or kinetic energy of relative motion would have the form Krel = 1 2µ12v2 12 + 1 2µ13v2 13 + . . . + 1 2µ23v2 23 + . . . (in this expression, something like µ23 means a reduced mass like the one in Eq. (4.14), only for masses m2 and m3, and so forth). When we get to the study of rotational motion, for instance, we will see that the total kinetic energy of an extended rigid object can be written as Kcm +Krot, where Krot, the rotational kinetic energy, is just the same kind of thing as what we have called the “convertible energy” here. All of the above still leaves unanswered the question of what happens to the convertible energy that is lost in an inelastic collision. Just what is it that it gets converted into? The answer to this question will be the subject of the following chapter. 4.2.1 Kinetic energy and momentum in different reference frames I have pointed out repeatedly before that all motion is relative, and so, to some extent, kinetic energy and momentum must be somewhat relative as well. A car in a freight train has a lot of momentum relative to an observer on the ground, but its momentum relative to another car on the same train is zero, since they are not moving relative to each other. The same could be said about its kinetic energy. In general, if you have a system with a total momentum ⃗psys and inertia M, its center of mass will have a velocity ⃗vcm = ⃗psys/M. Then, if you were to move alongside the system with a velocity exactly equal to ⃗vcm, the total momentum of the system relative to you would be zero. If the system was a solid object, it would not “hit” you if you made contact; there would be no collision. It may help here to think, for instance, of aircraft refueling in flight: if the two planes’ velocities are exactly matched, they can make contact without any damage, just as if they were at rest. A reference frame moving at a system’s center of mass velocity is, for this reason, called a zero-momentum frame for the system in question. Clearly, in such a reference frame, the translational kinetic energy of the system, Kcm = 1 2Mv2 cm, will also be zero (since, in that frame, the center of mass is not moving at all). However, the relative motion term, Kconv, would be completely unaffected by the change in reference frame. This is because, as you may have noticed by now, to convert velocities from one frame of reference to | University Physics I Classical Mechanics_Page_97_Chunk4295 |
80 CHAPTER 4. KINETIC ENERGY another we just add or subtract from all the velocities the relative velocity of the two frames. This operation, however, will not change any of the relative velocities of the parts of the system, since these are all differences to begin with. Mathematically, (v2 + v′) −(v1 + v′) = v2 −v1 regardless of the value of v′. So there something we might call absolute (as opposed to “relative”) about the convertible kinetic energy: it is the same, it will have the same value, for any observer, regardless of how fast or in what direction that observer may be moving relative to the system as a whole. We may think of it as an intrinsic (meaning, observer-independent) property of the system. 4.3 In summary 1. The kinetic energy of a particle of mass m moving with velocity v is defined as K = 1 2mv2. It is a scalar quantity, and it is always positive. For a system of particles or an extended object, we define Ksys as the sum of the kinetic energies of all the particles making up the system. 2. For any system, the total kinetic energy can be written as the sum of the translational (or center of mass) kinetic energy, Kcm, and another term that involves the motion of the parts of the system relative to each other. (See Eq. (4.11) above.) The translational kinetic energy is constant for an isolated system, and is always given by Kcm = 1 2Mv2 cm. 3. The kinetic energy of relative motion (which, in the context of collisions, is called the con- vertible energy) is given, for the special case of a system consisting of two particles (or two non-rotating extended objects), by Kconv = 1 2µv2 12, where µ = m1m2/(m1+m2) is the reduced mass, and v12 = v2 −v1 is the relative velocity of the two objects. 4. In a one-dimensional collision between two objects that do not pass through each other, the convertible energy always drops to zero at some point, as a result of the interaction; that is, it is converted entirely into some other form of energy. At the end of the interaction, all the convertible energy may be recovered (elastic collision), or only part of it (inelastic collision), or none of it (completely inelastic collision). 5. In terms of the coefficient of restitution e, defined as e = −v12,f/v12,i, elastic collisions have e = 1, totally inelastic collisions have e = 0, and inelastic collisions 0 < e < 1. The total change in kinetic energy in the collision can be written as ∆Ksys = ∆Kconv = (e2 −1)Kconv,i. 6. Another way to say the above is that in an elastic collision in one dimension, the two objects move apart after the collision at the same rate (relative speed) at which they approached each other initially. In a totally inelastic collision, conversely, the two objects do not move apart at all after the collision—they become “stuck together.” | University Physics I Classical Mechanics_Page_98_Chunk4296 |
4.3. IN SUMMARY 81 7. Besides the cases considered above, one may have collisions where the objects pass through each other, giving e < 0, and “explosive collisions,” where e > 1. In these latter collisions some internal source of energy is converted into additional kinetic energy when the objects interact. The extreme case of this is an explosive separation, which is the reverse of a totally inelastic collision—two objects initially moving together fly apart, with a net increase in the system’s kinetic energy. 8. The translational kinetic energy of a system will, in general, have different values for observers moving with different velocities. The convertible kinetic energy, on the other hand, is seen by all observers to have the same value, regardless of their relative state of motion. | University Physics I Classical Mechanics_Page_99_Chunk4297 |
82 CHAPTER 4. KINETIC ENERGY 4.4 Examples 4.4.1 Collision graph revisited Look again at the collision graph from example 3.5.1 from the point of view of the kinetic energy of the two carts. (a) What is the initial kinetic energy of the system? (b) How much of this is in the center of mass motion, and how much of is convertible? (c) Does the convertible kinetic energy go to zero at some point during the collision? If so, when? Is it fully recovered after the collision is over? (d) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution? Solution (a) From the solution to example 3.5.1 we know that v1i = −1 m s v2i = 0.5 m s v1f = 1 m s v2f = −0.5 m s and m1 = 1 kg and m2 = 2 kg. So the initial kinetic energy is Ksys,i = 1 2m1v2 1i + 1 2m2v2 2i = 0.5 J + 0.25 J = 0.75 J (4.17) (b) To calculate Kcm = 1 2(m1 + m2)v2 cm, we need vcm, which in this case is equal to vcm = m1v1i + m2v2i m1 + m2 = −1 + 2 × 0.5 3 = 0 so Kcm = 0, which means all the kinetic energy is convertible. We can also calculate that directly: Kconv,i = 1 2µv2 12,i = 1 2 $1 × 2 1 + 2 kg % × & 0.5 m s −(−1) m s '2 = 1.52 3 J = 0.75 J (4.18) (c) If we look at figure 3.5, we can see that the carts do not pass through each other, so their relative velocity must be zero at some point, and with that, the convertible energy. In fact, the figure makes it quite clear that both v1 and v2 are zero at t = 5 s, so at that point also v12 = 0, and the convertible energy Kconv = 0. (And so is the total Ksys = 0 at that time, since Kcm = 0 throughout.) | University Physics I Classical Mechanics_Page_100_Chunk4298 |
4.4. EXAMPLES 83 On the other hand, it is also clear that Kconv is fully recovered after the collision is over, since the relative velocity just changes sign: v12,i = v2i −v1i = 0.5 m s −(−1) m s = 1.5 m s v12,f = v2f −v1f = −0.5 m s −1 m s = −1.5 m s (4.19) Therefore Kconv,f = 1 2µv2 12,f = 1 2µv2 12,i = Kconv,i (d) Since the total kinetic energy (which in this case is only convertible energy) is fully recovered when the collision is over, the collision is elastic. Using equation (4.19), we can see that the coefficient of restitution is e = −v12,f v12,i = −−1.5 1.5 = 1 as it should be. 4.4.2 Inelastic collision and explosive separation Analyze example 3.5.2 from the point of view of the system’s kinetic energy. In particular, answer the following questions: (a) What is the total kinetic energy of the system (i) before the players collide, (ii) right after the collision, when they are holding to one another, and (iii) after they separate. How much of this energy is translational (that is, center-of-mass kinetic energy), and how much is convertible? (b) Answer the same questions from the point of view of the player who is skating at a constant 1.5 m/s to the right (player 3) (To avoid needless repetition, you may use already established results, such as conservation of momentum.) Solution (a) Before the players collide, we have Ksys,i = 1 2m1v2 1i + 1 2m2v2 2i = 1 2 (80 kg) × & 3 m s '2 + 1 2 (90 kg) × & −2 m s '2 = 540 J (4.20) While they are still holding to each other, we know from the solution to example 3.5.2 that their joint velocity is 0.353 , and that this has to be also the velocity of their center of mass, which is unchanged by the collision. So, we have Kcm = 1 2(m1 + m2)v2 cm = 1 2 (170 kg) & 0.353 m s '2 = 10.6 J (4.21) | University Physics I Classical Mechanics_Page_101_Chunk4299 |
84 CHAPTER 4. KINETIC ENERGY This is Kcm throughout, as well as Ksys right after the collision, since the collision is totally inelastic and that means that Kconv drops to zero. Also, subtracting this from (4.20) will give us the initial value of the convertible energy, without the need for a separate calculation, so Kconv,i = Ksys,i −Kcm = 540 J −10.6 J = 529.4 J ≃529 J (4.22) After the separation, the new total kinetic energy (for which I will use the subscript f) is Ksys,i = 1 2m1v2 1f + 1 2m2v2 2f = 1 2 (80 kg) × & −0.176 m s '2 + 1 2 (90 kg) × & 0.824 m s '2 = 31.8 J (4.23) where I have gotten the values for v1f and v2f from the solution to part (d) of Example 3.5.2. Subtracting Kcm from this will give us the final value of the convertible energy: Kconv,f = Ksys,f −Kcm = 31.8 J −10.6 J = 21.2 J (4.24) To summarize, then, we have: • Before the collision: Ksys,i = 540 J, Kcm = 10.6 J, Kconv,i = 529.4 J • Right after the collision (players still holding to each other): Ksys = Kcm = 10.6 J, Kconv = 0 • After the (explosive) separation: Ksys,f = 31.8 J, Kcm = 10.6 J, Kconv,i = 21.2 J So, in the collision, approximately 529 J of kinetic energy “disappeared” from the system (or, we could say, were “converted into some form of internal energy”), whereas the players’ pushing on each other managed to put about 21 J of kinetic energy back into the system; we will explore these kinds of processes in more detail in the following chapter! (b) We need to repeat all the above calculations with all the velocities shifted down by 1.5 m/s, to bring them to the reference frame of player 3. Instead of putting a subscript “3” on all the quantities, since we already have tons of subscripts to worry about, I’m going to follow an alternative convention and use a “prime” superscript (′) to denote all the quantities in this frame of reference. In brief, we have K′ sys,i = 1 2m1(v′ 1i)2 + 1 2m2(v′ 2i)2 = 1 2 (80 kg) × & 1.5 m s '2 + 1 2 (90 kg)× & −3.5 m s '2 = 641.3 J (4.25) | University Physics I Classical Mechanics_Page_102_Chunk4300 |
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