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4.4. EXAMPLES 85 K′ cm = 1 2(m1 + m2)(v′ cm)2 = 1 2 (170 kg) & 0.353 m s −1.5 m s '2 = 111.8 J (4.26) K′ conv,i = K′ sys,i −K′ cm = 641.3 J −111.8 J = 529.5 J ≃529 J (4.27) This shows explicitly that the convertible energy, as I pointed out earlier in this chapter, is the same in every reference frame! (The equality is... | University Physics I Classical Mechanics_Page_103_Chunk4301 |
86 CHAPTER 4. KINETIC ENERGY 4.5 Problems Problem 1 A 71-kg man can throw a 1-kg ball with a maximum speed of 6 m/s relative to himself. Imagine that one day he decides to try to do that on roller skates. Starting from rest, he throws the ball as hard as he can, so it ends up moving at 6 m/s relative to him, but he him... | University Physics I Classical Mechanics_Page_104_Chunk4302 |
4.5. PROBLEMS 87 (a) What is the initial kinetic energy of the system? How much of this is center of mass energy, and how much is convertible? (b) What is the maximum amount of kinetic energy that could be “lost” (converted to other forms of energy) in this collision? (c) If 60% of the amount you calculated in part (b)... | University Physics I Classical Mechanics_Page_105_Chunk4303 |
88 CHAPTER 4. KINETIC ENERGY | University Physics I Classical Mechanics_Page_106_Chunk4304 |
Chapter 5 Interactions and energy 5.1 Conservative interactions Let me summarize the physical concepts and principles we have encountered so far in our study of classical mechanics. We have “discovered” one important quantity, the inertia or inertial mass of an object, and introduced two different quantities based on th... | University Physics I Classical Mechanics_Page_107_Chunk4305 |
90 CHAPTER 5. INTERACTIONS AND ENERGY We are going to take the above description literally, and use the name conservative interaction for any interaction that can “store and restore” kinetic energy in this way. The “stored energy” itself—which is not actually kinetic energy while it remains stored, since it is not give... | University Physics I Classical Mechanics_Page_108_Chunk4306 |
5.1. CONSERVATIVE INTERACTIONS 91 Then we see from Eq. (5.2) that K + U G = constant (5.4) This is a statement of conservation of energy under the gravitational interaction. For any interaction that has a potential energy associated with it, the quantity K + U is called the (total) mechanical energy. Figure 5.1 shows h... | University Physics I Classical Mechanics_Page_109_Chunk4307 |
92 CHAPTER 5. INTERACTIONS AND ENERGY from any height I wanted to—for instance, taking the initial height of my hand to correspond to y = 0. This would shift the blue curve in Fig. 5.1 down by 4.9 J, but it would not change any of the physics. The only important thing I really want the potential energy for is to calcul... | University Physics I Classical Mechanics_Page_110_Chunk4308 |
5.1. CONSERVATIVE INTERACTIONS 93 As a result of the collision, the spring compresses and undergoes “half a cycle” of oscillation with an “angular frequency” ω = ! k/µ (where µ is, as in previous chapters, the “reduced mass” of the system, µ = m1m2/(m1 + m2)). That is, the spring is compressed and then pushes out until... | University Physics I Classical Mechanics_Page_111_Chunk4309 |
94 CHAPTER 5. INTERACTIONS AND ENERGY The main point is that this kind of physical setup (a cart fitted with a spring) would indeed give us an elastic collision, and a kinetic energy curve very much like the ones I used, for illustration purposes, in Chapter 4; only now we also have a potential energy curve to go with i... | University Physics I Classical Mechanics_Page_112_Chunk4310 |
5.1. CONSERVATIVE INTERACTIONS 95 dimension, then, we have a situation where, once the initial conditions (the particle’s initial position and velocity) are known, the motion of the particle can be completely determined from the function U(x), where x is the particle’s position at any given time. This can be done, usin... | University Physics I Classical Mechanics_Page_113_Chunk4311 |
96 CHAPTER 5. INTERACTIONS AND ENERGY At that point, the particle stops and turns around, just like an object thrown vertically upwards. As it moves “down the potential energy hill,” it recovers the kinetic energy it used to have, so that when it again reaches the starting point x = −2 m, its speed is again 2 m/s, but ... | University Physics I Classical Mechanics_Page_114_Chunk4312 |
5.2. DISSIPATION OF ENERGY AND THERMAL ENERGY 97 energy of the system, Emech = K + U, that is constant throughout the interaction. However, we already know from our study of inelastic collisions that this is rarely the case. Essential to the concept of potential energy is the idea of “storage and retrieval” of the kine... | University Physics I Classical Mechanics_Page_115_Chunk4313 |
98 CHAPTER 5. INTERACTIONS AND ENERGY of mechanical energy lost always resulted in the same predictable increase in the system’s thermal energy. Thermal energy is largely “invisible” at the macroscopic level, but we detect it indirectly through an object’s temperature. The crucial experiments to establish what at the t... | University Physics I Classical Mechanics_Page_116_Chunk4314 |
5.4. CONSERVATION OF ENERGY 99 level (and you will learn more about them next semester!). Many others, however, are more subtle and involve quantum mechanical effects (such as the exclusion principle) in a fundamental way. Among the most important of these is chemical energy, which is an extremely important source of en... | University Physics I Classical Mechanics_Page_117_Chunk4315 |
100 CHAPTER 5. INTERACTIONS AND ENERGY system is not necessarily the same thing as an “isolated” system: the former relates to the total energy, the latter to the total momentum. A parked car getting hotter in the sun is not a closed system (it is absorbing energy all the time) but, as far as its total momentum is conc... | University Physics I Classical Mechanics_Page_118_Chunk4316 |
5.5. IN SUMMARY 101 gravitational potential energy, and kinetic energy (diagram (a)). Note that we could write the total kinetic energy as Kcm + Kconv, as we did in the previous chapter, but because of the large mass of the earth, the center of mass of the system is essentially the center of the earth, which, in our ea... | University Physics I Classical Mechanics_Page_119_Chunk4317 |
102 CHAPTER 5. INTERACTIONS AND ENERGY the graph of the function U(x). The idea, elaborated in Section 5.1.2 above, is to imagine the equivalent motion of an object sliding without friction over the same landscape, under the influence of gravity. 6. The fundamental interactions currently known in physics are gravity, th... | University Physics I Classical Mechanics_Page_120_Chunk4318 |
5.6. EXAMPLES 103 5.6 Examples 5.6.1 Inelastic collision in the middle of a swing Tarzan swings on a vine to rescue a helpless explorer (as usual) from some attacking animal or another. He begins his swing from a branch a height of 15 m above the ground, grabs the explorer at the bottom of his swing, and continues the ... | University Physics I Classical Mechanics_Page_121_Chunk4319 |
104 CHAPTER 5. INTERACTIONS AND ENERGY (note that this is just the familiar result (2.10) for free fall! This is because, as I pointed out above, the vine does no work on the system.). Substituting, we get vbot1 = *& 5 m s '2 + 2 & 9.8 m s2 ' × (15 m) = 17.9 m s • Second part: the completely inelastic collision. The ex... | University Physics I Classical Mechanics_Page_122_Chunk4320 |
5.6. EXAMPLES 105 problem involves the conversion of kinetic energy into elastic potential energy, and back. In the absence of dissipation, Eq. (5.8), specialized to this system (the spring and the block) reads: K + U spr = constant (5.13) For part (a), we consider the whole process where the spring starts relaxed and ... | University Physics I Classical Mechanics_Page_123_Chunk4321 |
106 CHAPTER 5. INTERACTIONS AND ENERGY 5.7 Advanced Topics 5.7.1 Two carts colliding and compressing a spring Unlike the example 5.6.2, which considered a stationary spring and asked only questions about initial and final states, this example is intended to show you how one can use “energy methods” to solve for the actu... | University Physics I Classical Mechanics_Page_124_Chunk4322 |
5.7. ADVANCED TOPICS 107 where the quantity ω = ! k/µ, and the time tc is the time cart 1 first makes contact with the spring: tc = (x2i −x0 −x1i)/v1i. The solution (5.19) is valid for as long as the spring is compressed, which is to say, for as long as x12(t) < x0, or sin[ω(t −tc)] > 0, which translates to the conditio... | University Physics I Classical Mechanics_Page_125_Chunk4323 |
108 CHAPTER 5. INTERACTIONS AND ENERGY ical functions. This looks complicated, but it just gives you the shapes you want for the velocity curves. The derivative of the above is v1(t) = 1 3 (1 + 2 erf(10 −2t)) v2(t) = 1 3 (1 −erf(10 −2t)) (5.21) and you may want to try plotting these for yourself; the result should be F... | University Physics I Classical Mechanics_Page_126_Chunk4324 |
5.8. PROBLEMS 109 5.8 Problems Problem 1 A particle is in a region where the potential energy has the form U = 5/x (in joules, if x is in meters). (a) Sketch this potential energy function for x > 0. (b) Assuming the particle starts at rest at x = 0.5 m, which way will it go if released? Why? (c) Under the assumption i... | University Physics I Classical Mechanics_Page_127_Chunk4325 |
110 CHAPTER 5. INTERACTIONS AND ENERGY energy. (a) What is the kinetic energy of the ball just before it hits the ground? (b) What is the kinetic energy of the ball just after it bounces up? (c) What is the coefficient of restitution for this collision? (d) What kind of collision is this (elastic, inelastic, etc.)? Why? ... | University Physics I Classical Mechanics_Page_128_Chunk4326 |
5.8. PROBLEMS 111 here? Explain. (f) After the skydiver reaches terminal speed (and before he opens his parachute), he falls for a while at constant speed. What kind of energy conversion is taking place during this time? (Consider the system to be the earth, the skydiver, and the air around him). Problem 5 You shoot a ... | University Physics I Classical Mechanics_Page_129_Chunk4327 |
112 CHAPTER 5. INTERACTIONS AND ENERGY | University Physics I Classical Mechanics_Page_130_Chunk4328 |
Chapter 6 Interactions, part 2: Forces 6.1 Force As we saw in the previous chapter, when an interaction can be described by a potential energy function, it is possible to use this to get a full solution for the motion of the objects involved, at least in one dimension. In fact, energy-based methods (known as the Lagran... | University Physics I Classical Mechanics_Page_131_Chunk4329 |
114 CHAPTER 6. INTERACTIONS, PART 2: FORCES change of each object’s momentum as a measure of the force exerted on it by the other object. Mathematically, this means we will write for the average force exerted by 1 on 2 over the time interval ∆t the expression (F12)av = ∆p2 ∆t (6.1) Please observe the notation we are go... | University Physics I Classical Mechanics_Page_132_Chunk4330 |
6.1. FORCE 115 however, somewhat more general, so it is technically preferred, even though this semester we will directly use F = ma throughout. If you want an example of a physical situation where F = dp/dt is not equivalent to F = ma, consider a system where object 1 is a rocket, including its fuel, and “object” 2 ar... | University Physics I Classical Mechanics_Page_133_Chunk4331 |
116 CHAPTER 6. INTERACTIONS, PART 2: FORCES just equal to Mvcm (compare Eq. (3.11), in the “Momentum” chapter). So we have Fext,all = dpsys dt = d dt(Mvcm) (6.10) This extends a previous result. We already knew that in the absence of external forces, the mo- mentum of a system remained constant. Now we see that the sys... | University Physics I Classical Mechanics_Page_134_Chunk4332 |
6.2. FORCES AND POTENTIAL ENERGY 117 simplified setup to show you a very interesting relationship between potential energies and forces. Suppose this is a closed system in which no dissipation of energy is taking place. Then the total mechanical energy is a constant: Emech = 1 2mv2 + U(x) = constant (6.13) (Here, m is t... | University Physics I Classical Mechanics_Page_135_Chunk4333 |
118 CHAPTER 6. INTERACTIONS, PART 2: FORCES I claim that in that case you can again get the force on object 1, F21, by taking the derivative of U(x2 −x1) with respect to x1 (leaving x2 alone), and reciprocally, you get F12 by taking the derivative of U(x2 −x1) with respect to x2. Here is how it works, again using the c... | University Physics I Classical Mechanics_Page_136_Chunk4334 |
6.2. FORCES AND POTENTIAL ENERGY 119 Figure 6.1 shows, in black, all the forces exerted by a spring with one fixed end, according as to whether it is relaxed, compressed, or stretched. I have assumed that it is pushed or pulled by a hand (not shown) at the “free” end, hence the subscript “h”, whereas the subscript “w” s... | University Physics I Classical Mechanics_Page_137_Chunk4335 |
120 CHAPTER 6. INTERACTIONS, PART 2: FORCES only meaningful at the macroscopic level, since at the microscopic level objects never really touch, and all forces are field forces, it is just that some are “long range” and some are “short range.” For our purposes, really, the word “contact” will just be a convenient, catch... | University Physics I Classical Mechanics_Page_138_Chunk4336 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 121 of the object on which it acts (since it is a reaction force, it can assume any value as required to adjust to any circumstance—up to the point where the rope snaps, anyway). Thus, for instance, in the picture below, which shows two blocks connected by a rope over a p... | University Physics I Classical Mechanics_Page_139_Chunk4337 |
122 CHAPTER 6. INTERACTIONS, PART 2: FORCES then, the following two equations: F t = m1a F t −m2g = −m2a (6.23) The system (6.23) can be easily solved to get a = m2g m1 + m2 F t = m1m2g m1 + m2 (6.24) 6.3.2 Normal forces Normal force is the reaction force with which a surface pushes back when it is being pushed on. Aga... | University Physics I Classical Mechanics_Page_140_Chunk4338 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 123 for the apparent weightlessness experienced by the astronauts in the space station, where the force of gravity is, in fact, not very much smaller than on the surface of the earth. (We will return to this effect after we have a good grip on two-dimensional, and in parti... | University Physics I Classical Mechanics_Page_141_Chunk4339 |
124 CHAPTER 6. INTERACTIONS, PART 2: FORCES does this have to be? If there is no acceleration (a = 0), the equivalent of system (6.23) will be F s s,1 + F t = 0 F t −m2g = 0 (6.27) where F s s,1 is the force of static friction exerted by the surface on block 1, and we are going to let the math tell us what sign it is s... | University Physics I Classical Mechanics_Page_142_Chunk4340 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 125 Note that, unlike for static friction, this is not the maximum possible value of |F k|, but its actual value; so if we know F n (and µk) we know F k without having to solve any other equations (its sign does depend on the direction of motion, of course). The coefficient... | University Physics I Classical Mechanics_Page_143_Chunk4341 |
126 CHAPTER 6. INTERACTIONS, PART 2: FORCES drag force is proportional to the object’s speed, whereas for high velocities it is proportional to the square of the speed. In principle, one can use the appropriate drag formula together with Newton’s second law to calculate the effect of air resistance on a simple object th... | University Physics I Classical Mechanics_Page_144_Chunk4342 |
6.5. IN SUMMARY 127 Fr,1 t Fs,1 n FE,1 G Fs,1 k a Figure 6.3: Free-body diagram for block 1 in Figure 6.2, with the friction force adjusted so as to be compatible with a nonzero acceleration to the right. Note that I have drawn F n and the force of gravity F G E,1 as having the same magnitude, since there is no vertica... | University Physics I Classical Mechanics_Page_145_Chunk4343 |
128 CHAPTER 6. INTERACTIONS, PART 2: FORCES 6. When dealing with macroscopic objects we introduce several “constraint” forces whose values need to be determined from the accelerations through Newton’s second law: the tension F t in ropes, strings or cables; the normal force F n exerted by a surface in response to appli... | University Physics I Classical Mechanics_Page_146_Chunk4344 |
6.6. EXAMPLES 129 6.6 Examples 6.6.1 Dropping an object on a weighing scale (Short version) Suppose you drop a 5-kg object on a spring scale from a height of 1 m. If the spring constant is k = 20, 000 N/m, what will the scale read? (Long version) OK, let’s break that up into parts. Suppose that a spring scale is just a... | University Physics I Classical Mechanics_Page_147_Chunk4345 |
130 CHAPTER 6. INTERACTIONS, PART 2: FORCES up, with the object momentarily at rest, with only spring potential energy: U G i + U spr i = U G f + U spr f mgyi + 0 = mgyf + 1 2kd2 max (6.35) where I have used the subscript “max” on the compression distance to distinguish it from what I calculated in part (a) (this kind ... | University Physics I Classical Mechanics_Page_148_Chunk4346 |
6.6. EXAMPLES 131 pushes on the brake pedal. Assume that the coefficient of static friction between the tires and the road is µs = 0.7, and that the wheels don’t “lock”: that is to say, they continue rolling without slipping on the road as they slow down. What is the car’s minimum stopping distance? (e) Draw a free-body ... | University Physics I Classical Mechanics_Page_149_Chunk4347 |
132 CHAPTER 6. INTERACTIONS, PART 2: FORCES (d) This is the opposite of part (a): the driver now relies on the force of static friction to slow down the car. The shortest stopping distance will correspond to the largest (in magnitude) acceleration, as per our old friend, Eq. (2.10): v2 f −v2 i = 2a∆x (6.38) In turn, th... | University Physics I Classical Mechanics_Page_150_Chunk4348 |
6.6. EXAMPLES 133 This is a huge distance, close to half a football field! If these numbers are accurate, you can see that locking your brakes in the rain can have some pretty bad consequences. | University Physics I Classical Mechanics_Page_151_Chunk4349 |
134 CHAPTER 6. INTERACTIONS, PART 2: FORCES 6.7 Problems Problem 1 (a) Draw a free-body diagram for the skydiver in Problem 4 of Chapter 5. (b) What is the magnitude of the air drag force on the skydiver, after he reaches terminal speed? Problem 2 A book is sent sliding along a table with an initial velocity of 2 m/s. ... | University Physics I Classical Mechanics_Page_152_Chunk4350 |
6.7. PROBLEMS 135 Problem 6 Draw a free-body diagram for a 70-kg person standing in an elevator carrying a 15-kg backpack (do not consider the backpack a part of the person!). (a) if the elevator is not moving, and (b) if the elevator is accelerating downwards at 2 m/s2. In each case, what is the magnitude of the norma... | University Physics I Classical Mechanics_Page_153_Chunk4351 |
136 CHAPTER 6. INTERACTIONS, PART 2: FORCES | University Physics I Classical Mechanics_Page_154_Chunk4352 |
Chapter 7 Impulse, Work and Power 7.1 Introduction: work and impulse In physics, “work” (or “doing work”) is what we call the process through which a force changes the energy of an object it acts on (or the energy of a system to which the object belongs). It is, therefore, a very technical term with a very specific mean... | University Physics I Classical Mechanics_Page_155_Chunk4353 |
138 CHAPTER 7. IMPULSE, WORK AND POWER impulse, usually denoted as ⃗J ⃗J = ⃗F ∆t (7.2) Clearly, the impulse given by a force to an object is equal to the change in the object’s momentum (by Eq. (7.1)), as long as it is the only force (or, alternatively, the net force) acting on it. If the force is not constant, we brea... | University Physics I Classical Mechanics_Page_156_Chunk4354 |
7.2. WORK ON A SINGLE PARTICLE 139 In three dimensions, the force will be a vector ⃗F with components (Fx, Fy, Fz), and the displacement, likewise, will be a vector ∆⃗r with components (∆x, ∆y, ∆z). The work will be defined then as W = Fx∆x + Fy∆y + Fz∆z (7.5) This expression is an instance of what is known as the dot p... | University Physics I Classical Mechanics_Page_157_Chunk4355 |
140 CHAPTER 7. IMPULSE, WORK AND POWER each of them, calculating all those (possibly very small) “pieces of work,” and adding them all together. In one dimension, the final result can be expressed as the integral W = " xf xi F(x)dx (variable force) (7.9) So the work is given by the “area” under the F-vs-x curve. In more... | University Physics I Classical Mechanics_Page_158_Chunk4356 |
7.3. THE “CENTER OF MASS WORK” 141 Before we go there, however, I would like to take a little detour to explore another “reasonable” extension of the result (7.11), as well as its limitations. 7.3 The “center of mass work” All the physics I used in order to derive the result (7.11) was F = ma, and the expression v2 f −... | University Physics I Classical Mechanics_Page_159_Chunk4357 |
142 CHAPTER 7. IMPULSE, WORK AND POWER Fs,2 spr Fs,1 spr Fh,2 c Fh,2 c ∆x1 ∆x2 ∆xcm 1 1 2 2 Figure 7.2: A system of two blocks connected by a spring. A constant external force, ⃗F c h,2, is applied to the block on the right. Initially the spring is relaxed, but as soon as block 2 starts to move it stretches, pulling ba... | University Physics I Classical Mechanics_Page_160_Chunk4358 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 143 start by considering what happens over a time interval so short that all the forces are approximately constant (the final result will hold for arbitrarily long time intervals, just by adding, or integrating, over many such short intervals). I will also work expli... | University Physics I Classical Mechanics_Page_161_Chunk4359 |
144 CHAPTER 7. IMPULSE, WORK AND POWER Adding up very many such “infinitesimal” displacements will lead to the same final result, where ∆U will be the change in the potential energy over the whole process. This can also be proved using calculus, without any approximations: W(1, 2) = " x12,f x12,i F12dx12 = − " x12,f x12,... | University Physics I Classical Mechanics_Page_162_Chunk4360 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 145 ball goes up a little while in contact with your hand), and the rest, which is typically most of it, goes into increasing the system’s kinetic energy (in this case, just the ball’s; the earth’s kinetic energy does not change in any measurable way!). A B C rising... | University Physics I Classical Mechanics_Page_163_Chunk4361 |
146 CHAPTER 7. IMPULSE, WORK AND POWER practice, all we have to do is see how high the ball rises. At the ball’s maximum height (point C), as the second diagram shows, all the energy in the system is gravitational potential energy, and (as long as the system stays closed), all that energy is still equal to the work you... | University Physics I Classical Mechanics_Page_164_Chunk4362 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 147 the throw, from A to B ∆K Wgrav Whand rising, from A to C ∆K Whand Wgrav } Wext } Wext ∆K Whand Wgrav } Wext falling, from C to B Figure 7.4: Work-energy balance diagrams for the same toss illustrated in Fig. 7.3, but now the system is taken to be the ball only.... | University Physics I Classical Mechanics_Page_165_Chunk4363 |
148 CHAPTER 7. IMPULSE, WORK AND POWER The second observation is that the work done by an external force on a system does not depend on where the force comes from—that is to say, what physical arrangement we use to produce the force. Only the value of the force at each step and the displacement of the point of applicat... | University Physics I Classical Mechanics_Page_166_Chunk4364 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 149 force ⃗F t r,2 that is doing negative work on system B. System B, outlined in magenta, consists of block 2 and the earth and thus it includes only one internal interaction, namely gravity, which is conservative. This means that we can immediately apply the theor... | University Physics I Classical Mechanics_Page_167_Chunk4365 |
150 CHAPTER 7. IMPULSE, WORK AND POWER 7.4.3 Energy dissipated by kinetic friction In the situation illustrated in Fig. 7.5, we might calculate the energy dissipated by kinetic friction by indirect means. For instance, we can use the fact that the energy of system A is of two kinds, kinetic and “dissipated,” and theref... | University Physics I Classical Mechanics_Page_168_Chunk4366 |
7.6. IN SUMMARY 151 which is equal to 1 J/s. The average power going into or coming out of a system by mechanical means, that is to say, through the action of a force applied at a point undergoing a displacement ∆x, will be Pav = ∆E ∆t = W ∆t = F ∆x ∆t (7.29) assuming the force is constant. Note that in the limit when ... | University Physics I Classical Mechanics_Page_169_Chunk4367 |
152 CHAPTER 7. IMPULSE, WORK AND POWER 5. The result in 4 above holds only provided the boundary of the system is not drawn at a physical surface on which dissipation occurs. Put otherwise, kinetic friction or other similar dissipative forces (drag, air resistance) must be included as internal, not external forces. 6. ... | University Physics I Classical Mechanics_Page_170_Chunk4368 |
7.7. EXAMPLES 153 7.7 Examples 7.7.1 Braking Suppose you are riding your bicycle and hit the brakes to come to a stop. Assuming no slippage between the tire and the road: (a) Which force is responsible for removing your momentum? (By “you” I mean throughout “you and the bicycle.”) (b) Which force is responsible for rem... | University Physics I Classical Mechanics_Page_171_Chunk4369 |
154 CHAPTER 7. IMPULSE, WORK AND POWER 7.7.2 Work, energy and the choice of system: dissipative case Consider again the situation shown in Figure 7.5. Let m1 = 1 kg, m2 = 2 kg, and µk = 0.3. Use the solutions provided in Section 6.3 to calculate the work done by all the forces, and the changes in all energies, when the... | University Physics I Classical Mechanics_Page_172_Chunk4370 |
7.7. EXAMPLES 155 To plot all this as energy bars, if you do not have access to a very precise drawing program, you typically have to make some approximations. In this case, we see that ∆K2 = 2∆K1 (exactly), whereas ∆K1 ≃2∆Ediss, so we can use one box to represent Ediss, two boxes for ∆K1, three for Wext,A, four for ∆K... | University Physics I Classical Mechanics_Page_173_Chunk4371 |
156 CHAPTER 7. IMPULSE, WORK AND POWER (b) If the system is the block alone, the only energy it has is kinetic energy, which, as stated above, does not see a net change in this process. This means the net work done on the block by the external forces must be zero. The external forces in this case are the spring force a... | University Physics I Classical Mechanics_Page_174_Chunk4372 |
7.7. EXAMPLES 157 (a) The external forces on your body are gravity, pointing down, and the normal force from the floor, pointing up. Initially, as you start lowering your center of mass, the normal force has to be slightly smaller than gravity, since your center of mass acquires a small downward acceleration. However, e... | University Physics I Classical Mechanics_Page_175_Chunk4373 |
158 CHAPTER 7. IMPULSE, WORK AND POWER 7.8 Problems Problem 1 In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just before it hits the mattress? (b) How much work does th... | University Physics I Classical Mechanics_Page_176_Chunk4374 |
7.8. PROBLEMS 159 Problem 5 A block of mass 3 kg slides on a horizontal, rough surface towards a spring with k = 500 N/m. The kinetic friction coefficient between the block and the surface is µk = 0.6. If the block’s speed is 5 m/s at the instant it first makes contact with the spring, (a) Find the maximum compression of ... | University Physics I Classical Mechanics_Page_177_Chunk4375 |
160 CHAPTER 7. IMPULSE, WORK AND POWER | University Physics I Classical Mechanics_Page_178_Chunk4376 |
Chapter 8 Motion in two dimensions 8.1 Dealing with forces in two dimensions We have been able to get a lot of physics from our study of (mostly) one-dimensional motion only, but it goes without saying that the real world is a lot richer than that, and there are a number of new and interesting phenomena that appear whe... | University Physics I Classical Mechanics_Page_179_Chunk4377 |
162 CHAPTER 8. MOTION IN TWO DIMENSIONS F F Fx Fy x y θ θ 180° − θ Fx Figure 8.1: The components of a vector that makes an angle θ with the positive x axis. Two examples are shown, for θ < 90◦(in which case Fx > 0) and for 90◦< θ < 180◦(in which case Fx < 0). In both cases, Fy > 0. The triangle will always have the vec... | University Physics I Classical Mechanics_Page_180_Chunk4378 |
8.1. DEALING WITH FORCES IN TWO DIMENSIONS 163 eventually solve) the equations Fnet,x = max Fnet,y = may (8.2) We can show that Eqs. (8.2) must hold for any choice of orthogonal x and y axes, based on the fact that we know ⃗Fnet = m⃗a holds along one particular direction, namely, the direction common to ⃗Fnet and ⃗a, a... | University Physics I Classical Mechanics_Page_181_Chunk4379 |
164 CHAPTER 8. MOTION IN TWO DIMENSIONS 8.2 Projectile motion Projectile motion is basically just free fall, only with the understanding that the object we are tracking was “projected,” or “shot,” with some initial velocity (as opposed to just dropped from rest). Unlike in the previous cases of free fall that we have s... | University Physics I Classical Mechanics_Page_182_Chunk4380 |
8.2. PROJECTILE MOTION 165 The overall motion, then, is a combination of motion with constant velocity horizontally, and motion with constant acceleration vertically, and we can write down the corresponding equations of motion immediately: vx = vx,i vy = vy,i −gt x = xi + vx,it y = yi + vy,it −1 2gt2 (8.5) where (xi, y... | University Physics I Classical Mechanics_Page_183_Chunk4381 |
166 CHAPTER 8. MOTION IN TWO DIMENSIONS For projectile motion, however, vx does not change, so any change in K will affect only the second term in Eq. (8.8). Conservation of energy between any two instants i and f gives K + U G = 1 2mv2 x,i + 1 2mv2 y,i + mgyi = 1 2mv2 x,i + 1 2mv2 y,f + mgyf (8.9) The 1 2mv2 x,i term c... | University Physics I Classical Mechanics_Page_184_Chunk4382 |
8.3. INCLINED PLANES 167 8.3 Inclined planes Back in Chapter 2, I stated without proof that the acceleration of an object sliding, without friction, down an inclined plane making an angle θ with the horizontal was g sin θ. I can show you now why this is so, and introduce friction as well. FG θ θ F n F k a θ F k FG Eb s... | University Physics I Classical Mechanics_Page_185_Chunk4383 |
168 CHAPTER 8. MOTION IN TWO DIMENSIONS Newton’s second law, as given by equations (8.4) applied to this system, then reads: F g x + F k x = max = F g sin θ −F k (8.15) for the motion along the plane, and F g y + F n y = may = −F g cos θ + F n (8.16) for the direction perpendicular to the plane. Of course, since there ... | University Physics I Classical Mechanics_Page_186_Chunk4384 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 169 sliding, and on the other hand the constraint F s ≤µsmg cos θ. Putting these together we conclude that the block will not slide as long as mg sin θ ≤µsmg cos θ (8.23) or tan θ ≤µs (8.24) In short, as long as θ is small enough to satisfy Eq. (8.24), the block will not mo... | University Physics I Classical Mechanics_Page_187_Chunk4385 |
170 CHAPTER 8. MOTION IN TWO DIMENSIONS Another way to see this is to go back to the definition of acceleration. If an object has a velocity vector ⃗v(t) at the time t, and a different velocity vector ⃗v(t + ∆t) at the later time t + ∆t, then its average acceleration over the time interval ∆t is the quantity ⃗vav = (⃗v(t... | University Physics I Classical Mechanics_Page_188_Chunk4386 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 171 very small, and we can use the so-called “small angle approximation,” which states that sin x ≃x when x is small and expressed in radians. Therefore, by Eq. (8.25), |⃗aav| = 2v sin(θ/2) ∆t ≃vθ ∆t = v2∆t/R ∆t (8.26) This expression becomes exact as ∆t →0, and then ∆t can... | University Physics I Classical Mechanics_Page_189_Chunk4387 |
172 CHAPTER 8. MOTION IN TWO DIMENSIONS changing its direction of motion. You will find this effect illustrated in some detail in an example in the “Advanced Topics” section, if you want to look at it in more depth. On the other hand, getting a car to safely negotiate a turn is actually an important example of a situatio... | University Physics I Classical Mechanics_Page_190_Chunk4388 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 173 + r v P (x,y) x y R sin θ θ R cos θ Figure 8.6: A particle moving on a circle. The position vector has length R, so the x and y coordinates are R cos θ and R sin θ, respectively. The conventional positive direction of motion is indicated. The velocity vector is always, ... | University Physics I Classical Mechanics_Page_191_Chunk4389 |
174 CHAPTER 8. MOTION IN TWO DIMENSIONS where ω is positive for counterclockwise motion, and negative for clockwise. (There is a sense in which it is useful to think of ω as a vector, but, since it is not immediately obvious how or why, I will postpone discussion of this next chapter, after I have introduced angular mo... | University Physics I Classical Mechanics_Page_192_Chunk4390 |
8.5. IN SUMMARY 175 As we shall see later, the product Rα is also a useful quantity. It is not, however, equal to the mag- nitude of the acceleration vector, but only one of its two components, the tangential acceleration: at = Rα (8.37) The sign convention here is that a positive at represents a vector that is tangent... | University Physics I Classical Mechanics_Page_193_Chunk4391 |
176 CHAPTER 8. MOTION IN TWO DIMENSIONS on whatever value is necessary to keep the object from moving, up to a maximum value of F s max = µsF n. 5. An object moving in an arc of a circle of radius R with a speed v experiences a centripetal acceleration of magnitude ac = v2/R. “Centripetal” means the corresponding vecto... | University Physics I Classical Mechanics_Page_194_Chunk4392 |
8.6. EXAMPLES 177 8.6 Examples You will work out a rather thorough example of projectile motion in the lab, and Section 8.3 above already has the problem of a block sliding down an inclined plane worked out for you. The following example will show you how to use the kinematic angular variables of section 8.4.2 to deal ... | University Physics I Classical Mechanics_Page_195_Chunk4393 |
178 CHAPTER 8. MOTION IN TWO DIMENSIONS The angular acceleration, therefore, is α = ∆ω ∆t = ωf −ωi ∆t = 3.49 rad/s 1.3 s = 2.68 rad s2 (8.40) (b) The way to answer this question is to find out the total angular displacement, ∆θ, of the penny over the time interval considered (from t = 0 to t = 1.3 s), and then convert t... | University Physics I Classical Mechanics_Page_196_Chunk4394 |
8.7. ADVANCED TOPICS 179 8.7 Advanced Topics 8.7.1 Staying on track (This example studies a situation that you could easily setup experimentally at home (you can use a whole sphere instead of a half-sphere!), although to get the numbers to work out you really need to make sure that the friction between the surface and ... | University Physics I Classical Mechanics_Page_197_Chunk4395 |
180 CHAPTER 8. MOTION IN TWO DIMENSIONS The next thing we need to do is find the value of the speed v for a given angle θ. If we treat the object as a particle, its only energy is kinetic energy, and ∆K = Wnet (Eq. (7.11)), where Wnet is the work done on the particle by the net force acting on it. The normal force is al... | University Physics I Classical Mechanics_Page_198_Chunk4396 |
8.7. ADVANCED TOPICS 181 Now we just use these results in Eqs. (8.5). Specifically, we want to know how long it takes for the object to reach the ground, so we use the last equation (8.5) with y = 0 and solve for t: 0 = yi + vy,it −1 2gt2 (8.53) The result is t = 0.697 ! R/g. (You do not need to carry the “g” throughout... | University Physics I Classical Mechanics_Page_199_Chunk4397 |
182 CHAPTER 8. MOTION IN TWO DIMENSIONS so this is the natural extension of that. In general, you should always try to imagine which way the object would slide if friction disappeared altogether: ⃗F s must point in the direction opposite that. Thus, for a car traveling at a reasonable speed, the direction in which it w... | University Physics I Classical Mechanics_Page_200_Chunk4398 |
8.7. ADVANCED TOPICS 183 Note that the second equation would have F s becoming negative if v2 < gr tan θ. This means that below that speed, the force of static friction must actually point up the slope, as discussed above. We can call this particular speed, for which F s becomes zero, vno friction: vno friction = ! gr ... | University Physics I Classical Mechanics_Page_201_Chunk4399 |
184 CHAPTER 8. MOTION IN TWO DIMENSIONS 8.7.3 Rotating frames of reference: Centrifugal force and Coriolis force Imagine you are inside a rotating cylindrical room of radius R. There is a metal puck on the floor, a distance r from the axis of rotation, held in place with an electromagnet. At some time you switch offthe e... | University Physics I Classical Mechanics_Page_202_Chunk4400 |
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