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8.7. ADVANCED TOPICS 185 The cyan angle in the picture, which we could call ∆θpart, has tangent equal to √ R2 −r2/r, so we have ∆θroom = tan (∆θpart) (8.64) This tells us the two angles are going to be pretty close if they are small enough, which is what happens if the puck starts close enough to the wall in the first p... | University Physics I Classical Mechanics_Page_203_Chunk4401 |
186 CHAPTER 8. MOTION IN TWO DIMENSIONS 8.8 Problems Problem 1 A pitcher throws a fastball horizontally at a speed of 42 m/s. Neglecting air resistance, (a) How long does it take for the ball to reach the batter, a distance of 18.4 m away? (b) How much does the ball drop vertically in this time? (c) What is the vertica... | University Physics I Classical Mechanics_Page_204_Chunk4402 |
8.8. PROBLEMS 187 (b) What is the work done by gravity on the skier for the process described above? (Thinking now of the skier only as the system.) (c) If you could neglect the friction between the skis and the snow, what would be the speed of the skier at the bottom of the slope? Why? (d) If the speed of the skier is... | University Physics I Classical Mechanics_Page_205_Chunk4403 |
188 CHAPTER 8. MOTION IN TWO DIMENSIONS (d) Which force component in your diagram provides this centripetal acceleration? (e) Based on your results above, what is the angle the string must make with the horizontal? Problem 6 A golf ball is hit in such a way that it travels 300 m horizontally and stays in the air a tota... | University Physics I Classical Mechanics_Page_206_Chunk4404 |
8.8. PROBLEMS 189 (e) The power is turned offand the platform slows down to a stop with a constant angular accel- eration of −0.02 rad/s2. How long does it take for it to stop completely? (f) What is the man’s tangential acceleration during that time? Does his centripetal acceleration change? Why? (g) How many turns doe... | University Physics I Classical Mechanics_Page_207_Chunk4405 |
190 CHAPTER 8. MOTION IN TWO DIMENSIONS | University Physics I Classical Mechanics_Page_208_Chunk4406 |
Chapter 9 Rotational dynamics Rotational motion, which involves an object spinning around an axis, or revolving around a point in space, is actually rather common in nature, so much so that Galileo thought (mistakenly) that circular motion, rather than motion on a straight line, was the “natural,” or “unforced” state o... | University Physics I Classical Mechanics_Page_209_Chunk4407 |
192 CHAPTER 9. ROTATIONAL DYNAMICS Now, consider the kinetic energy of an extended object that is rotating around some axis. We may treat the object as being made up of many “particles” (small parts) of masses m1, m2 . . .. If the object is rigid, all the particles move together, in the sense that they all rotate throu... | University Physics I Classical Mechanics_Page_210_Chunk4408 |
9.2. ANGULAR MOMENTUM 193 9.2 Angular momentum Back in Chapter 3 we introduced the momentum of an object moving in one dimension as p = mv, and found that it had the interesting property of being conserved in collisions between objects that made up an isolated system. It seems natural to ask whether the corresponding r... | University Physics I Classical Mechanics_Page_211_Chunk4409 |
194 CHAPTER 9. ROTATIONAL DYNAMICS velocity. Yet, we would like to define L in such a way that it will remain constant when, in fact, nothing in the particle’s actual state of motion is changing. The way to do this, for a particle moving on a straight line, is to define L as the product of mv times, not the distance of t... | University Physics I Classical Mechanics_Page_212_Chunk4410 |
9.2. ANGULAR MOMENTUM 195 thing is that, with this definition, the angular momentum will be conserved in an important kind of process, namely, a collision that converts linear to rotational motion, as illustrated in Fig. 9.2 below. v r θ O 1 2 Figure 9.2: Collision between two particles, 1 and 2, of equal masses. Partic... | University Physics I Classical Mechanics_Page_213_Chunk4411 |
196 CHAPTER 9. ROTATIONAL DYNAMICS thin rod of mass m and length l pivoted at one end. What happens now when particle 1 strikes the rod? vi r θ O 1 vf Figure 9.3: Collision between a particle initially moving with velocity ⃗v1 and a rod of length l pivoted at an endpoint. The particle strikes the rod perpendicularly, a... | University Physics I Classical Mechanics_Page_214_Chunk4412 |
9.3. THE CROSS PRODUCT AND ROTATIONAL QUANTITIES 197 Note that this does reduce to our previous results for the collision of two particles if we make I = ml2 (which one could always do, by choosing the mass of the rod, M, appropriately). On the other hand, as indicated at the end of the previous subsection, for general... | University Physics I Classical Mechanics_Page_215_Chunk4413 |
198 CHAPTER 9. ROTATIONAL DYNAMICS B A × B B × A A A B Figure 9.4: The “right-hand rule” to determine the direction of the cross product. Line up the first vector with the fingers, and the second vector with the flat of the hand, and the thumb will point in the correct direction. In the first drawing, we are looking at the... | University Physics I Classical Mechanics_Page_216_Chunk4414 |
9.3. THE CROSS PRODUCT AND ROTATIONAL QUANTITIES 199 product ⃗r × ⃗v will always point upwards, along the positive z axis. Furthermore, since ⃗r and ⃗v always stay perpendicular, the magnitude of ⃗L, by Eq. (9.9), will always be |⃗L| = mR|⃗v|. Taking note of I = mR2 and of Eq. (8.36), we see we have then |⃗L| = ( mR2) ... | University Physics I Classical Mechanics_Page_217_Chunk4415 |
200 CHAPTER 9. ROTATIONAL DYNAMICS For the motion depicted in Fig. 9.5, the vector ⃗α will point along the positive z axis if the vector ⃗ω is growing (which means the particle is speeding up), and along the negative z axis if ⃗ω is decreasing. One important property the cross product does have is the distributive prop... | University Physics I Classical Mechanics_Page_218_Chunk4416 |
9.4. TORQUE 201 momentum involves the particle’s distance to a point. For particles at different “heights” along the axis of rotation, these quantities are different. It can be shown that, in the general case, all we can say is that Lz = Iωz, if we call z the axis of rotation and calculate ⃗L relative to a point on that ... | University Physics I Classical Mechanics_Page_219_Chunk4417 |
202 CHAPTER 9. ROTATIONAL DYNAMICS The quantity ⃗r × ⃗F is called the torque of a force around a point (the origin from which ⃗r is calculated, typically a pivot point or center of rotation). It is denoted with the Greek letter τ, “tau”: ⃗τ = ⃗r × ⃗F (9.21) For an extended object or system, the rate of change of the an... | University Physics I Classical Mechanics_Page_220_Chunk4418 |
9.4. TORQUE 203 wrenches). It is also perpendicular to the rod, for maximum effect (sin θ = 1). The force ⃗F2, by contrast, although also applied at the point A is at a disadvantage because of the relatively small angle it makes with ⃗rA. If you imagine breaking it up into components, parallel and perpendicular to the r... | University Physics I Classical Mechanics_Page_221_Chunk4419 |
204 CHAPTER 9. ROTATIONAL DYNAMICS Coming back to Eq. (9.22), the main message of this section (other, of course, than the definition of torque itself), is that the rate of change of an object or system’s angular momentum is equal to the net torque due to the external forces. Two special results follow from this one. Fi... | University Physics I Classical Mechanics_Page_222_Chunk4420 |
9.5. STATICS 205 important in engineering (particularly in mechanical engineering). In an introductory physics course, we can only deal with it at a very elementary level, by ignoring altogether the deformation of extended objects such as planks and beams (and the associated stresses), and just imposing two simple cond... | University Physics I Classical Mechanics_Page_223_Chunk4421 |
206 CHAPTER 9. ROTATIONAL DYNAMICS any of these complications, just to keep the example simple, but they could be dealt with in exactly the same way. θ ϖ/2– θ l Fwl n Fgl n FEl G Fgl s Figure 9.7: A ladder leaning against a frictionless wall: sketch and extended free-body diagram. With the convention that a vector quan... | University Physics I Classical Mechanics_Page_224_Chunk4422 |
9.6. ROLLING MOTION 207 9.6 Rolling motion As a step up from a statics problem, we may consider a situation in which the sum of the external forces is zero, as well as the sum of the external torques, yet the system is moving. We call this “unforced motion.” The first condition, 1 ⃗Fext = 0, means that the center of mas... | University Physics I Classical Mechanics_Page_225_Chunk4423 |
208 CHAPTER 9. ROTATIONAL DYNAMICS when the condition for rolling without slipping is satisfied: |vcm| = R|ω| (9.34) At this point, the object will start rolling without slipping, and losing speed at a much slower rate. The origin of the condition (9.34) is fairly straightforward. You can imagine an object that is rolli... | University Physics I Classical Mechanics_Page_226_Chunk4424 |
9.6. ROLLING MOTION 209 contact with the surface have zero instantaneous velocity. This means that, even if there was a force acting on the object at that point (such as the force of static friction), it would do no work, since the instantaneous power Fv for a force applied there would always be equal to zero. We do no... | University Physics I Classical Mechanics_Page_227_Chunk4425 |
210 CHAPTER 9. ROTATIONAL DYNAMICS For the situation shown in Fig. 9.9, if we take down the plane as the positive direction for linear motion, and clockwise torques as negative, we have to write acm = −Rα. In the direction perpen- dicular to the plane, we conclude from (9.35) that F n = Mg cos θ, an equation we will no... | University Physics I Classical Mechanics_Page_228_Chunk4426 |
9.7. IN SUMMARY 211 (Note how I have made use of the dot product to calculate the magnitude squared of a vector.) On the last line, the quantity ⃗p′ is the momentum of that particle in the CM frame. Adding those momenta for all the particles should give zero, since, as we saw in an earlier chapter, the center of mass f... | University Physics I Classical Mechanics_Page_229_Chunk4427 |
212 CHAPTER 9. ROTATIONAL DYNAMICS that the relation ⃗v = ⃗ω ×⃗r always holds, where ⃗r is the (instantaneous) position vector of the particle on the circle. 2. The particle’s kinetic energy can be written as Krot = 1 2Iω2, where I = mR2 is the rotational inertia or moment of inertia. For an extended object rotating ab... | University Physics I Classical Mechanics_Page_230_Chunk4428 |
9.8. EXAMPLES 213 9.8 Examples This was a long chapter, in part because it contains a number of useful worked-out examples; so please make sure not to overlook them! Section 9.2.2 showed a couple of examples of problems that can be solved using conservation of angular momentum. Section 9.3 shows you how to set up and s... | University Physics I Classical Mechanics_Page_231_Chunk4429 |
214 CHAPTER 9. ROTATIONAL DYNAMICS For the front wheel, this is in fact the only external force, and the only force of any sort that exerts a torque on that wheel (there are forces acting at the axle, but they exert no torque around the axle). Since the torque has to be clockwise, then, the force of static friction on ... | University Physics I Classical Mechanics_Page_232_Chunk4430 |
9.8. EXAMPLES 215 Solution The figure below shows the setup, plus free-body diagrams for the two blocks (the vertical forces on block 1 have been left out to avoid cluttering the figure, since they are not relevant here), and an extended free-body diagram for the pulley. (You can see from the pulley diagram that there ha... | University Physics I Classical Mechanics_Page_233_Chunk4431 |
216 CHAPTER 9. ROTATIONAL DYNAMICS which is easily solved for a: a = m2g m1 + m2 + I/R2 (9.55) If you look at the structure of this equation, it all makes sense. The numerator is the force of gravity on block 2, which is, ultimately, the force responsible for setting the whole thing in motion. The denominator is, essen... | University Physics I Classical Mechanics_Page_234_Chunk4432 |
9.9. PROBLEMS 217 9.9 Problems Problem 1 An ice skater has a moment of inertia equal to 1.9 kg·m2 when she is rotating with her arms stretched out, at a rate of 2 revolutions per second. She then brings her arms in, lined up with her axis of rotation, so her moment of inertia becomes 1.1 kg·m2. (a) What is her new angu... | University Physics I Classical Mechanics_Page_235_Chunk4433 |
218 CHAPTER 9. ROTATIONAL DYNAMICS (c) If the worker were to let go of the plank, what would its angular acceleration be as it starts swinging down? The moment of inertia is I = 1 3Ml2. (Note: assume the toolbox stops pressing down on the plank immediately. This is a good approximation, as you shall see below.) (d) Con... | University Physics I Classical Mechanics_Page_236_Chunk4434 |
9.9. PROBLEMS 219 A 20-kg plank of length l = 4 m is supported at both ends as shown in the figure. A 60-kg man is standing a distance l/3 from the right end of the plank. 1.33 m 4 m (a) Draw an extended free-body diagram for the plank. Try to get the scale of the forces at least qualitatively right. (b) Find the upward... | University Physics I Classical Mechanics_Page_237_Chunk4435 |
220 CHAPTER 9. ROTATIONAL DYNAMICS | University Physics I Classical Mechanics_Page_238_Chunk4436 |
Chapter 10 Gravity 10.1 The inverse-square law Up to this point, all I have told you about gravity is that, near the surface of the Earth, the gravitational force exerted by the Earth on an object of mass m is F G = mg. This is, indeed, a pretty good approximation, but it does not really tell you anything about what th... | University Physics I Classical Mechanics_Page_239_Chunk4437 |
222 CHAPTER 10. GRAVITY Equation (10.1), as stated, applies to particles, that is to say, in practice, to any objects that are very small compared to the distance between them. The net force between extended masses can be obtained using calculus, by breaking up the two objects into very small pieces and adding (vectori... | University Physics I Classical Mechanics_Page_240_Chunk4438 |
10.1. THE INVERSE-SQUARE LAW 223 near the surface of any other planet or moon, just replacing ME and RE by the mass and radius of the planet or moon in question. Thus, we could speak of gmoon, gMars, etc., and in some homework problems you will be asked to calculate these quantities. Clearly, besides telling you how fa... | University Physics I Classical Mechanics_Page_241_Chunk4439 |
224 CHAPTER 10. GRAVITY 10.1.1 Gravitational potential energy Ever since I introduced the concept of potential energy in Chapter 5, I have been using U G = mgy for the gravitational potential energy of the system formed by the Earth and an object of mass m a height y above the Earth’s surface. This works well as long a... | University Physics I Classical Mechanics_Page_242_Chunk4440 |
10.1. THE INVERSE-SQUARE LAW 225 it increases (since a number like, say, −0.1 is, in fact, greater than a number like −10). So as the particles are moved farther and farther apart, the potential energy of the system does increase—all the way up to a maximum value of zero! Still, even if it makes sense mathematically, t... | University Physics I Classical Mechanics_Page_243_Chunk4441 |
226 CHAPTER 10. GRAVITY time, and hence does not need to be included in most energy calculations involving gravitational forces between extended objects. One thing that you may be wondering about, regarding Eq. (10.6) for the potential energy of a pair of particles (or, for that matter, Eq. (10.1) for the force), is wh... | University Physics I Classical Mechanics_Page_244_Chunk4442 |
10.1. THE INVERSE-SQUARE LAW 227 initially on the surface of the earth, and then we move it to a height h above the earth. The change in potential energy, according to (10.6), is U G f −U G i = −GMEm RE + h + GMEm RE (10.9) If we write both terms with a common denominator, we get U G f −U G i = GMEm (RE + h)RE h ≃GMEm ... | University Physics I Classical Mechanics_Page_245_Chunk4443 |
228 CHAPTER 10. GRAVITY So, if we want to, say, put a satellite in a circular orbit around a central body of mass M and at a distance R from the center of that body, we can do it, but only provided we give the satellite an initial velocity v = ! GM/R in a direction perpendicular to the radius. But what if we were to re... | University Physics I Classical Mechanics_Page_246_Chunk4444 |
10.1. THE INVERSE-SQUARE LAW 229 and the distance of each focus to the center of the ellipse is given by the product ea, that is, the product of the eccentricity and the semimajor axis. (This explains why the “eccentricity” is called that: it is a measure of how “off-center” the focus is.) For an object moving in an ell... | University Physics I Classical Mechanics_Page_247_Chunk4445 |
230 CHAPTER 10. GRAVITY Since an ellipse has only two parameters, and we have two constants of the motion (the total energy, E, and the angular momentum, L), we should be able to determine what the orbit will look like based on just those two quantities. Under the assumption we are making here, that the very massive ob... | University Physics I Classical Mechanics_Page_248_Chunk4446 |
10.1. THE INVERSE-SQUARE LAW 231 elliptical orbit (of which a circle is a special case, if you give it the precise speed v = ! GM/r in the right direction). If you give it precisely the escape speed (10.15), the total energy of the system will be zero and the trajectory of the object will be a parabola; and if you give... | University Physics I Classical Mechanics_Page_249_Chunk4447 |
232 CHAPTER 10. GRAVITY For elliptical orbits, one can prove the result e = * 1 − L2 aGMm2 (10.16) which shows how the eccentricity increases as L decreases, for a given value of a (which is to say, for a given total energy). I should at least sketch how to obtain this result, since it is a variant of a procedure that ... | University Physics I Classical Mechanics_Page_250_Chunk4448 |
10.1. THE INVERSE-SQUARE LAW 233 10.1.3 Kepler’s laws The first great success of Newton’s theory was to account for the results that Johannes Kepler had extracted from astronomical data on the motion of the planets around the sun. Kepler had managed to find a number of regularities in a mountain of data (most of which we... | University Physics I Classical Mechanics_Page_251_Chunk4449 |
234 CHAPTER 10. GRAVITY instance, if the particle starts out at B instead, then in the same time interval ∆t it will move to a point B′ such that the area of the “curved triangle” OBB′ equals the area of OAA′. Qualitatively, this means that the particle needs to move more slowly when it is farther from the center of at... | University Physics I Classical Mechanics_Page_252_Chunk4450 |
10.1. THE INVERSE-SQUARE LAW 235 From this some simple manipulation gives you T2 = T1 $R2 R1 %3/2 (10.20) Note you can express R1 and R2 in any units you like, as long as you use the same units for both, and similarly T1 and T2. For instance, if you use the Earth as your reference “planet 1,” then you know that T1 = 1 ... | University Physics I Classical Mechanics_Page_253_Chunk4451 |
236 CHAPTER 10. GRAVITY solar system actually forms an isolated system with the sun, since all the planets are really pulling gravitationally on each other all the time. Particularly, Jupiter and Saturn have a non-negligible influence on each other’s orbits, and on the orbits of every other planet, which can only be per... | University Physics I Classical Mechanics_Page_254_Chunk4452 |
10.2. WEIGHT, ACCELERATION, AND THE EQUIVALENCE PRINCIPLE 237 (a) (b) Figure 10.8: If you are holding something while in free fall (a) and let go, since you are all accelerating at the same rate, it stays in the same position relative to you (b), so it appears to be weightless. The familiar sensation of weight, on the ... | University Physics I Classical Mechanics_Page_255_Chunk4453 |
238 CHAPTER 10. GRAVITY Newton’s second law). (a) gravity g, a = – g (b) gravity g, a = 0 (c) no gravity, a = g Fgl n Fgl n Fsl spr Fsu spr Fsu spr Fsl spr FEu G FEu G FEl G FEl G Figure 10.9: (a) In free fall, your skeleton (represented here by a relaxed spring) does not need to support your upper body, so there is no... | University Physics I Classical Mechanics_Page_256_Chunk4454 |
10.2. WEIGHT, ACCELERATION, AND THE EQUIVALENCE PRINCIPLE 239 illustrates what happens when you drop something while traveling in the upwardly accelerating rocket, in the absence of gravity. From an inertial observer’s point of view, the object you drop merely keeps the upward velocity it had the moment it left your ha... | University Physics I Classical Mechanics_Page_257_Chunk4455 |
240 CHAPTER 10. GRAVITY the object would “fall towards the wall,” just like the object considered in Example 9.6.3 (previous chapter). Unfortunately, while the idea might work for a space station, it would probably be impractical for a spaceship, since one would need a fairly large R and/or a fairly large rotation rate... | University Physics I Classical Mechanics_Page_258_Chunk4456 |
10.3. IN SUMMARY 241 shifted and/or distorted by the gravity of the galaxies that lie in between them and us. It has even become possible to imagine an object so dense that it would “capture” light, attracting it so strongly that it could not leave the object’s neighborhood. Such an object has come to be called a black... | University Physics I Classical Mechanics_Page_259_Chunk4457 |
242 CHAPTER 10. GRAVITY 6. The solutions to the Kepler problem are of two types, depending on the system’s total energy E: bound, elliptical orbits (including circles as a special case), if E < 0; and unbound hyperbolic trajectories, if E > 0. The special trajectory obtained when E = 0 is a parabola. 7. For the ellipti... | University Physics I Classical Mechanics_Page_260_Chunk4458 |
10.4. EXAMPLES 243 10.4 Examples 10.4.1 Orbital dynamics In the early days of space flight, astronauts sometimes mentioned the counterintuitive aspects of orbital flight. For example, if, from a circular orbit around the Earth, they wanted to move to a lower orbit, the way to do it was to slow down their capsule (by firin... | University Physics I Classical Mechanics_Page_261_Chunk4459 |
244 CHAPTER 10. GRAVITY To get the answer, recall that we found in (a) that the total mechanical energy E has gone down. But, since E is a negative number, this means the magnitude of E has gone up. Then, in the formula (10.14), E = −GMm 2a the semimajor axis a must have gone down. For the original circular orbit, we h... | University Physics I Classical Mechanics_Page_262_Chunk4460 |
10.4. EXAMPLES 245 The diagram of the situation is above (previous page). The long-dash circle is the original orbit; the solid line is the elliptical orbit resulting from the speed reduction at point A; the short-dash circle is the circular orbit that would result from another speed reduction at the point P. Note: the... | University Physics I Classical Mechanics_Page_263_Chunk4461 |
246 CHAPTER 10. GRAVITY equations (10.12) says as well.). So we have e = a −rmin a = 1 −rmin a = 1 −0.59 17.9 = 0.967 (10.23) Note that we did not even have to convert AU to kilometers. In these types of problems, particularly, where you have to manipulate very large numbers, it really pays offto do all the calculations... | University Physics I Classical Mechanics_Page_264_Chunk4462 |
10.5. ADVANCED TOPICS 247 10.5 Advanced Topics 10.5.1 Tidal Forces Throughout this chapter we have treated the objects interacting gravitationally as if they were particles, that is to say, as if they were non-deformable and their shape and relative orientation did not matter. However, these conditions are never quite ... | University Physics I Classical Mechanics_Page_265_Chunk4463 |
248 CHAPTER 10. GRAVITY forces from the primary will result in a net torque on the satellite that will tend to slow down its rotation; conversely, if it is rotating too slowly, the torque will tend to speed up the rotation. A torque-free situation will only happen when the satellite’s period of rotation exactly matches... | University Physics I Classical Mechanics_Page_266_Chunk4464 |
10.6. PROBLEMS 249 10.6 Problems Problem 1 Suppose you fire a projectile straight up from the Earth’s North Pole with a speed of 10.5 km/s. Ignore air resistance. (a) How far from the center of the Earth does the projectile rise? How high above the surface of the Earth is that? (The radius of the Earth is RE = 6.37 × 10... | University Physics I Classical Mechanics_Page_267_Chunk4465 |
250 CHAPTER 10. GRAVITY from the sun). What is then its angular momentum? (f) Draw a sketch of an elliptical orbit. On your sketch, indicate (1) the semimajor axis, and (2) qualitatively, where the sun might be. (g) The point in its orbit where the asteroid is farthest away from the sun is called aphelion. Use conserva... | University Physics I Classical Mechanics_Page_268_Chunk4466 |
10.6. PROBLEMS 251 an orbit around the sun, in earth years? (Do not look it up! You need to show how you can calculate it using what you have learned in this chapter.) | University Physics I Classical Mechanics_Page_269_Chunk4467 |
252 CHAPTER 10. GRAVITY | University Physics I Classical Mechanics_Page_270_Chunk4468 |
Chapter 11 Simple harmonic motion 11.1 Introduction: the physics of oscillations It is probably not an exaggeration to suggest that we are all introduced to oscillatory motion from our first moments of life. Babies, it seems, are constantly rocked to sleep, in many cases using devices, such as cradles and rocking chairs... | University Physics I Classical Mechanics_Page_271_Chunk4469 |
254 CHAPTER 11. SIMPLE HARMONIC MOTION In fact, oscillatory motion is extremely common, both in natural systems and in human-made structures. It essentially requires only two things: a stable equilibrium configuration, where the stability is ensured by what we call a restoring force; and inertia, which, of course, every... | University Physics I Classical Mechanics_Page_272_Chunk4470 |
11.2. SIMPLE HARMONIC MOTION 255 is to say, in one dimension, if x0 is the equilibrium position, the restoring force has the form F = −k(x −x0) (11.2) We are familiar with this from Hooke’s “law” for an ideal spring (see Chapter 6). So, an object attached to an ideal, massless spring, as in the figure below, should perf... | University Physics I Classical Mechanics_Page_273_Chunk4471 |
256 CHAPTER 11. SIMPLE HARMONIC MOTION harmonic motion only happens for relatively small oscillations, but “relatively small” can still be fairly large sometimes, and even as an approximation it is often an extremely valuable one. The other distinctive characteristic of simple harmonic motion is that the position funct... | University Physics I Classical Mechanics_Page_274_Chunk4472 |
11.2. SIMPLE HARMONIC MOTION 257 The answer is that there is a very close relationship between simple harmonic motion and circular motion with constant speed, as Figure 11.3 illustrates: as the point P rotates with constant angular velocity ω, its projection onto the x axis (the red dot in the figure) performs simple ha... | University Physics I Classical Mechanics_Page_275_Chunk4473 |
258 CHAPTER 11. SIMPLE HARMONIC MOTION The expression (11.4) for ω is typical of what we find for many different kinds of oscillators: the restoring force (here represented by the spring constant k) and the object’s inertia (m) together determine the frequency of the motion, acting in opposite directions: a larger restor... | University Physics I Classical Mechanics_Page_276_Chunk4474 |
11.2. SIMPLE HARMONIC MOTION 259 If we stick to using cosines, for definiteness, then the most general equation for the position of a simple harmonic oscillator is as follows: x(t) = A cos(ωt + φ) (11.10) where φ is what we call a “phase angle,” that allows us to match the function to the initial conditions—by which I m... | University Physics I Classical Mechanics_Page_277_Chunk4475 |
260 CHAPTER 11. SIMPLE HARMONIC MOTION so the total energy of the system is constant (independent of time), at it should be, in the absence of dissipation. Figure 11.5 shows how the potential and kinetic energies oscillate in opposition, so one is maximum whenever the other is minimum. It also shows that they oscillate... | University Physics I Classical Mechanics_Page_278_Chunk4476 |
11.2. SIMPLE HARMONIC MOTION 261 This is a remarkable result, because the force of gravity has disappeared completely from the final expression. Basically, the system behaves as if it consisted of just a spring of constant k with equilibrium length l′ = l + y0 −y′ 0, and no gravity. In other words, the only thing gravit... | University Physics I Classical Mechanics_Page_279_Chunk4477 |
262 CHAPTER 11. SIMPLE HARMONIC MOTION external force is constant, and does not change direction, this work will be positive half the time, and negative half the time. If it is kinetic friction, then of course it will change direction every half cycle, and the work will be negative all the time. In the case shown in Fi... | University Physics I Classical Mechanics_Page_280_Chunk4478 |
11.3. PENDULUMS 263 F t F G l θ o Figure 11.7: A simple pendulum. The mass of the bob is m, the length of the string is l, and torques are calculated around the point of suspension O. The counterclockwise direction is taken as positive. Instead, I will take advantage of the obvious fact that the bob moves on an arc of ... | University Physics I Classical Mechanics_Page_281_Chunk4479 |
264 CHAPTER 11. SIMPLE HARMONIC MOTION Equation (11.21) is an example of what is known as a differential equation. The problem is to find a function of time, θ(t), that satisfies this equation; that is to say, when you take its second derivative the result is equal to −(g/l) sin[θ(t)]. Such functions exist and are called ... | University Physics I Classical Mechanics_Page_282_Chunk4480 |
11.3. PENDULUMS 265 11.3.2 The “physical pendulum” By a “physical pendulum” one means typically any pendulum-like device for which the moment of inertia is not given by the simple expression I = ml2. This means that the mass is not concentrated into a single point-like particle a distance l away from the point of suspe... | University Physics I Classical Mechanics_Page_283_Chunk4481 |
266 CHAPTER 11. SIMPLE HARMONIC MOTION As an example, consider the oscillations of a uniform, thin rod of length l and mass m pivoted at one end. We then have I = ml2/3, and d = l/2, so Eq. (11.28) gives ω = * 3g 2l (11.29) This is about 22% larger than the result (11.25) for a simple pendulum of the same length, imply... | University Physics I Classical Mechanics_Page_284_Chunk4482 |
11.4. IN SUMMARY 267 9. A simple pendulum (a point particle of mass m suspended from a massless, inextensible string of length l) will perform harmonic oscillations around the vertical provided the small angle, approximation, sin θ ≃θ, holds. The angular frequency of these oscillations is ω = ! g/l. 10. A rigid object ... | University Physics I Classical Mechanics_Page_285_Chunk4483 |
268 CHAPTER 11. SIMPLE HARMONIC MOTION 11.5 Examples 11.5.1 Oscillator in a box (a basic accelerometer!) Consider a block-spring system inside a box, as shown in the figure. The block is attached to the spring, which is attached to the inside wall of the box. The mass of the block is 0.2 kg. For parts (a) through (f), a... | University Physics I Classical Mechanics_Page_286_Chunk4484 |
11.5. EXAMPLES 269 (b) The amplitude will be 10 cm, since it is released at that point with no kinetic energy. (c) The velocity is minimum (largest in magnitude, but with a negative sign) as the object passes through the equilibrium position moving to the left. vmin = −ωA = − $ 30 rad s % × 0.1 m = −3 m s (d) The accel... | University Physics I Classical Mechanics_Page_287_Chunk4485 |
270 CHAPTER 11. SIMPLE HARMONIC MOTION system subject to a gravitational interaction that pulls any object with mass m with a force equal to ma in the direction opposite the acceleration. Therefore, inside the box, which is accelerating towards the left, the block behaves as if there was a force of gravity of magnitude... | University Physics I Classical Mechanics_Page_288_Chunk4486 |
11.5. EXAMPLES 271 As shown in Section 11.3.2, we have then ω = * Mgw 2I (11.30) Squaring this, and solving for I/M, I M = gw 2ω2 = 9.8 m/s2 × 0.025 m 2 × (2π/5 s)2 = 0.0776 m2 (11.31) The moment of inertia is to be calculated around the point O, that is to say, the point of suspension (where the knot is in the figure).... | University Physics I Classical Mechanics_Page_289_Chunk4487 |
272 CHAPTER 11. SIMPLE HARMONIC MOTION 11.6 Advanced Topics 11.6.1 Mass on a spring damped by friction with a surface Consider the system depicted in Figure 11.2 in the presence of friction between the block and the surface. Let the coefficient of kinetic friction be µk and the coefficient of static friction be µs. As usua... | University Physics I Classical Mechanics_Page_290_Chunk4488 |
11.6. ADVANCED TOPICS 273 Figure 11.9: Damped oscillations. Note that, in general, the oscillator does not stop at the equilibrium position. Rather, its final position will be at the end of the last half-swing, which is either x′ 0 −An (if the number n of half-periods is odd), or −x′ 0 + An, if the number n is even. Eit... | University Physics I Classical Mechanics_Page_291_Chunk4489 |
274 CHAPTER 11. SIMPLE HARMONIC MOTION d (a) (b) (c) Figure 11.10: (a) Torsion balance. The extremes of the oscillation are drawn in black and gray, respectively. (b) The view from the top. The dashed line indicates the equilibrium position. (c) In the presence of two nearby large masses, the equilibrium position is ti... | University Physics I Classical Mechanics_Page_292_Chunk4490 |
11.6. ADVANCED TOPICS 275 the oscillations will not change, but the equilibrium position will. In Eq. (11.15) we found that y′ 0 −y0 = Fext/k for a spring of spring constant k, where y0 was the old and y′ 0 the new equilibrium position (the force was equal to −mg; the displacement of the equilibrium position will be in... | University Physics I Classical Mechanics_Page_293_Chunk4491 |
276 CHAPTER 11. SIMPLE HARMONIC MOTION 11.7 Problems Problem 1 A block of mass m is sliding on a frictionless, horizontal surface, with a velocity vi. It hits an ideal spring, of spring constant k, which is attached to the wall. The spring compresses until the block momentarily stops, and then starts expanding again, s... | University Physics I Classical Mechanics_Page_294_Chunk4492 |
11.7. PROBLEMS 277 (that is, with reference to the figure, y0 −y′ 0 = 0.02 m). (a) What is the value of the spring constant k? (b) If you stretch the spring by an additional 2 cm downward from this equilibrium position, and release it, what will be the frequency of the oscillations? (c) Now consider the system formed by... | University Physics I Classical Mechanics_Page_295_Chunk4493 |
278 CHAPTER 11. SIMPLE HARMONIC MOTION | University Physics I Classical Mechanics_Page_296_Chunk4494 |
Chapter 12 Waves in one dimension 12.1 Traveling waves In our study of mechanics we have so far dealt with particle-like objects (objects that have only translational energy), and extended, rigid objects, which may also have rotational energy. We have, however, implicitly assumed that all the objects we studied had som... | University Physics I Classical Mechanics_Page_297_Chunk4495 |
280 CHAPTER 12. WAVES IN ONE DIMENSION pulse” traveling down the slinky, with very little distortion; you may even be able to see it being reflected at the other end, and coming back, before all its energy is dissipated away. direction of propagation of the wave (x) direction of displacement of the medium (x) Figure 12.... | University Physics I Classical Mechanics_Page_298_Chunk4496 |
12.1. TRAVELING WAVES 281 Perhaps the most important (and remarkable) property of wave motion is that it can carry energy and momentum over relatively long distances without an equivalent transport of matter. Again, think of the slinky: the “pulse” can travel through the slinky’s entire length, carrying momentum and en... | University Physics I Classical Mechanics_Page_299_Chunk4497 |
282 CHAPTER 12. WAVES IN ONE DIMENSION of that slice, along the x axis, at the time t will be given by x + ξ(x, t). In both of these cases, the displacement vector ξ reduces to a single nonzero component (along the y or x axis, respectively), which can, of course, be positive or negative. I will restrict myself implici... | University Physics I Classical Mechanics_Page_300_Chunk4498 |
12.1. TRAVELING WAVES 283 0 2 4 6 8 10 -1 -0.5 0 0.5 1 0 2 4 6 8 10 0 2 4 6 10 t = 0 t = ∆t λ cΔt x ξ t = 0 t = ∆t x x Figure 12.3: Top: two snapshots of a traveling harmonic wave at t = 0 (solid) and at t = ∆t (dashed). The quantity ξ is the displacement of a typical particle of the medium at each point x (the wave is... | University Physics I Classical Mechanics_Page_301_Chunk4499 |
284 CHAPTER 12. WAVES IN ONE DIMENSION as ξ(x, t) = ξ0 sin 22π λ (x −ct) 3 (12.5) This suggests that if we want to have a wave moving to the left instead, all we have to do is change the sign of the term proportional to c, which is indeed the case. In contrast to the wave speed, which is a constant, the speed of any pa... | University Physics I Classical Mechanics_Page_302_Chunk4500 |
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