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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 place. The picture shows, for clarity, the case when r = 0.7R, which gives ∆θroom = 1.02 rad, and ∆θpart = tan−1(1.02) = 0.8 rad. For r = 0.9R, on the other hand, one finds ∆θroom = 0.48 rad, and ∆θpart = tan−1(0.48) = 0.45 rad. In terms of pseudoforces (forces that do not, physically, exist, but may be introduced to describe mathematically the motion of objects in non-inertial frames of reference), the non-inertial observer would say that the puck heads towards the wall because of a centrifugal force (that is, a force pointing away from the center of rotation), and while doing so it drifts to the right because of the so-called Coriolis force. | 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 vertical component of the ball’s velocity as it reaches the batter? (Just to set the record straight, a real fastball would probably not drop that much, because of a lift force, called the Magnus effect, due to the interaction of the air with the backspin of the ball!) Problem 2 Two blocks are connected by a massless string threaded over a massless, frictionless pulley, as shown in the picture. The mass of block 1 is 2 kg and the mass of block 2 is 1.5 kg. The angle of the incline is 30 degrees. There is friction between the block and the inclined surface. 1 2 30º (a) Start by assuming that the coefficient of static friction is strong enough to keep the system from moving, and draw free-body diagrams for the two blocks. Try to get all the forces approximately to scale. (The following questions may be helpful.) (b) If the system is not moving, what is the magnitude of the tension? (c) What is the magnitude of the normal force? (d) How large does the coefficient of static friction have to be to keep the system from moving? (e) Now suppose the system is moving, and the coefficient of sliding (kinetic) friction is 0.2. What is the acceleration of the system? (f) What is the tension now? (g) What is the rate at which energy is dissipated (instantaneous dissipated power) when the system’s velocity is 3 m/s? Problem 3 A 60-kg skier starts sliding from rest from the top of a slope that makes an angle of 30◦ with the horizontal. Assume the bottom of the slope is 100 m below the top (measured vertically). (a) What is the change in the gravitational potential energy of the system formed by the skier and the Earth, as the skier slides from the top to the bottom of the slope? | 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 only 30 m/s as she reaches the bottom of the slope, how much energy was dissipated by friction? (e) Draw a free-body diagram of the skier as she is sliding down the slope. Make sure you include friction, and indicate the direction of the acceleration. Use your diagram to answer the next couple of questions. (f) What is the magnitude of the normal force exerted by the ground on the skier? (g) Under the same assumptions as in part (d) above, what is the coefficient of friction between the skis and the snow? (h) Again under the assumption that her final speed is 30 m/s, what is the acceleration of the skier along the slope? Problem 4 A child is dragging a 2 kg sled through a flat patch of snow (coefficient of kinetic friction: 0.1) with a constant force, by pulling on a rope at an angle of 25◦to the horizontal. The sled is moving at a constant speed. (a) Draw a free body diagram for the sled as it is being pulled. (b) Find the magnitude of all the forces acting on the sled. (c) If the child were to suddenly release the rope, what would be the new value of the friction force? What would be the sled’s acceleration? (Take the initial direction of motion to be positive.) Problem 5 A man is swinging an object, attached to a string, in a circle over his head (see the sketches). v Top view Side view (a) Draw a free-body diagram for the object. Notice the string is not drawn horizontal in the “side view” diagram above. Why is that? (b) If the mass of the object is 1 kg, what is the vertical component of the tension? (c) If the object makes 3 turns per second, and the radius of the circle (as seen in the top view) is 0.8 m, what is the centripetal acceleration of the object? | 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 total of 6 s. What was its initial velocity? (give horizontal and vertical components, and also magnitude and direction). Problem 7 A 2 kg block is initially at rest at the top of a 35◦incline, and is then allowed to slide down the incline. The coefficient of kinetic friction is µk = 0.25. (a) Draw a free-body diagram for the block. (b) Find the components of the gravitational force in the given coordinate system. (c) Find the normal force and the force of kinetic friction. (d) Find the magnitude of the acceleration of the block. (e) What is the final speed of the block after sliding 0.75 m? (f) If the coefficient of static friction is µs = 0.4, what would the maximum value of the angle θ be for the block to not slide when it is released? Problem 8 You throw a ball for your dog to fetch. The ball leaves your hand with a speed of 2 m/s, at an angle of 30◦to the horizontal, and from a height of 1.5 m above the ground. The mass of the ball is 0.5 kg. Neglect air resistance in what follows. (a) What is the acceleration of the ball while it is in flight? Report it as a vector, that is, specify magnitude and direction (or vertical and horizontal components; if the latter, specify which direc- tion(s) you take as positive). (b) What is the kinetic energy of the ball as it leaves your hand? (c) Consider the Earth as being in the system. What is the potential energy of the Earth-ball system (1) as the ball leaves your hand, (2) at its maximum height, and (3) as it finally hits the ground? (d) How high does the ball rise above the ground? (e) What is the kinetic energy of the ball as it hits the ground? (f) Now let the system be the ball alone. How much work does the Earth do on the ball while it is in flight? (from start to finish) (g) What is the velocity of the ball as it hits the ground? Report it as a vector. (h) How far away from you (horizontally) does the ball land? Problem 9 A man is standing on the platform of a merry-go-round, not holding on to anything. The merry-go-round is turning at a constant rate, and makes a complete turn every 10 s. (a) What is the merry-go-round’s angular velocity? (b) If the man is standing at a distance of 2 m from the center of the merry-go-round, what is his centripetal acceleration? (c) Which actual force acting on the man is responsible for this acceleration? (d) What is the minimum value of µs, the static friction coefficient, between the soles of the man’s shoes and the platform? | 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 does the platform make before coming to a complete stop? | 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 of motion for any body. Galileo was wrong, but there is at least one sense in which it is true that rotational motion, once started, can go on forever in the absence of external forces. The underlying principle is the conservation of angular momentum, which I will introduce later in this chapter. As pointed out in the previous chapter, rotational motion is also extremely important in mechanical devices. In every case, the rotation of an extended, rigid body can be mathematically described as a collection of circular motions by the particles making up the body. Two very important quantities for dealing with such collections of particles in rotation are the rotational kinetic energy, and the angular momentum. These will both be introduced, and their properties explored, in this chapter. 9.1 Rotational kinetic energy, and moment of inertia If a particle of mass m is moving on a circle of radius R, with instantaneous speed v, then its kinetic energy is Krot = 1 2mv2 = 1 2mR2ω2 (9.1) using |⃗v| = R|ω|, Eq. (8.36). Note that, at this stage, there is no real reason for the subscript “rot”: equation (9.1) is all of the particle’s kinetic energy. The distinction will only become important later in the chapter, when we consider extended objects whose motion is a combination of translation (of the center of mass) and rotation (around the center of mass). 191 | 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 through the same angle in the same time, which means they all have the same angular velocity. However, the particles that are farther away from the axis of rotation are actually moving faster—they have a larger v, according to Eq. (8.36). So the expression for the total kinetic energy in terms of all the particles’ speeds is complicated, but in terms of the (common) angular velocity is simple: Krot = 1 2m1v2 1 + 1 2m2v2 2 + 1 2m3v2 3 + . . . = 1 2 ( m1r2 1 + m2r2 2 + m3r2 3 + . . . ) ω2 = 1 2Iω2 (9.2) where r1, r2, . . . represent the distance of the 1st, 2nd. . . particle to the axis of rotation, and on the last line I have introduced the quantity I = / all particles mr2 (9.3) which is usually called the moment of inertia of the object about the axis considered. In general, the expression (9.3) is evaluated as an integral, which can be written symbolically as I = 0 r2 dm; the “mass element” dm can be expressed in terms of the local density as ρdV , where V is a volume element. The integral is a multidimensional integral that may require somewhat sophisticated calculus skills, so we will not be calculating any of these this semester; rather, we will rely on the tabulated values for I for objects of different, simple, shapes. For instance, for a homogeneous cylinder of total mass M and radius R, rotating around its central axis, I = 1 2MR2; for a hollow sphere rotating through an axis through its center, I = 2 3MR2, and so on. As you can see, the expression (9.2) for the kinetic energy of a rotating body, 1 2Iω2, parallels the expression 1 2mv2 for a moving particle, with the replacement of v by ω, and m by I. This suggests that I is some sort of measure of a solid object’s rotational inertia, by which we mean the resistance it offers to being set into rotation about the axis being considered. We will see later on, when we introduce the torque, that this interpretation for I is indeed correct. It should be stressed that the moment of inertia depends, in general, not just on the shape and mass distribution of the object, but also on the axis of rotation. In general, the formula (9.3) shows that, the more mass you put farther away from the axis of rotation, the larger I will be. Thus, for instance, a thin rod of length l has a moment of inertia I = 1 12Ml2 when rotating around a perpendicular axis through its midpoint, whereas it has the larger I = 1 3Ml2 when rotating around a perpendicular axis through one of its endpoints. | 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 rotational quantity, formed by multiplying the “rotational inertia” I and the angular velocity ω, has any interesting properties as well. We are tentatively1 going to call the quantity Iω angular momentum (to distinguish it from the “ordinary,” or linear momentum, mv), and build towards a better understanding, and a formal definition of it, in the remainder of this section. Two things are soon apparent: one is that, unlike the rotational kinetic energy, which was just plain kinetic energy, the quantity Iω really is different from the ordinary momentum, since it has different dimensions (see the “extra” factor of R in Eq. (9.4) below). The other is that there are, in fact, systems in nature where this quantity appears to remain constant to a good approximation. For instance, the Earth spinning around its axis, at a constant rate of 2π radians every 24 hours, has, by virtue of that, a constant angular momentum (a constant I and ω). It is best, however, to start by thinking about how we would define “angular momentum” for a system consisting of a single particle, and then building up from there, as we have been doing all semester with every new concept. For a particle moving in a circle, according to the previous section, the moment of inertia is just I = mR2, and therefore Iω is just mR2ω. Using Eq. (8.36), we can write (the letter L is the conventional symbol for angular momentum; do not mistake it for a length!) L = Iω = ±mR|v| (particle moving in a circle) (9.4) where, for consistency with our sign convention for ω, we should use the positive sign if the rotation is counterclockwise, and the negative sign if it is clockwise. We can again readily think of examples in nature where the quantity (9.4) is conserved: for instance, the moon, if we treat it as a particle orbiting the Earth in an approximately circular orbit2, has then an approximately constant angular momentum, as given by Eq. (9.4). This example is also interesting because it offers an inkling of an important difference between ordinary momentum, and angular momentum: it appears that the latter can be conserved even in the presence of some kinds of external forces (in this case, the force of gravity due to the Earth that keeps the moon on its orbit). On the other hand, it is not immediately obvious how to generalize the definition (9.4) to other kinds of motion. If we simply try something like L = mvr, where r is the distance to a fixed axis, or to a fixed point, then we find that this yields a quantity that is constantly changing, even for the simplest possible physical system, namely, a particle moving on a straight line with constant 1As we shall see below, in general Iω is equal to only one component of the angular momentum vector. 2You can peek ahead at Figure 10.1, in the next chapter, to see how good an approximation this might be! | 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 the particle to a point, but the distance of the particle’s line of motion to the point considered. The “line of motion” is just a straight line that contains the velocity vector at any given time. The distance between a line and a point O is, by definition, the shortest distance from O to any point on the line; it is given by the length of a segment drawn perpendicular to the line through the point O. For a particle moving in a circle, the line of motion at any time is tangent to the circle, and so the distance between the line of motion and the center of the circle is just the radius R, so we recover the definition Eq. (9.4). For the general case, on the other hand, we have the situation shown in Fig. 9.1: if the instantaneous velocity of the particle is ⃗v, and we draw the position vector of the particle, ⃗r, with the point O as the origin, then the distance between O and the line of motion (sometimes also called the perpendicular distance between O and the particle) is given by r sin θ, where θ is the angle between the vectors ⃗r and ⃗v. v r θ r sin θ = r' sin θ' v O r' θ' ϖ–θ' Figure 9.1: For a particle moving on a straight line, the distance from the point O to the particle’s line of motion (dashed line) is equal to r sin θ at every point in the trajectory (two possible points are shown in the figure). The distance is the length of the blue segment. The figure shows there is some freedom in choosing how the angle θ between ⃗r and ⃗v is to be measured, since the sine of θ is the same as the sine of π −θ. We therefore try and define angular momentum relative to a point O as the product L = ±mr|v| sin θ (9.5) with the positive sign if the point O is to the left of the line of motion as the particle passes by (which corresponds to counterclockwise motion on the circle), and negative otherwise. There are two very good things about the definition (9.5): the first one is that there is already in existence a mathematical operation between vectors, called the vector, or cross, product, according to which |L|, as defined by (9.5), would just be given by |⃗r × ⃗p|, where ⃗r × ⃗p is the cross product of ⃗r and ⃗p; I will have a lot more to say about this in the next subsection. The second good | 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. Particle 2 is tied by a massless string to the point O. After the collision, particle 1 is at rest and particle 2 moves in the circle shown. In the picture, particle 1 is initially moving with constant velocity ⃗v and particle 2 is initially at rest, tied by a massless string to the point O. If the particles have the same mass, conservation of (ordinary) momentum and kinetic energy means that when they collide they exchange velocities: particle 1 comes to a stop and particle 2 starts to move to the right with the same speed v, but immediately the string starts pulling on it and bending its path into a circle. Assuming negligible friction between the string and the pivoting point (and all the surfaces involved), the speed of particle 2 will remain constant as it rotates, by conservation of kinetic energy, because the tension in the string is always perpendicular to the particle’s displacement vector, so it does no work on it. All of the above means that angular momentum is conserved: before the collision it was equal to −mvr sin θ = −mvR for particle 1, and 0 for particle 2; after the collision it is zero for particle 1 and Iω = mR2ω = −mRv for particle 2 (note ω is negative, because the rotation is clockwise). Kinetic energy is also conserved, for the reasons argued above. On the other hand, ordinary momentum is only conserved until just after the collision, when the string starts pulling on particle 2, since this represents an external force on the system. So we have found a situation in which linear motion is converted to circular motion while angular momentum, as defined by Eq. (9.5), is conserved, and this despite the presence of an external force. It is true that, in fact, we were able to solve for the final motion without making explicit use of conservation of angular momentum: we only had to invoke conservation of ordinary momentum for the short duration of the collision, and conservation of kinetic energy. But it is easy to generalize the problem depicted in Figure 9.2 to one that cannot be solved by these methods, but can be solved if angular momentum is constant. Suppose that we replace particle 2 and the string by a | 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, at the other end. After the collision, the particle is moving along the same line with velocity ⃗vf, and the rod is rotating around the point O with an angular velocity ω. If the particle strikes the rod perpendicularly, and the collision happens very fast (that is, it is over before the rod has time to move a significant distance), we may assume the force exerted by the rod on the particle (a normal force) is along the original line of motion, so the particle will continue to move on that line. On the other hand, this time we cannot neglect the force exerted by the pivot on the bar during the collision time, since the bar is one single rigid object. This means the system is not isolated during the collision, and we cannot rely on ordinary momentum conservation. Suppose, however, that the angular momentum L is conserved, as well as the total energy (the pivot does no work on the system, since there is no displacement of that point). The initial angular momentum is Li = −mvil. The final angular momentum is −mvfl for the particle (note that, if the particle bounces back, vf will be negative) and Iω for the rod. The initial kinetic energy is Ki = 1 2mv2 i , and the final kinetic energy is 1 2mv2 f for the particle and 1 2Iω2 for the rod (recall Eq. (9.2), so we have to solve the system −mvil = −mvfl + Iω 1 2mv2 i = 1 2mv2 f + 1 2Iω2 (9.6) The general solution, for arbitrary values of all the constants, is vf = ml2 −I ml2 + I vi ω = − 2ml ml2 + I vi (9.7) | 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 M we have I = 1 3Ml2 for the rod, so we can cancel l2 almost everywhere and end up with vf = 3m −M 3m + M vi ω = − 6m 3m + M vi l (9.8) In particular, we see that if m = M the particle continues to move forward with 1/2 of its initial velocity, and the rod spins with ω = −(3/2)vi/l, which is actually a larger angular velocity than what we found for the system in Fig. 9.2. This example shows how useful conservation of angular momentum can be, but, of course, we do not really know yet whether angular momentum is actually conserved in this problem! I will address this very important question—when is angular momentum conserved—in the section after next, which is to say, after I have properly developed angular momentum as a vector quantity. 9.3 The cross product and rotational quantities The cross, or vector, product of two vectors ⃗A and ⃗B is denoted by ⃗A × ⃗B. It is defined as a vector perpendicular to both ⃗A and ⃗B (that is to say, to the plane that contains them both), with a magnitude given by | ⃗A × ⃗B| = AB sin θ (9.9) where A and B are the magnitudes of ⃗A and ⃗B, respectively, and θ is the angle between ⃗A and ⃗B, when they are drawn either with the same origin or tip-to-tail. The specific direction of ⃗A × ⃗B depends on the relative orientation of the two vectors. Basically, if ⃗B is counterclockwise from ⃗A, when looking down on the plane in which they lie, assuming they are drawn with a common origin, then ⃗A × ⃗B points upwards from that plane; otherwise, it points downward (into the plane). One can also use the so-called right-hand rule, illustrated in Figure 9.4 (next page) to figure out the direction of ⃗A× ⃗B. Note that, by this definition, the direction of ⃗A× ⃗B is the opposite of the direction of ⃗B × ⃗A (as also illustrated in Fig. 9.4). Hence, the cross-product is non-commutative: the order of the factors makes a difference. ⃗A × ⃗B = −⃗B × ⃗A (9.10) | 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 plane formed by ⃗A and ⃗B from above; in the second drawing, we are looking at the plane from below, and calculating ⃗B × ⃗A. It follows from Eq. (9.10) that the cross-product of any vector with itself must be zero. In fact, according to Eq. (9.9), the cross product of any two vectors that are parallel to each other is zero, since in that case θ = 0, and sin 0 = 0. In this respect, the cross product is the opposite of the dot product that we introduced in Chapter 7: it is maximum when the vectors being multiplied are orthogonal, and zero when they are parallel. (And, of course, the result of ⃗A × ⃗B is a vector, whereas ⃗A · ⃗B is a scalar.) Besides not being commutative, the cross product also does not have the associative property of ordinary multiplication: ⃗A×( ⃗B × ⃗C) is different from ( ⃗A× ⃗B)× ⃗C. You can see this easily from the fact that, if ⃗A = ⃗B, the second expression will be zero, but the first one generally will be nonzero (since ⃗A × ⃗C is not parallel, but rather perpendicular to ⃗A). In spite of these oddities, the cross product is extremely useful in physics. We will use it to define the angular momentum vector ⃗L of a particle, relative to a point O, as follows: ⃗L = ⃗r × ⃗p = m⃗r × ⃗v (9.11) where ⃗r is the position vector of the particle, relative to the point O. This definition gives us a constant vector for a particle moving on a straight line, as discussed in the previous section: the magnitude of ⃗L, according to Eq. (9.9) will be mrv sin θ, which, as shown in Fig. 9.1, does not change as the particle moves. As for the direction, it is always perpendicular to the plane containing ⃗r and ⃗v (the plane of the paper, in Fig. 9.1), and if you imagine moving ⃗v to point O, keeping it parallel to itself, and apply the right-hand rule, you will see that ⃗L in Fig. 9.1 should point into the plane of the paper at all times. To see how the definition (9.11) works for a particle moving in a circle, consider again the situation shown in Figure 8.6 in the previous chapter, but now extend it to three dimensions, as in Fig. 9.5, on the next page. It is straightforward to verify that, for the direction of motion shown, the cross | 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) |⃗v| R = I|ω| (9.12) + r vP (x,y) x y R sin θ θ R cos θ z L ω Figure 9.5: A particle moving on a circle in the x-y plane. For the direction of rotation shown, the vectors ⃗L = m⃗r × ⃗v and ⃗ω lie along the z axis, in the positive direction. This suggests that we should define the angular velocity vector, ⃗ω, as a vector of magnitude |ω|, pointing along the positive z axis if the motion in the x-y plane is counterclockwise as seen from above (and in the opposite direction otherwise). Then this will hold as a vector equation: ⃗L = I⃗ω (9.13) It may seem a very strange choice to have the angular velocity point along the z axis, when the particle is moving in the x-y plane, but in a certain way it makes sense. Suppose the particle is moving with constant angular velocity: the directions of ⃗r and ⃗v are constantly changing, but ⃗ω is pointing along the positive z direction, which does remain fixed throughout. There are some other neat things we can do with ⃗ω as defined above. Consider the cross product ⃗ω ×⃗r. Inspection of Figure 9.5 and of Eq. (8.36) shows that this is nothing other than the ordinary velocity vector, ⃗v: ⃗v = ⃗ω × ⃗r (9.14) We can also take the derivative of ⃗ω to obtain the angular acceleration vector ⃗α, so that Eq. (8.33) will hold as a vector equation: ⃗α = lim ∆t→0 ⃗ω(t + ∆t) −⃗ω(t) dt = d⃗ω dt (9.15) | 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 property with respect to the sum: & ⃗A + ⃗B ' × ⃗C = ⃗A × ⃗C + ⃗B × ⃗C (9.16) This, it turns out, is all that’s necessary in order to be able to apply the product rule of differenti- ation to calculate the derivative of a cross product; you just have to be careful not to change the order of the factors in doing so. We can then take the derivative of both sides of Eq. (9.14) to get an expression for the acceleration vector: ⃗a = d⃗v dt = d⃗ω dt × ⃗r + ⃗ω × d⃗r dt = ⃗α × ⃗r + ⃗ω × ⃗v (9.17) The first term on the right-hand side, ⃗α × ⃗r, lies in the x-y plane, and is perpendicular to ⃗r; it is, therefore, tangential to the circle. In fact, looking at its magnitude, it is clear that this is just the tangential acceleration vector, which I introduced (as a scalar) in Eq. (8.37). As for the second term in (9.17), ⃗ω × ⃗v, noting that ⃗ω and ⃗v are always perpendicular, it is clear its magnitude is |ω||⃗v| = Rω2 = v2/R (making use of Eq. (8.36) again). This is just the magnitude of the centripetal acceleration we studied in the previous chapter (section 8.4). Also, using the right-hand rule in Fig. 9.5, you can see that ⃗ω ×⃗v always points inwards, towards the center of the circle; that is, along the direction of −⃗r. Putting all of this together, we can write this vector as just −ω2⃗r, and the whole acceleration vector as the sum of a tangential and a centripetal (radial) component, as follows: ⃗a = ⃗at + ⃗ac ⃗at = ⃗α × ⃗r ⃗ac = −ω2⃗r (9.18) To conclude this section, let me return to the angular momentum vector, and ask the question of whether, in general, the angular momentum of a rotating system, defined as the sum of Eq. (9.11) over all the particles that make up the system, will or not satisfy the vector equation ⃗L = I⃗ω. We have seen that this indeed works for a particle moving in a circle. It will, therefore, also work for any object that is essentially flat, and rotating about an axis perpendicular to it, since in that case all its parts are just moving in circles around a common center. This was the case for the thin rod we considered in connection with Figure 9.3 in the previous subsection. However, if the system is a three-dimensional object rotating about an arbitrary axis, the result ⃗L = I⃗ω does not generally hold. The reason is, mathematically, that the moment of inertia I is defined (Eq. (9.3)) in terms of the distances of the particles to an axis, whereas the angular | 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 axis. On the other hand, if the axis of rotation is an axis of symmetry of the object, then ⃗L has only a z component, and the result ⃗L = I⃗ω holds as a vector equation. Most of the systems we will consider this semester will be covered under this clause, or under the “essentially flat” clause mentioned above. In what follows we will generally assume that I has only a z component, and we will drop the subscript z in the equation Lz = Iωz, so that L and ω will not necessarily be the magnitudes of their respective vectors, but numbers that could be positive or negative, depending on the direction of rotation (clockwise or counterclockwise). This is essentially the same convention we used for vectors in one dimension, such as ⃗a or ⃗p, in the early chapters; it is fine for all the cases in which the (direction of the) axis of rotation does not change with time, which are the only situations we will consider this semester. 9.4 Torque We are finally in a position to answer the question, when is angular momentum conserved? To do this, we will simply take the derivative of ⃗L with respect to time, and use Newton’s laws to find out under what circumstances it is equal to zero. Let us start with a particle and calculate d⃗L dt = d dt (m⃗r × ⃗v) = md⃗r dt × ⃗v + m⃗r × d⃗v dt (9.19) The first term on the right-hand side goes as ⃗v×⃗v, which is zero. The second term can be rewritten as m⃗r × ⃗a. But, according to Newton’s second law, m⃗a = ⃗Fnet. So, we conclude that d⃗L dt = ⃗r × ⃗Fnet (9.20) So the angular momentum, like the ordinary momentum, will be conserved if the net force on the particle is zero, but also, and this is an important difference, when the net force is parallel (or antiparallel) to the position vector. For motion on a circle with constant speed, this is precisely what happens: the force acting on the particle is the centripetal force, which can be written as ⃗Fc = m⃗ac = −mω2⃗r (using Eq. (9.18)), so ⃗r × ⃗Fc = 0, and the angular momentum is constant. | 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 angular momentum vector would be given by the sum of the torques of all the forces acting on all the particles. For each torque one needs to use the position vector of the particle on which the force is acting. As was the case when calculating the rate of change of the ordinary momentum of an extended system (Section 6.1.1), Newton’s third law, with a small additional assumption, leads to the cancellation of the torques due to the internal forces3, and so we are left with only ⃗τext,all = d⃗Lsys dt (9.22) It goes without saying that all the torques and angular momenta need to be calculated relative to the same point. The torque of a force around a point is basically a measure of how effective the force would be at causing a rotation around that point. Since |⃗r × ⃗F| = rF sin θ, you can see that it depends on three things: the magnitude of the force, the distance from the center of rotation to the point where the force is applied, and the angle at which the force is applied. All of this can be understood pretty well from Figure 9.6 below, especially if you have ever had to use a wrench to tighten or loosen a bolt: A B O F1 F4 F3 F2 rA rB θ Figure 9.6: The torque around the point O of each of the forces shown is a measure of how effective it is at causing the rod to turn around that point. Clearly, the force ⃗F1 will not cause a rotation at all, and accordingly its torque is zero (since it is parallel to ⃗rA). On the other hand, of all the forces shown, the most effective one is ⃗F3: it is applied the farthest away from O, for the greatest leverage (again, think of your experiences with 3The additional assumption is that the force between any two particles lies along the line connecting the two particles (which means it is parallel or antiparallel to the vector ⃗r1 −⃗r2). In that case, ⃗r1 × ⃗F12 + ⃗r2 × ⃗F21 = (⃗r1 −⃗r2) × ⃗F12 = 0. Most forces in nature satisfy this condition. | 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 rod, only the perpendicular component (whose magnitude is F2 sin θ) would be effective at causing a rotation; the other component, the one parallel to the rod, would be wasted, like ⃗F1. In order to calculate torques, then, we basically need to find, for every force, the component that is perpendicular to the position vector of its point of application. Clearly, for this purpose we can no longer represent an extended body as a mere dot, as we did for the free-body diagrams in Chapter 6. What we need is a more careful sketch of the object, just detailed enough that we can tell how far from the center of rotation and at what angle each force is applied. That kind of diagram is called an extended free-body diagram. Figure 9.6 could be an example of an extended free-body diagram, for an object being acted on by four forces. Typically, though, instead of drawing the vectors ⃗rA and ⃗rB we would just indicate their lengths on the diagram (or maybe even leave them out altogether, if we do not want to overload the diagram with detail). I will show a couple of examples of extended free-body diagrams in the next couple of sections. As indicated above, to calculate the torque of each force acting on an extended object you should use the position vector ⃗r of the point where the force is applied. This is typically unambiguous for contact forces4, but what about gravity? In principle, the force of gravity would act on all of the particles making up the body, and we would have to add up all the corresponding torques: ⃗τ G = ⃗r1 × ⃗F G E,1 + ⃗r2 × ⃗F G E,2 + . . . (9.23) We can, however, simplify this substantially by noting that (near the surface of the Earth, at any rate), all the forces F G E,1, F G E,2, . . . point in the same direction (which is to say, down), and they are all proportional to each particle’s mass. If I let the total mass of the object be M, and the total force due to gravity on the object be ⃗F G E,obj, then I have ⃗F G E,1 = m1 ⃗F G E,obj/M, ⃗F G E,2 = m2 ⃗F G E,obj/M, . . ., and I can rewrite Eq. (9.23) as ⃗τ G = m1⃗r1 + m2⃗r2 + . . . M × ⃗F G E,obj = ⃗rcm × ⃗F G E,obj (9.24) where ⃗rcm is the position vector of the object’s center of mass. So to find the torque due to gravity on an extended object, just take the total force of gravity on the object (that is to say, the weight of the object) to be applied at its center of mass. (Obviously, then, the torque of gravity around the center of mass itself will be zero, but in some important cases an object may be pivoted at a point other than its center of mass.) 4Actually, friction forces and normal forces may be “spread out” over a whole surface, but, if the object has enough symmetry, it is usually OK to have them “act” at the midpoint of that surface. This can be proved along the lines of the derivation for gravity that follows. | 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. First, if the net external torque is zero, angular momentum will be conserved, as was the case, in particular, for the collision illustrated earlier, in Fig. 9.3, between a particle and a rod pivoted at one end. The only external force in that case was the force exerted on the rod, at the pivot point, by the pivot itself, but the torque of that force around that point is obviously zero, since ⃗r = 0, so our assumption that the total angular momentum around that point was conserved was legitimate. Secondly, if ⃗L = I⃗ω holds, and the moment of inertia I does not change with time, we can rewrite Eq. (9.22) as ⃗τext,all = I⃗α (9.25) which is basically the rotational equivalent of Newton’s, second law, ⃗F = m⃗a. We will use this extensively in the remainder of this chapter. Finally, note that situations where the moment of inertia of a system, I, changes with time are relatively easy to arrange for any deformable system. Especially interesting is the case when the external torque is zero, so L is constant, and a change in I therefore brings about a change in ω = L/I: this is how, for instance, an ice-skater can make herself spin faster by bringing her arms closer to the axis of rotation (reducing her I), and, conversely, slow down her spin by stretching out her arms. This can be done even in the absence of a contact point with the ground: high-board divers, for instance, also spin up in this way when they curl their bodies into a ball. Note that, throughout the dive, the diver’s angular momentum around its center of mass is constant, since the only force acting on him (gravity, neglecting air resistance) has zero torque about that point. In all the cases just mentioned, the angular momentum is constant, but the rotational kinetic energy changes. This is due to the work done by the internal forces (of the ice-skater’s or the diver’s body), converting some internal energy (such as elastic muscular energy) into rotational kinetic energy, or vice-versa. A convenient expression for a system’s rotational kinetic energy when ⃗L = I⃗ω holds is Krot = 1 2Iω2 = L2 I (9.26) which shows explicitly how K would change if I changed and L remained constant. 9.5 Statics Statics is the branch of mechanics concerned with the forces and stresses5 needed to keep a system at rest, in a stable equilibrium—so that it will not move, bend or collapse. It is, obviously, extremely 5A “stress” is a “distributed” force in an extended object, varying continuously from one point to another. | 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 conditions for static equilibrium: first, the net (external) force on the system must be zero, to make sure its center of mass stays at rest; and second, the net (external) torque on the system must also be zero, so that it does not rotate. These conditions can be symbolically expressed as / ⃗Fext = 0 / ⃗τext = 0 (9.27) You may ask about which point one should calculate the torques. The answer is that, as long as the first condition is satisfied (sum of the forces is zero), it does not matter! The proof is simple, but you are welcome to skip it if you are not interested. Suppose you have two points, A and B, around which to calculate the torques. Let ⃗rA1,⃗rA2, . . . be the position vectors of the points of application of the forces ⃗F1, ⃗F2 . . ., relative to point A, and ⃗rB1,⃗rB2, . . ., the same, but relative to point B. If you go all the way back to Figure 1.6 (in Chapter 1), you can see that these vectors only differ from the first set by the single constant vector ⃗rAB that gives the position of point B relative to point A: ⃗rA1 = ⃗rAB + ⃗rB1, etc. Then, for the sum of torques around A we have ⃗rA1 × ⃗F1 + ⃗rA2 × ⃗F2 + . . . = (⃗rAB + ⃗rB1) × ⃗F1 + (⃗rAB + ⃗rA2) × ⃗F2 + . . . = ⃗rAB × & ⃗F1 + ⃗F2 . . . ' + ⃗rB1 × ⃗F1 + ⃗rB2 × ⃗F2 + . . . (9.28) The first term on the last line is zero if the sum of all the forces is zero, and what is left is the sum of all the torques around B. This, indeed, for statics problems, as long as we are enforcing 1 ⃗Fext = 0, it does not matter about which point we choose to calculate the torque. A natural choice is the system’s center of mass, since that is typically a point of high symmetry, but we may also choose a point where there are many applied forces, and so get rid of them quickly (since their torques about that point will be zero). The way all this works is probably best illustrated with an example. Figure 9.9 (next page) shows a classic one, a ladder leaning against a wall. The sketch on the left shows the angles and dimensions involved, whereas the proper extended free-body diagram, showing all the forces and their points of application, is on the right. The minimum number of forces needed to balance the system is four: the weight of the ladder (acting at the center of mass), a normal force from the ground, another normal force from the wall, and a force of static friction from the ground that prevents the ladder from slipping. In real life there should also be a force of static friction from the wall, pointing upwards (also to prevent slippage); and, of course, if there is a person on the ladder she will exert an additional force down on it (equal to her weight), applied at whatever point she is standing. I am not going to consider | 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 quantity without an arrow on top represents that vector’s magnitude, the equation for the balance of the vertical forces reads F N gl −mg = 0 (9.29) For the horizontal forces, we have F N wl −F s gl = 0 (9.30) Then, taking torques around the point where the ladder is in contact with the ground, neither of the two forces applied at that point will contribute, and the condition that the sum of the torques equal zero becomes F N wll sin θ −mg l 2 cos θ = 0 (9.31) This is because the angle made by the force of gravity with the position vector of its point of application is π 2 −θ, and sin(π 2 −θ) = cos θ. From the first equation we get that F N gl = mg; from the second we get that the other normal force, F N wl = F s gl. If we substitute this in (9.31), and cancel out l, the length of the ladder, we get the condition F s gl = 1 2mg cot θ (9.32) But the force of static friction cannot exceed µsF N gl = µsmg, so, setting the right-hand side of (9.32) to be lower than or equal to µsmg, and canceling the common factor mg, we get the condition cot θ ≤2µs, or tan θ ≥ 1 2µs (9.33) for the minimum angle θ at which we can lean the ladder before it slips and falls. | 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 mass of the system must be moving with constant velocity; the second condition means that the total angular momentum must be constant. For a rigid body, this means that the most general kind of unforced motion can be described as a translation of the center of mass with constant velocity, accompanied by a rotation with constant angular velocity around the center of mass. For an extended, deformable system, on the other hand, the presence of internal forces can make the general motion a lot more complicated. Just think, for instance, of the solar system: although everything is, loosely speaking, revolving around the sun, the motions of individual planets and (especially) moons can be fairly complicated. A simple example of (for practical purposes) unforced motion is provided by a symmetric, rigid object (such as a ball, or a wheel) rolling on a flat surface. The normal and gravity forces cancel each other, and since they lie along the same line their torques cancel too, so both ⃗vcm and ⃗L remain constant. In principle, you could imagine removing the ground and gravity and nothing would change: the same motion (in the absence of air resistance) would just continue forever. In practice, there is energy dissipation associated with rolling motion, primarily because, if the rolling object is not perfectly rigid6, then, as it rolls, different parts of it get compressed under the combined pressure of gravity and the normal force, expand again, get compressed again. . . This kind of constant “squishing” ends up converting macroscopic kinetic energy into thermal energy: you may have noticed that the tires on a car get warm as you drive around, and you may also be familiar with the fact that you get a better gas mileage (less energy dissipation) when your tires are inflated to the right pressure than when they are low (because they are more “rigid,” less deformable, in the first case). This conversion of mechanical energy into thermal energy can be formally described by introducing another “friction” force that we call the force of rolling friction. Eventually, rolling friction alone would bring any rolling object to a stop, even in the absence of air resistance. It is, however, usually much weaker than sliding friction, so we will continue to ignore it from now on. You may have noticed already that typically an object can roll on a surface much farther than it can slide without rolling on the same surface. In fact, what happens often is that, if you try to send the object (for instance, a billiard ball) sliding, it will lose kinetic energy rapidly to the force of kinetic friction, but it will also start spinning under the influence of the same force, until a critical point is reached 6There is a simple argument, based on Einstein’s theory of relativity, that shows an infinitely rigid object cannot exist: if it did, you could send a signal instantly from one end of it to the other, just by pushing or pulling on your end. In practice, such motion cannot reach the other end faster than the speed of a sound wave (that is to say, a compression wave) in the material. We will study such waves (which imply the medium is not infinitely rigid) in Chapter 12. | 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 rolling without slipping as “measuring the surface” as it rolls (or vice-versa, the surface measuring the circumference of the object as its different points are pressed against it in succession). So, after it has completed a revolution (2π radians), it should have literally “covered” a distance on the surface equal to 2πR, that is, advanced a distance 2πR. But the same has to be true, proportionately, for any rotation angle ∆θ other that 2π: since the length of the corresponding arc is s = R|∆θ|, in a rotation over an angle |∆θ| the center of mass of the object must have advanced a distance |∆xcm| = s = R|∆θ|. Dividing by ∆t as ∆t →0 then yields Eq. (9.34). θ vcm Figure 9.8: Left: illustrating the rolling without slipping condition. The cyan line on the surface has the same length as the cyan-colored arc, and will be the distance traveled by the disk when it has turned through an angle θ. Right: velocities for four points on the edge of the disk. The pink arrows are the velocities in the center of mass frame. In the Earth reference frame, the velocity of the center of mass, ⃗vcm, in green, has to be added to each of them. The resultant is shown in blue for two of them. Note that, unlike Eq. (8.36), which it very much resembles, Eq. (9.34) is not a “vector identity in disguise”: there is nothing like Eq. (9.14) that we could substitute for it in order to make the signs automatically come out right. You should just treat it as a relationship between the magnitudes of ⃗vcm and ⃗ω and just pick the signs appropriately for each circumstance, based on your convention for positive directions of translation and rotation. In fact, we could use Eq. (9.14) to find the velocity of any point on the circle, if we go to a reference frame where the center is at rest—which is to say, the center of mass (CM) reference frame; then, to go back to the Earth frame, we just have to add ⃗vcm (as a vector) to the vector we obtained in the CM frame. Fig. 9.8 shows the result. Note, particularly, that the point at the very bottom of the circle has a velocity −R|ω| in the CM frame, but when we go back to the Earth frame, its velocity is −R|ω| + vcm = −R|ω| + R|ω| = 0 (by the condition (9.34)). Thus, as long as the condition for rolling without slipping holds, the point (or points) on the rolling object that are momentarily in | 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 not actually need the force of static friction to keep an object rolling on a flat surface (as I mentioned above, the motion could in principle go on “unforced” forever), but things are different on an inclined plane. Fig. 9.9 shows an object rolling down an inclined plane, and the corresponding extended free-body diagram. vcm x y θ FG Fn Fs Figure 9.9: An object rolling down an inclined plane, and the extended free-body diagram. Note that neither gravity (applied at the CM) nor the normal force (whose line of action passes through the CM) exert a torque around the center of mass; only the force of static friction, ⃗F s, does. The basic equations we use to solve for the object’s motion are the sum of forces equation: / ⃗Fext = M⃗acm (9.35) the net torque equation, with torques taken around the center of mass7 / ⃗τext = I⃗α (9.36) and the extension of the condition of rolling without slipping, (9.34), to the accelerations: |acm| = R|α| (9.37) 7You may feel a little uneasy about the fact that the CM frame is now an accelerated, and therefore non-inertial, frame. How do we know Eq. (9.36) even applies there? This is, indeed, a non-trivial point. However, as we shall see in a later chapter, being in a uniformly-accelerated reference frame is equivalent to being in a uniform gravitational field, and we have just shown that, for all torque-related purposes, one may treat such a field as a single force applied to the center of mass of an object. Such a force (or “pseudoforce” in this case) clearly does not contribute to the torque about the center of mass, and so Eq. (9.36) applies in the CM frame. | 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 not actually need8; in the direction along the plane, we have Macm = Mg sin θ −F s (9.38) and the torque equation just gives −F sR = Iα, which with acm = −Rα becomes F sR = I acm R (9.39) We can eliminate F s in between these two equations and solve for acm: acm = g sin θ 1 + I/(MR2) (9.40) Now you can see why, earlier in the semester, we were always careful to assume that all the objects we sent down inclined planes were sliding, not rolling! The acceleration for a rolling object is never equal to simply g sin θ. Most remarkably, the correction factor depends only on the shape of the rolling object, and not on its mass or size, since the ratio of I to MR2 is independent of m and R for any given geometry. Thus, for instance, for a disk, I = 1 2MR2, so acm = 2 3g sin θ, whereas for a hoop, I = MR2, so acm = 1 2g sin θ. So any disk or solid cylinder will always roll down the incline faster than any hoop or hollow cylinder, regardless of mass or size. This rather surprising result may be better understood in terms of energy. First, let me show (a result that is somewhat overdue) that for a rigid object that is rotating around an axis passing through its center of mass with angular velocity ω we can write the total kinetic energy as K = Kcm + Krot = 1 2Mv2 cm + 1 2Iω2 (9.41) This is because for every particle the velocity can be written as ⃗v = ⃗vcm+ ⃗v′, where ⃗v′ is the velocity relative to the center of mass (that is, in the CM frame). Since in this frame the motion is a simple rotation, we have |v′| = ωr, where r is the particle’s distance to the axis. Therefore, the kinetic energy of that particle will be 1 2mv2 = 1 2⃗v · ⃗v = 1 2m & ⃗vcm + ⃗v′ ' · & ⃗vcm + ⃗v′ ' = 1 2mv2 cm + 1 2mv′2 + m⃗vcm · ⃗v′ = 1 2mv2 cm + 1 2mr2ω2 + ⃗vcm · ⃗p′ (9.42) 8Unless we were trying to answer a question such as “how steep does the plane have to be for rolling without slipping to become impossible?” | 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 frame is the zero momentum frame. Then, adding the contributions of all particles to the first and second terms in 9.42 gives Eq. (9.41). Coming back to our rolling body, using Eq. (9.41) and the condition of rolling without slipping (9.34), we see that the ratio of the translational to the rotational kinetic energy is Kcm Krot = mv2 cm Iω2 = mR2 I (9.43) The amount of energy available to accelerate the object initially is just the gravitational potential energy of the object-earth system, and that has to be split between translational and rotational in the proportion (9.43). An object with a proportionately larger I is one that, for a given angular velocity, needs more rotational kinetic energy, because more of its mass is away from the rotation axis. This leaves less energy available for its translational motion. Resources Unfortunately, we will not really have enough time this semester to explore further the many inter- esting effects that follow from the vector nature of Eq. (9.20), but you are at least subconsciously familiar with some of them if you have ever learned to ride a bicycle! A few interesting Internet references (some of which could perhaps inspire a good Honors project!) are the following: • Walter Lewin’s lecture on gyroscopic motion (and rolling motion): https://www.youtube.com/watch?v=N92FYHHT1qM • A “Veritasium” video on “antigravity”: https://www.youtube.com/watch?v=GeyDf4ooPdo https://www.youtube.com/watch?v=tLMpdBjA2SU • And the old trick of putting a gyroscope (flywheel) in a suitcase: https://www.youtube.com/watch?v=zdN6zhZSJKw If any of the above links are dead, try googling them. (You may want to let me know, too!) 9.7 In summary (Note: this summary makes extensive use of cross products, but does not include a summary of cross product properties. Please refer to Section 9.3 for that!) 1. The angular velocity and acceleration of a particle moving in a circle can be treated as vectors perpendicular to the plane of the circle, ⃗ω and ⃗α, respectively. The direction of ⃗ω is such | 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 about an axis, Krot = 1 2Iω2 also applies if I is defined as the sum of the quantities mr2 for all the particles making up the object, where r is the particle’s distance to the rotation axis. 3. For a rigid object that is rotating around an axis passing through its center of mass with angular velocity ω the total kinetic energy can be written as K = Kcm+Krot = 1 2Mv2 cm+ 1 2Iω2. This would apply also to a non-rigid system, provided all the particles have the same angular velocity. 4. The angular momentum, ⃗L, of a particle about a point O is defined as ⃗L = ⃗r × ⃗p = m⃗r × ⃗v, where ⃗r is the position vector of the particle relative to the origin O, and ⃗v and ⃗p its velocity and momentum vectors. For an extended object or system, ⃗L is defined as the sum of the quantities m⃗r × ⃗v for all the particles making up the system. 5. For a solid object rotating around a symmetry axis, ⃗L = I⃗ω. This applies also to an essentially flat object rotating about a perpendicular axis, even if it is not an axis of symmetry. 6. The torque, ⃗τ, of a force about a point O is defined as τ = ⃗r × ⃗F, where ⃗r is the position vector of the point of application of the force relative to the origin O. It is a measure of the effectiveness of the force at causing a rotation around that point. 7. The rate of change of a system’s angular momentum about a point O is equal to the sum of the torques, about that same point, of all the external forces acting on the system: 1⃗τext = d⃗Lsys/dt. Hence, angular momentum is constant whenever all the external torques vanish (conservation of angular momentum). 8. For the cases considered in point 7. above, if the moment of inertia I is constant, the equation 1⃗τext = d⃗Lsys/dt can be written in the form 1⃗τext = I⃗α, which strongly resembles the familiar 1 ⃗Fext = m⃗a. Note, however, that deformable systems where I may change with time as a result of internal forces are relatively common, and for those systems this simpler equation would not apply. 9. For an object to be in static equilibrium, we require that both the sum of the external forces and of the external torques be equal to zero: 1 ⃗Fext = 0 and 1⃗τext = 0. Note that if the first condition applies, it does not matter about which point we calculate the torque, so we are free to choose whichever is most convenient. 10. For a rigid object of radius R rolling without slipping on some surface, the relations |vcm| = R|ω| and |acm| = R|α| hold. The relative signs of, for instance, vcm and ω (understood here as the relevant components of their respective vectors) need to be chosen so as to be consistent with whatever convention one has adopted for the positive direction of motion and the positive direction of rotation (typically, a counterclockwise rotation is considered positive). | 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 solve the equilibrium equations for a leaning ladder, which is the archetype of all statics problems; and Section 9.4 also solves for you the problem of a generic object rolling down an inclined plane. The first couple of additional examples in this section show you have to set up and solve the equations of motion for somewhat more complicated systems, and you should study them carefully. The third one is slightly more sophisticated and you may treat it as optional reading. 9.8.1 Torques and forces on the wheels of an accelerating bicycle Consider an accelerating bicycle. The rider exerts a torque on the pedals, which is transmitted to the rear wheel by the chain (possibly amplified by the gears, etc). How does this “drive” torque on the rear wheel (call it τd) relate to the final acceleration of the center of mass of the bicycle? Fr,ft Fr,rt s s Solution We need first to figure out how many external forces, at a minimum, we have to deal with. As the bicycle accelerates, two things happen: the wheels (both wheels) turn faster, so there must be a net torque (clockwise in the picture, if the bicycle is accelerating to the right) on each wheel; and the center of mass of the system accelerates, so there must be a net external force on the whole system. The system is only in contact with the road, and so, as long as no slippage happens, the only external source of torques or forces on the wheels has to be the force of static friction between the tires snd the road. | 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 the front wheel, applied as it is at the point of contact with the road, must point backwards, that is, opposite the direction of motion. We get then one equation of motion (of the type (9.25)) for that wheel: −F s r,ftR = Iα (9.44) where the subscript “ft” stands for “front tire”, and the wheel is supposed to have a radius R and moment of inertia I. For the rear wheel, we have the “drive torque” τd, exerted by the chain, and another torque exerted by the force of static friction, ⃗F s r,rt, between that tire and the road. However, now the force ⃗F s r,rt needs to point forward. This is because the net external force on the whole bicycle-rider system is ⃗F s r,rt + ⃗F s r,ft, and that has to point forward, or the center of mass could never accelerate in that direction. Since we have established that ⃗F s r,ft has to point backwards, it follows that ⃗F s r,rt needs to be larger, and in the forward direction. This means we get, for the center of mass acceleration, the equation (Fnet = Macm) F s r,rt −F s r,ft = Macm (9.45) and for the rear wheel, the torque equation F s r,rtR −τd = Iα (9.46) I am following the convention that clockwise torques are negative, and also that a force symbol without an arrow on top represents the magnitude of the force. If a clockwise angular acceleration is likewise negative, the condition of rolling without slipping [Eq. (9.37)] needs to be written as acm = −Rα (9.47) These are all the equations we need to relate the acceleration to τd. We can start by solving (9.44) for F s r,ft and substituting in (9.45), then likewise solving (9.46) for F s r,rt and substituting in (9.45). The result is Iα + τd R + Iα R = Macm (9.48) then use Eq (9.47) to write α = −acm/R, and solve for acm: acm = τd MR + 2I/R (9.49) 9.8.2 Blocks connected by rope over a pulley with non-zero mass Consider again the setup illustrated in Figure 6.2, but now assume that the pulley has a mass M and radius R. For simplicity, leave the friction force out. What is now the acceleration of the system? | 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 has to be another force acting on it, to balance the two forces shown. This would be a contact force at the axle, directed upwards and to the left. If this was a statics problem, I would have to include it, but since it does not exert a torque around the axis of rotation, it does not contribute to the dynamics of the system, so I have left it out as well.) 1 2 Fr,1 t Fr,2 t FE,2 G a1 a2 1 2 Fr,pl t Fr,pd t The key new feature of this problem is that the tension on the string has to have different values on either side of the pulley, because there has to be a net torque on the pulley. Hence, the leftward force on the pulley (F t r,pl) has to be smaller than the downward force (F t r,pd). On the other hand, as long as the mass of the rope is negligible, it will still be the case that the horizontal part of the rope will pull with equal strength on block 1 and on the pulley, and similarly the vertical part of the rope will pull with equal strength on the pulley and on block 2. (To make this point clearer, I have “color-coded” these matching forces in the figure.) This means that we can write F t r,pl = F t r,1 and F t r,pd = F t r,2, and write the torque equation (9.25) for the pulley as F t r,1R −F t r,2R = Iα (9.50) We also have F = ma for each block: F t r,1 = m1a (9.51) F t r,2 −m2g = −m2a (9.52) where I have taken a to be a = |⃗a1| = |⃗a2|. The condition of rolling without slipping, Eq. (9.37), applied to the pulley, gives then −Rα = a (9.53) since, in the situation shown, α will be negative, and a has been defined as positive. Substituting Eqs. (9.51), (9.52), and (9.53) into (9.50), we get m1aR −(m2g −m2a)R = −Ia R (9.54) | 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, essentially, the inertia of the system: ordinary inertia for the blocks, and rotational inertia for the pulley. Note further that, if we treat the pulley as a flat, homogeneous disk of mass M, then I = 1 2MR2, and the denominator of (9.55) becomes just m1 + m2 + M/2. | 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 angular velocity? (You may use radians per second, or revolutions per second if you prefer.) (b) What is the change in her kinetic energy? (c) Where did this energy come from? Problem 2 Two identical pucks, each of inertia m, are connected to a rod of length 2r and negligible inertia that is pivoted about its center (that is, there is some sort of pin though its center, around which it can rotate without friction). A third puck of inertia m/2 strikes one of the connected pucks perpendicular to the rod with a speed vi. Assume the collision is elastic. (a) Draw a diagram of the situation, clearly labeling the direction of vi and what direction the connected pucks will rotate. (b) Is the total momentum of the system (the three pucks and the rod) conserved throughout the interaction? Why? Is the system isolated, or can you identify an external force acting on it? (c) Is the total kinetic energy of the system conserved? Why? If you found an external force in part (b), explain why it does or does not do work on the system. (d) Is the total angular momentum of the system conserved? Why? If you found an external force in part (b), explain why it does or does not exert a torque on the system. (e) Write down an expression for the moment of inertia (rotational inertia) of the system formed by the connected pucks. (f) Based on all of the above, write down equations expressing the conservation of the two quantities that are in fact conserved. These equations should involve only the given data (masses, length of rod); the initial and final velocities, vi and vf, of the free puck; and ω, the angular speed of the connected pucks after the collision. (Assume the final velocity of the free puck lies along the same line as its initial velocity, that is, it does not bounce offat some random angle.) Problem 3 A plank of length l = 2 m is hinged at one end to a wall. The other end is being (temporarily) supported by a worker who is holding it up with his hand, keeping the plank horizontal. The plank has a mass of 20kg, and there is also a toolbox of mass 5 kg sitting on it, 50 cm away from the worker (1.5 m away from the wall). (a) Draw a free body diagram and an extended free-body diagram for the plank. (b) What are the magnitudes of (1) the upwards force exerted by the worker on the plank and (2) the force at the hinge? | 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) Consider a point on the plank located immediately below the toolbox. As the plank swings, this point moves in a circle of radius 1.5 m. What is its linear (tangential) acceleration as it starts going down, and how does it compare to the acceleration of gravity? Problem 4 A solid sphere of radius 5 cm and mass 0.2 kg is rolling without slipping, on level ground, at a constant speed of 0.5 m/s. It is heading toward a ramp that makes an angle of 30◦with the horizontal. (a) What is the angular velocity of the sphere, in radians per second? (b) If the moment of inertia of a solid, homogeneous sphere is I = 2 5mR2, what is the initial angular momentum of this sphere? (c) What is the translational kinetic energy of the sphere? (d) What is its rotational kinetic energy? (e) How high (vertically) will the sphere rise as it goes up the ramp, still rolling without slipping, before it stops and turns around? (f) Draw an extended free-body diagram for the sphere as it rolls up the ramp. On the diagram, indicate the direction of rotation, and the direction of the acceleration of the center of mass. (g) Referring to the extended free-body diagram in the previous question, which force is responsible for the change in the angular momentum of the sphere as it rolls up the ramp? Explain briefly: Why that force and not another one? Does the direction you have assumed for this force agree with the torque it has to provide? (If not, you’d better go back and fix that right now!) (h) Calculate the acceleration of the sphere as it rolls up the ramp. Problem 5 A very light, inextensible string is wrapped around a cylindrical spool. The end of the string is held fixed, and the spool is released so it starts falling, as the string unwinds. Because the spool is basically hollow, you can take its moment of inertia to be I = mR2. (a) Draw an extended free-body diagram for the spool as it unrolls. (b) Find the linear acceleration of the spool as it accelerates toward the ground. (c) Let the mass of the spool be 0.1 kg. What is its translational kinetic energy after it has fallen for 0.5 s? (d) What is its rotational kinetic energy at that time? (e) What is the tension in the string? Does this change as the spool falls? (Remember the mass of the string is negligible.) (f) If the radius of the spool is 3 cm, what is the magnitude of the torque (around the center of mass) exerted by the tension? Problem 6 | 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 upwards force on the plank exerted by each of the two supports. | 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 the gravitational force is where other objects or distances are involved. The first comprehensive theory of gravity, formulated by Isaac Newton in the late 17th century, postulates that any two “particles” with masses m1 and m2 will exert an attractive force (a “pull”) on each other, whose magnitude is proportional to the product of the masses, and inversely pro- portional to the square of the distance between them. Mathematically, we write F G 12 = Gm1m2 r2 12 (10.1) Here, r12 is just the magnitude of the position vector of particle 2 relative to particle 1 (so r12 is, indeed, the distance between the two particles), and G is a constant, known as “Newton’s constant” or the gravitational constant, which at the time of Newton still had not been determined experimentally. You can see from Eq. (10.1) that G is simply the magnitude, in newtons, of the attractive force between two 1-kg masses a distance of 1 m apart. This turns out to have the ridiculously small value G = 6.674×10−11 N m2/kg2 (or, as is more commonly written, m3kg−1s−2). It was first measured by Henry Cavendish in 1798, in what was, without a doubt, an experimental tour de force for that time (more on that later, but you may peek at the “Advanced Topics” section of next chapter if you are curious already). As you can see, gravity as a force between any two ordinary objects is absolutely insignificant, and it takes the mass of a planet to make it into something you can feel. 221 | 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 (vectorially!) the force exerted by every small part of one object on every small part of the other object. This requires the use of integral calculus at a fairly advanced level, and for irregularly- shaped objects can only be computed numerically. For spherically-symmetric objects, however, it turns out that the result (10.1) still holds exactly, provided the quantity r12 is taken to be the distance between the center of the spheres. The same result also holds for the force between a finite-size sphere and a “particle,” again with the distance to the particle measured from the center of the sphere. For the rest of this chapter, we will simply assume that Eq. (10.1) is a good approximation to be used in any of the problems we will encounter involving extended objects. You can estimate visually how good it may be, for instance, when applied to the Earth-moon system (as we will do in a moment), from a look at Figure 10.1 below: Figure 10.1: The moon, the earth, and the distance between them, all approximately to scale. Based on this picture it seems that it might be OK to treat the moon as a “particle,” but that it would not do, in general, to neglect the radius of the Earth; that is to say, we should use for r12 the distance from the moon to the center of the Earth, not just to its surface. Before we get there, however, let us start closer to home and see what happens if we try to use Eq. (10.1) to calculate the force exerted by the Earth on an object if mass m near its surface—say, at a height h above the ground. Clearly, if the radius of the Earth is RE, the distance of the object to the center of the Earth is RE + h, and this is what we should use for r12 in Eq. (10.1). However, noting that the radius of the Earth is about 6, 000 km (more precisely, RE = 6.37×106 m), and the tallest mountain peak is only about 9 km above sea level, you can see that it will almost always be a very good approximation to set r12 equal to just RE, which results in a force F G E,o ≃GMEm R2 E = m 6.674 × 10−11 m3kg−1s−2 × 5.97 × 1024 kg (6.37 × 106 m)2 = m × 9.82m s2 (10.2) where I have used the currently known value ME = 5.97 × 1024 kg for the mass of the Earth. As you can see, we recover the familiar result F G = mg, with g ≃9.8 m/s2, which we have been using all semester. We can rewrite this result (canceling the mass m) in the form gE = GME R2 E (10.3) Here I have put a subscript E on g to emphasize that this is the acceleration of gravity near the surface of the Earth, and that the same formula could be used to find the acceleration of gravity | 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 fast things fall on a given planet, the quantity gplanet allows you to figure out how much something would weigh on that planet’s surface (just multiply gplanet by the mass of the object); alternatively, the ratio of gplanet to gE will be the ratio of the object’s weight on that planet to its weight here on Earth. Of course, historically, this is not what Newton and his contemporaries would have done: they had measurements of objects in free fall (or sliding on inclined planes) that would have given them the value of g, and they even had a pretty good idea of the radius of the Earth1, but they did not know either G or the mass of the Earth, so all they could get from Eq. (10.3) was the value of the product GME. It was only a century later, when Cavendish measured G, that they could get from that the mass of the earth (as a result of which, he became known as “the man who weighed the earth”!) What Newton could do, however, with just this knowledge of the value of GME, was something that, for its time, was even more dramatic and far-reaching: namely, he could “prove” his intuition that the same fundamental interaction—gravity—that causes an apple to fall near the surface of the Earth, reaching out hundreds of thousands of miles away into space, also provides the force needed to keep the moon on its orbit. This brought together Earth-bound science (physics) and “celestial” science (astronomy) in a way that no one had ever dreamed of before. To see how this works, let us start by assuming that the moon does move in a circle, with the Earth motionless at the center (these are all approximations, as we shall see later, but they give the right order of magnitude at the end, which is all that Newton could have hoped for anyway). The force F G E,m then has to provide the centripetal force Fc = Mmoonω2re,m, where ω is the moon’s angular velocity. We can cancel the moon’s mass from both terms and write this condition as GME r2e,m = ω2re,m (10.4) The moon revolves around the earth once about every 29 days, which is about 29 × 24 × 3600 = 2.5 × 106 s. So ω is 2π radians per 2.5 million seconds, or ω = 2.5 × 10−6 rad/s. Substituting this into Eq. (10.4), as well as the result GME = gR2 E (note that, as stated above, we do not need to know G and ME separately), we get re,m = 3.99 × 108 m, pretty close to today’s accepted value of the average Earth-moon distance, which is 3.84 × 108 km. Newton would not have known re,m to such an accuracy, but he would still have had a pretty good idea that this was, indeed, the correct order of magnitude2. 1The radius of the earth was known, at least approximately, since ancient times. The Greeks first figured it out by watching the way tall objects disappear below the horizon when viewed from a ship at sea. Next time somebody tries to tell you that people used to believe the earth was flat, ask them just which “people” they are talking about! 2And, yes, the first estimates of the distance between the earth and the moon also go back to the ancient Greeks! You can figure it out, if you know the radius of the earth, by looking at how long it takes the moon to transit across the earth’s shadow during a lunar eclipse. | 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 as the force of gravity is approximately constant, which is to say, as long as y is much smaller than the radius of the earth, but obviously it must break down at some point. Recall that, if the interaction between two objects can be described by a potential energy function of the objects’ coordinates, U(x1, x2), then (in one dimension) the force exerted by object 1 on object 2 could be written as F12 = −dU/dx2. Since the force of gravity does lie along the line joining the two particles, we can cheat a bit and treat this as a one-dimensional problem, with (F G 12)x = −Gm1m2/(x1 −x2)2 (I’ve put a minus sign there under the assumption that particle 1 is to the left of particle 2, that is, x1 < x2, and the force on 2 is to the left), and find a potential energy function whose derivative gives that. The answer is clearly U G(x1, x2) = −Gm1m2 x2 −x1 + C (10.5) where C is an arbitrary constant. (Please take a moment to verify for yourself that, indeed, −dU G/dx2 = −(F G 12)x, and also −dU G/dx1 = −(F G 21)x.) Since I have assumed x2 > x1, the denominator in Eq. (10.5) is just the distance between the two particles, and the potential energy function could be written, in three dimensions, as U G(r12) = −Gm1m2 r12 (10.6) where r12 = |⃗r2 −⃗r1|, and I have set the constant C equal to zero. This means that the potential energy of the system is always negative, which is, on the face of it, a strange result. However, there is no way to choose the constant C in (10.5) that will prevent that: no matter how big and positive C might be, the first term in (10.5) can always become larger (in magnitude) and negative, if the particles are very close together. So we might as well choose C = 0, which, at least, gives us the somewhat comforting result that the potential energy of the system is zero when the particles are “infinitely” distant from each other—that is to say, so far apart that they do not feel a force from each other any more. But Eq. (e10.5) also makes sense in a different way: namely, it shows that the system’s potential energy increases as the particles are moved farther apart. Indeed, we expect, physically, that if you separate the particles by a great distance and then release them, they will pick up a lot of speed as they approach each other; or, put differently, that the force doing work over a large distance will give them a large amount of kinetic energy—which must come from the system’s potential energy. But, in fact, mathematically, Eq. (10.6) agrees with this expectation: for any finite distance, U G is negative, and it gets smaller in magnitude as the distance increases, which means algebraically | 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, the notion of a “negative energy” is hard to wrap your mind around. I can only offer you two possible ways to look at it. One is to simply not think of potential energy as being anything like a “substance,” but just an accounting device that we use to keep track of the potential that a system has to do work for us—or (more or less equivalently) to give us kinetic energy, which is always positive and hence may be thought of as the “real” energy. From this point of view, whether U is positive or negative does not matter: all that matters is the change ∆U, and whether this change has a sign that makes sense. This, at least, is the case here, as I have argued in the paragraph above. The other perspective is almost opposite, and based on Einstein’s theory of relativity: in this theory, the total energy of a system is indeed “something like a substance,” in that it is directly related to the system’s total inertia, m, through the famous equation E = mc2. From this point of view, the total energy of a system of two particles, interacting gravitationally, at rest, and separated by a distance r12, would be the sum of the gravitational potential energy (negative), and the two particles’ “rest energies,” m1c2 and m2c2: Etotal = m1c2 + m2c2 −Gm1m2 r12 (10.7) and this quantity will always be positive, unless one of the “particles” is a black hole and the other one is inside it3! Please note that we will not use equation (10.7) this semester at all, since we are concerned only with nonrelativistic mechanics here. In other words, we will not include the “rest energy” in our calculations of a system’s total energy at all. However, if we did, we would find that a system whose rest energy is given by Eq. (10.7) does, in fact, have an inertia that is less than the sum m1 + m2. This strongly suggests that the negativity of the potential energy is not just a mathematical convenience, but rather it reflects a fundamental physical fact. For a system of more than two particles, the total gravitational potential energy would be obtained by adding expressions like (10.6) over all the pairs of particles. Thus, for instance, for three particles one would have U G(⃗r1,⃗r2,⃗r3) = −Gm1m2 r12 −Gm1m3 r13 −Gm2m3 r23 (10.8) A large mass such as the earth, or a star, has an intrinsic amount of gravitational potential energy that can be calculated by breaking it up into small parts and performing a sum like (10.8) over all the possible pairs of “parts.” (As usual, this sum is usually evaluated as an integral, by taking the limit of an infinite number of infinitesimally small parts.) This “self-energy” does not change with 3For a justification of this statement, please see the definition of the Schwarzschild radius, later on in this chapter. | 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 what happens when the distance r12 goes to zero, since the mathematical expression appears to become infinite. This is technically true, but, in practice, it would only be a problem for a pair of true point particles—objects that would literally be mathematical points, with no dimensions at all. Such things may exist in some sense—electrons may well be an example—but they need to be described by quantum mechanics, which is an altogether different mathematical theory. For finite-sized objects, you cannot continue to use an equation like (10.6) (or (10.1), for the force) when you are under the surface of the object. If you could dig a tunnel all the way down to the center of a hypothetical “earth” that had a constant density, the potential energy of the system formed by this “earth” and a particle of mass m, a distance r from the center, would look as shown in Fig. 10.2. Notice how U G becomes “flat,” indicating an equilibrium position (zero force), as r →0. It stands to reason that the net gravitational force at the center of this model “earth” should be zero, since one would be pulled equally strongly in all directions by all the mass around. r/RE UG/mgRE Figure 10.2: Gravitational potential energy of a system formed by a particle of mass m and a hypothetical earth with uniform density, a mass M, and a radius RE, as a function of the distance r between the particle and the center of the “earth” (solid line). The dashed line shows the result for a system of two (point-like) particles. The energy U G is expressed in units of mgRE, where g = GM/R2 E. Finally, let me show you that the result (10.6) is fully consistent with the approximation U G = mgy that we have been using up till now near the surface of the earth. (If you are not interested in mathematical derivations, feel free to skip this next bit.) Consider a particle of mass m that is | 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 R2 E h = mgh (10.10) The only approximation here has been to set RE + h ≃RE in the denominator of this expression. Since RE is of the order of thousands of kilometers, this is an excellent approximation, as long as h is less than, say, a few hundred meters. 10.1.2 Types of orbits under an inverse-square force Consider a system formed by two particles (or two perfect, rigid spheres) interacting only with each other, through their gravitational attraction. Conservation of the total momentum tells us that the center of mass of the system is either at rest or moving with constant velocity. Let us assume that one of the objects has a much greater mass, M, than the other, so that, for practical purposes, its center coincides with the center of mass of the whole system. This is not a bad approximation if what we are interested in is, for instance, the orbit of a planet around the sun. The most massive planet, Jupiter, has only about 0.001 times the mass of the sun. Accordingly, we will assume that the more massive object does not move at all (by working in its center of mass reference frame, if necessary—note that, by our assumptions, this will be an inertial reference frame to a good approximation), and we will be concerned only with the motion of the less massive object under the force F = GMm/r2, where r is the distance between the centers of the two objects. Since this force is always pulling towards the center of the more massive object (it is what is often called a central force), its torque around that point is zero, and therefore the angular momentum, ⃗L, of the less massive body around the center of mass of the system is constant. This is an interesting result: it tells us, for instance, that the motion is confined to a plane, the same plane that the vectors ⃗r and ⃗v defined initially, since their cross-product cannot change. In spite of this simplification, the calculation of the object’s trajectory, or orbit, requires some fairly advanced mathematical techniques, except for the simplest case, which is that of a circular orbit of radius R. Note that this case requires a very precise relationship to hold between the object’s velocity and the orbit’s radius, which we can get by setting the force of gravity equal to the centripetal force: GMm R2 = mv2 R (10.11) | 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 release the satellite at the same distance R, but with a different velocity, either in magnitude or direction? Too much speed would pull it away from the circle, so the distance to the center, r, would temporarily increase; this would increase the system’s potential energy and accordingly reduce the satellite’s velocity, so eventually it would get pulled back; then it would speed up again, and so on. You may experiment with this kind of thing yourself using the PhET demo at this link: https://phet.colorado.edu/en/simulation/gravity-and-orbits You will find that, as long as you do not give the satellite—or planet, in the simulation—too much speed (more on this later!) the orbit you get is, in fact, a closed curve, the kind of curve we call an ellipse. I have drawn one such ellipse for you in Fig. 10.3. a b e a r P A O Figure 10.3: An elliptical orbit. The semimajor axis is a, the semiminor axis is b, and the eccentricity e = ! 1 −b2/a2 = 0.745 in this case.. The “center of attraction” (the sun, for instance, in the case of a planet’s or comet’s orbit) is at the point O. As a geometrical curve, any ellipse can be characterized by a couple of numbers, a and b, which are the lengths of the semimajor and semiminor axes, respectively. These lengths are shown in the figure. Alternatively, one could specify a and a parameter known as the eccentricity, denoted by e (do not mistake this “e” for the coefficient of restitution of Chapter 4!), which is equal to e = ! 1 −b2/a2. If a = b, or e = 0, the ellipse becomes a circle. The most striking feature of the elliptical orbits under the influence of the 1/r2 gravitational force is that the “central object” (the sun, for instance, if we are interested in the orbit of a planet, asteroid or comet) is not at the geometric center of the ellipse. Rather, it is at a special point called the focus of the ellipse (labeled “O” in the figure, since that is the origin for the position vector of the orbiting body). There are actually two foci, symmetrically placed on the horizontal (major) axis, | 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 elliptical orbit around the sun, the distance to the sun is minimal at a point called the perihelion, and maximal at a point called the aphelion. Those points are shown in the figure and labeled “P” and “A”, respectively. For an object in orbit around the earth, the corresponding terms are perigee and apogee; for an orbit around some unspecified central body, the terms periapsis and apoapsis are used. There is some confusion as to whether the distances are to be measured from the surface or from the center of the central body; here I will assume they are all measured from the center, in which case the following relationships follow directly from Figure 10.3: rmax = (1 + e)a rmin = (1 −e)a rmin + rmax = 2a e = rmax −rmin 2a (10.12) The ellipse I have drawn in Fig. 10.3 is actually way too eccentric to represent the orbit of any planet in the solar system (although it could well be the orbit of a comet). The planet with the most eccentric orbit is Mercury, and that is only e = 0.21. This means that b = 0.978a, an almost imperceptible deviation from a circle. I have drawn the orbit to scale in Fig. 10.4, and as you can see the only way you can tell it is an ellipse is, precisely, because the sun is not at the center. Figure 10.4: Orbit of Mercury, with the sun approximately to scale. | 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 object does not move at all, the total energy of the system is just E = 1 2mv2 −GMm r (10.13) For a circular orbit, the radius R determines the speed (as per Eq. (10.11)), and hence the total energy, which is easily seen to be E = −GMm 2R . It turns out that this formula holds also for elliptical orbits, if one substitutes the semimajor axis a for R: E = −GMm 2a (10.14) Note that the total energy (10.14) is negative. This means that we have a bound orbit, by which I mean, a situation where the orbiting object does not have enough kinetic energy to fly arbitrarily far away from the center of attraction. Indeed, since U G →0 as r →∞, you can see from Eq. (10.13) that if the two objects could be infinitely far apart, the total energy would eventually have to be positive, for any nonzero speed of the lighter object. So, if E < 0, we have bound orbits, which are ellipses (of which a circle is a special case), and conversely, if E > 0 we have “unbound” trajectories, which turn out to be hyperbolas4. These trajectories just pass near the center of attraction once, and never return. The special borderline case when E = 0 corresponds to a parabolic trajectory. In this case, the particle also never comes back: it has just enough kinetic energy to make it “to infinity,” slowing down all the while, so v →0 as r →∞. The initial speed necessary to accomplish this, starting from an initial distance ri, is usually called the “escape velocity” (although it really should be called the escape speed), and it is found by simply setting Eq. (10.13) equal to zero, with r = ri, and solving for v: vesc = * 2GM ri (10.15) In general, you can calculate the escape speed from any initial distance ri to the central object, but most often it is calculated from its surface. Note that vesc does not depend on the mass of the lighter object (always assuming that the heavier object does not move at all). The escape velocity from the surface of the earth is about 11 km/s, or 1.1×104 m/s; but this alone would not be enough to let you leave the attraction of the sun behind. The escape speed from the sun starting from a point on the earth’s orbit is 42 km/s. To summarize all of the above, suppose you are trying to put something in orbit around a much more massive body, and you start out a distance r away from the center of that body. If you give the object a speed smaller than the escape speed at that point, the result will be E < 0 and an 4There is apparently a way to describe a hyperbola as an ellipse with eccentricity e > 1, but I’m definitely not going to go there. | 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 it more speed than vesc, the total energy will be positive and the trajectory will be a hyperbola. This is illustrated in Fig. 10.5 below. Figure 10.5: Possible trajectories for an object that is “released” with a sideways velocity at the lowest point in the figure, under the gravitational attraction of a large mass represented by the black circle. Each trajectory corresponds to a different value of the object’s initial kinetic energy: if Kcirc is the kinetic energy needed to have a circular orbit through the point of release, the figure shows the cases Ki = 0.5Kcirc (small ellipse), Ki = Kcirc (circle), Ki = 1.5Kcirc (large ellipse), Ki = 2Kcirc (escape velocity, parabola), and Ki = 2.5Kcirc (hyperbola). Note that all the trajectories shown in Figure 10.5 have the same potential energy at the “point of release” (since the distance from that point to the center of attraction is the same for all), so increasing the kinetic energy at that point also means increasing the total energy (10.13) (which is constant throughout). So the picture shows different orbits in order of increasing total energy. For a given total energy, the total angular momentum does not change the fundamental nature of the orbit (bound or unbound), but it can make a big difference on the orbit’s shape. Generally speaking, for a given energy the orbits with less angular momentum will be “narrower,” or “more squished” than the ones with more angular momentum, since a smaller initial angular momentum at the point of insertion means a smaller sideways velocity component. In the extreme case of zero initial angular momentum (no sideways velocity at all), the trajectory, regardless of the total energy, reduces to a straight line, either straight towards or straight away from the center of attraction. | 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 you may have to use for some homework problems this semester. You start by writing the angular momentum as L = mrPvP (or mrAvA), where A and P are the special points shown in Figure 10.2, where ⃗v and ⃗r are perpendicular. Then, you note that rP = rmin = (1 −e)a (or, alternatively, rA = rmax = (1+ e)a), so vP = L/[m(1−e)a]. Then substitute these expressions for rP and vP in Eq. (10.13), set the result equal to the total energy (10.14), and solve for e. ri vi vi vi Figure 10.6: Effect of the “angle of insertion” on the orbit. Figure 10.6 illustrates the effect of varying the angular momentum, for a given energy. All the initial velocity vectors in the figure have the same magnitude, and the release point (with position vector ⃗ri) is the same for all the orbits, so they all have the same energy; indeed, you can check that the semimajor axis of the two ellipses is the same as the radius of the circle, as required by Eq. (10.14). The difference between the orbits is their total angular momentum. The green orbit has the maximum angular momentum possible at the given energy, since the green velocity vector is perpendicular to ⃗ri. Note that this (maximizing L for a given E < 0) always results in a circle, in agreement with Eq. (e10.15): the eccentricity is zero when L = Lcirc ≡ √ aGMm2, which is the largest value of L allowed in Eq. (e10.15). For the other two orbits, ⃗vi and ⃗ri make angles of 45◦and 135◦, and so the angular momentum L has magnitude L = Lcirc sin 45◦= Lcirc/ √ 2. The result are the red and blue ellipses, with eccentricities e = ! 1 −sin2(45◦) = 0.707. | 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 were observations by his mentor, the Danish astronomer Tycho Brahe), and expressed them in a succinct way in mathematical form. These results have come to be known as Kepler’s laws, and they are as follows: 1. The planets move around the sun in elliptical orbits, with the sun at one focus of the ellipse. 2. (Law of areas) A line that connects the planet to the sun (the planet’s position vector) sweeps equal areas in equal times. 3. The square of the orbital period of any planet is proportional to the cube of the semimajor axis of its orbit (the same proportionality constant holds for all the planets). I have discussed the first “law” at length in the previous section, and also pointed out that the math necessary to prove it is far from trivial. The second law, on the other hand, while it sounds com- plicated, turns out to be a straightforward consequence of the conservation of angular momentum. To see what it means, consider Figure 10.7. O A B A’ B’ rA rB vA∆t vB∆t Figure 10.7: Illustrating Kepler’s law of areas. The two gray “curved triangles” have the same area, so the particle must take the same time to go from A to A′ as it does to go from B to B′. Suppose that, at some time tA, the particle is at point A, and a time ∆t later it has moved to A′. The area “swept” by its position vector is shown in grey in the figure, and Kepler’s second law states that it must be the same, for the same time interval, at any point in the trajectory; so, for | 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 attraction, and faster when it is closer. Quantitatively, this actually just means that its angular momentum is constant! To see this, note that the straight distance from A to A′ is the displacement vector ∆⃗rA, which, for a sufficiently short interval ∆t, will be approximately equal to ⃗vA∆t. Again, for small ∆t, the area of the curved triangle will be approximately the same as that of the straight triangle OAA′. It is a well-known result in trigonometry that the area of a triangle is equal to 1/2 the product of the lengths of any two of its sides times the sine of the angle they make. So, if the two triangles in the figures have the same areas, we must have |⃗rA| |⃗vA|∆t sin θA = |⃗rB| |⃗vB|∆t sin θB (10.17) and we recognize here the condition |⃗LA| = |⃗LB|, which is to say, conservation of angular momen- tum. (Once the result is established for infinitesimally small ∆t, we can establish it for finite-size areas by using integral calculus, which is to say, in essence, by breaking up large triangles into sums of many small ones.) As for Kepler’s third result, it is easy to establish for a circular orbit, and definitely not easy for an elliptical one. Let us call T the orbital period, that is, the time it takes for the less massive object to go around the orbit once. For a circular orbit, the angular velocity ω can be written in terms of T as ω = 2π/T, and hence the regular speed v = Rω = 2πR/T. Substituting this in Eq. (10.11), we get GM/R2 = 4π2R/T 2, which can be simplified further to read T 2 = 4π2 GM R3 (10.18) Again, this turns out to work for an elliptical orbit if we replace R by a. Note that the proportionality constant in Eq. (10.18) depends only on the mass of the central body. For the solar system, that would be the sun, of course, and then the formula would apply to any planet, asteroid, or comet, with the same proportionality constant. This gives you a quick way to calculate the orbital period of anything orbiting the sun, if you know its distance (or vice-versa), based on the fact that you know what these quantities are for the Earth. More generally, suppose you have two planets, 1 and 2, both orbiting the same star, at distances R1 and R2, respectively. Then their orbital periods T1 and T2 must satisfy T 2 1 = (4π2/GM)R3 1 and T 2 2 = (4π2/GM)R3 2. Divide one equation by the other, and the proportionality constant cancels, so you get $T2 T1 %2 = $R2 R1 %3 (10.19) | 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 (in years), and R1 = 1, in AU (an AU, or “astronomical unit,” is the distance from the Earth to the sun). A hypothetical planet at a distance of 4 AU from the sun should then have an orbital period of 8 Earth-years, since 43/2 = √ 43 = √ 64 = 8. A formula just like (10.18), but with a different proportionality constant, would apply to the satellites of any given planet; for instance, the myriad of artificial satellites that orbit the Earth. Again, you could introduce a “reference satellite” labeled 1, with known period and distance to the Earth (the moon, for instance?), and derive again the result (10.20), which would allow you to get the period of any other satellite, if you knew how its distance to the earth compares to the moon’s (or, conversely, the distance at which you would need to place it in order to get a desired orbital period). For instance, suppose I want to place a satellite on a “geosynchronous” orbit, meaning that it takes 1 day for it to orbit the Earth. I know the moon takes 29 days, so I can write Eq. (10.20) as 1 = 29(R2/R1)3/2, or, solving it, R2/R1 = (1/29)2/3 = 0.106, meaning the satellite would have to be approximately 1/10 of the Earth-moon distance from (the center of) the Earth. In hindsight, it is somewhat remarkable that Kepler’s laws are as accurate, for the solar system, as they turned out to be, since they can only be mathematically derived from Newton’s theory by making a number of simplifying approximations: that the sun does not move, that the gravitational force of the other planets has no effect on each planet’s orbit, and that the planets (and the sun) are perfect spheres, for instance. The first two of these approximations work as well as they do because the sun is so massive; the third one works because the sizes of all the objects involved (including the sun) are much smaller than the corresponding orbits. Nevertheless, Newton’s work made it clear that Kepler’s laws could only be approximately valid, and scientists soon set to work on developing ways to calculate the corrections necessary to deal with, for instance, the trajectories of comets or the orbit of the moon. Of the main approximations I have listed above, the easiest one to get rid of (mathematically) is the first one, namely, that the sun does not move. Instead, what one finds is that, as long as the sun and the planet are still treated as an isolated system, they will both revolve around the system’s center of mass. Of course, the sun’s motion (a slight “wobble”) is very small, but not completely negligible. You can even see it in the simulation I mentioned earlier, at https://phet.colorado.edu/en/simulation/gravity-and-orbits. What is much harder to deal with, mathematically, is the fact that none of the planets in the | 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 perceived over centuries. Basically, the orbits still look like ellipses to a very good degree, but the ellipses rotate very, very slowly (so they fail to exactly close in on themselves). This effect, known as orbital precession, is most dramatic for Mercury, where the ellipse’s axes rotate by more than one degree per century. Nevertheless, the Newtonian theory is so accurate, and the calculation techniques developed over the centuries so sophisticated, that by the early 20th century the precession of the orbits of all planets except Mercury had been calculated to near exact agreement with the best observational data. The unexplained discrepancy for Mercury amounted only to 43 seconds of arc per century, out of 5600 (an error of only 0.8%). It was eventually resolved by Einstein’s general theory of relativity. 10.2 Weight, acceleration, and the equivalence principle Whether we write it as mg or as GMm/r2, the force of gravity on an object of mass m has the remarkable property—not shared by any other known force—of being proportional to the inertial mass of the object. This means that, if gravity is the only force acting on a system made up of many particles, when you divide the force on each particle by the particle’s mass in order to find the particle’s acceleration, you get the same value of a for every particle (at least, assuming that they are all at about the same distance from the object exerting the force in the first place). Thus, all the parts making up the object will accelerate together, as a whole. Suppose that you are holding an object, while in free fall (remember that “free fall” means that gravity is the only force acting on you), and you let go of it, as in Fig. 10.8 below. Since gravity will give you and the object the same acceleration, you’ll find that it does not “fall” relative to you—that is, it will not fall any faster nor more slowly than yourself. From your own reference frame, you will just see it hovering motionless in front of you, in the same position (relative to you) that it occupied before you let go of it. This is exactly what you see in videos shot aboard the International Space Station. The result is an impression of weightlessness, or “zero gravity”—even though gravity is very much nonzero; the space station, and everything inside it, is constantly “falling” to the earth, it just does not hit it because it has some sideways velocity (or angular momentum) to begin with, and the earth’s pull just bends its trajectory around enough to keep it moving in a circle. But gravity is the only force acting on it, and on everything in it (at least until somebody pushes himself against a wall, or something like that). So, the kind of acceleration you get from gravity is, paradoxically, such that, if you give in to it completely, you feel like there is no gravity. | 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 other hand, comes precisely from not giving in, and rather, enlisting other forces to fight against gravity. When you do this—when you simply stand on the surface of the earth, for instance—your feet are supported by the ground below you, but every other part of your body is supported by some other part of your body, immediately above or below it, that you can think of as a sort of spring that is either somewhat stretched or somewhat compressed. It is primarily your skeleton, and mostly your spine, that bears most of the compressive load. (See Fig. 10.9, next page.) The sensation of weight is your response to this load. Interestingly, even though this constant squishing may actually result in your losing a little height in the course of a day (which you recover at night, when you lie horizontally), it is not a bad thing, rather the contrary: your bones have evolved so that they need this constant pressure to grow and replace the mass that they would otherwise lose in a “weightless” environment. On the other hand, as shown in Fig. 10.9 (c), the same compression (or extension—for instance, for the muscles in your arms, as they hang by your side) would result from a situation in which you were, say, standing motionless inside a rocket that is accelerating upwards with a = g, but very far away from any gravity source. In Fig. 10.9 (b), the “spring” that represents your skeleton needs to be compressed so it can exert an upward force F spr = mug to support the weight of your upper body (simplified here as just a single mass mu). In Fig. 10.9 (c), it needs to be compressed by the same amount, so it can exert the upward force F spr = mua needed to give your upper body an acceleration a = g. The equality of the two expressions is a direct consequence of the fact that the force of gravity is proportional to an object’s inertial mass (since the second expression is just | 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 sensation of weight. When standing on the ground motionless under the influence of gravity, however (b), every part of your body needs to compress a little in order to support the weight of the parts above it (as shown here by the compressed spring). The same compression, and hence the same subjective sensation of weight, results if you are moving upwards with an acceleration a = g, but in the absence of gravity (c). (The subscripts u and l on the forces stand for “upper” and “lower” body, respectively.) In general, then, when your whole body is subjected to an upward acceleration a, it feels like your weight is increased by an amount ma. The same thing holds regardless of direction—a forward acceleration a on a jet pilot’s body feels like a “weight” ma pushing her against her seat. This is why these “effective forces” (or, more precisely, the accelerations that cause them) are measured in g’s: a “force” of, say, 5g, means that the pilot feels pushed against her seat with a “force” equal to 5 times her weight. What’s really happening, of course, is the opposite—her seat is pushing her forward, but her internal organs are being compressed (in order to provide that same forward acceleration) the way they would under a gravity force five times stronger than at the earth’s surface. The parallels between being in a constantly accelerating frame of reference and being at rest under the influence of a constant gravity force go beyond the subjective sensation of weight. Figure 10.10 | 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 hand; but, since you are in contact with the rocket, your own velocity is constantly increasing, and as a result of that you see the object fall—relative to you. (a) (b) Figure 10.10: “Dropping” an object inside a constantly accelerating rocket, away from any gravity. From a practical point of view, this suggests a couple of ways to provide an “artificial gravity” for astronauts who might one day have to spend a long time in space, either under extremely weak gravity (for instance, during a trip to Mars), or, what amounts to essentially the same thing, in free fall (as in a space station orbiting a planet). The one most often seen in movies consists in having the space station (or spaceship) constantly spin around an axis with some angular velocity ω. Then any object that is moving with the station, a distance R away from the axis, will experience a centripetal acceleration of magnitude ω2R, which will feel like a gravity force mω2R directed in the opposite direction, that is to say, away from the center. People would then basically “walk on the walls” (that is to say, sideways as seen from above, with their feet away from the rotation axis and their heads towards the rotation axis). If somebody let go of something they were holding, | 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 to get ω2R ≃g. (On the other hand, probably even something like 1 5g is better than nothing, so who knows. . .) On a fundamental level, the equivalence between a constantly accelerated reference frame, and an inertial frame with a uniform gravitational field (such as, approximately, the surface of the earth), was elevated by Einstein to a basic principle of physics, which became the foundation of his general theory of relativity. This equivalence principle asserts that it is absolutely impossible to distinguish, by any kind of physics experiment, between the two situations just mentioned: a constantly accelerated reference frame is postulated to be completely equivalent in every way to an inertial frame with a uniform gravitational field. A remarkable consequence of the equivalence principle is that light, despite having technically “zero rest mass,” must bend its trajectory under the influence of gravity. This can be seen as follows. Imagine shooting a projectile horizontally inside the rocket in Figure 10.10. Although an inertial observer, looking from the outside, would see the projectile travel in a straight line, the observer inside the rocket would see its path bend down, just as for the projectiles we studied back in Chapter 8. This is for the same reason he would see the object fall, relative to him, in Figure 10.10: the projectile has a constant velocity, so it travels the same distance in every equal time interval, but the rocket is accelerating, so the distance it travels in equal time intervals is constantly increasing. In basically the same way, then, a beam of light sent horizontally inside the rocket, and traveling with constant velocity (and, therefore, in a straight line) in an inertial frame, would be seen as bending down in the rocket’s reference frame. However, if the equivalence principle is true, and physical phenomena look the same in a constantly accelerating frame as in an inertial frame with a constant gravitational field, it follows that light must also bend its path in the latter system, in much the same way as a projectile would. (I say “much the same way” because the effect is not just as simple as giving light an “effective mass”; there are other relativistic effects, such as space contraction and time dilation, that must also be reckoned with.) This gravitational bending was one of the most important early predictions of Einstein’s General Relativity theory, and certainly the most spectacular. Since one needs the light rays to pass vary close to a large mass to get an observable effect, the way the prediction was verified was by looking at the apparent position of the stars that can be seen close to the edge of the sun’s disk during a solar eclipse. The slight (apparent) shift in position predicted by Einstein was observed by Sir Arthur Eddington during the solar eclipse of 1919 (two expeditions were sent to remote corners of the earth for this purpose), and it was primarily responsible for Einstein’s sudden fame among the general public of his day. Today, with modern telescopes, this so-called “gravitational lensing” effect has become an important tool in astronomy, allowing us to interpret the pictures taken of distant galaxies, which are often | 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 hole. If you set the escape velocity of Eq. (10.15) equal to the speed of light in vacuum, c, and solve for ri, you obtain what is called the Schwarzschild radius, rs, for a black hole of mass M; the idea being that, in order to be a black hole, the object has to be so dense that all its mass M is inside a sphere of radius smaller than rs. Physicists today believe in the existence (and even what one might call the ubiquity) of black holes, of which the Schwarzschild solution was only the first calculated example. Note that rs does not define the actual, physical surface of the object; it does, however, locate what is known as the black hole’s event horizon. Nothing can be known, through observation, about anything that might happen closer to the black hole’s center than the distance rs, since no information can be transmitted faster than light, and no light can escape from a distance ri < rs. 10.3 In summary 1. In Newton’s theory of gravity, two particles of inertial masses m1 and m2, separated by a distance r12, exert a gravitational force on each other which is attractive, along the line joining the two particles, and has magnitude F G 12 = F G 21 = Gm1m2/r2 12, with G = 6.674 × 10−11 m3kg−1s−2. 2. The gravitational force between two extended objects is found by adding (vectorially) the forces between all the pairs of particles that make up the objects. For objects with spherical symmetry, the result has the same form as above, with r12 being now the distance between the centers of the spheres. 3. The gravitational potential energy of a system of two particles of masses m1 and m2 is U G = −Gm1m2/r12. For systems of more particles, one should just add the corresponding energies for all the possible pairs. For a pair of spheres, one may use the same result as for two particles, as long as one is not interested in the spheres’ gravitational self-energy. 4. The expressions given above for F G and U G reduce, respectively, to mg and mgy+C (where C is an unimportant constant) near the surface of the earth, to a good approximation, provided the distance y to the surface is much smaller than the radius of the earth, RE. If ME is the mass of the earth, one has g = GME/R2 E. 5. A good first approximation to many astronomical problems is obtained by considering the motion of a particle (or sphere) of mass m under the gravitational pull of an object (also treated as a particle or sphere) of much larger mass M, which is assumed to not move at all. This is sometimes called the Kepler problem. | 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 elliptical orbits, one has E = −GMm/2a, where a is the ellipse’s semimajor axes. The large mass object is not at the center, but at one of the foci of the ellipse. The distance from the focus to the center is equal to ea, where e is called the eccentricity of the ellipse. 8. The escape speed of an object bound gravitationally to a mass M, a distance ri away from that mass’s center, is obtained by setting the total energy of the system equal to zero. It is the speed the object needs in order to be able to just escape to “infinity” and “stop there” (mathematically, v →0 as r →∞, which makes E = K + U G = 0). 9. The angular momentum of a particle in a Kepler trajectory (circle, ellipse, parabola or hy- perbola), relative to the point where the large mass M is located, is constant. For a given energy, orbits with less angular momentum are more eccentric. 10. A consequence of conservation of angular momentum is Kepler’s second law, or “law of areas”: The orbiting object’s position vector (with the origin at the location of the large mass), sweeps equal areas in equal times. 11. The square of the orbital period of any object in a Kepler elliptical orbit is proportional to the cube of the semimajor axis of the ellipse. This is Kepler’s third law. Specifically, one finds, from Newton’s theory, T 2 = (4π2/GM)a3. 12. According to Einstein’s principle of equivalence, a constant acceleration a of a reference frame is experienced by every object in that reference frame as an “extra weight,” or gravitational force, equal to −ma (that is, of magnitude ma and in the direction opposite the acceleration). | 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 firing a thruster in the direction opposite their motion). This would take them to a lower orbit, but then the capsule would start speeding up, on its own. Use the concepts introduced in this chapter to explain what is going on in this scenario. Let R be the radius of the initial orbit. For simplicity, assume the thruster is on only for a very short time, so you can neglect the motion of the capsule during this time. In other words, treat it as an instantaneous reduction in velocity, and discuss: (a) What happens to the system’s potential and kinetic energy, and angular momentum? (b) Is the new orbit circular or elliptic? How do you know? What is the new orbit’s rmax (maximum distance to the center of the Earth)? (c) Why does the capsule speed up in its new orbit? (d) If the new orbit is not circular, what would the astronauts need to do to make it so? (Without getting any closer to the Earth, that is, keeping rmin the same.) Make sure to draw a diagram of the situation. Make it as accurate as you can. Solution (a) Under the assumption that the capsule barely changes position during the thruster firing, the potential energy of the system, which is equal to U G = −GMm/R, will not change: U G f = U G i . The kinetic energy, on the other hand, will go down, since the capsule’s speed is reduced: Kf < Ki. Hence, the total mechanical energy of the system, E = K + U G, will decrease: Ef < Ei. The angular momentum will go down, since v goes down. (b) The new orbit has to be elliptical, since to have a circular orbit at a distance R requires a precise velocity (given just below Eq. (10.11) by v = ! GM/R), and now we have changed that. However, since the orbit must still be a closed curve, it will contain the starting point, which is, by our assumption, a distance R away from the Earth. Also, if the direction of the velocity vector does not change as a result of the thruster firing (only the magnitude of v is supposed to change), it follows that at this point the velocity and the position vectors are perpendicular. For a circular orbit, this is the case everywhere. For an elliptical orbit, this is only true at the two extreme points labeled P and A in Figure 10.3 (the perigee and apogee, respectively). So, the initial position of the capsule becomes either the perigee or the apogee of the new orbit. Which is it? | 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 had a = R; now, we must have a < R. This means that the starting point, a distance R away from the (center of the) Earth, cannot be the perigee (the point of closest approach), since at that point r = rmin, and rmin is always less than a (check again Fig. 10.3, or Eqs. (10.12)). Instead, the starting point has to be the apogee of the new orbit, and therefore the distance at that point is also the maximum distance: rmax = R. (c) The capsule speeds up in its new orbit because, as we just saw, it starts as far away from the Earth as it’s going to get; therefore, as it moves it will start getting closer to the Earth, and we know from Kepler’s second law that as it gets closer it has to speed up. (You can also say that, as it gets closer, the gravitational potential energy of the system will go down, and therefore its kinetic energy must increase.) (d) The easiest way to change the new orbit to a circular orbit with radius rmin would be to perform another speed-reduction maneuver, but this time at perigee. At perigee, the distance to the Earth is already rmin, which is what you want it to be, but the capsule is moving too fast to stay on a circular orbit (put differently, the gravitational force of the Earth at that point is too weak to bend the orbit into a circle): that is why it eventually ends up “overshooting” the Earth on the other side. Reducing v will further reduce E and, by the same argument as above, it will result in an orbit with a smaller a, which is what you want (since, at the moment, a > rmin, and you want the new a to be equal to rmin). A P | 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 size of the orbits is greatly exaggerated compared to those in the early space flights, which were much closer to the Earth! The way to draw this kind of figure is to first draw an accurate ellipse, making sure you know where the focus is; then draw the circles centered at the focus and touching the ellipse at the right points. An ellipse’s equation in polar form, with the origin at one focus, is r = a + ae cos φ. 10.4.2 Orbital data from observations: Halley’s comet Halley’s comet follows an elliptical orbit around the sun. At its closest approach, it is a distance of 0.59 AU from the sun (an astronomical unit, AU, is defined as the average distance from the earth to the sun: 1 AU = 1.496 × 1011 m), and it is moving at 5.4 × 104 m/s. We know its period is approximately 76 years. Ignoring the forces exerted on the comet by the other solar system objects (a rather rough approximation): (a) Use the appropriate Kepler law to infer the value of a (the semimajor axis) for the comet’s orbit. (b) What is the eccentricity of the comet’s orbit? (c) Using the result in (a) and conservation of angular momentum, find the speed of the comet at aphelion (the point in its orbit when it is farthest away from the sun). Solution (a) The “appropriate Kepler law” here is the third one. For any two objects orbiting, for instance, the sun, the square of their orbital periods is proportional to the cube of their orbits’ semimajor axes, with the same proportionality constant (4π/GMsun; see Eq. (10.18)). We do not even need to calculate the proportionality constant; we can divide the equation for Halley’s comet by the equation for the earth, and get T 2 Halley T 2 earth = a3 Halley a3 earth (10.21) where T 2 earth = 1 yr2, and a3 earth = 1 AU3, so we get immediately aHalley = ( 762)1/3 AU = 17.9 AU (10.22) (b) We can get this one from a look at Figure 10.3: the product ea, plus the minimum distance between the comet and the sun (0.59 AU) is equal to a. (This is just what the second of 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 symbolically and not substitute the numbers in until the very end, to see if something cancels out, and to prevent mistakes when copying large numbers form one line to the next; and sometimes, like here, you do not even have to convert to other units! (c) At the point of closest approach (perihelion), the velocity and the position vector of the comet are perpendicular, and so the magnitude of the comet’s angular momentum is just equal to L = mrv. The same happens at the farthest point in the orbit (aphelion), and since angular momentum is conserved for the Kepler problem, we can write mrminvmax = mrmaxvmin (10.24) (the reason for this choice of subscripts is that we know that when r is maximum, v is minimum, and vice-versa). Solving for vmin, the speed at aphelion, we get vmin = rmin rmax vmax = rmin 2a −rmin vmax = 0.59 2 · 17.9 −0.59 5.4 × 104 m s = 905 m s (10.25) Here again the equation I used to find rmax can be derived directly from Fig. 10.3 (and it is also one of the equations (10.12): rmin + rmax = 2a). Once again, I was able to use AU throughout, since the units of distance cancel out in the fraction rmin/rmax. | 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 realized in real life. Some planets, like our own Earth, are particularly susceptible to deformation, because of the large amount of fluid matter on their surface, and even rocky planets and moons are sensitive to tidal forces, which are the differences on the gravitational pull by the central attractor on different parts of the object considered. If you look back on Figure 10.1, for instance, it is easy to see that the moon must be pulling more strongly on the side of the Earth that is closer to it (the left side, on that picture) than on the farther side. This is, of course, because the force of gravity depends on the distance between the interacting objects, and is stronger when the objects are closer. You can easily calculate that the force by the moon on a given volume of Earth (say, a cubic meter) is about 7% stronger on the near side than on the far side. A deformable object subject to such a pair of forces will naturally be stretched along the direction of the pull: in the case of the Earth, this “stretching” affects primarily the water in the oceans that cover most of the surface, resulting in two “tidal bulges” that account for the well-known phenomenon of tides: as the earth rotates around its axis, each point on the surface passes through one of the bulges once a day, resulting in two high tides each day (and, in between, a comparatively lower water level, or low tide, twice a day as well).5 For many objects in the solar system, this tidal stretching has, over millions of years, resulted in a permanent deformation. The tidal forces by the Earth on the moon are weaker than those by the moon on the Earth (since the moon is much smaller, the difference between the Earth’s pull on the moon’s near and far side is less than 1.5%), but at a time when the moon was more malleable than at present, it was enough to produce an elongation along the Earth-moon axis that is now pretty much frozen in place. Once a satellite (I will use the term generically to refer either to a planet orbiting the sun, or a moon orbiting a planet) becomes permanently deformed, a new phenomenon, known as tidal locking, can happen. Suppose the satellite is rotating around its own axis, in addition to orbiting around the primary body. As you can see from the figure below, if the rotation is too fast, the gravitational 5You may ask, besides being stretched, shouldn’t the net pull of the moon cause the Earth to fall towards it? The answer is that the Earth does not fall straight towards the moon, for the same reason the moon does not fall straight towards the Earth: both have a sideways velocity that keeps them in a steady orbit, for which the net gravitational pull provides the centripetal force. Both actually orbit around the center of mass of the system, which is located about 1, 690 km under the Earth’s surface; the Earth’s orbit (and its orbital velocity) is much smaller than the moon’s, by virtue of its much larger mass, of course. | 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 its orbital period, so that it always shows the same side to the primary body. This is the situation with the Earth’s moon, and indeed for most of the major moons of the giant planets. (a) (b) τ τ Figure 10.11: An elongated moon revolving around a planet in a clockwise orbit, and at the same time rotating clockwise around an axis through its center. In (a), the rotation is too fast, resulting in a counter- clockwise “tidal torque.” In (b), the rotation is too slow, and the “tidal torque” is clockwise. In both cases, the torque is due to the moon’s misalignment, and to the gravitational force on its near side being stronger than on its far side, as shown by the blue force vectors. Note that, by the same argument, we would expect the tidal forces on the Earth due to the moon to try and bring the Earth into tidal locking with the moon—that is, to try to bring the duration of an Earth day closer to that of a lunar month. Indeed, the moon’s tidal forces have been slowing down the Earth’s rotation for billions of years now, and continue to do so by about 15 microseconds every year. This process requires dissipation of energy, (which is in fact associated with the ocean tides: think of the frictional forces caused by the waves, as the tide comes in and out); however, to the extent that the Earth-moon system may be treated as isolated, its total angular momentum cannot change, and so the slowing-down of the Earth is accompanied by a very gradual increase in the radius of the moon’s orbit—about 3.8 cm per year, currently—to keep the total angular momentum constant. | 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 × 106 m, and the mass of the Earth is M = 5.97 × 1024 kg.) (b) How different is the result you got in part (a) above from what you would have obtained if you had treated the Earth’s gravitational force as a constant (independent of height), as we did in previous chapters? (c) Using the correct expression for the gravitational potential energy, what is the total energy of the projectile-Earth system, if the projectile’s mass is 1, 000 kg? Now assume the projectile is fired horizontally instead, with the same speed. This time, it actually goes into orbit! (Well, it would, if you could neglect things like air resistance, and mountains and stufflike that. Assume it does, anyway, and answer the following questions:) (d) What is the projectile’s angular momentum around the center of the Earth? (e) How far from the center of the Earth does it make it this time? (You will need to use conservation of energy and angular momentum to answer this one, unless you can think of a shortcut. . .) (f) Draw a sketch of the Earth and the projectile’s trajectory. Problem 2 You want to put a satellite in a geosynchronous orbit around the earth. (This means the asteroid takes 1 day to complete a turn around the earth.) (a) At what height above the surface do you need to put it? (b) How fast is it moving? (c) How does the answer to (b) compare to the escape speed from the earth, for an object at this height? Problem 3 Suppose that one day astronomers discover a new asteroid that moves on a very elliptical orbit around the sun. At the point of closest approach (perihelion), the asteroid is 1.61 × 108 km away from the (center of the) sun, and its speed is 38.9 km/s. (a) What is the escape velocity from the sun at this distance? The mass of the sun is 2 × 1030 kg. (b) The astronomers estimate the mass of the asteroid as 1012 kg. What is its kinetic energy at perihelion? (c) What is the gravitational potential energy of the sun-asteroid system at perihelion? (d) What is the total energy of the sun-asteroid system? Is it positive or negative? Is this consistent with the assumption that the orbit is an ellipse? What would a positive total energy mean? (e) At perihelion, the asteroid’s velocity vector is perpendicular to its position vector (as drawn | 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 conservation of energy and angular momentum to figure out the asteroid’s distance to the sun at aphelion. (Hint: if solving simultaneous equations does not appeal to you, there is a formula in this chapter which you can use to answer this question fairly quickly, based on something you have calculated already.) (h) How fast is the asteroid moving at aphelion? Problem 4 The mass of the moon is 7.34 × 1022 kg, and its radius is about 1.74 × 106 m (a) What is the value of “gmoon”, that is, the acceleration of gravity for a falling object near the surface of the moon? (b) What is the escape speed (from the moon) for an object on the surface of the moon? (c) What is the escape speed from the earth for an object that is as far from the earth as the orbit of the moon? (d) At some point between the earth and the moon, an object would be pulled with equal strength towards both bodies. How far from the earth is that point? Problem 5 On August 17, 2017, the LIGO observatory reported the detection of gravitational waves from the merger of two neutron stars. Neutron stars are extremely dense (“a teaspoon of neutron star material has a mass of about a billion tons”) and very small—only about 10 or 20 km in diameter. The stars were estimated to have been separated by about 300 km when the merger signal became detectable. Let us start a little before that. Suppose the stars have the same mass, M = 2.6 × 1030 kg (approximately 1.3 times the mass of the sun), and are separated (center-to-center distance) by 1000 km. They pull on each other gravitationally, and as a result each one moves in a circular orbit around their common center of mass. What is then (a) their period of revolution, and (b) their speed? (Hint: what is the centripetal force in this case?) Problem 6 (a) Consider two possible circular orbits for a satellite around a planet, with radii R1 and R2. If R1 < R2, which of the two orbits has (i) the largest total energy, and (ii) the largest total angular momentum? Explain. (b) For an object in a circular orbit around a planet, how does the orbital velocity compare to the escape velocity from the same orbit? Problem 7 Jupiter’s distance to the sun is 5.2 astronomical units. How long does it take for Jupiter to complete | 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, that exemplify the kind of mechanical oscillator with which this chapter is concerned. And then, of course, there are swings, which function essentially like the pendulum depicted below. (a) (b) (c) (d) F t F G Fnet v F t F G Fnet Figure 11.1: A simple pendulum. In (a), the equilibrium position, the tension and gravity forces balance out. In (b), they combine to produce a restoring force (in blue) pointing back towards equilibrium. In (c), the bob is passing through equilibrium and the net force on it at that instant is again zero, but its momentum keeps it going. At (d) we have the mirror image of (b). 253 | 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 physical system has. The pendulum in Figure 11.1 illustrates how these things combine to produce an oscillation. As the pendulum bob is displaced from its equilibrium position, a net force on it appears (a combination of gravity and the tension in the string), pointing back towards the vertical. When the bob is released, it accelerates under the influence of this force, with the result that when it reaches back the equilibrium position, its inertia (or, if you prefer, its momentum) causes it to overshoot it. Once this happens, the restoring force changes direction, always trying to bring the mass back to equilibrium; as a result, the bob slows down, and eventually reverses course, accelerates again towards the vertical, overshoots it again. . . the process will repeat itself, until all the energy we initially put in the system (gravitational potential energy, in this case) is dissipated away (or damped), mostly through friction at the pivot point, though air resistance plays a small part as well. That the motion, in the absence of dissipation, must be symmetric around the equilibrium position follows from conservation of energy: the speed of the bob at any given height must be the same on either side, in order for the sum of its potential and kinetic energies to be the same. In particular, if released from rest from some height, it will stop when it reaches the same height on the other side. In the presence of dissipation, the motion is neither exactly symmetric, nor exactly periodic (that is to say, it does not repeat itself exactly—the maximum height gets lower every time, the speed as it passes through the equilibrium position gets also smaller and smaller), but when the dissipation is not very large one can always define an approximate period (which we will denote with the letter T) as the time it takes to complete one full swing. The inverse of the period is the frequency, f, which tells us how many full swings the pendulum completes per second. These two quantities, T and f, can be defined for any type of periodic (or approximately periodic) motion, and will always satisfy the relationship f = 1 T (11.1) The units of frequency are, of course, inverse seconds (s−1). In this context, however, this unit is called a ”hertz,” and abbreviated Hz. 11.2 Simple harmonic motion A particularly important kind of oscillatory motion is called simple harmonic motion. This is what happens when the restoring force is linear in the displacement from the equilibrium position: that | 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 perform simple harmonic motion. This kind of oscillation is distinguished by the following characteristics: • The position as a function of time, x(t), is a sinusoidal function. • The period of the oscillations does not depend on their amplitude (by “amplitude” we mean the maximum displacement from the equilibrium position). What this second property means is that, for instance, with reference to Fig. 11.2, you can displace the mass a distance A, or A/2, or A/3, or whatever you choose, and the period (and frequency) of the resulting oscillations will be the same regardless. (This means, actually, that if you displace it farther it has to end up moving faster, to cover the larger distance in the same time.) x x0 (a) (b) (c) A A Figure 11.2: A mass attached to a spring and sliding on a frictionless surface. Figure (a) shows the spring in its relaxed state (the “equilibrium” position for the mass, at coordinate x0). If displaced from equilibrium a distance A and released (b), the mass will perform simple harmonic oscillations with amplitude A. Since we know that “Hooke’s law” is actually just an approximation, valid only provided that the spring is not compressed or stretched too much, we expect that in real life the “ideal” simple harmonic motion properties I have listed above will only hold approximately, as well; so, in fact, if you stretch a spring too much you will get a different period, eventually, than if you stay in the “linear,” Hooke’s law regime. This is a general characteristic of most physical systems: simple | 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 function is sinusoidal, by which I mean a sine or a cosine. Thus, for example, if the mass in Fig. 11.2 is released from rest at t = 0, and the position x is measured from the equilibrium position x0 (that is, the point x = x0 is taken as the origin of coordinates), the function x(t) will be x(t) = A cos(ωt) (11.3) where the quantity ω, known as the oscillator’s angular frequency, is given by ω = * k m (11.4) Here, k is the spring constant, and m the mass of the object (remember the spring is assumed to be massless). I will prove that Eq. (11.3), together with (11.4), satisfy Newton’s second law of motion for this system in a moment; first, however, I need to say a couple of things about ω. You’ll recall that we have used this symbol before, in Chapter 9, to represent the angular velocity of a particle moving in a circle (or, more generally, of any rotating object). Why bring it up again now for an apparently completely different purpose? + r v P x y R sin(ωt) θ R cos(ωt) Figure 11.3: A particle moving on a circle with constant angular velocity ω. Assuming θ = 0 at t = 0, we have θ = ωt, and therefore the particle’s x coordinate is given by the function x(t) = R cos(ωt). This means the corresponding point on the x axis (the red dot) performs simple harmonic motion with angular frequency ω as the particle rotates. | 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 harmonic motion with angular frequency ω (and amplitude R). (Of course, there is nothing special about the x axis; the projection on any other axis will also perform simple harmonic motion with the same angular frequency; for example, the blue dot on the figure.) If the angular velocity of the particle in Fig. 11.3 is constant, then its “orbital period” (the time needed to complete one revolution) will be T = 2π/ω, and this will also be the period of the associated harmonic motion (the time it takes for the motion to repeat itself). You can see this directly from Eq. (11.3): if you increase the time t by 2π/ω, you get the same value of x: x $ t + 2π ω % = A cos 2 ω $ t + 2π ω %3 = A cos(ωt + 2π) = A cos(ωt) = x(t) (11.5) Since the frequency f of an oscillator is equal to 1/T, this gives us the following relationship between f and ω: f = 1 T = ω 2π (11.6) One way to tell whether one is talking about an oscillator’s frequency (f) or its angular frequency (ω)—apart from the different symbols, of course—is to pay attention to the units. The frequency f is usually given in hertz, whereas the angular frequency ω is always given in radians per second. Apart from the factor of 2π, they are, of course, completely equivalent; sometimes one is just more convenient than the other. On the other hand, the only way to tell whether ω is a harmonic oscillator’s angular frequency or the angular velocity of something moving in a circle is from the context. (In this chapter, of course, it will always be the former). Let us go back now to Eq. 11.3 for our block-on-a-spring system. The derivative with respect to time will give us the block’s velocity. This is a simple application of the chain rule of calculus: v(t) = dx dt = −ωA sin(ωt) (11.7) Another derivative will then give us the acceleration: a(t) = dv dt = −ω2A cos(ωt) (11.8) Note that the acceleration is always proportional to the position, only with the opposite sign. The proportionality constant is ω2. Since the force exerted by the spring on the block is F = −kx (because we are measuring the position from the equilibrium position x0), Newton’s second law, F = ma, gives us ma = −kx (11.9) and you can check for yourself that this will be satisfied if x is given by Eq. (11.3), a is given by Eq. (11.8), and ω is given by Eq. (11.4). | 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 restoring force means a higher frequency (faster oscillations) whereas a larger inertia means a lower frequency (slower oscillations—a more “sluggish” response). The position, velocity and acceleration graphs for the motion (11.3) are shown in Fig. 11.4 below. You may want to pay attention to some of their main features. For instance, the position and the velocity are what we call “90◦out of phase”: one is maximum (or minimum) when the other one is zero. The acceleration, on the other hand, is “180◦out of phase” (that is to say, in complete opposition) with the position. As a result of that, all combinations of signs for a and v are possible: the object may be moving to the right with positive or negative acceleration (depending on which side of the origin it’s on), and likewise when it is moving to the left. A –A 0 ωA –ωA 0 ω2A –ω2A 0 x(t) v(t) a(t) T T/2 3T/2 2T 0 Figure 11.4: Position, velocity and acceleration as a function of time for an object performing simple harmonic motion according to Eq. (11.3). Since the time we choose as t = 0 is arbitrary, the function in Eq. (11.3) (which assumes that t = 0 is when the object’s displacement is maximum and positive) is clearly not the most general formula for simple harmonic motion. Another way to see this is to realize that we could have started the motion differently. For instance, we could have hit the block with a sharp, “impulsive” force, lasting only a very short time, so it would have acquired a substantial velocity before it could have moved very far from its initial (equilibrium) position. In such a case, the motion would be better described by a sine function, such as x(t) = A sin(ωt), which is zero at t = 0 but whose derivative (the object’s velocity) is maximum at that time. | 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 mean, the object’s initial position and velocity. Specifically, you can see, by setting t = 0 in Eq. (11.10) and its derivative, that the initial position and velocity of the motion described by Eq. (11.10) are xi = A cos φ vi = −ωA sin φ (11.11) Conversely, if you are given xi and vi, you can use Eqs. (11.11) to determine A and φ, which is what you need to know in order to use Eq. (11.10) (note that the angular frequency, ω, does not depend on the initial conditions—it is always the same regardless of how you choose to start the motion). Specifically, you can verify that Eqs. (11.11) imply the following: A2 = x2 i + v2 i ω2 (11.12) and then, once you know A, you can get φ from either xi = A cos φ or vi = −ωA sin φ (in fact, since the inverse sine and inverse cosine are both multivalued functions, you should use both equations, to make sure you get the correct sign for φ). 11.2.1 Energy in simple harmonic motion Equation (11.11) above actually follows from the conservation of energy principle for a harmonic oscillator. Consider again the mass on the spring in Figure 11.3. Its kinetic energy is clearly K = 1 2mv2, whereas the potential energy in the spring is 1 2kx2. Using Eq. (11.10) and its derivative, we have U spr = 1 2kA2 cos2(ωt + φ) K = 1 2mω2A2 sin2(ωt + φ) (11.13) Recalling Eq. (11.4), note that ω2 = k/m, so if you substitute this in the second equation above, you can see that the amplitude of both the potential and the kinetic energy is the same, namely, 1 2kA2. Since, for any angle θ, it is always true that cos2 θ + sin2 θ = 1, we find Esys = U spr + K = 1 2kA2 = 1 2mω2A2 (11.14) | 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 twice as fast as the oscillator itself: for example, the potential energy is maximum both when the displacement is maximum (spring maximally stretched) and when it is minimum (spring maximally compressed). Similarly, the kinetic energy is maximum when the oscillator passes through the equilibrium posi- tion, regardless of whether it is moving to the left or to the right. T T/2 3T/2 2T 0 Esys Uspr K 0 Figure 11.5: Kinetic (red), potential (blue) and total (black) energy for the oscillator shown in Fig. 11.4. 11.2.2 Harmonic oscillator subject to an external, constant force Consider a mass hanging from an ideal spring suspended from the ceiling, as in Fig. 11.6 below (next page). Supposed the relaxed length of the spring is l, such that, in the absence of gravity, the object’s equilibrium position would be at the height y0 shown in figure 11.6(a). In the presence of gravity, of course, the spring needs to stretch, to balance the object’s weight, and so the actual equilibrium position for the system will be y′ 0, as shown in figure 11.6(b). The upwards force from the spring at that point will be −k(y′ 0 −y0), and to balance gravity we must have −k(y′ 0 −y0) −mg = 0 (11.15) Suppose that we now stretch the spring beyond this new equilibrium position, so the mass is now at a height y (figure 11.6(c)). What happens then? The net upwards force will be −k(y −y0)−mg, but using Eq. (11.15) this can be rewritten as Fnet = −k(y −y0) −mg = −k(y −y′ 0) −−k(y′ 0 −y0) −mg = −k(y −y′ 0) (11.16) | 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 gravity does is to change the equilibrium position, so that if you now displace the mass, it will oscillate around y′ 0 instead of around y0. The oscillation’s period and frequency are the same as if the spring was horizontal. y0 y0' y l l' (a) (b) (c) Figure 11.6: (a) An ideal (massless) spring hanging from the ceiling, in its relaxed position. (b) With a mass m hanging from its end, the spring stretches to a new length l′, so that k(l′ −l) = mg. (c) If the mass is now displaced from this equilibrium position (labeled y′ 0 in the figure) it will perform harmonic oscillations symmetrically around the point y′ 0, with the same frequency as if the spring was horizontal. Although I have established this here for the specific case where the oscillator involves a spring, and the external force is gravity, this is a completely general result, valid for any simple harmonic oscillator, since for such a system the restoring force will always be a linear function of the displace- ment (which is all that is required for the math to work). As long as the external force is constant, the frequency of the oscillations will not be affected, and only the equilibrium position will change. In an example at the end of the chapter (under “Advanced Topics”) I will show you how you can make use of this to calculate the effect of friction on the horizontal mass-spring combination in Fig. 11.2. One thing you need to keep in mind, however, is that when the oscillator is subjected to an external force, as was the case here, its energy will not, in general, remain constant (unlike what we saw in Section 11.2.1), since the external force will be doing work on the system as it oscillates. If the | 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 Figure 11.6, the external force is gravity, which we know to be a conservative force, so the energy that will be conserved will be the total energy of the system that includes the oscillation and the Earth, and hence also the gravitational potential energy (for which we can use here the familiar form U G = mgy): Eosc+earth = U spr + K + U G = 1 2(y −y0)2 + 1 2mv2 + mgy = const (11.17) The reason it is no longer possible to combine the terms U spr + K into the constant 1 2kA2, as in Eq. (11.14), is that now we have y(t) = y′ 0 + A cos(ωt + φ) v(t) = −ωA sin(ωt + φ) (11.18) so the oscillations are centered around the new equilibrium position y′ 0, but the spring energy is not zero at that point: it is zero at y = y0 instead. You can check for yourself, however, that if you substitute Eqs. (11.18) into Eq. (11.17), and make use of the fact that k(y′ 0 −y0) = −mg (Eq. (11.15)), you do indeed get a constant, as you should. 11.3 Pendulums 11.3.1 The simple pendulum Besides masses on springs, pendulums are another example of a system that will exhibit simple harmonic motion, at least approximately, as long as the amplitude of the oscillations is small. The simple pendulum is just a mass (or “bob”), approximated here as a point particle, suspended from a massless, inextensible string, as in Fig. 11.7 on the next page. We could analyze the motion of the bob by using the general methods introduced in Chapter 8 to deal with motion in two dimensions—breaking down all the forces into components and applying ⃗Fnet = m⃗a along two orthogonal directions—but this turns out to be complicated by the fact that both the direction of motion and the direction of the acceleration are constantly changing. Although, under the assumption of small oscillations, it turns out that simply using the vertical and horizontal directions is good enough, this is not immediately obvious, and arguably it is not the most pedagogical way to proceed. | 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 a circle, and that we have developed already in Chapter 9 a whole set of tools to deal with that kind of motion. Let us, therefore, describe the position of the pendulum by the angle it makes with the vertical, θ, and let α = d2θ/dt2 be the angular acceleration; we can then write the equation of motion in the form τnet = Iα, with the torques taken around the center of rotation—which is to say, the point from which the pendulum is suspended. Then the torque due to the tension in the string is zero (since its line of action goes through the center of rotation), and τnet is just the torque due to gravity, which can be written τnet = −mgl sin θ (11.19) The minus sign is there to enforce a consistent sign convention for θ and τ: if, for instance, I choose counterclockwise as positive for both, then I note that when θ is positive (pendulum to the right of the vertical), τ is clockwise, and hence negative, and vice-versa. This is characteristic of a restoring torque, that is to say, one that will always try to push the system back to its equilibrium position (the vertical in this case). As for the moment of inertia of the bob, it is just I = ml2 (if we treat it as just a point particle), so the equation τnet = Iα takes the form ml2 d2θ dt2 = −mgl sin θ (11.20) The mass and one factor of l cancel, and we get d2θ dt2 = −g l sin θ (11.21) | 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 elliptic functions; they are included in many modern mathematical packages, but they are still not easy to use. More importantly, the oscillations they describe, in general, are not of the simple harmonic type. On the other hand, if the amplitude of the oscillations is small, so that the angle θ, expressed in radians, is a small number, we can make an approximation that greatly simplifies the problem, namely, sin θ ≃θ (11.22) This is known as the small angle approximation, and requires θ to be in radians. As an example, if θ = 0.2 rad (which corresponds to about 11.5◦), we find sin θ = 0.199, to three-figure accuracy. With this approximation, the equation to solve becomes much simpler: d2θ dt2 = −g l θ (11.23) We have, in fact, already solved an equation completely equivalent to this one in the previous section: that was equation (11.9) for the mass-on-a-spring system, which can be rewritten as d2x dt2 = −k m x (11.24) since a = d2x/dt2. Just like the solutions to (11.24) could be written in the form x(t) = A cos(ωt + φ), with ω = ! k/l, the solutions to (11.23) can be written as θ(t) = A cos(ωt + φ) ω = *g l (11.25) This tells us that if a pendulum is not pulled too far away from the vertical (say, about 10◦or less) it will perform approximate simple harmonic oscillations, with a period of T = 2π ω = 2π . l g (11.26) This depends only on the length of the pendulum, and remains constant even as the oscillations wind down, which is why it became the basis for time-keeping devices, beginning with the invention of the pendulum clock by Christiaan Huygens in 1656. In particular, a pendulum of length l = 1 m will have a period of almost exactly 2 s, which is what gives you the familiar “tick-tock” rhythm of a “grandfather’s clock,” once per second (that is to say, once every half period). | 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 suspension; rather, for example, the bob could have a size that is not negligible compared to l (as in Fig. 11.1), or the “string” could have a substantial mass of its own—it could, for instance, be a chain, like in a playground swing, or a metal rod, as in most pendulum clocks. F G o θ d F G θ (a) (b) cm l Figure 11.8: The “physical pendulum.” Figure (a) shows an arbitrary distribution of mass, pivoted at point O, with center of mass at CM, oscillating under the restoring torque provided by gravity. Figure (b) shows the special case of a thin rod of length l pivoted at one end (the distance d = l/2 in this case). In both cases, there is an additional force (not shown) acting at the pivot point, to balance gravity. Regardless of the reason, having to deal with a distributed mass means also that one needs to use the center of mass of the system as the point of application of the force of gravity. When this is done, the motion of the pendulum can again be described by the angle between the vertical and a line connecting the point of suspension and the center of mass. If the distance between these two points is d, then the torque due to gravity is −mgd sin θ, and the only other force on the system, the force at the pivot point, exerts no torque around that point, so we can write the equation of motion in the form I d2θ dt2 = −mgd sin θ (11.27) Under the small-angle approximation, this will again lead to simple harmonic motion, only now with an angular frequency given by ω = * mgd I (11.28) | 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, implying a correspondingly shorter period. 11.4 In summary 1. Most stable physical systems will oscillate when displaced from their equilibrium position. The oscillations are due to a restoring force (or a restoring torque) and the system’s own inertia. 2. The period T and frequency f of an oscillatory motion are related by f = 1/T. The units of frequency are s−1 or hertz (abbreviated Hz). 3. A special (but very common) kind of oscillation is simple harmonic motion. This happens whenever the restoring force is a linear function of (that is, it is proportional to) the system’s displacement from the equilibrium position. 4. The angular frequency, ω, of a simple harmonic oscillator is related to the regular frequency by ω = 2πf. One of the properties of simple harmonic motion is that its frequency does not depend on the initial conditions, that is, on the velocity or displacement with which the motion is started. 5. If the equilibrium position is chosen to correspond to x = 0, the most general form of the position function for a simple harmonic oscillator is x(t) = A cos(ωt+φ), where The amplitude A and phase angle φ are determined by the initial conditions. The velocity function is then v(t) = −ωA sin(ωt + φ), and the acceleration a(t) = −ω2A cos(ωt + φ). 6. The total energy (potential plus kinetic) in a simple harmonic oscillator is equal to Esys = 1 2mω2A2. The kinetic and potential energies oscillate (in opposition, that is to say, each being maximum when the other is minimum) between this value and zero. 7. A mass attached to an ideal (massless) spring is an example of a simple harmonic oscillator. If the spring constant is k, the angular frequency of this system is ω = ! k/m. 8. An external, constant force acting on a harmonic oscillator does not change its period (or frequency), only its equilibrium position. For the mass-on-a-spring system, a force F will cause a displacement of the equilibrium position equal to F/k, in the direction of the force. | 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 of mass m pivoted around some point and performing oscillations under the influence of gravity is sometimes called a physical pendulum. Just as for the simple pendu- lum, the oscillations will be harmonic if the small-angle approximation holds. The angular frequency will then be ! mgd/I, where I is the moment if inertia around the pivot point, and d the distance from the pivot point to the center of mass. | 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), assume that the box does not move. Suppose you pull the block 10 cm to the right and release it. The angular frequency of the oscilla- tions is 30 rad/s. Neglect friction between the block and the bottom of the box. (a) What is the spring constant? (b) What will be the amplitude of the oscillations? (c) Taking to the right to be positive, at what point in the oscillation is the velocity minimum and what is its minimum value? (d) At what point in the oscillation is the acceleration minimum, and what is its minimum value? (e) What is the total energy of the spring-block system? (f) If you take t = 0 to be the instant when you release the block, write an equation of motion for the oscillation, x(t) =?, identifying the values of all constants that you use. (g) Imagine now that the box, with the spring and block in it, starts moving to the left with an acceleration a = −4 m/s2. By how much does the equilibrium position of the block shift (relative to the box), and in what direction? Solution Most of this is really pretty straightforward, since it is just a matter of using the equations intro- duced in this chapter properly: (a) Since we know that for this kind of situation, the angular frequency, the mass and the spring constant are related by ω = * k m we can solve this for k: k = mω2 = 0.2 kg × $ 30 rad s %2 = 180 N m | 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 acceleration is minimum (again, largest in magnitude, but with a negative sign) when the spring is maximally stretched (block is farthest to the right), since this gives you the maximal force in the negative direction: amin = −ω2A = − $ 30 rad s %2 × 0.1 m = −90 m s2 (e) The total energy is given by the formula (either one is acceptable) E = 1 2mω2A2 = 1 2kA2 = 1 2 (180 N/m) × (0.1 m)2 = 0.9 J (You could also use E = 1 2mv2 max.) (f) The result is x(t) = A cos(ωt) = A sin & ωt + π 2 ' with A = 0.1 m and ω = 30 rad/s. You could also just write the numbers directly in the formula, but in that case you need to include the units implicitly or explicitly. What I mean by “implicitly” is to say something like: “x(t) = 0.1 cos(30t), with x in meters and t in seconds.” (g) The equilibrium position is where the block could sit at rest relative to the box. In that case, relative to the ground outside the box, it would be moving with an acceleration a = −4 m/s2, and the spring force (which is the only actual force acting on the block) would have to provide this acceleration: F spr x = −k∆x = ma so ∆x = −ma k = 0.2 kg × 4 m/s2 180 N/m = 0.00444 m or 4.44 mm. This is positive, so the spring stretches—the equilibrium position for the block is shifted to the right, relative to the box’s walls. Another way to see this is the following. As we saw in the previous chapter (section 10.2), an accelerated reference system, with acceleration a, appears “from the inside” as an inertial reference | 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 ma, pulling it to the right. In other words, we have a situation like the one illustrated in Fig. 11.6, only sideways. As in that case, we find the equilibrium position is shifted just enough for the force of the stretched spring to match the “force of gravity,” and in this way we get again the equation F spr x = ma. To get an accelerometer, we provide the box with some readout mechanism that can tell us the change in the oscillator’s equilibrium position. This basic principle is one of the ways accelerometers— and so-called “inertial navigation systems”—work. 11.5.2 Meter stick as a physical pendulum While working on the lab on torques, you notice that a meter stick suspended from the middle behaves a little like a pendulum, in that it performs very slow oscillations when you tilt it slightly. Intrigued, you notice that it is suspended by a simple loop of string tied in a knot at the top (see figure). You measure the period of the oscillations to be about 5 s, and the width of the stick to be about 2.5 cm. (a) What does this tell you about the quantity I/M, where M is the mass of the stick, and I its moment of inertia around a certain point? (b) What is the “certain point” mentioned in (a)? Solution As the picture below shows, the stick will behave like a physical pendulum, oscillating around the point of suspension O, which in this case is just next to the stick, where the knot is. As seen in the blown-up detail, if the width of the stick is w, the center of mass of the stick is located a distance d = w/2 away from the point of suspension: w cm O | 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). For reference, the moment of inertia of a thin rod of length l around its midpoint is Ml2/12 = 0.083l2. The length of the meter stick is, of course, 1 m, so the result I/M ∼0.08 m2 obtained above seems reasonable. | 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 usual, we will assume that µs ≥µk. As the mass oscillates, it will experience a kinetic friction force of magnitude F k = µkmg, in the direction opposite the direction of motion; that is to say, a force that changes direction every half period. As shown in section 11.2.2, this force does not change the frequency of the motion, but it displaces the equilibrium position in the direction of the force, which is to say, closer to the starting point for each half-swing. As a result of that, the amplitude for each half-swing is reduced from the previous one. Let the original equilibrium position (in the absence of friction) be x0 = 0. Suppose we displace the mass a distance A to the right (call this position, the starting point for the first half-swing, x1 = A), and let go. In the presence of friction, the equilibrium position for this first half-swing becomes the point x′ 0 = F k/k = µkmg, so the real amplitude of this first half-oscillation will be A1 = x1 −x′ 0 = A −x′ 0, and the resulting motion will be x(t) = x′ 0 + A1 cos(ωt) (first half-period, 0 ≤t ≤π/ω) (11.32) The mass then stops, momentarily, at t = π/ω, at the position x2 = x′ 0 −A1, and turns around for the second half-swing. However, now the external force has reversed direction, so the new equilibrium position is at −x′ 0, and the new amplitude is A2 = −x′ 0 −x2 = A1 −2x′ 0 = A −3x′ 0. The motion for this next half-period is then x(t) = −x′ 0 + A2 cos(ωt) (second half-period, π/ω ≤t ≤2π/ω) (11.33) Continuing the process, we see that A1 = A−x′ 0, A2 = A−3x′ 0, A3 = A−5x′ 0 . . . An = A−(2n−1)x′ 0. Of course, this can’t go on forever, since we require the amplitude to be a positive quantity; so the motion will consist of only n half-periods, where n is an integer such that A −(2n −1)x′ 0 > 0 but A −(2n + 1)x′ 0 < 0. (That is to say, n is equal to the integer part of (A/x′ 0 + 1)/2.) The figure shows an example of how this would go, for the following choice of parameters: period T = 1 s, µk = 0.1, and A = 0.18 m. Note that, since x′ 0 depends only on the ratio m/k = 1/ω2, there is no need to specify m and k separately. We get ω = 2π/T = 2π rad/s, x′ 0 = µkg/ω2 = 0.0248 m, and (A/x′ 0+1)/2 = 4.13, which means that the motion will go on for 4 half-periods before stopping. | 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. Either way, at that point the spring will be exerting a force of magnitude F spr = k|x′ 0 −An| = k|x′ 0 −A + (2n −1)x′ 0| = k|A −2nx′ 0| < kx′ 0 = F k (11.34) Since we expect the force of static friction, F s, to be greater than F k, this tells us that at this point the spring is not exerting enough force to get the mass to move again. Note: Just for the record, this is not the way dissipation in simple harmonic motion is usually handled. The conventional thing is to assume a damping force that is proportional to the oscillator’s velocity. You will almost certainly see this more standard approach (which leads to a relatively simple differential equation) in some later course. 11.6.2 The Cavendish experiment: how to measure G with a torsion balance Suppose that you want to try and duplicate Cavendish’s experiment to measure directly the grav- itational force between two masses (and hence, indirectly, the value of G). You take two relatively small, identical objects, each of mass m, and attach them to the ends of a rod of length l (let us say the mass of the rod is negligible, for simplicity), making a sort of dumbbell; then you suspend this from the ceiling, by the midpoint, using a nylon line. | 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 tilted very slightly; the light blue lines in the background show the oscillation in the absence of the masses, for reference. You have now made a torsion balance similar to the one Cavendish used. You will probably find out that it it is very hard to keep it motionless: the slightest displacement causes it to oscillate around an equilibrium position. The way it works is that an angular displacement θ from equilibrium puts a small twist on the line, which results in a restoring torque τ = −κθ, where κ is the torsion constant for your setup. If your dumbbell has moment of inertia I, then the equation of motion τ = Iα gives you I d2θ dt2 = −κθ (11.35) If you compare this to Eq. (11.21), and follow the derivation there, you can see that the period of oscillation is T = 2π * I κ (11.36) so if you measure T you can get κ, since I = 2m(l/2)2 = ml2/2 for the dumbbell. Now suppose you bring two large masses, a distance d each from each end of the dumbbell, per- pendicular to the dumbbell axis, and one on either side, as in the figure. The gravitational force F G = GmM/d2 between the large and small mass results in a net “external” torque of magnitude τext = 2F G × l 2 = F Gl (11.37) This torque will cause a very small displacement, so small that the change in d will be practically negligible, so you can treat F G, and hence τext, as a constant. Then the situation is analogous to that of an oscillator subjected to a constant external force (section 11.2.2): the frequency of | 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 the direction of the force). For the torsion balance, the equivalent result is θ′ 0 −θ0 = τext κ = F Gl κ (11.38) So, if you measure the angular displacement of the equilibrium position, you can get F G. This displacement is going to be very small, but you can try to monitor the position of the dumbbell by, for instance, reflecting a laser from it (or, one or both of your small masses could be a small laser). Tracking the oscillations of the point of laser light on the wall, you might be able to detect the very small shift predicted by Eq. (11.38). | 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, so the block ultimately bounces off(see Example 5.6.2). (a) Write down an equation of motion (a function x(t)) for the block, which is valid for as long as it is in contact with the spring. For simplicity, assume the block is initially moving to the right, take the time when it first makes contact with the spring to be t = 0, and let the position of the block at that time to be x = 0. Make sure that you express any constants in your equation (such as A or ω) in terms of the given data, namely, m, vi, and k. (b) Sketch the function x(t) for the relevant time interval. Problem 2 For this problem, imagine that you are on a ship that is oscillating up and down on a rough sea. Assume for simplicity that this is simple harmonic motion (in the vertical direction) with amplitude 5 cm and frequency 2 Hz. There is a box on the floor with mass m = 1 kg. (a) Assuming the box remains in contact with the floor throughout, find the maximum and mini- mum values of the normal force exerted on it by the floor over an oscillation cycle. (b) How large would the amplitude of the oscillations have to become for the box to lose contact with the floor, assuming the frequency remains constant? (Hint: what is the value of the normal force at the moment the box loses contact with the floor?) Problem 3 Imagine a simple pendulum swinging in an elevator. If the cable holding the elevator up was to snap, allowing the elevator to go into free fall, what would happen to the frequency of oscillation of the pendulum? Justify your answer. Problem 4 Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring constant k2. Each spring is attached on the other side to a wall, and the block slides without friction on a horizontal surface. When the block is sitting at x = 0, both springs are relaxed. Write Newton’s second law, F = ma, as a differential equation for an arbitrary position x of the block. What is the period of oscillation of this system? Problem 5 Consider the block hanging from a spring shown in Figure 11.6. Suppose the mass of the block is 1.5 kg and the system is at rest when the spring has been stretched 2 cm from its original length | 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 the spring, the block, and the earth. Take the “zero” of gravitational potential energy to be at the height y′ 0 (the equilibrium point; you may also use this as the origin for the vertical coordinate!), and calculate all the energies in the system (kinetic, spring/elastic, and gravitational) at the highest point in the oscillation, the equilibrium point, and the lowest point. Verify that the sum is indeed constant. | 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 some internal structure, or were to some extent deformable, whenever we allowed for the possibility of their storing other forms of energy, such as chemical or thermal. This chapter deals with a very common type of organized (as opposed to incoherent) motion exhibited by extended elastic objects, namely, wave motion. (Often, the “object” in which the wave motion takes place is called a “medium.”) Waves can be “traveling” or “standing,” and we will start with the traveling kind, since they are the ones that most clearly exhibit the characteristics typically associated with wave motion. A traveling wave in a medium is a disturbance of the medium that propagates through it, in a definite direction and with a definite velocity. By a “disturbance” we typically mean a displacement of the parts that make up the medium, away from their rest or equilibrium position. The idea here is to regard each part of an elastic medium as, potentially, an oscillator, which couples to the neighboring parts by pushing or pulling on them (for an example of how to model this mathematically, see Advanced Topic 12.6.1 at the end of this chapter). When the traveling wave reaches a particular location in the medium, it sets that part of the medium in motion, by giving it some energy and momentum, which it then passes on to a neighboring part, and so on down the line. You can see an example of how this works in a slinky. Start by stretching the slinky somewhat, then grab a few coils, bunch them up at one end, and release them. You should see a “compression 279 | 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 12.1: A longitudinal (compression) wave pulse traveling down a slinky. The compression pulse in the slinky in Fig. 12.1 is an example of what is called a longitudinal wave, because the displacement of the parts that make up the medium (the rings, in this case) takes place along the same spatial dimension along which the wave travels (the horizontal direction, in the figure). The most important examples of longitudinal waves are sound waves, which work a bit like the longitudinal waves on the slinky: a region of air (or some other medium) is compressed, and as it expands it pushes on a neighboring region, causing it to compress, and passing the disturbance along. In the process, regions of rarefaction (where the density drops below its average value) are typically produced, alongside the regions of compression (increased density). The opposite of a longitudinal wave is a transverse wave, in which the displacement of the medium’s parts takes place in a direction perpendicular to the wave’s direction of travel. It is actually also relatively easy to produce a transverse wave on a slinky: again, just stretch it somewhat and give one end a vigorous shake up and down. It is, however, a little hard to draw the resulting pulse on a long spring with all the coils, so in Figure 12.2 below I have instead drawn a transverse wave pulse on a string, which you can produce in the same way. (Strings have other advantages: they are also easier to describe mathematically, and they are very relevant, particularly to the production of musical sounds.) direction of propagation of the wave (x) x direction of displacement of the medium (y) y Figure 12.2: A transverse wave pulse traveling down a string. This pulse can be generated by giving an end of the string a strong shake, while holding the string taut. (You can do this on a slinky, too.) | 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 energy with it, but each individual ring does not move very far away from its equilibrium position. Ideally, after the pulse has passed through a particular location in the medium, the corresponding part of the medium returns to its equilibrium position and does not move any more: all the energy and momentum it momentarily acquired is passed forward. The same is (ideally) true for the transverse wave on the string in Fig. 12.2. Since this is meant to be a very elementary introduction to waves, I will consider only this case of “ideal” (technically known as “linear and dispersion-free”) wave propagation, in which the speed of the wave does not depend on the shape or size of the disturbance. In that case, the disturbance retains its “shape” as it travels, as I have tried to illustrate in figures 12.1 and 12.2. 12.1.1 The “wave shape” function: displacement and velocity of the medium. In a slinky, what I have been calling the “parts” of the medium are very clearly seen (they are, naturally, the individual rings); in a “homogeneous” medium (one with no visible parts), the way to describe the wave is to break up the medium, in your mind, into infinitely many small parts or “particles” (as we have been doing for extended systems all semester), and write down equations that tell us how each part moves. Physically, you should think of each of these “particles” as being large enough to contain many molecules, but small enough that its position in the medium may be represented by a mathematical point. The standard way to label each “particle” of the medium is by the position vector of its equilibrium position (the place where the particle sits at rest in the absence of a wave). In the presence of the wave, the particle that was initially at rest at the point ⃗r will undergo a displacement that I am going to represent by the vector ⃗ξ (where ξ is the Greek letter “xi”). This displacement will in general be a function of time, and it may also be different for different particles, so it will also be a function of ⃗r, the equilibrium position of the particle we are considering. The particle’s position under the influence of the wave becomes then ⃗r + ⃗ξ(⃗r, t) (12.1) This is very general, and it can be given a simpler form for simple cases. For instance, for a transverse wave on a string, we can label each part of the string at rest by its x coordinate, and then take the displacement to lie along the y axis; the position vector, then, could be written in component form as (x, ξ(x, t), 0). Similarly, we can consider a “plane” sound wave as a longitudinal wave traveling in the x direction, where the density of the medium is independent of y and z (that is, it is constant on planes perpendicular to the direction of propagation). In that case, the equilibrium coordinate x can be used to refer to a whole “slice” of the medium, and the position | 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 implicitly to these simple cases and treat ξ as a scalar from this point on. Under these conditions, the function ξ(x, t) (which is often called the wave function) gives us the shape of the “displacement wave,” that is to say, the displacement of every part of the medium, labeled by its equilibrium x-coordinate, at any instant in time. Accordingly, taking the derivative of ξ gives us the velocity of the corresponding part of the medium: vmed = dξ dt (12.2) This is also, in general, a vector (along the direction of motion of the wave, if the wave is longitu- dinal, or perpendicular to it if the wave is transverse). It is also a function of time, and in general will be different from the speed of the wave itself, which we have taken to be constant, and which I will denote by c instead. 12.1.2 Harmonic waves An important class of waves are those for which the wave function is sinusoidal. This means that the different parts of the medium execute simple harmonic motion, all with the same frequency, but each (in general) with a different phase. Specifically, for a sinusoidal wave we have ξ(x, t) = ξ0 sin 22πx λ −2πft 3 (12.3) In Eq. (12.3), f stands for the frequency, and plays the same role it did in the previous chapter: it tells us how often (that is, how many times per second) the corresponding part of the medium oscillates around its equilibrium position. The constant ξ0 is just the amplitude of the oscillation (what we used to call A in the previous chapter). The constant λ, on the other hand, is sometimes known as the “spatial period,” or, most often, the wavelength of the wave: it tells you how far you have to travel along the x axis, from a given point x, to find another one that is performing the same oscillation with the same amplitude and phase. A couple of snapshots of a harmonic wave are shown in Fig. 12.3 (next page). The figure shows the displacement ξ, at two different times, and as a function of the coordinate x used to label the parts into which we have broken up the medium (as explained in the previous subsection). As such, the wave it represents could equally well be longitudinal or transverse. If it is transverse, like a wave on a string, then you can think of ξ as being essentially just y, and then the displacement curve (the blue line) just gives you the shape of the string. If the wave is longitudinal, however, then it is a bit harder to visualize what is going on just from the plot of ξ(x, t). This is what I have tried to do with the density plots at the bottom of the figure. | 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 traveling in the positive x direction). Units for both x and ξ are arbitrary. Bottom: The corresponding densities, for the case of a longitudinal wave. Imagine the wave is longitudinal, and consider the x = π point on the t = 0 curve (the first zero, not counting the origin). A particle of the medium immediately to the left of that point has a positive displacement, that is, it is pushed towards x = π, whereas a slice on the right has a negative displacement—which means it is also pushed towards x = π. We therefore expect the density of the medium to be highest around that point, whereas around x = 2π the opposite occurs: particles to the left are pushed to the left and those to the right are pushed to the right, resulting in a low-density region. The density plot labeled t = 0 attempts to show this using a grayscale where darker and lighter correspond to regions of higher and smaller density, respectively. At the later time t = ∆t the high and low density regions have moved a distance c∆t to the right, as shown in the second density plot. Regardless of whether the wave is longitudinal or transverse, if it is harmonic, the spatial pattern will repeat itself every wavelength; you can think of the wavelength λ as the distance between two consecutive crests (or two consecutive troughs) of the displacement function, as shown in the figure. If the wave is traveling with a speed c, an observer sitting at a fixed point x would see the disturbance pass through that point, the particles move up and down (or back and forth), and the motion repeat itself after the wave has traveled a distance λ, that is, after a time λ/c. This means the period of the oscillation at every point is T = λ/c, and the corresponding frequency f = 1/T = c/λ: f = c λ (12.4) This is the most basic equation for harmonic waves. Making use of it, Eq. (12.3) can be rewritten | 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 part of the medium, with equilibrium position x, at the time t, can be calculated from Eqs. (12.2) and (12.3) to be vmed(x, t) = 2πfξ0 cos 22πx λ −2πft 3 = ωξ0 cos 22πx λ −2πft 3 (12.6) (where I have introduced the angular frequency ω = 2πf). Again, this is a familiar result from the theory of simple harmonic motion: the velocity is “90 degrees out of phase” with the displacement, so it is maximum or minimum where the displacement is zero (that is, when the particle is passing through its equilibrium position in one direction or the other). Note that the result (12.6) implies that, for a longitudinal wave, the “velocity wave” is in phase with the “density wave”: that is, the medium velocity is large and positive where the density is largest, and large and negative where the density is smallest (compare the density plots in Fig. 12.3). If we think of the momentum of a volume element in the medium as being proportional to the product of the instantaneous density and velocity, we see that for this wave, which is traveling in the positive x direction, there is more “positive momentum” than “negative momentum” in the medium at any given time (of course, if the wave had been traveling in the opposite direction, the sign of vmed in Eq. (12.6) would have been negative, and we would have found the opposite result). This confirms our expectation that the wave carries a net amount of momentum in the direction of propagation. A detailed calculation (which is beyond the scope of this book) shows that the time-average of the “momentum density” (momentum per unit volume) can be written as p V = 1 2cρ0ω2ξ2 0 (12.7) where ρ0 is the medium’s average mass density (mass per unit volume). Interestingly, this result applies also to a transverse wave! As mentioned in the introduction, the wave also carries energy. Equation (12.6) could be used to calculate the kinetic energy of a small region of the medium (with volume V and density ρ0, and therefore m = ρ0V ), and its time average. This turns out to be equal to the time average of the elastic potential energy of the same part of the medium (recall that we had the same result for harmonic oscillators in the previous chapter). In the end, the total time-averaged energy density (energy per unit volume) in the region of the medium occupied by the wave is given by E V = 1 2ρ0ω2ξ2 0 (12.8) | University Physics I Classical Mechanics_Page_302_Chunk4500 |
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