Ph1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004

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1 Ph1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004 Problem 1 (10 points) - The Delivery A crate of mass M, which contains an expensive piece of scientific equipment, is being delivered to Caltech. The delivery truck has a freight bed of length L (see the figure), with a coefficient of static friction µ s and a coefficient of kinetic friction µ k. ather than move the heavy crate himself, the driver tilts the truck bed by an angle θ (see figure) and then drives the truck forward with increasing acceleration a, until the crate begins to slide. y x L M µ s, µ k g θ a For this problem, use the x, y coordinate system shown in the figure. These coordinates are fixed with respect to the truck s bed, not to the ground. (a) (2 points) Draw a free-body force diagram for the crate in the truck s frame of reference. Solution: See Figure 1. N is the normal force that the truck s bed exerts on the crate, and f is the corresponding friction. N y x Ma f θ Mg Figure 1: Free-body diagram for the crate in the truck s frame of reference

2 (b) (3 points) Write down Newton s second law for the motion of the crate in the x and y directions, just before it begins to slip. Solution: While the crate remains on the truck s bed, a y = 0. Thus: Ma y = N Mg cos θ + Ma sin θ = 0. The maximum static friction is given by f max = µ s N. This must be the magnitude of the friction just before the crate begins to slip. Therefore: Ma x = f max Mg sin θ Ma cos θ = µ s N Mg sin θ Ma cos θ = 0. (c) (2 points) Determine the minimum acceleration a min for which the crate will begin to slip. Express your answer in terms of the constants shown in the figure. Solution: If we solve for a in the equations obtained in part (b) for the crate just before it begins to slip, the result must correspond to a min. Therefore a min = g µ s cos θ sin θ µ s sin θ + cos θ. When the truck reaches a min and the driver notices the crate beginning to slide, he continues at that constant acceleration. (d) (3 points) Find the speed of the crate along the x direction when the crate leaves the truck bed. Neglect the size of the crate. You may leave your answer in terms of a min. Solution: If the crate is sliding, then the friction is kinetic rather than static. By Newton s second law, the acceleration is: a x = µ k (g cos θ a min sin θ) g sin θ a min cos θ. Since this acceleration is constant, we can find the final velocity of the crate by using the equation v 2 f = 2 a x L. Notice that, in our coordinate system, a x for the sliding crate is negative, so that a x = a x. Thus we have that v f = 2L [g sin θ + a min cos θ µ k (g cos θ a min sin θ)]. Another way of arriving at this result is to set the initial energy of the crate E i = MgL sin θ equal to its final energy E f = Mv 2 f/2 minus the work done by the frictional force W = L µ k (g cos θ a min sin θ).

3 Problem 2 (10 points) - A Cylindrical Child A playground merry-go-round of uniformly distributed mass M and radius is initially at rest. A kid with uniformly distributed mass 2M (whose shape is remarkably close to that of a disk of radius /2) runs at a constant velocity v towards point A on the edge of the merry-go-round. The angle between v and the line that runs from A to the center of the merry-go-round is 45 degrees, as shown in the figure. /2 45 v A ω before after At the instant when he jumps aboard, the child grabs hold of the edge so that he is fixed with respect to the merry-go-round, with his center of mass at point A. The merry-go-round then begins turning with angular velocity ω. (a) (2 points) Find the total moment of inertia I tot of the system about the axis of the merry-go-round, after the child has jumped on. Solution: The moment of inertia of the merry-go-round is M 2 /2. The moment of inertia of the child around around his own center of mass is 2M(/2) 2 /2 = M 2 /4. But after jumping on the merry-go-round, he will be spinning on an axis displaced from his center of mass by a distance. By the parallel axis theorem, his moment of inertia will therefore by M 2 /4 + 2M 2 = 9M 2 /4. Therefore the total moment of inertia of the system after the child has jumped onto the merry-go-round is I = 11 4 M2. (b) (3 points) Find the angular velocity ω of the system after the jump. Express your answer in terms of v, M, and. Solution: Since there are no external torques in this problem (because the torque exerted by the child on the merry-go-round during the jump is equal to the torque that the merry-go-round exerts on the child), total angular momentum is conserved. The initial angular momentum with respect to the axis of the merry-go-round comes entirely from the child s running, and is given by L = p r = 2Mv sin 45 = 2Mv.

4 This must be equal to the angular momentum of the spinning system after the jump, L = Iω = 11 4 M2 ω. Solving for ω we obtain that ω = v. (c) (2 points) Find the fraction of the initial kinetic energy that is lost in the collision between the kid and the merry-go-round. Solution: The initial energy is simply the kinetic energy of the child s running E i = Mv 2. The final energy is: E f = 1 2 Iω2 = 4 11 Mv2, which implies that seven elevenths of the energy is dissipated into heat during the child s collision with the the merry-go-round. (d) (3 points) After a few revolutions at a constant angular velocity ω, the kid decides to move to the center of the merry-go-round with a constant radial speed v r = dr/dt. Find the angular acceleration α of the merry-go-round as a function of r. Express your answer in terms of M,, v r and the initial angular velocity ω. Hint: You might want to save yourself some computation by first finding an expression for ω as a function of r, and then using the chain rule α = dω dt = dω dr dr dt. Solution: Following the hint, we begin by expressing the moment of inertia as a function of r: I(r) = 1 2 M2 + 1 ( ) 3 4 M2 + 2Mr 2 = M r 2. By conservation of angular momentum, which gives us that: L = I(r)ω(r) = 11 4 M2 ω 0, ω(r) = 112 ω r 2. The first derivative of ω as a function or r is therefore dω dr = 176r2 ω 0 ( r 2 ) 2. Using the chain rule and substituting dr/dt = v r we have that: α(r) = 176r2 ω 0 v r ( r 2 ) 2.

5 Problem 3 (10 points) - Underwater Seesaw Two solid cubes of edge length a and uniform density ρ are partly immersed in an ideal fluid of density ρ 0. The cubes are connected at their centers by a rigid, massless rod of length L, with a fixed pivot point that lies on the surface of the liquid, as shown in the figure. The system is built so that the cubes remain upright even as the rod turns on its pivot. a θ ρ ρ 0 L Suppose that the system is tilted until the rod makes an angle θ with the surface of the fluid. Assume that for this value of θ both cubes remain in contact with the fluid. (a) (2 points) Calculate the net force on the rod s pivot. For what value of ρ/ρ 0 is this force zero? Hint: The value is independent of θ. Solution: egardless of the angle θ, the volume of fluid displaced by the cubes is always a 3. The corresponding weight of the displaced fluid is gρ 0 a 3, which gives the buoyant force acting on the cubes. The weight of the two cubes together is 2gρa 3. Therefore, for ρ/ρ 0 = 1/2 the buoyant force will equal the weight of the cubes, and the force on the pivot will vanish. (b) (2 points) Calculate the torque τ about the pivot as a function of θ and the other parameters specified in the problem. Solution: The weight of each cube is the same, and since the cubes at the same distance from the pivot on opposite sides of it, the torque that their weight exerts is zero. If the rod is tilted by an angle θ, the buoyant force on the cube that was pushed down will be gρ 0 (a 3 + a 2 L sin θ)/2. The buoyant force on the cube that was raised will be gρ 0 (a 3 a 2 L sin θ)/2. Each of these buoyant forces exerts a torque of opposite sign on the pivot. Thus, the net torque will be τ = gρ 0 a 2 L sin θ L 2 cos θ = gρ 0a 2 L 2 sin 2θ. 4

6 Notice that this torque always acts against the tilting, which is why we have have given τ with an overall minus sign. (c) (3 points) If the system is tilted by an angle θ 0 and then released, it will oscillate. Assume that a L, so that the cubes may be treated as point masses for the purpose of computing the moment of inertia. Find an expression for the angular acceleration θ of the rod. Solution: If we treat the cubes as point masses, then the moment of inertia of the system is I = a 3 ρl 2 /2. The rotational acceleration of the system is given by which implies that I θ = τ = gρ 0a 2 L 2 sin 2θ, 4 ( ) gρ0 θ = sin 2θ. 2aρ (d) (3 points) Now assume that θ 0 is very small. Express the period T of the oscillation in terms of the parameters in the problem. Solution: In the small angle approximation sin 2θ 2θ, so that the angular acceleration is approximately given by ( ) gρ0 θ = θ. aρ This has the form of the equation for simple harmonic motion, since θ is proportional to θ, with the proportionality constant being negative. In simple harmonic motion, the proportionality constant (without the minus sign) is equal to the square of the angular frequency ω. In this case, then, we have gρ0 ω = aρ. The period T is then given by: T = 2π ω = 2π aρ gρ 0.

7 Problem 4 (10 points) - Exploding Satellite A small spherical satellite of mass m orbits a planet of much larger mass M at a speed u on a circular orbit of radius. At a time t = 0, an internal explosion breaks the satellite into two equal hemispheres A and B, each of mass m/2. Immediately after the explosion, the speed of piece B is 5u/4, and it is moving along the same direction as it was before the explosion. The figures shows the system immediately before and immediately after the satellite s explosion. u m 5u/4 m/2 A B m/2 M M Before After (a) (3 points) What is the speed u, angular momentum L, and total energy E (kinetic plus gravitational potential) of the satellite just before the explosion? Choose your energy scale so that a stationary object infinitely far away from the planet will have zero energy. Express your answers in terms of G, m, M, and. Solution: In the energy scale indicated, the gravitational potential energy for the satellite is GM m/. Meanwhile, the gravitational force that pulls the satellite towards the planet is GMm/ 2, which must be equal to the centripetal force mu 2 / necessary to keep the satellite on its circular orbit. Therefore u = GM, which corresponds to a kinetic energy mu 2 /2 = GMm/2. The total energy of the satellite is E = 1 2 mu2 GMm = GMm 2. The angular momentum is simply given by L = mu = m GM. (b) (2 points) What are the angular momenta L A and L B and the total energies E A and E B of parts A and B for times t > 0? You may leave your answer in terms of u if you did not complete part (a).

8 Solution: Immediately after the explosion the two hemispheres are still a distance away from the center of the planet, and their velocities are still tangential. Therefore L B = m 5u 2 4 = 5 8 mu. In order to compute L A we must first find the velocity of A after the explosion. Since the explosion was internal, total linear momentum should have been conserved. This implies that the velocity of A must also be tangential to the original orbit of the satellite, and that its magnitude must be 3u/4. Then L A = m 3u 2 4 = 3 8 mu. Notice that total angular momentum is also conserved, as expected. The energy of A is E A = 1 ( m 3u The energy of B is E B = 1 ( m 5u ) 2 GMm 2 ) 2 GMm 2 = GMm = GMm ( ) = 23 GMm 64. ( ) = 7 GMm (c) (2 points) How much mechanical work was done by the explosion that broke the satellite apart? Solution: Only the kinetic energy of the satellite parts changes during the explosion. The initial kinetic energy is simply mu 2 /2, while the final energy is (5u/4) 2 m/4+(3u/4) 2 m/4 = 17mu 2 /32. Therefore the energy of the system has increased by mu 2 32 = GMm 32, which is the amount of work done by the explosion. The same result may be obtained from W = E A + E B E, using the results from parts (a) and (b). (d) (1 point) Find the length of the semimajor axes a A and a B for the orbits of the pieces A and B for t > 0. Solution: Using the energies calculated in part (b) and the equation for the length of the semi-major axis length in terms of the energy of the orbiting body, we obtain that and a B = 16/7. a A = GMm/2 2E A = 16/23,

9 (e) (2 points) Sketch the orbits of pieces A and B for t > 0. Clearly label the point at which the explosion occurred. Solution: Since the velocity of both A and B at the point of the explosion is entirely tangential, the orbits for the hemispheres must have either their apogee or their perigee at that location. Notice that A has less energy than is necessary to keep it in the circular orbit of radius. It will then describe a smaller ellipse with semi-major axis a A = 16/23. The planet will be at the focus furthest from the point of the explosion. (Notice that if the perigee distance of the orbit is smaller than the radius of the planet, A would crash into it.) Meanwhile, B has more energy than necessary to keep it on the circular orbit, but not enough for it to escape, because the energy is still negative. Therefore it will describe an elliptical orbit with the planet at the focus nearest from the point of the explosion. The semi-major axis of this orbit is the a B computed in part (d). Boom! 16/23 A 16/7 B Figure 2: Sketch of the orbits of the hemispheres of the satellite after the explosion

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