P442 Analytical Mechanics - II

Transcription

1 Solutions to Problem Set Wednesday - January 8, 6 Written or last updated: January 8, 6 P44 Analytical Mechanics - II Alex R. Dzierba Problem - Problem 9. in the text. Solution Assume that the base of the hemisphere lies on the (x, y) plane where z = and that the hemisphere is centered on the z-axis. By symmetry x = ȳ =. Also assume that the mass density is ρ. The total mass of the hemisphere is then: M = π π/ r r ρr sin θdφdθdr () The first integral is over φ - the azimuthal angle, the second is over θ - the polar angle and the third integral is over r. The z position of the center of mass is: z = M π π/ r r zρr sin θdφdθdr () But z = r cos θ. The result is: z = 3 r 4 r 4 8 r 3 r3 (3) Problem - Problem 9.8 in the text. Solution Referring to Figure, we can see that the differential (shaded) area da is given by: ( ) a da = y dy Assuming a constant areal density σ we see that the total mass is M = σ a/ ( ) a y dy = σ a

2 y y a x Figure : Problems and 3 The center of mass is given by: ȳ = a/ M σ ( ) a y ydy = a 3 Problem 3 with y. - Problem 9.8 in the text with the added complication that the areal density increases linearly Solution In this case we assume σ(y) = σ y where σ is a constant. The mass is: a/ M = σ ( ) a y a 3 ydy = σ 6 The center of mass is given by: ȳ = M σ a/ ( ) a y y dy = M σ a 4 4 = a Problem 4 - Problem 9. in the text. Solution The time required for the projectile to reach the apex of the trajectory (where the projectile has has no vertical component of velocity) is t = v sin θ/g. The horizontal component is v cos θ. The projectile of mass M explodes into two parts of masses m and m, which immediately after the explosion have only horizontal velocities v and v respectively. Using conservation of momentum and energy we have: (m + m )v cos θ = m v + m v (4) (m + m )v cos θ + E = m v + m v (5)

3 From these equations you can solve for v and v : m E v = v cos θ ± m (m + m ) m E v = v cos θ m (m + m ) The time for the projectile elements to hit the water from the time of the explosion is just t and their separation is just given by v = v t. Problem 5 Assume that two cannon balls of masses 3m and m are each shot from guns at 45 with respect to the horizontal and with and initial speed of v. The guns are aimed and positioned so that the cannon balls collide at the highest point in their trajectories. The collision is perfectly elastic. Please answer the following: (a) Does one cannon ball hit the ground before the other and if so which one and by how much time? (b) With what speeds do each of the cannon balls hit the ground? (c) How far apart are the cannon balls after they have hit the ground? Solution Recall from the famous colliding pendula where one has a mass 3 that of the other that when they approach each other with equal velocities, each with v, after the collision the heavier mass comes to a stop and the lighter mass rebounds with velocity v. In this case the horizontal velocity of the balls before the collision is v / so the horizontal velocity of the lighter ball after the collision is v. After the collision both balls are at the same height and have zero vertical velocity so they must hit the ground at the same time. The collision occurs at height h = v /4g. The heavier ball hits the ground with speed v /. The lighter ball has this same vertical velocity and a horizontal velocity of v so the speed is.5v. The time after the collision that the balls hit the ground is the same as the time from launch to the collision - which is t = v /( g). The distance between the balls when they hit the ground is the horizontal distance traveled by the lighter ball which is v /g. Problem 6 A long, thin, pliable carpet is laid on the floor. One of the carpet is bent back and then pulled with constant unit velocity just above the carpet which is still at rest on the floor. See Figure. Find the speed of the center of mass of the moving part. What is the minimum force needed to pull the moving part if the carpet has unit mass and unit length. This problem came from Puzzling Physics Problems by Peter Gnädig, Gyula Honyek and Ken Riley, Cambridge Press () 3

4 Solution When the position of the end of the carpet is given by x, the other end of the moving end is at x/ (see Figure ). The coordinate of the center of mass of the moving part is given by 3x/4. Thus, if the end of the carpet is pulled so that dx/dt =, the speed of the center of mass of the moving part is 3/4. v = x/ x Figure : Problem 6 The linear momentum of the moving part of the carpet is mv and the force needed to move the carpet is: F = dp dt = dm dt v + mdv dt = dm dt + (6) To find dm/dt consider this: when the end of the carpet reaches x = at time t = the whole carpet is moving so dm/dt = /. By the way, when the end of the carpet is at x = the center of mass is at x = 3/ and it took t = to get there so indeed the velocity of the center of mass of the moving part is 3/4. Now getting back to the force the force needed is F = /. Now here is a nice twist on all this. You might think that the work done by the force F in moving the end of the carpet to x = L = (L is the length of the carpet) would be W = F L. Then using W = mv / you might conclude that F = /4. How do you get around this problem? Problem 7 Two steel balls are touching each other and released from rest. Initially the center of the lower ball is a distance h from the ground as shown in Figure 3. The radius of the upper ball is a and that of the lower ball is a. After the balls are released they move so that their centers are always on the same vertical line. Any collisions that occur are elastic. What is the maximum height reached by upper ball. a h Figure 3: Problem 7 Solution Given their radii the mass of the smaller sphere is m and that of the larger sphere is m = 8m. The larger sphere lands with velocity v where: v = g(h a) (7) This problem and the next one came from Problems and Solutions on Mechanics Compiled by the Physics Coaching Class of Science and Technology of China, edited by Lim Yung-kuo, World Scientific (994) 4

5 The large sphere immediately rebounds with this velocity traveling upwards with velocity v upwards and immediately collides with the smaller sphere which has velocity v = v downwards. After the collision of these two spheres they have velocities u and u. Momentum and kinetic energy conservation lead to: m v m v = m u + m u m v + m v = m u + m u We know that m = 8m and v = v. Solving: u = 5 9 v u = 3 9 v So the maximum height (measured from the ground) reached by the smaller sphere is: H = 3a + u g Problem 8 A railroad car of mass M is initially at rest. The car can move without friction on a horizontal rail. N women, each of mass m, are initially standing at rest with respect to the car. The situation is shown in Figure 4. (a) All N women run in unison to one end of the car and jump off at the same time. They each reach velocity v r relative to the just before jumping off. After they all jump off what is the speed of the car? (b) Now consider a different situation. Each woman runs to the end of the car and jumps off but they do it one at a time. In each case, as a woman jumps off, her speed relative to the car is v r. After the Nth woman jumps off, what is the speed of the car? (c) In which of the above two cases is the the final speed of the car larger? m m M Figure 4: Problem 8 Solution First we deal with the case of all women jumping off at the same time. The center of mass of the car plus women system does not move, so ẋ cm =. The location of the center of mass is given by: 5

6 x cm = Mx c + Nmx w M + Nm and therefore: ẋ cm = = Mẋ c + Nmẋ w The velocity of the car is ẋ c = V and the velocity of any woman is ẋ w = V v r. This allows us to solve for the velocity of the car: V = Nmv r M + Nm Now the women jump off one at a time. Consider the transition in going from n women on the car to n women. The velocity of the car with n women on it is V n and the momentum of the system is P n : P n = MV n + nmv n When a woman jumps off the momentum becomes: P n = MV n + (n )mv n + m(v n v r ) Conservation of momentum implies: P n = P n or: (M + nm)v n = (M + nm)v n mv r or V n = V n + mv r M + nm and that would imply: V n s = V n + s i= mv r M + (n i + )m 6

7 We want n = N and we know V N = so the final velocity when N = is: V = N i= N mv r M + (N i + )m = mv r M + nm i= This leads to a larger velocity than the previous case. Problem 9 - Problem 9.5 in the text. Solution The mass per length of the rope is λ. At any given instant the mass of the hanging part of the rope is m and the force on it due to gravity is mg. We then have: F = mg = ṗ = ṁv + m v But m = λx and ṁ = λẋ. So we get: xg = ẋv + x dv dt = v + xv dv dx Try a solution of the form v = ax n that leads to: xg = a (n + )x n For this to hold for all x we need n = / and that leads to a = g/3 finally giving us: v = gx 3 Problem - Problem 9.7 in the text. Solution - see the next page. 7

8 9-7. As the problem states, we need to perform the following integral α τ = dα () α ε ( α) 4 Our choice of ε is for this calculation, and the results are shown in the figure. We plot the natural velocity dα dτ = x gb vs. the natural time τ..5 dα dτ τ