# 3 problems on Rotational Dynamics

by chavic
Tags: dynamics, rotational
 P: 9 On this assignment I could do most of the work easily, I just have some questions on three problems 3. The problem statement, all variables and given/known data A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 270 kg·m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? 3. Relevant equations L=rmvsin[\theta] I=.5MR$$^{2}$$ $$\omega$$= v/r 3. The attempt at a solution I attempted to use conservation of momentum. I=.5MR$$^{2}$$ Solve for M (mass of merry-go-round) I/.5R$$^{2}$$=M M=135 kg Angular velocity Circumference of merry-go-round=12.5664 m 12.5664 * 9 Initial velocity=113.0976 Initial Momentum L=rmv =2(135)(133.09760) Li=30536.352 So final momentum should be the same L=rmv v=L/rm 30536.352/ 2(162)=94.248 This is wrong -------------------------------------------------------------------- 4. The problem statement, all variables and given/known data A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it. (a) What is the angular momentum of the bullet- block system? (For the following answers, use M for the mass M, m for the mass m, and L for the length .) (b) What fraction of the original kinetic energy is converted into internal energy in the collision? 4. Relevant equations ? If someone could just point me in the right direction 4. The attempt at a solution -------------------------------------------------------------------- 5. The problem statement, all variables and given/known data Two astronauts, each having a mass of 70.0 kg, are connected by a 9.5 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.50 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. (b) Calculate the rotational energy of the system. (c) By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. What is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut do in shortening the rope? 5. Relevant equations L=mrvsin[\theta] KE=.5I$$\omega$$$$^{2}$$ Work=$$\Delta$$KE I=MR$$^{2}$$ $$\omega$$= v/r 5. The attempt at a solution I have the answers for a-e, but I'm having a hard time with f A=3657.5 kg·m$$^{2}$$/s B=2117.5193 J C=3657.5 kg·m$$^{2}$$/s D=10.5 m/s E=7717.5 J *fixed momentums So F should be Work=$$\Delta$$KE or 7717.5-2117.5193=5599.9807 But this is the wrong answer *I figured it out, its supposed to be in Kj
 Quote by chavic On this assignment I could do most of the work easily, I just have some questions on three problems 3. The problem statement, all variables and given/known data A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 270 kg·m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 26.0 kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed of the merry-go-round? 3. Relevant equations L=rmvsin[\theta] Consider the angular momentum of a rigid body, $$L=I \omega$$. $$L = mvrsin\theta$$ is for a particle I=.5MR$$^{2}$$ $$\omega$$= v/r 3. The attempt at a solution I attempted to use conservation of momentum. I=.5MR$$^{2}$$ Solve for M (mass of merry-go-round) You don't need to find the mass of the merry-go-round since you've been given the moment of inertia of the merry-go-round. I/.5R$$^{2}$$=M M=135 kg Angular velocity Circumference of merry-go-round=12.5664 m 12.5664 * 9 Initial velocity=113.0976 Initial Momentum L=rmv =2(135)(133.09760) Li=30536.352 So final momentum should be the same L=rmv v=L/rm 30536.352/ 2(162)=94.248 Remember that I(final) will be the sum of the two objects. So I(final) = I(merry-go-round) + I(child). Assume the child is a point mass. This is wrong -------------------------------------------------------------------- 4. The problem statement, all variables and given/known data A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it. (a) What is the angular momentum of the bullet- block system? (For the following answers, use M for the mass M, m for the mass m, and L for the length .) (b) What fraction of the original kinetic energy is converted into internal energy in the collision? 4. Relevant equations ? If someone could just point me in the right direction Start with the conservation of momentum of the collision between the bullet and the block. You need to figure out an expression for the velocity after the collision. Then consider the angular momentum of the block + bullet about the pivot point. 4. The attempt at a solution -------------------------------------------------------------------- 5. The problem statement, all variables and given/known data Two astronauts, each having a mass of 70.0 kg, are connected by a 9.5 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.50 m/s. (a) Treating the astronauts as particles, calculate the magnitude of the angular momentum. (b) Calculate the rotational energy of the system. (c) By pulling on the rope, one of the astronauts shortens the distance between them to 5.00 m. What is the new angular momentum of the system? (d) What are the astronauts' new speeds? (e) What is the new rotational energy of the system? (f) How much work does the astronaut do in shortening the rope? 5. Relevant equations L=mrvsin[\theta] KE=.5I$$\omega$$$$^{2}$$ Work=$$\Delta$$KE I=MR$$^{2}$$ $$\omega$$= v/r I would have used $$L=I \omega$$. Remember that I will be the sum of the two astronauts. Although your way should work for this. 5. The attempt at a solution I have the answers for a-e, but I'm having a hard time with f A=3675 kg·m$$^{2}$$/s B=2117.5193 J C=3675 kg·m$$^{2}$$/s D=10.5 m/s E=7717.5 J I get slightly different numbers than you. Can you show how you got the 3675? So F should be Work=$$\Delta$$KE This should work. or 7717.5-2117.5193=5599.9807 But this is the wrong answer