32- In the figure below, the hanging object has a mass of m 1 = 0.460 kg; the sl
ID: 2250640 • Letter: 3
Question
32-In the figure below, the hanging object has a mass of m1 = 0.460 kg; the sliding block has a mass of m2 = 0.765 kg; and the pulley is a hollow cylinder with a mass ofM = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
32-In the figure below, the hanging object has a mass of m1 = 0.460 kg; the sliding block has a mass of m2 = 0.765 kg; and the pulley is a hollow cylinder with a mass ofM = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.Explanation / Answer
The total energy in the system is given by
T = 1/2 m2 Vo^2 + 1/2 m1 Vo^2 + 1/2 I wo^2
= 1/2 m2 Vf^2 + 1/2 m1 Vf^2 + 1/2 I wf^2 - m1gd + uk m2g d
where Vo,Vf are the initial and final velocity of the blocks, wo/wf are the initial and final angular velocity of the pulley, I is the moment of inertia of the pulley. w is related to V as
Vo = R2 wo and Vf = R2 wf.
Also, the moment of inertia is given by
I = 1/2 m (R2^2 + R1^2)
You have all the data now, you can go ahead and solve for Vf in the energy equation above. After that, you can calculate wf.