In the figure below, the hanging object has a mass of m1 = 0.390 kg; the sliding
ID: 2304667 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.390 kg; the sliding block has a mass of m2 = 0.835 kg; and the pulley is a hollow cylinder with a mass of M = 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.
(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. (in m/s)
(b) Find the angular speed of the pulley at the same moment. (in rad/s)
Explanation / Answer
a)
Moment of inertia
I=(1/2)M(R12+R22) =(1/2)(0.35)(0.022+0.032)
I=2.275*10-4 Kg-m2
Initial angular velocity
Wi=Vi/R2 =0.82/0.03 =27.33 rad/s
Final angular velocity
Wf=Vf/0.03
By Conservation of energy
(1/2)(m1+m2)Vi2+m1gx +(1/2)IWi2 -ukm2gx =(1/2)(m1+m2)Vf2+(1/2)IWf2
(1/2)(0.39+0.835)*0.822+0.39*9.81*0.7+(1/2)*(2.275*10-4)(27.332)-0.25*0.835*9.81*0.7 =(1/2)(0.39+0.835)Vf2+(1/2)(2.275*10-4)(Vf/0.03)2
3.483 =1.47778Vf2
Vf =1.535 m/s =1.54 m/s (approx)
b)
Final angular speed
Wf=Vf/R2 =1.54/0.03
Wf =51.2 rad/s