In the figure below, the hanging object has a mass of m 1 = 0.465 kg; the slidin

Question

In the figure below, the hanging object has a mass of m1 = 0.465 kg; the sliding block has a mass of m2 = 0.845 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.
m/s

(b) Find the angular speed of the pulley at the same moment.
rad/s

Explanation / Answer

a)

Moment of inertia of pulley

I=(1/2)M(R12+R22)=(1/2)(0.35)(0.022+0.032) =2.275*10-4 Kg-m2

By Conservation of energy

(1/2)m1Vi2+(1/2)m2Vi2+m1gh+(1/2)IWi2-ukm2g =(1/2)m1Vf2+(1/2)m2Vf2+(1/2)IWf2

Since W=V/r

(1/2)m1Vi2+(1/2)m2Vi2+m1gh+(1/2)I(Vi/R2)2-ukm2g =(1/2)m1Vf2+(1/2)m2Vf2+(1/2)I(Vf/R2)2

(1/2)*0.465*0.822 +( 1/2)*0.845*0.822 + 0.465*9.81*0.7 + (1/2)(2.275*10-4)(0.822/0.032)- 0.25*0.845*9.81*0.7 = (1/2)(0.465)Vf2 + (1/2)(0.845)Vf2 + (1/2)(2.275*10-4)(Vf/0.03)2

Vf=1.7036 m/s

b)

Angular speed

Wf=1.7036/0.03 =56.788 rad/s

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