In the figure below, the hanging object has a mass of m -0.370 kg; the sliding b

Question

In the figure below, the hanging object has a mass of m -0.370 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 is0.250. 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 = 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 Need Help?Readt

Explanation / Answer

The change in kinetic energy of the system {block m1+block m2+pulley} is equal to the net work done on the

system. Only friction on the block m2 and gravitation force on block m1 have non-zero work. On the other hand,

angular speed of the pulley is related to the sped of the objects = v/ R^2 and a pulley of a hollow cylinder shape

has a moment of inertia of : 1/2M(R1^2 +R2^2)

1/2(m1+m2)(vf^2 - vi^2) + 1/2*I*(wf^2 - wi^2) = m1gh - mu*m2*g

1/2(m1+m2)(vf^2 - vi^2) + 1/2*1/2*M*(R1^2+R2^2)(vf^2 -vi^2)/R2^2 = m1gh - mu*m2*g

so

Vf = sqrt (vi^2 + {m1gh - mu*m2*g} / {1/2(m1+m2)+1/2M(1+R1^2/R2^2)}

Vf = sqrt (0.820^2 + {0.370*9.8*0.7 - 0.835*0.250*9.8}/{0.5(0.370+0.835)+0.5*0.350(1+0.02^2/0.03^2)}

Vf = 1.11 m/s

the angular sped of the pulley

w = vf/R2 = 1.11/0.03 = 37.24 rad/s

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