in the figure below, the hanging object has a mass of m_1 = 0.460 kg; the slidin
ID: 1438694 • Letter: I
Question
in the figure below, the hanging object has a mass of m_1 = 0.460 kg; the sliding block has a mass of m_2 0.880 kg, and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner R_1 =0.020 0 m, and an outer radius of R_2 = 0.030 0 m. Assume the mass of the spokes eggible. The coefficient of kinetic friction between the block and the horizontal surface is muk.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 v_1 = 0.820 m/s toward the pulley when It passes a reference point on the table. Use energy methods to predict its speed after it has moved to a second point, 0.700 n away. Find the angular speed of the pulley at the same moment.Explanation / Answer
a)
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 / R2
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)+1/2I((f^2-(i^2) = m1gh- µkm2g
1/2(m1 + m2)(vf^2-vi^2)+1/2*1/2M(R1^2+R2^2*(vf^2-vi^2)/R2^2 = m1gh- µkm2g
Rearranging should give
vf = sqrt vi^2 + m1gh- µkm2g/1/2(m1+m2)+1/2M(1+R1^2/R2^2)
With the given numerical values we get :
vf = 1.378m/s
b)
The angular speed of the pulley :
f = vf/R2
= 1.378 / 0.030
= 45.96rad/s
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