In the figure below, the hanging object has a mess of m 0.465 kg; the sliding bl

Question

In the figure below, the hanging object has a mess of m 0.465 kg; the sliding block has a mass of m2 0.840 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 frictian between the block and the horizontal surface is x-0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the puley. The block has a velocity of v 0.820 m/s toward the pulley when it passes a reference point on the table. mi (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

Given,

m1 = 0.465 kg ; m2 = 0.840 kg ; M = 0.35 kg ; R1 = 0.02 m ; R2 = 0.03 m ; uk = 0.25 ; vi = 0.82 m/s

a)The moment of inertia of the pulley is:

I = 1/2 M (R1^2 + R2^2)

I = 0.5 x 0.35 x [0.02^2 + 0.03^2] = 0.0002275 kg-m^2

KE of pulley

KE = 1/2 I w^2

KE-p = 0.5 x 0.0002275 x 0.82/0.03 = 0.085 J

KE of m1 is:

KE = 1/2 m1 vi^2

KE1 = 0.5 x 0.465 x 0.82^2 = 0.156 J

PE of m2

U = m g x

U1 = 0.465 x 9.81 x 0.7 = 3.19 J

KE of m2 is:

KE2 = 0.5 x 0.84 x 0.82^2 = 0.282 J

U2 = 0 J

force of friction of f2 is:

Ff = uk N = 0.25 x 0.84 x 9.81 = 2.06 N

Wf = 2.06 x 0.7 = 1.44 J

from conservation of energy

KE1 + KE2 + U1 + U2 + KE-p = Wf + KE1' + KE2' + KE-p'

0.156 + 0.282 + 3.19 + 0 + 0.085 = 2.06 + 0.5 (0.465 x vf^2 + 0.84 vf^2 + 0.000275 x vf^2/0.03^2)

3.713 - 2.06 = 1.61 vf^2 => vf = 1.05 m/s

Hence, vf = 1.05 m/s

b)w = vf/R2

w = 1.01/0.03 = 33.67 rad/s

Hence, w = 33.67 rad/s

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