(b) Find the angular speed of the pulley at the same moment. In the figure below

Question

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

In the figure below, the hanging object has a mass of m1 = 0.500 kg; the sliding block has a mass of m2 = 0.770 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 mu 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. (b) Find the angular speed of the pulley at the same moment.

Explanation / Answer

we are given with mass of the sliding block m1 = 0.850kg mass of the counterweight m2 = 0.420 kg mass of the pulley mp = 0.350kg outer radius of the pulleyis Rp = 0.0300 m the coefficient of kinetic friction between theblock and the horizontal surface is µk = 0.250 the initial velocity of the block is vi =0.820 m / s distance moved by the blocks betwwen the photogatesis s = 0.700 m the moment of inertia of the pulley is I = (1 /2) mp Rp2 the angular velocity of the pulley is?p = v / Rp the frictional force retarding the sliding block isgiven by fk = µk n = µk(m1 g) (a) here to get the result we choose that thegravitational potential energy to be zero at the level of the counter weight when the slidingmass reachesthe second photogate applying the work energy theorem we get Wnc = (KEtrans+ KErot + PEg)f -(KEtrans + KErot +PEg)i - fk s = (1 / 2) (m1 +m2) vf2 + (1 / 2) [(1 / 2)mp Rp2] (vf2/ Rp2) + 0 - (1 / 2) (m1 + m2) vi2+ (1 / 2) [(1 / 2) mp Rp2](vi2 / Rp2) -m2 g s on solving we get the final velocity as vf = v{vi2+ [2 (m2 - µk m1) gs / (m1 + m2 + (1 / 2) mp)]} = v{[(0.820 m /2)2] + [2 (0.208 kg) (9.80 m / s2) (0.700 m)/ (1.45 kg)]} = 1.625 m / s (b) the angular speed of the pulley will be ?f = vf / Rp = 1.625 m / s / 0.0300m = 54.16 rad / s

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