the figure below, the hanging object has a mass of m 1 = 0.360 kg; the sliding b

Question

the figure below, the hanging object has a mass of m1 = 0.360 kg; the sliding block has a mass of m2 = 0.900 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.
Ans: ??


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

Explanation / Answer

m1 =0.36 kg , m2 =0.9 kg, M =0.35 kg , R1 =0.02m

R2 =0.03 m , uk = 0.25, vi =0.82 m/s

a) Moment of inertia = (1/2)M (R1^2+R2^2)

from conservation of energy

(1/2)(m1+m2)(vf^2 - vi^2) +(1/2)I(wf^2 -wi^2) = m1gh - ukm2g

(1/2)(m1+m2)(vf^2 - vi^2) +(1/2)(1/2)M(R1^2+R2^2)(wf^2 -wi^2) = m1gh - ukm2g

vf = (vi^2 +((m1gh - ukm2g)/((1/2)(m1+m2)+(1/2)M(1+(R1^2/R2^2))))^0.5

vf^2 = (0.82)^2 +(((0.36*9.8*0.914) - (0.25*0.9*9.8))/((0.5*1.26)+(0.5*0.35(1+(0.02^2/0.03^2)))))

vf = 1.35 m/s

(b) the angular speed of the pulley

wf = vf/R2 = 1.35/0.03

wf = 45 rad/s

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