In the figure below, the hanging object has a mass of m 1 = 0.440 kg; the slidin
ID: 1349854 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.440 kg; the sliding block has a mass of m2 = 0.820 kg; and the pulley is a hollow cylinder with a mass ofM = 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.
___m/s
(b) Find the angular speed of the pulley at the same moment.
___rad/s
Explanation / Answer
a) Moment of interia of Cylinder I = 0.5 M (R1^2 + R2^2)
I(cylinder) = 0.5*0.35*(0.02^2+0.03^2) =2.275*10^-4
now By the law of conservation of energy:
inital KE = 0.5 m1v1^2 + 0.5 m2v2^2 + 0.5 IW^2 - umgh
initial KE = 0.5*0.82*0.82^2+0.5*0.365*0.82^2- 0.25*0.82*9.8*0.7+0.5*2.275*10^-4*(0.82/0.03)^2
Intiai energy = -0.93 JOules
simlilary final KE = 0.5*0.82*V^2+0.5*0.365*V^2-0.365*9.8*0.7+0.5*2.275*10^-4*(V/0.03)^2
KEf = 0.72 V^2-2.5
Thus: 0.72 V^2-2.5= -0.93
V=1.48 m/s
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(b) Use the relation for angular velocity
W = v/sqrt(R1^2+R2^2)
w = 1.48/ sqrt(.02^2 + .03^2)
W =41.13 rad/s