A 5.0-kg solid cylinder of radius 0.17m is free to rotate about an axle that run
ID: 1769396 • Letter: A
Question
A 5.0-kg solid cylinder of radius 0.17m is free to rotate about an axle that runs along the cylinder length and passes through its center. A thread wrapped around the cylinder is pulled straight from the cylinder so as to unwrap with a steady tensile force of 20 N. As the thread unwinds, the cylinder rotates and there is no slippage between thread and cylinder. The cylinder starts from rest at ti = 0. Ignore any friction in the axle.
A- Calculate its rotational velocity at tf = 5.0 s .
B- Calculate the angle through which it has turned at tf = 5.0 s .
Explanation / Answer
torque = F*R = 20*0.17 = 3.4 Nm
MOI = I = (1/2)*M*R^2 = (1/2)*5*0.17*0.17 = 0.07225 kg m^2
A) torque = I*alfa = I*w/t
w = Torque *t/I
w = (3.4*5)/0.07225
w = 235.29 rad/s
--------
theta = average rotational velocity * time
theta = (w + 0)/2 *t
theta = (235.29/2)*5
theta = 588.225 radians
theta = 93.7 revolutions
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