A 5.0-kg solid cylinder of radius 0.27m is free to rotate about an axle that run

Question

A 5.0-kg solid cylinder of radius 0.27m 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 atti = 0. Ignore any friction in the axle.

Part A

Calculate its rotational velocity at tf = 5.0 s .

Part B

Calculate the angle through which it has turned at tf= 5.0 s .

?f??i = radians

Explanation / Answer

net torque = F*R = 20*0.27 = 5.4 Nm

net torque = I*alfa

I = (1/2)*M*R^2 = (1/2)*5*0.27*0.27 = 0.18225 kg m^2

I*alfa= 5.4

alfa = 5.4/0.18225 = 29.63 rad/s^2

#A)

w = wo + alfa*t

W = 0 + (29.63*5)

W = 148.15 s^-1

#B)

?f??i = wo*t + (1/2)*alfa*t^2 = 0 + ((1/2)*29.63*5*5)

?f??i = 370.375 radians

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