A 5.0-kg solid cylinder of radius 0.33m is free to rotate about an axle that run

Question

A 5.0-kg solid cylinder of radius 0.33m 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.

Calculate its rotational velocity at tf = 5.0 s .

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

Explanation / Answer

Moment of inertia, I = mr2/2 = 5*0.332/2 = 0.27225 kg m2

Torque, T = r x F = 0.33*20 = 6.6 Nm

Angular acceleration, a = T/I =24.24 /s2

Angular velocity, w = at = 24.24 * 5 = 121.21 /s

Angle traveresed, s = 0.5*a*t2 = 0.5*24.24*52 = 303 radians

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