First slide

Torque

Question

A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s^-2 is

Moderate

A

25 N
B

50 N
C

78.5 N
D

157 N

Solution

Here $α = 2$ revolutions/s² $= 4 πrad / s^{2}$
$I = \frac{1}{2} {MR}^{2} = \frac{1}{2} (50) (0 .5)^{2} = \frac{25}{4} kg - m^{2}$
As $τ = Iα so TR = Iα$
$\Rightarrow T = \frac{Iα}{R} = \frac{(\frac{25}{4}) (4 π)}{(0 .5)} N = 50 πN = 157 N$

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