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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

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78.5 N

157 N

25 N

50 N

answer is D.

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Detailed Solution

Torque is produced due to the tension (T) in the string and torque $(τ) = rF = rT and, τ = lα$

where I the moment of inertia $α$ = angular acceleration= $2 \times 2 πrad / s^{2}$

$rT = l α$

$T = \frac{lα}{r} = \frac{{Mr}^{2}}{2} \frac{α}{r} = \frac{Mrα}{2} = \frac{50 \times 0.5 \times 4 π}{2}$

$= 50 π = 157 N$

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