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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:
A
25 N
B
50 N
C
78.5 N
D
157 N
Detailed Solution
$\alpha = 2$ rev/$s^2$ $= 4\pi$ rad/$s^2$
$I = \frac{1}{2}MR^2 = \frac{1}{2}(50)(0.5)^2 = \frac{25}{4}$ kg $m^2$
$\tau = I\alpha \Rightarrow TR = I\alpha$
$T = \frac{I\alpha}{R} = \frac{\frac{25}{4}\times 4\pi}{0.5} = 50\pi$ N = 157 N
$I = \frac{1}{2}MR^2 = \frac{1}{2}(50)(0.5)^2 = \frac{25}{4}$ kg $m^2$
$\tau = I\alpha \Rightarrow TR = I\alpha$
$T = \frac{I\alpha}{R} = \frac{\frac{25}{4}\times 4\pi}{0.5} = 50\pi$ N = 157 N