A solid cylinder of mass $50\text{ kg}$ and radius $0.5\text{ m}$ 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 $2\text{ rev/s}^{2}$ is: (2014)
Solution:
Moment of inertia $I = \frac{1}{2}MR^{2} = 6.25\text{ kg m}^{2}$. Angular acceleration $\alpha = 2\text{ rev/s}^{2} = 4\pi\text{ rad/s}^{2}$. Torque $\tau = I\alpha = 25\pi = 78.5\text{ N m}$. Tension $T = \frac{\tau}{R} = \frac{78.5}{0.5} = 157\text{ N}$.
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