A solid cylinder of mass $50 ext{ kg}$ and radius $0.5 ext{ m}$ is free to rotate about the horizontal axi

Question

A solid cylinder of mass $50 	ext{ kg}$ and radius $0.5 	ext{ m}$ 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 $2 	ext{ rev/s}^2$ is:

Accepted Answer

Given:
Mass of the cylinder, $M = 50 	ext{ kg}$
Radius of the cylinder, $R = 0.5 	ext{ m}$
Angular acceleration, $\alpha = 2 	ext{ rev/s}^2 = 2 	imes 2\pi 	ext{ rad/s}^2 = 4\pi 	ext{ rad/s}^2$

The moment of inertia of a solid cylinder about its central axis is given by:
$I = \frac{1}{2}MR^2$

The torque $	au$ produced by the tension $T$ in the string acting at the rim of the cylinder is:
$	au = T 	imes R$

According to Newton's second law for rotational motion, we know that:
$	au = I\alpha$

Equating the two expressions for torque:
$T 	imes R = \left(\frac{1}{2}MR^2ight)\alpha$
$T = \frac{1}{2}MR\alpha$

Substituting the given values into the equation:
$T = \frac{1}{2} 	imes 50 	imes 0.5 	imes 4\pi$
$T = 12.5 	imes 4\pi = 50\pi 	ext{ N}$

Taking $\pi \approx 3.14$:
$T = 50 	imes 3.14 = 157 	ext{ N}$

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