A solid cylinder on moving with constant speed $v_{0}$ reaches the bottom of an incline of 30${^\circ}$. A hollow cylinder of same mass and radius moving with the same constant speed $v_{0}$ reaches the bottom of a different incline of $ heta$. There is no slipping and both of them go through the same distance in the same time ; $ heta$ is then equal to

Question

A solid cylinder on moving with constant speed $v_{0}$ reaches the bottom of an incline of 30${^\circ}$. A hollow cylinder of same mass and radius moving with the same constant speed $v_{0}$ reaches the bottom of a different incline of $	heta$. There is no slipping and both of them go through the same distance in the same time ; $	heta$ is then equal to

Dr. Abubacker · Accepted Answer

(c) For solid cylinder, $	heta = 30{^\circ},\ K^{2} = \frac{1}{2}R^{2}$
For hollow cylinder, $	heta = ?,\ K^{2} = R^{2}$
Using we find,
$\frac{\left( 1 + \frac{1}{2} ight)}{\sin{30{^\circ}}} = \frac{1 + 1}{\sin	heta}$
$	herefore\ \sin	heta = \frac{2}{3} = 0.6667$
$	heta = 42{^\circ}$

Explanation

Reviewed by Dr. Abubacker

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