A bob of mass ' m ' is tied by a massless string whose other end is… | MHT CET 2025 (25 Apr, Shift 2)

Q19MHT CET 2025 · 25 APR · SHIFT 2Medium

Rotational Kinematics & Dynamics

A bob of mass ' $m$ ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' $R$ ' and mass ' $m$ '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' $h$ ', then angular speed of wheel will be (There is no slipping between string and wheel, g - acceleration due to gravity)

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Answer: (a)

Energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\omega^2$ with $v = \omega R$ .
$mgh = \frac{3}{4}m\omega^2R^2 \Rightarrow \omega^2 = \frac{4gh}{3R^2}$
$\omega = \frac{1}{R}\sqrt{\frac{4gh}{3}} = \frac{2}{R}\sqrt{\frac{gh}{3}}$
So the answer is (a).

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