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JEE Main 2021 (26 Feb, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from JEE Main 2021 (26 Feb, Shift 2)

Q43JEE Main 2021 · 26 FEB · SHIFT 2Medium

Rotational Kinematics & Dynamics

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be:

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Answer: (b)

Energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ with $v = \omega r$ .
$mgh = \frac{1}{2}\omega^2(I + mr^2)$
So the answer is (b) $\omega^2 = \frac{2mgh}{I + mr^2}$ .

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