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MHT CET 2026 (19 Apr, Shift 1) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2026 (19 Apr, Shift 1)

Q11MHT CET 2026 · 19 APR · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A thin uniform rod AB of mass ' $m$ ' and length ' $l$ ' is hinged at one end A to the ground level. Initially the rod stands vertically and is allowed to fall freely to the ground in the vertical plane. The angular velocity of the rod when its B end strikes the ground is

( $g$ = acceleration due to gravity)

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Answer: (b)

Centre of mass falls by $\frac{l}{2}$ ; $I_A = \frac{ml^2}{3}$
$\frac{1}{2}\cdot\frac{ml^2}{3}\omega^2 = mg\frac{l}{2}$
$\omega^2 = \frac{3g}{l}$
So the answer is (b) $\sqrt{\frac{3g}{l}}$ .

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