Physics · Rotational Dynamics

JEE Main 2026 — 24 January, Evening Shift — Question 31

A thin uniform rod $(\mathrm{X})$ of mass M and length L is pivoted at a height $\left(\frac{\mathrm{L}}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is $\_\_\_\_$ . ( $\mathrm{g}=$ gravitational acceleration)

Option A:
$\sqrt{\frac{3}{2} \frac{\mathrm{~g}}{\mathrm{~L}}}$
Option B:
$\frac{3}{\sqrt{2}} \sqrt{\frac{\mathrm{~g}}{\mathrm{~L}}}$
Option C:
$\frac{1}{\sqrt{2}} \sqrt{\frac{\mathrm{~g}}{\mathrm{~L}}}$
Option D:
$\sqrt{\frac{3 \mathrm{~g}}{\mathrm{~L}}}$
Correct

Answer: D

Step-by-step solution

$\operatorname{mg} \frac{\ell}{6}=\frac{1}{2} \mathrm{I} \omega^{2}$ Here $I=\frac{m \ell^{2}}{12}+\frac{m \ell^{2}}{36}=\frac{m \ell^{2}}{9}$ $\mathrm{mg} \frac{\ell}{6}=\frac{\mathrm{m} \ell^{2}}{18} \omega^{2} \Rightarrow \omega^{2}=\frac{3 \mathrm{~g}}{\ell}$ $\omega=\sqrt{\frac{3 \mathrm{~g}}{\ell}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Rotational Dynamics
Topic: Moment of Inertia

A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to I 1 . A small co-centric cylinder of length…

Answer: D

Step-by-step solution

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