JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

A thin uniform rod (X) of mass $M$ and length $L$ is pivoted at a height $(\frac{L}{3})$ 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 ______.

( $g$ = gravitational acceleration)

A
$\frac{3 g}{2 L}$
B
$\frac{3}{2} \frac{g}{L}$
C
$\frac{1}{2} \frac{g}{L}$
D
$\frac{3 g}{L}$

Solution

The pivot is at $L /3$ from the lower end, so $L /3$ below the centre of mass (the rod's centre is at $L /2$ from the lower end). When the rod falls from vertical to horizontal, the centre of mass drops by $h = L /2 - L /3 = L /6$ .

Moment of inertia about the pivot (parallel-axis theorem):

$I = \frac{M L ^{2}}{12} + M (\frac{L}{6})^{2} = \frac{M L ^{2}}{12} + \frac{M L ^{2}}{36} = \frac{M L ^{2}}{9}$ .

Energy conservation:

$M g \cdot \frac{L}{6} = \frac{1}{2} I ω^{2} = \frac{1}{2} \cdot \frac{M L ^{2}}{9} \cdot ω^{2}$

$ω^{2} = \frac{3 g}{L} \Rightarrow ω = \frac{3 g}{L}$ .

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