JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Rotational Mechanics

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Two small balls with masses m and 2m are attached to both ends of a rigid rod of length d and negligible mass. If angular momentum of this system is L about an axis (A) passing through its centre of mass and perpendicular to the rod then angular velocity of the system about A is :

(A) $\frac{3}{2} \frac{L}{m d ^{2}}$ ✓ Correct
(B) $\frac{2 L}{m d ^{2}}$
(C) $\frac{4}{3} \frac{L}{m d ^{2}}$
(D) $\frac{2 L}{5 m d ^{2}}$

Answer: (A)

Solution

The centre of mass lies at

\frac{2 d}{3}

from the mass

m

and

\frac{d}{3}

from the mass

2 m

. The moment of inertia about an axis through the centre of mass, perpendicular to the rod, is

I = m (\frac{2 d}{3})^{2} + 2 m (\frac{d}{3})^{2} = \frac{4 m d ^{2}}{9} + \frac{2 m d ^{2}}{9} = \frac{2 m d ^{2}}{3}

. From

L = I ω

ω = \frac{L}{I} = \frac{3 L}{2 m d ^{2}} = \frac{3}{2} \frac{L}{m d ^{2}}

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Rotational Mechanics

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