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Physics · Rotational Dynamics

JEE Main 2026 — 23 January, Morning Shift — Question 40

Two small balls with masses m and 2 m 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 :

  1. Option A:

    32 Lmd2\frac{3}{2} \frac{\mathrm{~L}}{\mathrm{md}^{2}}

    Correct
  2. Option B:

    2 Lmd2\frac{2 \mathrm{~L}}{\mathrm{md}^{2}}

  3. Option C:

    43 Lmd2\frac{4}{3} \frac{\mathrm{~L}}{\mathrm{md}^{2}}

  4. Option D:

    2 L5md2\frac{2 \mathrm{~L}}{5 \mathrm{md}^{2}}

Answer: A

Step-by-step solution

L=Iω\mathrm{L}=\mathrm{I} \omega and ω=LI\omega=\frac{\mathrm{L}}{\mathrm{I}} ω=Lm(2 d3)2+2 m( d3)2=L49md2+29md2=L6md29\omega=\frac{\mathrm{L}}{\mathrm{m}\left(\frac{2 \mathrm{~d}}{3}\right)^{2}+2 \mathrm{~m}\left(\frac{\mathrm{~d}}{3}\right)^{2}}=\frac{\mathrm{L}}{\frac{4}{9} \mathrm{md}^{2}+\frac{2}{9} \mathrm{md}^{2}}=\frac{\mathrm{L}}{\frac{6 \mathrm{md}^{2}}{9}} ω=3 L2md2\omega=\frac{3 \mathrm{~L}}{2 \mathrm{md}^{2}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Morning Shift
Subject
Physics
Chapter
Rotational Dynamics
Topic
Moment of Inertia

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