Conservation Of Angular Momentum (numerical)
A child of negligible spin sits on a rotating stool that turns at 2 rad/s with moment of inertia 4 kg·m²; he then extends weights so the inertia rises to 8 kg·m². What is the new angular velocity?
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About This Question
- Subject
- physics
- Chapter
- rotational motion
- Topic
- conservation of angular momentum (numerical)
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1rad/s
Drawing on NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), when no external torque acts about the rotation axis the total angular momentum L = Iω of the child-and-stool system is conserved. This is the same principle that governs spinning skaters and divers, and it holds whenever internal rearrangements of mass occur without any outside twisting influence. Extending the weights outward is purely an internal action that increases the moment of inertia but cannot change the angular momentum, so the conservation statement I₁ω₁ = I₂ω₂ applies. Given I₁ = 4 kg·m², ω₁ = 2 rad/s, and I₂ = 8 kg·m², we solve for the new angular velocity as ω₂ = I₁ω₁/I₂ = (4 × 2)/8 = 8/8 = 1 rad/s. The angular speed halves precisely because the moment of inertia doubled. The option 4 rad/s wrongly increases the speed, which would occur only if the inertia had decreased. The option 2 rad/s ignores the change in inertia entirely. The option 0.5 rad/s over-reduces the speed by an extra unjustified factor. A consistency check confirms that L stays fixed at 8 kg·m²/s both before and after, so doubling I must exactly halve ω, matching the computed result of 1 rad/s.
This medium difficulty physics question is from the chapter rotational motion, covering the topic of conservation of angular momentum (numerical). It appeared in the 2025 exam.
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