AIPMT MAINS 2009 Physics Conservation of Angular Momentum MCQ Question
A thin circular ring of mass M and radius R is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity. If two objects each of mass m be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity:
ωM/(M + 2m)
(ω(M + 2m))/M
ωM/(M + m)
(ω(M - 2m))/(M + 2m)
Correct Answer
Detailed Explanation
The conservation of angular momentum applies here. The initial angular momentum is I₁ω = MR²ω. After attaching the masses, the new moment of inertia is I₂ = MR² + 2mR². Solving for the new angular velocity gives ω' = ωM/(M + 2m).
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