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A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be

1. (a) (M − 4m)ω / M + 4m
2. (b) Mω / 4m
3. (c) Mω / M + 4m
4. (d) (M + 4m)ω / M

Correct answer: (a) (M − 4m)ω / M + 4m

Solution

Using the conservation of angular momentum, the initial angular momentum of the ring (I₁ω₁) equals the final angular momentum (I₂ω₂). The initial moment of inertia is I₁ = Mr², and the final moment of inertia is I₂ = Mr² + 4m(r²). Solving for ω₂ gives (M − 4m)ω / (M + 4m).

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