A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :

Solution

Physics sir · 3 steps

No outside torque acts, so Iω stays fixed: the extra mass on the rim must slow the ring down.

1. No external torque. Attaching the particles involves only internal forces between them and the ring. So the angular momentum about the axis is conserved: I₁ω₁ = I₂ω₂.
2. Moments of inertia. Before, the ring alone has I₁ = Mr². After, each particle sits at distance r from the axis and adds mr², so I₂ = Mr² + 2mr² = (M + 2m)r².

   Mr² ω = (M + 2m)r² ω' ⇒ ω' = ω M/(M + 2m)

3. So. The new angular speed is ωM/(M + 2m), which is option (c). It is less than ω, as it should be once the moment of inertia grows.

The idea to keep. When mass is added or moved with no outside torque, keep Iω constant. Bigger I means smaller ω, and every point mass at distance r adds mr².

Answer from the NTA final answer key. Solution by PrepWiser.