A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads⁻¹ in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads⁻¹).

Solution

Physics sir · 3 steps

No outside torque acts, so Iω stays fixed: add 2mR² to the ring's MR² and solve.

1. Why it is conserved. The masses are attached gently, so no external torque acts about the vertical axis. Angular momentum Iω about that axis stays the same before and after.
2. Moments of inertia. Before: the ring alone, I₁ = MR², with ω₁ = 2 rad s⁻¹. After: each mass m sits at distance R from the axis and adds mR², so I₂ = MR² + 2mR² = (M + 2m)R².

   MR² × 2 = (M + 2m)R² ω₂ ⇒ ω₂ = 2M/(M + 2m)

3. So. R² cancels and the ring slows to ω₂ = 2M/(M + 2m) rad s⁻¹, which is option (c).

The idea to keep. When mass is added to a spinning body with no outside torque, use I₁ω₁ = I₂ω₂. More mass far from the axis means a larger I and a slower spin.

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