A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre _________.

Solution

Physics sir · 3 steps

No outside torque acts, so as the person walks in and I drops, the platform has to spin faster.

1. Why L is conserved. The person and the platform push on each other only through internal forces. No external torque acts about the axis, so the total angular momentum Iω stays the same.
2. Moments of inertia. Treat the platform as a uniform disc: Iₚ = 1/2(200)R² = 100R². At the rim the person adds 80R², so I₁ = 180R². At the centre the person is at r = 0 and adds nothing, so I₂ = 100R².

   I₁ω₁ = I₂ω₂ ⇒ 180R² × 5 = 100R² × ω₂

3. So. ω₂ = (180 × 5)/100 = 9 rpm. The answer is 9.

The common slip: 7. Taking the platform as a ring with I = MR² instead of a disc with 1/2MR² gives (280 × 5)/200 = 7 rpm, not 9.

The idea to keep. When mass moves towards the axis with no external torque, Iω stays fixed, so a smaller I means a larger ω. Since only the ratio matters, you can keep the rpm units as they are.

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