A circular plate is rotating in horizontal plane, about an axis passing through its center and perpendicular to the plate, with an angular velocity ω. A person sits at the center having two dumbbells in his hands. When he stretches out his hands, the moment of inertia of the system becomes triple. If E be the initial Kinetic energy of the system, then final Kinetic energy will be E/x. The value of x is

Solution

Physics sir · 3 steps

No outside torque, so L stays fixed; write KE as L²/2I and tripling I cuts KE to a third.

1. Angular momentum conserved. Stretching his arms is an internal action. No external torque acts about the axis, so L = Iω stays the same. When I becomes 3I, the angular velocity drops to ω/3.
2. Energy using L. Rotational kinetic energy is 1/2Iω². Since ω = L/I, this is the same as L²/2I, which is handy because L does not change.

   E' = L²/2(3I) = 1/3·L²/2I = E/3

3. So. The final kinetic energy is E/3, so x = 3. The missing energy is absorbed by his muscles as the dumbbells move outward.

The common slip: 0.333. Keeping ω unchanged and using 1/2Iω² makes the energy triple to 3E, so x comes out as 1/3, not 3.

The idea to keep. When I changes with no external torque, hold L fixed and use KE = L²/2I. KE goes as 1/I, not as I or 1/I².

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