Q161 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
A child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of a system is K. The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is
- A$\dfrac{K}{4}$
- Correct answerB$\dfrac{K}{2}$
- C$2K$
- D$4K$
Solution
By the principle of conservation of angular momentum, the angular momentum $L$ of the system remains constant because no external torque acts on it.
The kinetic energy of a rotating body is given by $K = \dfrac{L^2}{2I}$, where $I$ is the moment of inertia.
Given that the initial kinetic energy is $K$ and the initial moment of inertia is $I$.
When the child stretches his arms, the new moment of inertia becomes $I' = 2I$.
The new kinetic energy $K'$ is given by $K' = \dfrac{L^2}{2I'} = \dfrac{L^2}{2(2I)} = \dfrac{1}{2} \left( \dfrac{L^2}{2I} \right) = \dfrac{K}{2}$.
Answer: $\dfrac{K}{2}$
The kinetic energy of a rotating body is given by $K = \dfrac{L^2}{2I}$, where $I$ is the moment of inertia.
Given that the initial kinetic energy is $K$ and the initial moment of inertia is $I$.
When the child stretches his arms, the new moment of inertia becomes $I' = 2I$.
The new kinetic energy $K'$ is given by $K' = \dfrac{L^2}{2I'} = \dfrac{L^2}{2(2I)} = \dfrac{1}{2} \left( \dfrac{L^2}{2I} \right) = \dfrac{K}{2}$.
Answer: $\dfrac{K}{2}$
Hence, the correct answer is option: 'B'.