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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA child stands at the centre of a turntable with outstretched arms rotating with an initial rotational kinetic energy \( K_0 \). When the child folds their arms, the moment of inertia of the system reduces to \( \frac{2}{5} \) of its initial value. Assuming no external torque acts, the ratio of the new rotational kinetic energy to the initial rotational kinetic energy (\( K_f / K_0 \)) is:
Correct answer · Option 2
\( 5/2 \)
Explanation
By conservation of angular momentum \( L = I\omega = \text{constant} \). Kinetic energy \( K = \frac{L^2}{2I} \). Therefore, \( \frac{K_f}{K_0} = \frac{I_0}{I_f} = \frac{1}{2/5} = \frac{5}{2} \).
