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(A) The initial kinetic energy of the system is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the initial moment of inertia and $\omega$ is the i

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$\frac{K}{2}$

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(A) The initial kinetic energy of the system is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the initial moment of inertia and $\omega$ is the initial angular velocity.According to the law of conservation of angular momentum,since no external torque acts on the system,the angular momentum $L$ remains constant: $L = I \omega = I^{\prime} \omega^{\prime}$.Given that the new moment of inertia is $I^{\prime} = 2I$,we have $2I \omega^{\prime} = I \omega$,which implies $\omega^{\prime} = \frac{\omega}{2}$.The new kinetic energy $K^{\prime}$ is given by $K^{\prime} = \frac{1}{2} I^{\prime} \omega^{\prime 2}$.Substituting the values,$K^{\prime} = \frac{1}{2} (2I) \left( \frac{\omega}{2} \right)^2 = I \left( \frac{\omega^2}{4} \right) = \frac{1}{2} \left( \frac{1}{2} I \omega^2 \right) = \frac{K}{2}$.

$A$ child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is $K$. The child now stretches his arms so that the moment of inertia of the system becomes double. The kinetic energy of the system now is

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