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(A) Let the initial moment of inertia be $I_1$ and the final moment of inertia be $I_2 = 0.75 I_1 = \frac{3}{4} I_1$.Since no external torque acts on

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increase by $33.3 \%$

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(A) Let the initial moment of inertia be $I_1$ and the final moment of inertia be $I_2 = 0.75 I_1 = \frac{3}{4} I_1$.Since no external torque acts on the system,the angular momentum $L$ is conserved,so $L = I_1 \omega_1 = I_2 \omega_2$.Thus,$\omega_2 = \frac{I_1}{I_2} \omega_1 = \frac{I_1}{0.75 I_1} \omega_1 = \frac{4}{3} \omega_1$.The initial rotational kinetic energy is $K_1 = \frac{L^2}{2 I_1}$.The final rotational kinetic energy is $K_2 = \frac{L^2}{2 I_2} = \frac{L^2}{2 (0.75 I_1)} = \frac{L^2}{1.5 I_1} = \frac{4}{3} K_1$.The percentage increase in kinetic energy is $\frac{K_2 - K_1}{K_1} 	imes 100 = (\frac{4}{3} - 1) 	imes 100 = \frac{1}{3} 	imes 100 \approx 33.3 \%$.

$A$ man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is $75 \%$ of that with outstretched arms,then his rotational kinetic energy will

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