Solution:
Since external torque is zero, angular momentum \(L = I\omega\) is conserved. Rotational kinetic energy is \(K = \frac{L^2}{2I}\). If \(I' = I/4\), then \(K' = 4K\), so \(K' : K = 4 : 1\).
If moment of inertia of a spinning object drops to \(\left(\frac{1}{4}\right)^{\text{th}}\) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)
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