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A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity $$\omega$$. A person sits at the centre having two dumbbells in his hands. When he stretched out his hands, the moment of inertia of the system becomes triple. If $$E$$ be the initial Kinetic energy of the system, then final Kinetic energy will be $$\frac{E}{x}$$. The value of $$x$$ is _______
Correct Answer: 3
When the person stretches his hands, the moment of inertia becomes $$3I$$ (triple).
By conservation of angular momentum (no external torque):
$$L = I\omega = 3I\omega'$$
$$\omega' = \frac{\omega}{3}$$
Initial kinetic energy: $$E = \frac{1}{2}I\omega^2$$
Final kinetic energy:
$$E_f = \frac{1}{2}(3I)\omega'^2 = \frac{1}{2}(3I)\frac{\omega^2}{9} = \frac{1}{6}I\omega^2 = \frac{E}{3}$$
Therefore $$x = 3$$.
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