Question 420: Centre of Mass and Rotational Motion

Question

A circular plate is rotating in horizontal plane, about an axis passing through its center and perpendicular to the plate, with an angular velocity $$\omega$$. A person sits at the center having two dumbbells in his hands. When he stretches 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

Accepted Answer

The conservation of angular momentum states that the angular momentum (L) remains constant. The relation between kinetic energy (KE), angular momentum (L), and moment of inertia (I) is given by:
$$
\mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}
$$
Using this relation, we can find the ratio of the final kinetic energy ($$\mathrm{KE}_{	ext{final}}$$) to the initial kinetic energy ($$\mathrm{KE}_{	ext{initial}}$$ or E):
$$
\frac{\mathrm{KE}_{	ext{final}}}{\mathrm{KE}_{	ext{initial}}}=\frac{\mathrm{I}_{	ext{initial}}}{\mathrm{I}_{	ext{final}}}
$$
Since the moment of inertia triples, we have $$\mathrm{I}_{	ext{final}} = 3\mathrm{I}_{	ext{initial}}$$. Therefore,
$$
\frac{\mathrm{KE}_{	ext{final}}}{\mathrm{E}}=\frac{\mathrm{I}_{	ext{initial}}}{3\mathrm{I}_{	ext{initial}}}=\frac{1}{3}
$$
This means that the final kinetic energy of the system is:
$$
\mathrm{KE}_{	ext{final}}=\frac{E}{3}
$$
So, the value of $$x$$ is 3.

Centre of Mass and Rotational Motion

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