Rotational Motion

JEE Main level, numerical answer, medium

155 of 280

Question 155 of 280

A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity ω\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 EE be the initial Kinetic energy of the system, then final Kinetic energy will be Ex\dfrac{E}{x}. The value of xx is

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Answer: 3

The kinetic energy is given by
E=12Iω2=(L22I)E=\dfrac{1}{2}I{\omega }^{2}=\left(\dfrac{{L}^{2}}{2I}\right)
It is given that the new moment of inertia
I′=3I{I}^{'}=3I
Clearly, the final kinetic energy is
Ef=L22I′=L26I⇒Ef=(E3){E}_{f}=\dfrac{{L}^{2}}{2I'}=\dfrac{{L}^{2}}{6I} \Rightarrow {E}_{f}=\left(\dfrac{E}{3}\right)

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