(B) The rotational kinetic energy $K$ of a rotating body is given by the formula $K = \frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and

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(B) The rotational kinetic energy $K$ of a rotating body is given by the formula $K = \frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.Initially, $K = \frac{1}{2} I \omega^2$.After the dancer stretches his arms, the new moment of inertia $I' = 3I$ and the new angular velocity $\omega' = \frac{1}{3} \omega$.The new kinetic energy $K'$ is given by:$K' = \frac{1}{2} I' (\omega')^2$$K' = \frac{1}{2} (3I) (\frac{1}{3} \omega)^2$$K' = \frac{1}{2} (3I) (\frac{1}{9} \omega^2)$$K' = \frac{1}{3} (\frac{1}{2} I \omega^2)$Since $K = \frac{1}{2} I \omega^2$, we have $K' = \frac{K}{3}$.

Class 11 Physics System of Particles and Rotational Motion Rotational Kinetic Energy

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$A$ kathak dancer is standing on a horizontal surface with folded hands. Initially, the dancer is rotating about his central axis, and his kinetic energy is $K$. The dancer then stretches his arms such that the moment of inertia becomes three times its initial value and the angular velocity becomes one-third of its initial value. The kinetic energy of the dancer now is:

MHT CET 2026 All MHT CET

A
$K/6$
B
$K/3$
C
$3K$
D
$6K$

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Class 11

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System of Particles and Rotational Motion

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Rotational Kinetic Energy

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MediumNEET 2016

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MediumAIPMT 2005

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