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JEE Main 2003 (AIEEE) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from JEE Main 2003 (AIEEE)

Q53JEE Main 2003 · AIEEEMedium

Rotational Kinetic Energy, Work & Power

A particle performing uniform circular motion has angular momentum $L$ . If its angular frequency is doubled & its kinetic energy halved, then the new angular momentum is

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Answer: (a)

$K = \frac{1}{2}L\omega \Rightarrow L = \frac{2K}{\omega}$
$\omega' = 2\omega$ , $K' = \frac{K}{2}$ : $L' = \frac{2(K/2)}{2\omega} = \frac{L}{4}$
So the answer is (a) $\frac{L}{4}$ .

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