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JEE Main 2021 (18 Mar, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2021 (18 Mar, Shift 1)

Q52JEE Main 2021 · 18 MAR · SHIFT 1Medium

Angular Momentum & its Conservation

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega$ . Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become:

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Answer: (c)

Conservation of $L$ : $Mr^2\omega = (M+2m)r^2\omega'$ .
$\omega' = \omega\frac{M}{M+2m}$
So the answer is (c) $\omega\frac{M}{M+2m}$ .

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