Question

A thin circular ring of mass \(M\) and radius \(r\) is rotating about its axis with an angular speed \(\omega\). Two particles having mass \(\mathrm{m}\) each are now attached at diametrically opposite points. The angular speed of the ring will become
(a) \(\frac{\omega M}{M+m}\)
(b) \(\frac{\omega M}{M+2 m}\)
(c) \(\frac{\omega(M-2 m)}{M+2 m}\)
(d) \(\frac{\omega(M+2 m)}{M}\).

Answer 1

The correct answer is (b)
Explanation:
Moment of inertia of the ring \(I=M r^{2}\)
Angular Momentum \(=1 \omega\)
When the masses are attached, the moment of Inertia I'= \(M r^{2}+2 m r^{2}\) \(=(M+2 m) r^{2}\)

Let the new angular speed be w'. So the angular momentum \(=l^{\prime} \omega^{\prime} .\)
Since the angular momentum is conserved. I' \(\omega\) ' \(=I \omega\)
\(\rightarrow \omega^{\prime}=\mid \omega / I^{\prime}=\omega M r^{2} /(M+2 m) r^{2}=\omega M /(M+2 m)\)

Answer 2

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Answer 3

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