Question:
A thin circular ring of mass $M$ and radius $r$ is rotating gabout 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:
Correct Option: , 3
Solution:
Using conservation of angular momentum
$\left(\mathrm{Mr}^{2}\right) \omega=\left(\mathrm{Mr}^{2}+2 m r^{2}\right) \omega^{\prime}$
$\omega^{\prime}=\frac{M \omega}{M+2 m}$
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