# A thin circular ring of mass M and radius r is rotating

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:

1. $\omega \frac{\mathrm{M}}{\mathrm{M}+\mathrm{m}}$

2. $\omega \frac{\mathrm{M}+2 \mathrm{~m}}{\mathrm{M}}$

3. $\omega \frac{M}{M+2 m}$

4. $\omega \frac{M-2 m}{M+2 m}$

Correct Option: , 3

Solution:

Using conservation of angular momentum

$\left(\mathrm{Mr}^{2}\right) \omega=\left(\mathrm{Mr}^{2}+2 m r^{2}\right) \omega^{\prime}$

$\omega^{\prime}=\frac{M \omega}{M+2 m}$