Question

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

(b) $\frac{\omega M}{M+2m}$

(c) $\frac{\omega \left(M-2m\right)}{M+2m}$

(d) $\frac{\omega \left(M+2m\right)}{M}$.

Answer 1

(b) $\frac{\omega M}{M+2m}$

No external torque is applied on the ring; therefore, the angular momentum will be conserved.
$$\begin{gathered} I\omega =I\text{'}\omega \text{'} \\ \Rightarrow \omega \text{'}=\frac{I\omega }{I\text{'}}...\left(i\right) \end{gathered}$$
$$\begin{gathered} I=M{r}^2 \\ I\text{'}=M{r}^2+2m{r}^2 \end{gathered}$$
On putting these values in equation (i), we get:
$\omega \text{'}=\frac{\omega M}{M+2m}$