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 $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:

• A. $\begin{array}{l}\omega \left(\frac{M}{M+2m}\right)\end{array}$
• B. $\begin{array}{l}\omega \left(\frac{M}{M+m}\right)\end{array}$
• C. $\begin{array}{l}\omega \left(\frac{M+2m}{M}\right)\end{array}$
• D. $\begin{array}{l}\omega \left(\frac{M-2m}{M+2m}\right)\end{array}$

Answer 1

The correct option is A

$\begin{array}{l}\omega \left(\frac{M}{M+2m}\right)\end{array}$

Step 1. Given Data,

A thin circular ring of mass, $M$

The radius of the thin circular ring is, $r$

Angular speed, $\omega$

Two particles having mass, $m$

Step 2. We have to find the angular speed ${\omega }_{final}$ of the ring will become,

Since there is no external force applied during the addition of mass:

$T_{net}=0$

Step 3. Conservation of angular momentum

So, angular momentum is conserved:

$$\begin{gathered} I_{intial}\omega =I_{final}{\omega }_{final} \\ {I}_{intial}=M{r}^2 \\ {I}_{final}=M{r}^2+m{r}^2+m{r}^2 \end{gathered}$$

Step 4. By substituting the value of $I_{initial}\&I_{final}$ above in the equation:

$$\begin{gathered} M{r}^2\omega =(M{r}^2+m{r}^2+m{r}^2){\omega }_{final} \\ \begin{array}{l}{\omega }_{final}=\frac{\left(M{r}^2\omega \right)}{M{r}^2+2m{r}^2}\end{array} \\ {\omega }_{final}=\frac{M\omega }{M+2m} \end{gathered}$$

Hence, the correct option is (A).