Question

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Two objects, each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity of

• A. $\frac{\omega M}{M+m}$
• B. $\frac{\omega (M-2m)}{M+2m}$
• C. $\frac{\omega M}{M+2m}$
• D. $\frac{\omega (M+2m)}{M}$

Answer 1

The correct option is C $\frac{\omega M}{M+2m}$

Mass of circular ring $=M$
Radius of circular ring $=r$
Angular velocity $=\omega$
Mass of each object $=m$


Initial angular velocity $=\omega$
Initial angular momentum $L_i=I\omega =M{r}^2\omega ...\left(i\right)$


After placing masses,final angular velocity of system $={\omega }^{\prime }$
$(1,3$ represents masses and $2$ for the ring)
Final angular momentum
$\Rightarrow L_f=(I_1+I_2+I_3){\omega }^{\prime }$
$L_f=(m{r}^2+M{r}^2+m{r}^2){\omega }^{\prime }$
$\therefore L_f=(M+2m)r^2{\omega }^{\prime }...\left(ii\right)$
As the masses were placed gently on ring, all interactions between ring and masses being internal for the system of (ring+masses) hence the external torque is zero.
${\tau }_{\text{ext}}=0$
Applying angular momentum conservation:
$\Rightarrow L_i=L_f$
Or, $M{r}^2\omega =(M+2m)r^2{\omega }^{\prime }$
$\therefore {\omega }^{\prime }=\frac{M\omega }{M+2m}$
New angular velocity ${\omega }^{\prime }$ of the ring is $\frac{M\omega }{M+2m}$