A thin circular ring of mass $$M$$ and radius $$r$$ is rotating about its axis with a constant angular velocity $$\omega .$$ Four objects each of mass $$m,$$ are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be

Solution :
External torque $${\tau _{{\text{ext}}}} = 0$$
So, $$\frac{{dL}}{{dt}} = 0$$
Angular momentum, $$L$$ = constant
or $$I\omega = {\text{constant}}$$
$$\therefore {I_1}{\omega _1} = {I_2}{\omega _2}\,......\left( {\text{i}} \right)$$
So, for two different cases
Here, $${I_1} = M{r^2},{\omega _1}$$
$$ = \omega ,{I_2} = M{r^2} + 4m{r^2}$$
Hence, Eq. (i) can be written as
$$\eqalign{ & M{r^2}\omega = \left( {M{r^2} + 4m{r^2}} \right){\omega _2} \cr & \therefore {\omega _2} = \frac{{M\omega }}{{M + 4m}} \cr} $$