Physics - Rotation Question with Solution | TestHub

Initial moment of inertia of the ring is $I_1=M{r}^2$

Initial angular momentum is $L_1=I_1\omega =M{r}^2\omega$

When six objects of mass $m$ are placed on the ring at the ends of three diameters, they are at a distance $r$ from the axis of rotation. The final moment of inertia becomes $I_2=M{r}^2+6m{r}^2=(M+6m)r^2$

Since no external torque acts on the system, angular momentum is conserved.

$L_1=L_2$

$M{r}^2\omega =(M+6m)r^2{\omega }^{\prime }$

${\omega }^{\prime }=\frac{M\omega }{M+6m}$

Answer: $\frac{M\omega }{M+6m}$