Question

A thin circular ring of mass $M$ and radius $R$ is rotating about its axis with an angular speed ${\omega }_0$. Two particles each of mass $m$ are now attached at diametrically opposite points. The new angular speed of the ring is $\frac{{\omega }_{0}M}{M+xm}$. Find $x$

Answer 1

Since , there is no external force applied during addition of mass, angular momentum is conserved.

$\Rightarrow I_{intial}{\omega }_0=I_{final}{\omega }_{final}$

$I_{intial}=M{R}^2$

$I_{final}=M{R}^2+m{R}^2+m{R}^2$

by substituting above in equation,

$M{R}^2{\omega }_0=(M{R}^2+m{R}^2+m{R}^2){\omega }_{final}$

$\Rightarrow {\omega }_{final}=\frac{M{\omega }_0}{M+2m}$