A thin circular ring of mass 'M' and radius 'r' is rotating about its axis with an angular speed ' $\omega$…
A thin circular ring of mass 'M' and radius 'r' is rotating about its axis with an angular speed ' $\omega$ '. Two particles each of mass 'm' are now attached at diametrically opposite points. The angular speed of the ring will become
$\frac{\omega M}{M+2 m}$
$\frac{\omega M}{M+m}$
$\frac{\omega(M-2 m)}{M}$
$\frac{\omega(M-2 m)}{M+2 m}$
Solution
Using angular momentum conservation:
$I_{\text {initial }} \cdot \omega=I_{\text {final }} \cdot \omega_{\text {final }}$
- Initial moment of inertia: $I_{\text {initial }}=M r^2$
- Final moment of inertia: $I_{\text {final }}=(M+2 m) r^2$
Substitute:
$\begin{gathered}
M r^2 \cdot \omega=(M+2 m) r^2 \cdot \omega_{\text {final }} \\
\omega_{\text {final }}=\frac{\omega M}{M+2 m}
\end{gathered}$