A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\mathrm{\omega }$ . Two particles having mass m each are now attached at diametrically opposite points.  The angular speed of the ring will become

Two particles mass ‘m’ each are now attached at diametrically opposite points
$\therefore$Moment of inertia of two particles$={\mathrm{mr}}^2+{\mathrm{mr}}^2=2{\mathrm{mr}}^2$
External torque being absent, net angular momentum should be conserved about the axis of ring
$\begin{array}{l}\Rightarrow {\mathrm{I}}_{\mathrm{ring}}\mathrm{\omega }=\left({\mathrm{I}}_{\mathrm{ring}}+{\mathrm{I}}_{\mathrm{particles}}\right){\mathrm{\omega }}_{\mathrm{f}}\\ \Rightarrow {\mathrm{\omega }}_{\mathrm{f}}=\frac{{\mathrm{Mr}}^2\mathrm{\omega }}{{\mathrm{Mr}}^2+2{\mathrm{mr}}^2}\\ \therefore {\mathrm{\omega }}_{\mathrm{f}}=\frac{\mathrm{M}}{\mathrm{M}+2\mathrm{m}}\mathrm{\omega }\end{array}$