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

Since no external torque acts on the system, angular momentum is conserved. The initial angular momentum is $$L_i = I_{\text{ring}}\,\omega = Mr^2\omega$$.

When two particles of mass $$m$$ each are attached at diametrically opposite points on the ring, the new moment of inertia becomes $$I_f = Mr^2 + 2mr^2 = (M + 2m)r^2$$.

By conservation of angular momentum, $$I_i\omega = I_f\omega'$$, which gives $$Mr^2\omega = (M + 2m)r^2\omega'$$. Solving for $$\omega'$$, we get $$\omega' = \frac{M\omega}{M + 2m}$$.