1. Use conservation of angular momentum
   Since the two masses are attached gently, there is no external torque about the rotation axis. Hence angular momentum is conserved.

2. Initial moment of inertia
   For a thin ring of mass $M$ and radius $R$ about its central axis perpendicular to the plane: $I_i = MR^2$ Initial angular velocity is: $\omega_i = 2\ \text{rad s}^{-1}$ So initial angular momentum is: $L_i = I_i\omega_i = MR^2 \cdot 2 = 2MR^2$

3. Final moment of inertia
   Two particles, each of mass $m$, are attached at opposite ends of a diameter. Each particle is at distance $R$ from the axis.

Moment of inertia of the two masses: $I_{masses} = mR^2 + mR^2 = 2mR^2$

Therefore total final moment of inertia: $I_f = MR^2 + 2mR^2 = (M+2m)R^2$

4. Apply conservation of angular momentum
   $L_i = L_f$ $2MR^2 = (M+2m)R^2\,\omega_f$ Cancel $R^2$: $2M = (M+2m)\omega_f$ Thus, $\omega_f = \frac{2M}{M+2m}\ \text{rad s}^{-1}$

5. Match with the options
   This corresponds to: $\boxed{\frac{2M}{M+2m}}$ which is Option C.