Solution:

We can solve this problem using the principle of conservation of angular momentum since no external torque acts on the system.

Step 1: Initial Angular Momentum

The moment of inertia of a thin circular ring about its axis is:

$$I_{\text{initial}} = M R^2$$

The initial angular momentum is given by:

$$L_{\text{initial}} = I_{\text{initial}} \cdot \omega = (M R^2) \cdot \omega$$

Step 2: Final Moment of Inertia

When two objects of mass m are attached to the ring, assuming they are symmetrically placed on the ring, their contribution to the moment of inertia is:

$$I_{\text{added}} = 2m R^2$$

Thus, the new total moment of inertia becomes:

$$I_{\text{final}} = M R^2 + 2m R^2 = (M + 2m) R^2$$

Step 3: Applying Conservation of Angular Momentum

Since no external torque acts on the system:

$$L_{\text{initial}} = L_{\text{final}}$$

$$(M R^2) \cdot \omega = (M + 2m) R^2 \cdot \omega'$$

Canceling

$R^2$

from both sides:

$$M \omega = (M + 2m) \omega'$$

Solving for the new angular velocity

$\omega'$

:

$$\omega' = \frac{M \omega}{M + 2m}$$

Final Answer:

$$\omega' = \frac{M \omega}{M + 2m}$$

This shows that the angular velocity decreases after attaching the masses, as expected due to an increase in the moment of inertia.