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Step-by-Step Detailed Solution
Step 1: Identify the Physical Principle
When the two objects each of mass m are gently placed on the ring, there is no external torque acting on the system in the horizontal plane. Consequently, the total angular momentum of the system must remain conserved.
Step 2: Write the Initial Angular Momentum
The initial moment of inertia of the thin circular ring (of mass M) about the axis of rotation is
$I = M R^2$.
If the initial angular velocity is $ \omega $, the initial angular momentum $L_{\text{initial}}$ is:
$$
L_{\text{initial}} = I \times \omega = (M R^2) \, \omega.
$$
Step 3: Find the Final Moment of Inertia
When two objects each of mass m are placed on the ring at opposite ends of a diameter, the new moment of inertia $I'$ becomes the original ring’s moment of inertia plus the contribution from the two masses:
$$
I' = M R^2 + 2 \, m \, R^2 = (M + 2m) R^2.
$$
Step 4: Apply Conservation of Angular Momentum
Since angular momentum is conserved,
$$
L_{\text{initial}} = L_{\text{final}} \quad \Rightarrow \quad I \omega = I' \omega',
$$
where $ \omega' $ is the new angular velocity. Substituting the values of $I$ and $I'$, we get:
$$
M R^2 \, \omega = (M + 2m) R^2 \, \omega'.
$$
Step 5: Solve for the New Angular Velocity
Cancelling out the common factor $R^2$ on both sides, we obtain:
$$
M \, \omega = (M + 2m)\,\omega' \quad \Rightarrow \quad \omega' = \frac{M \, \omega}{M + 2m}.
$$
Final Answer
The new angular velocity of the ring after gently placing the two masses each of mass m is
$$
\omega' = \frac{M \, \omega}{M + 2m}.
$$
