A thin circular ring of mass `m` and radius `R` is rotating about its axis with a constant angular velocity `omega`. Two objects each of mass `M` are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity `omega' =`

Options

A. `(omega(m + 2 M))/(m)`
B. `(omega(m-2M))/((m + 2M))`
C. `(omega m)/((m +M))`
D. `(omega m)/((m + 2M))`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. Let's break down the solution step by step.

### Step 1: Understand the Initial Conditions
Initially, we have a thin circular ring of mass `m` and radius `R` rotating with an angular velocity `ω`. The angular momentum of the system can be calculated using the formula:
\[ L_i = I \cdot \omega \]
where \( I \) is the moment of inertia of the ring.

### Step 2: Calculate the Moment of Inertia of the Ring
The moment of inertia \( I \) of a thin circular ring about its axis is given by:
\[ I = mR^2 \]

### Step 3: Calculate the Initial Angular Momentum
Using the moment of inertia calculated in Step 2, the initial angular momentum \( L_i \) can be expressed as:
\[ L_i = I \cdot \omega = mR^2 \cdot \omega \]

### Step 4: Analyze the Final Conditions
When two objects of mass `M` are attached to the opposite ends of a diameter of the ring, the total moment of inertia of the system changes. We need to calculate the new moment of inertia \( I' \).

### Step 5: Calculate the Moment of Inertia After Adding the Masses
The moment of inertia of each mass \( M \) located at a distance \( R \) from the axis of rotation is given by:
\[ I_{mass} = M R^2 \]
Since there are two masses, the total contribution from the masses is:
\[ I_{masses} = 2 \cdot M R^2 \]

Thus, the new moment of inertia \( I' \) of the system becomes:
\[ I' = I + I_{masses} = mR^2 + 2MR^2 = (m + 2M) R^2 \]

### Step 6: Apply Conservation of Angular Momentum
According to the conservation of angular momentum, the initial angular momentum must equal the final angular momentum:
\[ L_i = L_f \]
\[ mR^2 \cdot \omega = I' \cdot \omega' \]

Substituting \( I' \) from Step 5:
\[ mR^2 \cdot \omega = (m + 2M) R^2 \cdot \omega' \]

### Step 7: Solve for the New Angular Velocity \( \omega' \)
We can simplify the equation by dividing both sides by \( R^2 \):
\[ m \cdot \omega = (m + 2M) \cdot \omega' \]

Now, solving for \( \omega' \):
\[ \omega' = \frac{m \cdot \omega}{m + 2M} \]

### Final Result
The new angular velocity \( \omega' \) of the ring after the masses are attached is:
\[ \omega' = \frac{m \cdot \omega}{m + 2M} \]

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