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 wheel now rotates with an angular velocity `omega=`

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have a thin circular ring of mass \( M \) and radius \( r \) rotating with an angular velocity \( \omega \). Two objects, each of mass \( m \), are attached to the opposite ends of a diameter of the ring.

### Step 2: Apply the Conservation of Angular Momentum
Since there is no external torque acting on the system, the angular momentum of the system is conserved. This means:
\[ L_{\text{initial}} = L_{\text{final}} \]
Where \( L \) is the angular momentum.

### Step 3: Calculate Initial Angular Momentum
The initial angular momentum \( L_{\text{initial}} \) of the ring can be calculated using the formula:
\[ L_{\text{initial}} = I \cdot \omega \]
Where \( I \) is the moment of inertia of the ring. For a ring, the moment of inertia is given by:
\[ I = M r^2 \]
Thus,
\[ L_{\text{initial}} = M r^2 \cdot \omega \]

### Step 4: Calculate Final Angular Momentum
After attaching the two masses, the new moment of inertia \( I' \) of the system (ring + two masses) is:
\[ I' = I + 2 \cdot (m \cdot r^2) \]
Where \( 2 \cdot (m \cdot r^2) \) accounts for the two masses located at a distance \( r \) from the center of the ring. Therefore:
\[ I' = M r^2 + 2 m r^2 = (M + 2m) r^2 \]
The final angular momentum \( L_{\text{final}} \) can be expressed as:
\[ L_{\text{final}} = I' \cdot \omega' \]
Where \( \omega' \) is the new angular velocity after the masses are added.

### Step 5: Set Initial and Final Angular Momentum Equal
Setting the initial and final angular momentum equal gives us:
\[ M r^2 \cdot \omega = (M + 2m) r^2 \cdot \omega' \]
We can simplify this equation by canceling \( r^2 \) (assuming \( r \neq 0 \)):
\[ M \omega = (M + 2m) \omega' \]

### Step 6: Solve for the New Angular Velocity
Rearranging the equation to solve for \( \omega' \):
\[ \omega' = \frac{M \omega}{M + 2m} \]

### Final Answer
The new angular velocity of the system after the two masses are attached is:
\[ \omega' = \frac{M \omega}{M + 2m} \]

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