A thin circular ring of mass M and radius r is rotating about its axis with an angular speed `omega`. Two particles having mass `m` each are now attached at diametrically opposite points. The angular speed of the ring will become

Options

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

Answer

To solve the problem of finding the new angular speed of a thin circular ring after attaching two particles at diametrically opposite points, we will use the principle of conservation of angular momentum. Here’s the step-by-step solution:

### Step 1: Understand the System
We have a thin circular ring of mass \( M \) and radius \( r \) rotating with an initial angular speed \( \omega \). Two particles, each of mass \( m \), are attached at diametrically opposite points on the ring.

### Step 2: Apply Conservation of Angular Momentum
Since no external torque acts on the system, the angular momentum before and after the particles are attached must be conserved. This can be expressed as:
\[ I_1 \omega_1 = I_2 \omega_2 \]
Where:
- \( I_1 \) is the moment of inertia of the ring before the particles are attached.
- \( \omega_1 \) is the initial angular speed (\( \omega \)).
- \( I_2 \) is the moment of inertia of the system after the particles are attached.
- \( \omega_2 \) is the new angular speed (\( \omega' \)).

### Step 3: Calculate the Initial Moment of Inertia \( I_1 \)
The moment of inertia of the ring about its axis is given by:
\[ I_1 = M r^2 \]

### Step 4: Calculate the New Moment of Inertia \( I_2 \)
After attaching the two particles, the new moment of inertia \( I_2 \) is the sum of the moment of inertia of the ring and the moment of inertia due to the two particles. Each particle contributes \( m r^2 \) to the moment of inertia (since they are at a distance \( r \) from the axis of rotation). Therefore, for two particles:
\[ I_2 = I_{\text{ring}} + I_{\text{particles}} = M r^2 + 2(m r^2) = M r^2 + 2m r^2 = (M + 2m) r^2 \]

### Step 5: Substitute into the Conservation of Angular Momentum Equation
Now substituting \( I_1 \), \( \omega_1 \), and \( I_2 \) into the conservation equation:
\[ M r^2 \cdot \omega = (M + 2m) r^2 \cdot \omega' \]

### Step 6: Simplify the Equation
We can cancel \( r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ M \omega = (M + 2m) \omega' \]

### Step 7: Solve for the New Angular Speed \( \omega' \)
Rearranging the equation to solve for \( \omega' \):
\[ \omega' = \frac{M \omega}{M + 2m} \]

### Final Answer
Thus, the new angular speed of the ring after attaching the two particles is:
\[ \omega' = \frac{M \omega}{M + 2m} \]

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