A thin smooth rod of length L and mass M is rotating freely with angular speed `omega_(0)` about an axis perpendicular ot the rod and passing through its centre. Two beads of mass m and negligible sizer are at the center of the rod intially. The beads are free to slide along the rod. the angular speed of the system, when the beads reach the oppsite ends of rod, will be :

Options

A. `(M omega_(0))/(M+m)`
B. `(Momega_(0))/(M+3m)`
C. `(Momega_(0))/(M+6m)`
D. `(Momega_(0))/(M+2m)`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Setup
We have a thin smooth rod of length \( L \) and mass \( M \) rotating with an initial angular speed \( \omega_0 \). Two beads, each of mass \( m \), are located at the center of the rod.

### Step 2: Calculate Initial Angular Momentum
The initial angular momentum \( L_i \) of the system can be calculated as the sum of the angular momentum of the rod and the beads.

1. **Angular Momentum of the Rod**:
The moment of inertia \( I \) of the rod about its center is given by:
\[ I_{rod} = \frac{1}{12} M L^2 \]
Therefore, the angular momentum of the rod is:
\[ L_{rod} = I_{rod} \cdot \omega_0 = \frac{1}{12} M L^2 \cdot \omega_0 \]

2. **Angular Momentum of the Beads**:
Initially, the beads are at the center of the rod, so their moment of inertia is zero, and thus their angular momentum is:
\[ L_{beads} = 0 \]

So, the total initial angular momentum \( L_i \) is:
\[ L_i = L_{rod} + L_{beads} = \frac{1}{12} M L^2 \cdot \omega_0 \]

### Step 3: Calculate Final Angular Momentum
When the beads slide to the opposite ends of the rod, their positions change. The final angular momentum \( L_f \) can be calculated as follows:

1. **Final Moment of Inertia**:
When the beads are at the ends of the rod, the moment of inertia of each bead is:
\[ I_{bead} = m \left(\frac{L}{2}\right)^2 = \frac{m L^2}{4} \]
Since there are two beads:
\[ I_{total\_beads} = 2 \cdot \frac{m L^2}{4} = \frac{m L^2}{2} \]

2. **Total Moment of Inertia of the System**:
The total moment of inertia of the system when the beads are at the ends is:
\[ I_f = I_{rod} + I_{total\_beads} = \frac{1}{12} M L^2 + \frac{m L^2}{2} \]

3. **Final Angular Momentum**:
The final angular momentum \( L_f \) can be expressed as:
\[ L_f = I_f \cdot \omega \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions we derived:
\[ \frac{1}{12} M L^2 \cdot \omega_0 = \left(\frac{1}{12} M L^2 + \frac{m L^2}{2}\right) \cdot \omega \]

### Step 5: Solve for Final Angular Speed \( \omega \)
1. Cancel \( L^2 \) from both sides:
\[ \frac{1}{12} M \cdot \omega_0 = \left(\frac{1}{12} M + \frac{m}{2}\right) \cdot \omega \]

2. Rearranging gives:
\[ \omega = \frac{\frac{1}{12} M \cdot \omega_0}{\frac{1}{12} M + \frac{m}{2}} \]

3. Simplifying further:
\[ \omega = \frac{M \cdot \omega_0}{M + 6m} \]

### Final Answer
The angular speed of the system when the beads reach the opposite ends of the rod is:
\[ \omega = \frac{M \cdot \omega_0}{M + 6m} \]