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

Options

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

Explanation

To solve the problem, we will apply the principle of conservation of angular momentum. Here are the step-by-step calculations:

### Step 1: Understand the Initial Conditions
Initially, we have a thin smooth rod of mass \( M \) and length \( L \) rotating about its center with an angular speed \( \omega_0 \). Two beads 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. Since the beads are at the center, their contribution to the angular momentum is zero.

The moment of inertia \( I_{rod} \) of the rod about its center is given by:
\[ I_{rod} = \frac{1}{12} M L^2 \]
Thus, the initial angular momentum is:
\[ L_i = I_{rod} \cdot \omega_0 = \frac{1}{12} M L^2 \cdot \omega_0 \]

### Step 3: Final Conditions
When the beads slide to the ends of the rod, they are at a distance of \( \frac{L}{2} \) from the axis of rotation. The final moment of inertia \( I_f \) of the system (rod + beads) is:
\[ I_f = I_{rod} + 2 \cdot m \left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + 2 \cdot m \cdot \frac{L^2}{4} = \frac{1}{12} M L^2 + \frac{1}{2} m L^2 \]

### Step 4: Calculate Final Angular Momentum
The final angular momentum \( L_f \) can be expressed as:
\[ L_f = I_f \cdot \omega \]
where \( \omega \) is the new angular speed we want to find.

### Step 5: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions we derived:
\[ \frac{1}{12} M L^2 \cdot \omega_0 = \left(\frac{1}{12} M L^2 + \frac{1}{2} m L^2\right) \cdot \omega \]

### Step 6: Simplify the Equation
We can cancel \( L^2 \) from both sides (assuming \( L \neq 0 \)):
\[ \frac{1}{12} M \cdot \omega_0 = \left(\frac{1}{12} M + \frac{1}{2} m\right) \cdot \omega \]

### Step 7: Solve for \( \omega \)
Rearranging gives:
\[ \omega = \frac{\frac{1}{12} M \cdot \omega_0}{\frac{1}{12} M + \frac{1}{2} m} \]
To simplify further, we can find a common denominator:
\[ \omega = \frac{M \cdot \omega_0}{M + 6m} \]

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

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