Moment of inertia of uniform rod of mass M and length L about an axis through its centre and perpendicular to its length is given by `(ML^(2))/(12)`. Now consider one such rod pivoted at its centre, free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass M moving horizontally at a speed v strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be

Options

A. `(v)/L`
B. `(2v)/L`
C. `(3v)/2L`
D. `(6v)/L`

Explanation

To solve the problem, we need to apply the principle of conservation of angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a uniform rod of mass \( M \) and length \( L \) that is pivoted at its center and is initially at rest in a vertical position. A bullet of mass \( M \) is moving horizontally with speed \( v \) and strikes one end of the rod, embedding itself in the rod.

### Step 2: Calculate Initial Angular Momentum
Before the collision, the rod is at rest, so its angular momentum is zero. The bullet has angular momentum about the pivot point (the center of the rod). The distance from the pivot to the point where the bullet strikes is \( \frac{L}{2} \).

The angular momentum \( L_i \) of the bullet before the collision is given by:
\[ L_i = M \cdot v \cdot \frac{L}{2} \]

### Step 3: Calculate Final Angular Momentum
After the collision, the bullet and the rod will rotate together about the pivot point with a common angular velocity \( \omega \). The total moment of inertia \( I_f \) of the system after the collision consists of the moment of inertia of the rod and the moment of inertia of the bullet treated as a point mass at a distance \( \frac{L}{2} \) from the pivot.

The moment of inertia of the rod about its center is:
\[ I_{\text{rod}} = \frac{ML^2}{12} \]

The moment of inertia of the bullet (considered as a point mass) is:
\[ I_{\text{bullet}} = M \left(\frac{L}{2}\right)^2 = \frac{ML^2}{4} \]

Thus, the total moment of inertia after the collision is:
\[ I_f = I_{\text{rod}} + I_{\text{bullet}} = \frac{ML^2}{12} + \frac{ML^2}{4} \]

### Step 4: Simplify the Total Moment of Inertia
To combine the terms, we need a common denominator:
\[ I_f = \frac{ML^2}{12} + \frac{3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3} \]

### Step 5: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
where \( L_f \) is the final angular momentum after the collision:
\[ L_f = I_f \cdot \omega \]

Substituting the values we have:
\[ M \cdot v \cdot \frac{L}{2} = \left(\frac{ML^2}{3}\right) \cdot \omega \]

### Step 6: Solve for Angular Velocity \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{M \cdot v \cdot \frac{L}{2}}{\frac{ML^2}{3}} = \frac{3v}{2L} \]

### Final Result
Thus, the angular velocity of the rod just after the collision is:
\[ \omega = \frac{3v}{2L} \]