A uniform rod of mass m and length / is hinged at upper end. Rod is free to rotate in vertical plane. A bail of mass m moving horizontally with velocity vo collides at lower end of rod perpendicular to it and sticks to it. The minimum velocity of the ball such that combined system just completes the vertical circle will be

Options

A. `2[sqrt(2gl]`
B. `sqrt(2gl)`
C. `2[sqrt(gl)]`
D. `sqrt(gl)`

Answer

To solve the problem step by step, we will analyze the situation involving the collision of a ball with a hinged rod and determine the minimum velocity required for the system to complete a vertical circle.

### Step 1: Understand the System
We have a uniform rod of mass \( m \) and length \( l \) that is hinged at its upper end. A ball of mass \( m \) is moving horizontally with velocity \( v_0 \) and collides with the lower end of the rod, sticking to it.

### Step 2: Apply Conservation of Angular Momentum
Since the hinge can exert forces, we cannot use linear momentum conservation. Instead, we will apply the conservation of angular momentum about the hinge point.

The initial angular momentum of the system before the collision is given by:
\[ L_{\text{initial}} = m v_0 l \]
where \( m \) is the mass of the ball, \( v_0 \) is its velocity, and \( l \) is the distance from the hinge to the point of collision.

After the collision, the moment of inertia \( I \) of the combined system (the rod and the ball) is:
\[ I = \frac{1}{3} m l^2 + m l^2 = \frac{4}{3} m l^2 \]
where \( \frac{1}{3} m l^2 \) is the moment of inertia of the rod about the hinge and \( m l^2 \) is the moment of inertia of the ball treated as a point mass at distance \( l \) from the hinge.

Let \( \omega \) be the angular velocity of the system after the collision. By conservation of angular momentum:
\[ m v_0 l = I \omega \]
Substituting for \( I \):
\[ m v_0 l = \frac{4}{3} m l^2 \omega \]
Cancelling \( m \) and rearranging gives:
\[ \omega = \frac{3 v_0}{4 l} \]

### Step 3: Determine Conditions for Completing a Vertical Circle
For the system to complete a vertical circle, we need to ensure that at the highest point of the circle, the centripetal force is sufficient to keep the system in circular motion. The minimum condition is that the gravitational force provides the necessary centripetal force.

At the highest point, the potential energy change and kinetic energy must be considered. The potential energy change for the center of mass of the rod (which moves from \( -\frac{l}{2} \) to \( +\frac{l}{2} \)) is:
\[ \Delta U_{\text{rod}} = mg \cdot l \]
For the ball, which moves from \( -l \) to \( +l \):
\[ \Delta U_{\text{ball}} = mg \cdot 2l \]
Thus, the total potential energy change is:
\[ \Delta U = mg \cdot l + mg \cdot 2l = 3mgl \]

### Step 4: Apply Conservation of Energy
At the bottom, the total mechanical energy is purely kinetic:
\[ \text{K.E.}_{\text{initial}} = \frac{1}{2} I \omega^2 \]
At the top, the energy is a combination of kinetic and potential energy:
\[ \text{K.E.}_{\text{final}} + \text{P.E.}_{\text{final}} = \frac{1}{2} I \omega^2 - 3mgl \]
Setting the initial kinetic energy equal to the final energy gives:
\[ \frac{1}{2} I \omega^2 = 3mgl \]
Substituting \( I \) and \( \omega \):
\[ \frac{1}{2} \cdot \frac{4}{3} m l^2 \cdot \left(\frac{3 v_0}{4 l}\right)^2 = 3mgl \]
Simplifying:
\[ \frac{2}{3} m l^2 \cdot \frac{9 v_0^2}{16 l^2} = 3mgl \]
Cancelling \( m \) and \( l^2 \):
\[ \frac{2 \cdot 9 v_0^2}{48} = 3g \]
\[ \frac{3 v_0^2}{8} = 3g \]
\[ v_0^2 = 8g \]
\[ v_0 = 2\sqrt{2g} \]

### Final Answer
Thus, the minimum velocity \( v_0 \) required for the ball is:
\[ v_0 = 2\sqrt{2g l} \]