A particle of mass `m` is fixed to one end of a light rigid rod of length `l` and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be

Options

A. zero
B. `sqrt(gl)`
C. `sqrt(1.5gl)`
D. `sqrt(2gl)`

Explanation

To find the minimum speed of a particle of mass \( m \) at its highest point when it is attached to a rigid rod of length \( l \) and rotates in a vertical circular path, we can analyze the forces acting on the particle at that point.

### Step-by-Step Solution:

1. **Identify the Forces at the Highest Point**:
At the highest point of the circular motion, the forces acting on the particle are:
- The gravitational force acting downward, which is \( mg \) (where \( g \) is the acceleration due to gravity).
- The tension in the rod (which we will denote as \( T \)) acting downward as well.

2. **Apply Newton's Second Law**:
At the highest point, the centripetal force required to keep the particle moving in a circular path is provided by the sum of the gravitational force and the tension in the rod:
\[ \frac{mv^2}{l} = mg + T \]
where \( v \) is the speed of the particle at the highest point.

3. **Determine the Minimum Speed Condition**:
The minimum speed occurs when the tension \( T \) in the rod is zero (i.e., the rod is just taut and not under any tension). Thus, we can set \( T = 0 \):
\[ \frac{mv^2}{l} = mg \]

4. **Solve for the Minimum Speed \( v \)**:
Rearranging the equation gives:
\[ mv^2 = mgl \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ v^2 = gl \]
Taking the square root of both sides, we find:
\[ v = \sqrt{gl} \]

5. **Conclusion**:
Therefore, the minimum speed of the particle at its highest point must be:
\[ v = \sqrt{gl} \]