In a vertical circle the minimum or critical velocity at highest point of path will be

Options

A. `sqrt(2rg)`
B. `sqrt(rg)`
C. `romega`
D. zero

Explanation

To find the minimum or critical velocity at the highest point of an object moving in a vertical circle, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces at the Highest Point**:
At the highest point of the circular path, two forces act on the object: the gravitational force (mg, acting downward) and the tension in the string (T, also acting downward if the string is taut).

2. **Applying Centripetal Force Equation**:
The centripetal force required to keep the object moving in a circle is provided by the net force acting towards the center of the circle. Therefore, we can write the equation:
\[ T + mg = \frac{mv^2}{r} \]
where:
- \( m \) is the mass of the object,
- \( v \) is the velocity at the highest point,
- \( r \) is the radius of the circular path.

3. **Finding Minimum Velocity**:
To find the minimum velocity at the highest point, we need to consider the case where the tension \( T \) is zero. This is because the minimum velocity occurs when the object is just about to lose contact with the circular path. Thus, we set \( T = 0 \):
\[ 0 + mg = \frac{mv^2}{r} \]

4. **Simplifying the Equation**:
By simplifying the equation, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ mg = \frac{mv^2}{r} \]
This simplifies to:
\[ g = \frac{v^2}{r} \]

5. **Solving for Velocity**:
Rearranging the equation to solve for \( v \):
\[ v^2 = rg \]
Taking the square root of both sides gives us the critical velocity:
\[ v = \sqrt{rg} \]

### Final Answer:
The minimum or critical velocity at the highest point of the path is:
\[ v = \sqrt{rg} \]

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