The ride is circus rides along a circular track in a vertical plane. The minimum velocity at the highest point of the track will be

Options

A. `sqrt((2g R))`
B. `sqrt((5 g R))`
C. `sqrt((3g R))`
D. `sqrt((g R))`

Answer

To find the minimum velocity at the highest point of a circular track in a vertical plane, we can analyze the forces acting on the object at that point.

### Step-by-Step Solution:

1. **Identify Forces at the Highest Point**:
At the highest point of the circular track, the forces acting on the object are:
- The gravitational force (weight) acting downwards: \( F_g = mg \)
- The centripetal force required to keep the object moving in a circular path, which is provided by the gravitational force at this point.

2. **Centripetal Force Requirement**:
The centripetal force needed to keep the object moving in a circular path is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
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. **Setting Forces Equal**:
At the highest point, the gravitational force must provide the necessary centripetal force. Therefore, we set the two forces equal:
\[ mg = \frac{mv^2}{r} \]

4. **Cancel Mass**:
Since \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)):
\[ g = \frac{v^2}{r} \]

5. **Rearranging the Equation**:
To find the minimum velocity \( v \) at the highest point, we rearrange the equation:
\[ v^2 = gr \]
Taking the square root of both sides gives:
\[ v = \sqrt{gr} \]

### Final Answer:
The minimum velocity at the highest point of the track is:
\[ v = \sqrt{gr} \]