A body of mass m is moving in a circle of radius r with a constant speed v, The force on the body is `(mv^(2))/(r )` and is directed towards the centre what is the work done by the from in moving the body over half the circumference of the circle?

Options

A. `(mv^2)/(pir^2)`
B. zero
C. `(mv^2)/(r^2)`
D. `(pir^2)/(mv^2)`

Explanation

To solve the problem, we need to determine the work done by the centripetal force on a body of mass \( m \) moving in a circle of radius \( r \) with a constant speed \( v \) as it moves over half the circumference of the circle.

### Step-by-Step Solution:

1. **Understanding the Forces Involved**:
- The body is moving in a circular path, and the force acting on it is the centripetal force, given by \( F = \frac{mv^2}{r} \).
- This force is always directed towards the center of the circle.

2. **Identifying the Direction of Motion**:
- As the body moves along the circular path, its displacement is tangential to the circle at any point.
- When the body moves from one point on the circle to the opposite point (half the circumference), the displacement is tangential to the circle.

3. **Calculating Work Done**:
- Work done \( W \) by a force is defined as:
\[ W = \int \vec{F} \cdot d\vec{r} \]
- Here, \( \vec{F} \) is the centripetal force directed towards the center, and \( d\vec{r} \) is the displacement vector along the path of motion.

4. **Angle Between Force and Displacement**:
- The angle \( \theta \) between the centripetal force (which is radial) and the displacement (which is tangential) is \( 90^\circ \).
- Therefore, the dot product \( \vec{F} \cdot d\vec{r} \) can be expressed as:
\[ W = F \cdot dr \cdot \cos(\theta) = F \cdot dr \cdot \cos(90^\circ) \]
- Since \( \cos(90^\circ) = 0 \), we have:
\[ W = F \cdot dr \cdot 0 = 0 \]

5. **Conclusion**:
- The work done by the centripetal force in moving the body over half the circumference of the circle is:
\[ W = 0 \]

### Final Answer:
The work done by the force in moving the body over half the circumference of the circle is **0**.