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 force in moving the body over half the circumference of the circle?

Options

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

Explanation

To solve the problem, we need to determine the work done by the centripetal force on a body moving in a circular path over half the circumference of the circle.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A body of mass \( m \) is moving in a circle of radius \( r \) with a constant speed \( v \). The force acting on the body is the centripetal force, which is given by the formula:
\[ F = \frac{mv^2}{r} \]
- This force is always directed towards the center of the circle.

2. **Displacement Over Half the Circumference**:
- The total circumference of the circle is given by:
\[ C = 2\pi r \]
- Therefore, half the circumference is:
\[ \text{Half Circumference} = \frac{C}{2} = \pi r \]

3. **Direction of Force and Displacement**:
- As the body moves along the circular path, the displacement \( ds \) at any instant is tangential to the circle, while the centripetal force \( F \) is directed radially inward (towards the center).
- The angle \( \theta \) between the force vector \( F \) and the displacement vector \( ds \) is always \( 90^\circ \) during circular motion.

4. **Calculating Work Done**:
- The work done \( W \) by a force is calculated using the formula:
\[ W = F \cdot d = F \cdot ds \cdot \cos(\theta) \]
- Since \( \theta = 90^\circ \), we have:
\[ \cos(90^\circ) = 0 \]
- Therefore, the work done in moving the body over half the circumference is:
\[ W = F \cdot ds \cdot 0 = 0 \]

5. **Conclusion**:
- The total 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 \( 0 \).

---