A particle undergoes uniform circular motion. About which point on the plane of the circle, will the angular momentum of the particle remain conserved?

Options

A. a. Centre of the circle
B. b. On the circumference of the circle
C. c. Inside the circle
D. d. Outside the circle

Explanation

To determine about which point on the plane of the circle the angular momentum of a particle undergoing uniform circular motion remains conserved, we can analyze the torque acting on the particle with respect to various points.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
Angular momentum (L) of a particle is given by the formula:
\[ L = r \times p \]
where \( r \) is the position vector from the point of reference to the particle, and \( p \) is the linear momentum of the particle. For a particle of mass \( m \) moving with velocity \( v \), \( p = mv \).

2. **Torque and Angular Momentum Relationship**:
The relationship between torque (\( \tau \)) and angular momentum is given by:
\[ \tau = \frac{dL}{dt} \]
If the torque is zero, then the angular momentum is conserved (\( \frac{dL}{dt} = 0 \)).

3. **Analyzing Torque about Different Points**:
- **About the Center of the Circle (Point O)**:
- The centripetal force (\( F_c \)) acts towards the center of the circle.
- The position vector \( r \) from the center to the particle is directed outward.
- The angle between \( F_c \) and \( r \) is 180 degrees.
- Therefore, the torque \( \tau \) about the center is:
\[ \tau = F_c \times r = F_c \cdot r \cdot \sin(180^\circ) = 0 \]
- Since the torque is zero, the angular momentum is conserved about the center of the circle.

- **About a Point on the Circumference**:
- Here, the position vector \( r \) from the point on the circumference to the particle forms a 90-degree angle with the centripetal force.
- The torque is:
\[ \tau = F_c \times r = F_c \cdot r \cdot \sin(90^\circ) = F_c \cdot r \neq 0 \]
- Since the torque is not zero, the angular momentum is not conserved about a point on the circumference.

- **About a Point Inside the Circle**:
- The angle between the centripetal force and the position vector is less than 90 degrees.
- The torque is:
\[ \tau = F_c \times r = F_c \cdot r \cdot \sin(\theta) \quad (\theta < 90^\circ) \]
- This torque is also not zero, indicating that angular momentum is not conserved about a point inside the circle.

- **About a Point Outside the Circle**:
- The angle between the centripetal force and the position vector is greater than 90 degrees.
- The torque is:
\[ \tau = F_c \times r = F_c \cdot r \cdot \sin(\theta_1) \quad (\theta_1 > 90^\circ) \]
- This torque is again not zero, so angular momentum is not conserved about a point outside the circle.

4. **Conclusion**:
The angular momentum of the particle undergoing uniform circular motion remains conserved only about the center of the circle.

### Final Answer:
The angular momentum of the particle remains conserved about the **center of the circle**.