A sphere is released on a smooth inclined plane from the top. When it moves down, its angular momentum is

Options

A. conserved about every point
B. conserved about the point of contact only
C. conserved about the centre of the sphere only
D. conserved about any point on a line parallel to the inclined plane and passing through the centre of the ball

Explanation

To determine the angular momentum of a sphere released on a smooth inclined plane, we need to analyze the forces acting on the sphere and the conditions under which angular momentum is conserved.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a sphere released from the top of a smooth inclined plane. The plane is smooth, meaning there is no friction acting on the sphere.

2. **Identifying Forces**:
- The forces acting on the sphere are:
- Gravitational force (weight) \( \mathbf{W} = mg \) acting downwards.
- Normal force \( \mathbf{N} \) acting perpendicular to the surface of the inclined plane.

3. **Torque and Angular Momentum**:
- Angular momentum \( \mathbf{L} \) is conserved when the net torque \( \tau \) acting on the system is zero.
- Torque is defined as \( \tau = \mathbf{R} \times \mathbf{F} \), where \( \mathbf{R} \) is the position vector from the point about which we are calculating torque to the point of application of the force \( \mathbf{F} \).

4. **Choosing a Point for Torque Calculation**:
- To find where the angular momentum is conserved, we can analyze the torque about different points.
- If we take the center of mass of the sphere as the reference point, the torque due to the gravitational force will be zero because the line of action of the weight passes through the center of mass.

5. **Torque Calculation at Different Points**:
- If we consider any point along the incline, the gravitational force \( mg \) can be resolved into two components:
- \( mg \sin \theta \) (parallel to the incline)
- \( mg \cos \theta \) (perpendicular to the incline)
- The torque due to the normal force \( \mathbf{N} \) and the component \( mg \cos \theta \) will also be zero at the center of mass since their lines of action pass through this point.

6. **Conclusion on Angular Momentum Conservation**:
- Since the net torque about the center of mass is zero, the angular momentum of the sphere is conserved as it rolls down the incline.
- However, if we consider points other than the center of mass, the torque due to \( mg \sin \theta \) will not be zero, and thus angular momentum will not be conserved at those points.

### Final Answer:
The angular momentum of the sphere is conserved at the center of mass of the sphere as it rolls down the smooth inclined plane.