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 solve the problem of the angular momentum of a sphere released on a smooth inclined plane, we will follow these steps:

### Step 1: Understand the System
We have a sphere released from the top of a smooth inclined plane. The plane is smooth, which means there is no friction acting on the sphere as it rolls down.

### Step 2: Identify Forces Acting on the Sphere
When the sphere is on the inclined plane, the forces acting on it are:
- The gravitational force (weight) \( mg \), acting vertically downwards.
- The normal force \( N \), acting perpendicular to the surface of the inclined plane.

### Step 3: Break Down the Weight into Components
The weight can be resolved into two components:
- A component parallel to the incline: \( mg \sin \theta \)
- A component perpendicular to the incline: \( mg \cos \theta \)

### Step 4: Analyze Torque
To determine if angular momentum is conserved, we need to analyze the torque about specific points. The condition for conservation of angular momentum is that the net torque about a point must be zero.

1. **About the Center of Mass**:
- The forces \( mg \sin \theta \) and \( mg \cos \theta \) act through the center of mass of the sphere.
- Since the torque due to these forces about the center of mass is zero, angular momentum is conserved about the center of mass.

2. **About the Point of Contact**:
- The torque due to the weight component \( mg \sin \theta \) will not be zero because it does not pass through the point of contact.
- Therefore, angular momentum is not conserved about the point of contact.

3. **About Any Point Parallel to the Inclined Plane**:
- For any point along a line parallel to the inclined plane and passing through the center of the sphere, the torque due to \( mg \sin \theta \) will be zero, as it acts through that line.
- However, the torque due to \( mg \cos \theta \) will still be present, which means angular momentum is not conserved about all such points.

### Step 5: Conclusion
The angular momentum of the sphere is conserved about its center of mass. Therefore, the correct answer to the question is:

**The angular momentum of the sphere is conserved about its center of mass.**

### Final Answer:
**Conserved about the center of the sphere only.**