A uniform ring of mass m and radius R is released from top of an inclined plane. The plane makes an angle `theta` with horizontal. The cofficent of friction between the ring and plane is `mu`. Initially, the point of contact of ring and plane is P. Angular momentum of ring about an axis passing from point P and perpendicular to plane of motion as a function of time t is

Options

A. `mgR(sintheta)t-mumgR(costheta)t`
B. `mgR(sintheta)t`
C. `mgR(sintheta)t+mumgR(costheta)t`
D. `mgR(1-mu^(2))(sintheta)t`

Answer

To solve the problem of finding the angular momentum of a uniform ring about an axis passing through the point of contact with an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Ring**:
- The forces acting on the ring include:
- Gravitational force \( mg \) acting downward.
- Normal force \( N \) acting perpendicular to the inclined plane.
- Frictional force \( f \) acting parallel to the inclined plane.

2. **Resolve the Gravitational Force**:
- The gravitational force can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos(\theta) \)
- Parallel to the inclined plane: \( mg \sin(\theta) \)

3. **Determine the Torque about Point P**:
- The torque \( \tau \) about point P due to the gravitational force component \( mg \sin(\theta) \) is calculated as:
\[ \tau = r \cdot F = R \cdot (mg \sin(\theta)) \]
- Here, \( R \) is the radius of the ring and \( F \) is the force causing the torque.

4. **Calculate the Angular Impulse**:
- The angular impulse is given by the product of torque and time \( t \):
\[ \text{Angular Impulse} = \tau \cdot t = (mg R \sin(\theta)) \cdot t \]

5. **Relate Angular Momentum to Angular Impulse**:
- The change in angular momentum \( \Delta L \) is equal to the angular impulse:
\[ L_f - L_i = \tau \cdot t \]
- Since the initial angular momentum \( L_i \) is zero (the ring starts from rest), we have:
\[ L_f = mg R \sin(\theta) \cdot t \]

6. **Final Expression for Angular Momentum**:
- Therefore, the final expression for the angular momentum of the ring about the point P as a function of time \( t \) is:
\[ L_f = mg R \sin(\theta) \cdot t \]

### Final Answer:
The angular momentum of the ring about an axis passing through point P and perpendicular to the plane of motion as a function of time \( t \) is:
\[ L_f = mg R \sin(\theta) \cdot t \]