A solid uniform disc of mass `m` rolls without slipping down a fixed inclined plank with an acceleration a. The frictional force on the disc due to surface of the plane is

Options

A. 2 ma
B. `(3)/(2)` ma
C. ma
D. `(1)/(2)` ma

Answer

To solve the problem of finding the frictional force on a solid uniform disc rolling down an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Disc**:
- The weight of the disc, \( mg \), acts vertically downward.
- This weight can be resolved into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

2. **Set Up the Equations of Motion**:
- The vertical component \( mg \cos \theta \) is balanced by the normal force \( N \).
- The net force acting down the incline is given by:
\[ F_{\text{net}} = mg \sin \theta - F \]
- Here, \( F \) is the frictional force acting up the incline.

3. **Relate Linear Acceleration to Angular Acceleration**:
- Since the disc rolls without slipping, the linear acceleration \( a \) is related to the angular acceleration \( \alpha \) by:
\[ a = R \alpha \]
- Thus, we can express \( \alpha \) as:
\[ \alpha = \frac{a}{R} \]

4. **Apply the Torque Equation**:
- The torque \( \tau \) acting on the disc due to the frictional force is given by:
\[ \tau = F \cdot R \]
- This torque is also equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
- For a solid disc, the moment of inertia \( I \) about its center is:
\[ I = \frac{1}{2} m R^2 \]

5. **Equate Torque Expressions**:
- Setting the two expressions for torque equal gives:
\[ F \cdot R = \frac{1}{2} m R^2 \cdot \frac{a}{R} \]
- Simplifying this, we get:
\[ F \cdot R = \frac{1}{2} m a R \]
- Dividing both sides by \( R \) (assuming \( R \neq 0 \)):
\[ F = \frac{1}{2} m a \]

6. **Conclusion**:
- The frictional force \( F \) acting on the disc due to the surface of the inclined plane is:
\[ F = \frac{1}{2} m a \]

### Final Answer:
The frictional force on the disc due to the surface of the plane is \( \frac{1}{2} m a \).