A circular disc of radius R rolls without slipping along the horizontal surface with constant velocity `v_0` . We consider a point A on the surface of the disc. Then, the acceleration of point A is

Options

A. Constant in magnitude
B. Constant in direction
C. Constant in magnitude as well as direction
D. None of the above

Answer

To solve the problem, we need to analyze the motion of a point A on the surface of a circular disc that rolls without slipping along a horizontal surface with a constant velocity \( v_0 \).

### Step-by-Step Solution:

1. **Understanding the Motion of the Disc**:
- The disc rolls without slipping, which means that the point of contact with the ground (let's call it point P) is momentarily at rest.
- The disc has a radius \( R \) and rolls with a constant translational velocity \( v_0 \).

2. **Relating Translational and Rotational Motion**:
- The angular velocity \( \omega \) of the disc is related to the translational velocity \( v_0 \) by the equation:
\[ v_0 = R \omega \]
- Since the disc rolls without slipping, the point A on the surface of the disc will have a velocity that is a combination of its translational motion and its rotational motion about point P.

3. **Analyzing the Point A**:
- Point A is located on the surface of the disc. As the disc rolls, point A will move in a circular path about point P.
- The acceleration of point A can be divided into two components:
- **Tangential Acceleration**: This is due to the change in the speed of point A along the circular path. Since the disc rolls with a constant velocity \( v_0 \), the tangential acceleration is zero.
- **Centripetal Acceleration**: This is due to the circular motion of point A around point P. The centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{R} \]
- Here, \( v \) is the linear speed of point A, which is equal to \( v_0 \) since point A is moving with the disc.

4. **Calculating the Centripetal Acceleration**:
- Substituting \( v = v_0 \) into the centripetal acceleration formula:
\[ a_c = \frac{v_0^2}{R} \]

5. **Conclusion**:
- Since there is no tangential acceleration (the speed is constant), the total acceleration of point A is purely centripetal:
\[ a_A = \frac{v_0^2}{R} \]

### Final Answer:
The acceleration of point A on the surface of the disc is:
\[ a_A = \frac{v_0^2}{R} \]