A ring rolls without slipping on the ground. Its centre `C` moves with a constant speed `u.P` is any point on the ring. The speed of `P` with respect to the ground is `v`.

Options

A. `0levle2u`
B. `v=u,` if `CP` is horizontal
C. `v=u`, if `CP` makes an angle of `30^@` with the horizontal and `P` is below the horizontal level of `C`.
D. `v=sqrt(2)u,` if `CP` is horizontal

Explanation

To solve the problem, we need to analyze the motion of a ring rolling without slipping and determine the speed of a point \( P \) on the ring with respect to the ground. Let's break down the steps involved in finding the speed \( v \) of point \( P \).

### Step-by-Step Solution:

1. **Understanding the Motion of the Ring**:
- The ring rolls without slipping on the ground, meaning that the point of contact with the ground has zero velocity relative to the ground.
- The center of the ring \( C \) moves with a constant speed \( u \).

2. **Identifying the Speed Components**:
- For any point \( P \) on the ring, the total velocity \( v_P \) can be expressed as:
\[ v_P = v_{P, \text{cm}} + v_{\text{cm}} \]
where \( v_{P, \text{cm}} \) is the velocity of point \( P \) relative to the center of mass \( C \), and \( v_{\text{cm}} \) is the velocity of the center of mass with respect to the ground, which is \( u \).

3. **Analyzing the Point of Contact**:
- The point of contact between the ring and the ground has zero velocity. Therefore, if we consider the ring's rotation, the velocity of the point at the bottom (point of contact) is zero.

4. **Calculating the Velocity of Point \( P \)**:
- The speed of point \( P \) will depend on its position on the ring.
- If \( P \) is at the top of the ring, it will have both the translational speed of the center \( u \) and the tangential speed due to rotation \( \omega r \) (where \( \omega \) is the angular velocity and \( r \) is the radius of the ring). Since \( u = \omega r \), the total speed at the top will be:
\[ v_{\text{max}} = u + u = 2u \]
- If \( P \) is at the bottom (point of contact), the speed will be:
\[ v_{\text{min}} = u - u = 0 \]

5. **Considering Other Positions**:
- For point \( P \) at a horizontal position (to the right or left of \( C \)), it will have a resultant speed of:
\[ v = \sqrt{u^2 + u^2} = u\sqrt{2} \]
- If point \( P \) makes an angle \( \theta \) with the horizontal, we can resolve the velocities into components and use trigonometric relations to find the resultant speed.

6. **Conclusion**:
- The maximum speed \( v \) of point \( P \) can vary from \( 0 \) to \( 2u \) depending on its position on the ring. The specific values depend on the angle or position of point \( P \) relative to the center \( C \).