The figure shows a system consisting of (i) a ring the outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed `omega` and (ii) an inner disc of radius 2R rotating anti clockwise with angular speed `omega//2.` The ring and disc are separted. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of `30^@` with the horizontal. Then with respect to the horizontal surface,

Options

A. The point `O` has a linear velocity `3Romegahati`
B. The point `P` has a linear velocity `11/4 Romegahati+(sqrt(3))/4Romegahatk`
C. The point `P` has a linear velocity `13/4Romegahati+(sqrt(3))/4Romegahatk`
D. The point `P` has a linear velocity `(3-(sqrt(3))/4)Romegahati+1/4Romegahatk`

Explanation

To solve the problem, we need to determine the velocity of point P on the inner disc with respect to the horizontal surface. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Components of the System
We have:
- A ring of outer radius \(3R\) rolling clockwise with angular speed \(\omega\).
- An inner disc of radius \(2R\) rotating anti-clockwise with angular speed \(\frac{\omega}{2}\).
- Point P is located at a distance \(R\) from the origin, making an angle of \(30^\circ\) with the horizontal.

### Step 2: Determine the Linear Velocity of the Ring
The linear velocity \(v\) of the center of the ring can be calculated using the formula:
\[ v = R \cdot \omega \]
Since the radius of the ring is \(3R\), the linear velocity of the center of the ring is:
\[ v_{\text{ring}} = 3R \cdot \omega \]
This velocity acts in the negative x-direction (since it rolls clockwise).

### Step 3: Calculate the Velocity of Point P on the Inner Disc
The point P on the inner disc is at a distance \(R\) from the center of the disc. The angular velocity of the inner disc is \(\frac{\omega}{2}\). The linear velocity of point P due to the rotation of the disc is given by:
\[ v_P = R \cdot \frac{\omega}{2} = \frac{R \omega}{2} \]
Since the disc rotates anti-clockwise, this velocity will have components in both the x and y directions.

### Step 4: Resolve the Velocity of Point P into Components
The angle OP makes with the horizontal is \(30^\circ\). Therefore, we can resolve the velocity of point P into its x and y components:
- The x-component of the velocity:
\[ v_{Px} = \frac{R \omega}{2} \cdot \cos(30^\circ) = \frac{R \omega}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} R \omega}{4} \]
- The y-component of the velocity:
\[ v_{Py} = \frac{R \omega}{2} \cdot \sin(30^\circ) = \frac{R \omega}{2} \cdot \frac{1}{2} = \frac{R \omega}{4} \]

### Step 5: Combine the Velocities
Now we combine the velocities of the ring and the point P:
- The total x-component of the velocity:
\[ v_{total_x} = -3R \omega + \frac{\sqrt{3} R \omega}{4} \]
- The total y-component of the velocity:
\[ v_{total_y} = \frac{R \omega}{4} \]

### Step 6: Final Velocity of Point P
The total velocity of point P with respect to the horizontal surface is:
\[ v_P = \left(-3R \omega + \frac{\sqrt{3} R \omega}{4}\right) \hat{i} + \frac{R \omega}{4} \hat{j} \]
This simplifies to:
\[ v_P = \left(-\frac{12R \omega}{4} + \frac{\sqrt{3} R \omega}{4}\right) \hat{i} + \frac{R \omega}{4} \hat{j} \]
\[ v_P = \left(-\frac{(12 - \sqrt{3})R \omega}{4}\right) \hat{i} + \frac{R \omega}{4} \hat{j} \]

### Conclusion
The final expression gives us the velocity of point P with respect to the horizontal surface.