A particle of mass 'm' is attached to the rim of a uniform disc of mass 'm' and radius R. The disc is rolling wihtout slipping on a stationery horizontal surface as shown in the figure. At a particular instant, the particle is at the topmost position and centre of the disc has speed `v_(0)` amd its angular speed is `omega`. Choose the correct option (s).

Options

A. `v_(0)=omegaR`
B. kinetic energy of the system is `(11)/(4) mv_(0)^(2)`.
C. speed of point mass m is less than `2v_(0)`
D. `|v_(C)-v_(B)|=|v_(B)-v_(D)|`

Answer

To solve the problem, we need to analyze the motion of the disc and the particle attached to it. Let's break down the steps:

### Step 1: Understand the System
We have a uniform disc of mass 'm' and radius 'R' rolling without slipping on a horizontal surface. A particle of mass 'm' is attached to the rim of the disc at the topmost position. The center of the disc has a linear speed \( v_0 \) and an angular speed \( \omega \).

### Step 2: Relate Linear and Angular Speeds
For a disc rolling without slipping, the relationship between the linear speed \( v_0 \) of the center of mass and the angular speed \( \omega \) is given by:
\[ v_0 = \omega R \]

### Step 3: Velocity of the Particle
At the topmost position, the velocity of the particle can be calculated as the sum of the linear velocity of the center of the disc and the tangential velocity due to the rotation of the disc. The velocity of the particle \( v_A \) at the topmost position is:
\[ v_A = v_0 + \omega R \]
Substituting \( v_0 = \omega R \):
\[ v_A = \omega R + \omega R = 2\omega R \]
Thus, the velocity of the particle at the topmost position is \( 2v_0 \).

### Step 4: Kinetic Energy Calculation
The total kinetic energy \( K \) of the system consists of the kinetic energy of the disc and the kinetic energy of the particle:
- Kinetic energy of the disc:
\[ K_{disc} = \frac{1}{2} I \omega^2 \]
where \( I \) (moment of inertia of the disc) is \( \frac{1}{2} m R^2 \):
\[ K_{disc} = \frac{1}{2} \left(\frac{1}{2} m R^2\right) \omega^2 = \frac{1}{4} m R^2 \omega^2 \]
Using \( v_0 = \omega R \):
\[ K_{disc} = \frac{1}{4} m \left(\frac{v_0^2}{R^2}\right) R^2 = \frac{1}{4} m v_0^2 \]

- Kinetic energy of the particle:
\[ K_{particle} = \frac{1}{2} m v_A^2 = \frac{1}{2} m (2v_0)^2 = 2 m v_0^2 \]

Thus, the total kinetic energy is:
\[ K = K_{disc} + K_{particle} = \frac{1}{4} m v_0^2 + 2 m v_0^2 = \frac{1}{4} m v_0^2 + \frac{8}{4} m v_0^2 = \frac{9}{4} m v_0^2 \]

### Step 5: Conclusion
Now we can summarize the findings:
1. The relationship between linear and angular speeds is \( v_0 = \omega R \).
2. The velocity of the particle at the topmost position is \( 2v_0 \).
3. The total kinetic energy of the system is \( \frac{9}{4} m v_0^2 \).