A uniform disc of mass m and radius R is thrown on horizontal lawn in such a way that it initially slides with speed `V_0` without rolling. The distance travelled by the disc till it starts pure rolling is (Coefficient friction between the contact is 0.5)

Options

A. ` v^2_0 / 9g`
B. `5 v^2_0 / 9g`
C. ` v^2_0 / 3g`
D. `3v^2_0 / 8g`

Answer

To solve the problem, we need to determine the distance traveled by a uniform disc until it starts pure rolling after being thrown with an initial speed \( V_0 \). The coefficient of friction between the disc and the surface is given as \( \mu = 0.5 \).

### Step-by-Step Solution:

1. **Understanding the System**:
- The disc is initially sliding with speed \( V_0 \) and not rolling.
- As it slides, friction will act on the disc, causing it to decelerate linearly and accelerate rotationally until it reaches pure rolling condition.

2. **Friction Force**:
- The friction force \( f \) acting on the disc can be calculated using the formula:
\[ f = \mu m g \]
- Here, \( g \) is the acceleration due to gravity.

3. **Linear Deceleration**:
- The linear deceleration \( a \) of the disc due to friction is given by:
\[ a = \frac{f}{m} = \mu g \]
- Substituting \( \mu = 0.5 \):
\[ a = 0.5 g \]

4. **Angular Acceleration**:
- The torque \( \tau \) caused by the friction force about the center of the disc is:
\[ \tau = f \cdot R = \mu m g R \]
- The angular acceleration \( \alpha \) can be found using the moment of inertia \( I \) of the disc:
\[ I = \frac{1}{2} m R^2 \]
- Thus, the angular acceleration is:
\[ \alpha = \frac{\tau}{I} = \frac{\mu m g R}{\frac{1}{2} m R^2} = \frac{2 \mu g}{R} \]

5. **Relation Between Linear and Angular Velocity**:
- The condition for pure rolling is given by:
\[ V = \omega R \]
- Initially, the linear velocity \( V_0 \) is reduced to \( V \) and the angular velocity \( \omega \) increases from 0.

6. **Using Kinematic Equations**:
- We can use the kinematic equation:
\[ V^2 = V_0^2 - 2a s \]
- Here, \( V = \omega R \) and \( \omega = \frac{V}{R} \).
- Since \( \alpha = \frac{d\omega}{dt} \) and \( a = \frac{dV}{dt} \), we can relate \( V \) and \( \omega \) through the distance \( s \).

7. **Finding the Distance \( s \)**:
- We know that:
\[ V = \frac{2}{3} V_0 \quad \text{(from conservation of angular momentum)} \]
- Substituting this into the kinematic equation:
\[ \left(\frac{2}{3} V_0\right)^2 = V_0^2 - 2 \cdot (0.5 g) \cdot s \]
- Simplifying gives:
\[ \frac{4}{9} V_0^2 = V_0^2 - g s \]
- Rearranging leads to:
\[ g s = V_0^2 - \frac{4}{9} V_0^2 = \frac{5}{9} V_0^2 \]
- Therefore, the distance \( s \) is:
\[ s = \frac{5 V_0^2}{9g} \]

### Final Answer:
The distance traveled by the disc until it starts pure rolling is:
\[ s = \frac{5 V_0^2}{9g} \]