A uniform disc of mass m and radius R is thrown on horizontal law 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_0^2 / (9g)`
B. `5v _0^2 /(9g)`
C. ` v_0^2 / (3g)`
D. `3v_0^2 / (8g)`

Answer

To solve the problem of a uniform disc of mass \( m \) and radius \( R \) that is thrown on a horizontal surface with an initial speed \( V_0 \) while sliding (not rolling), we need to determine the distance it travels until it starts pure rolling. The coefficient of friction between the disc and the surface is given as \( \mu = 0.5 \).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- The disc is sliding with an initial speed \( V_0 \).
- The initial angular velocity \( \omega_0 = 0 \) (since it is not rolling).

2. **Frictional Force:**
- The frictional force \( f \) acting on the disc is given by:
\[ f = \mu mg = 0.5mg \]
- This frictional force will cause a linear deceleration \( a \) of the disc:
\[ a = \frac{f}{m} = \frac{0.5mg}{m} = 0.5g \]

3. **Linear Velocity After Time \( t \):**
- The linear velocity \( V \) of the disc after time \( t \) can be expressed as:
\[ V = V_0 - at = V_0 - 0.5gt \]

4. **Angular Acceleration:**
- The torque \( \tau \) due to friction about the center of mass is:
\[ \tau = f \cdot R = 0.5mg \cdot R \]
- The moment of inertia \( I \) of the disc is:
\[ I = \frac{1}{2}mR^2 \]
- The angular acceleration \( \alpha \) is given by:
\[ \tau = I \alpha \implies 0.5mgR = \frac{1}{2}mR^2 \alpha \implies \alpha = \frac{g}{R} \]

5. **Angular Velocity After Time \( t \):**
- The angular velocity \( \omega \) after time \( t \) is:
\[ \omega = \omega_0 + \alpha t = 0 + \frac{g}{R}t = \frac{gt}{R} \]

6. **Condition for Pure Rolling:**
- Pure rolling starts when the condition \( V = \omega R \) is satisfied:
\[ V_0 - 0.5gt = \left(\frac{gt}{R}\right) R \]
- Simplifying gives:
\[ V_0 - 0.5gt = gt \implies V_0 = 1.5gt \]

7. **Solving for Time \( t \):**
- Rearranging gives:
\[ t = \frac{V_0}{1.5g} = \frac{2V_0}{3g} \]

8. **Distance Traveled Until Pure Rolling Starts:**
- The distance \( s \) traveled by the disc during this time can be calculated using the equation of motion:
\[ s = V_0 t - \frac{1}{2} a t^2 \]
- Substituting for \( t \) and \( a \):
\[ s = V_0 \left(\frac{2V_0}{3g}\right) - \frac{1}{2} (0.5g) \left(\frac{2V_0}{3g}\right)^2 \]
- Simplifying:
\[ s = \frac{2V_0^2}{3g} - \frac{1}{2} \cdot 0.5g \cdot \frac{4V_0^2}{9g^2} \]
\[ s = \frac{2V_0^2}{3g} - \frac{2V_0^2}{9g} \]
\[ s = \frac{6V_0^2}{9g} - \frac{2V_0^2}{9g} = \frac{4V_0^2}{9g} \]

### Final Result:
The distance traveled by the disc until it starts pure rolling is:
\[ s = \frac{4V_0^2}{9g} \]