A heavy disc is thrown on a horizontal surface in such a way that it slides with a speed `V_(0)` intially without rollind . It will start rolling without slipping when its speed is reduced

Options

A. `1/3V_(0)`
B. `2/3V_(0)`
C. `3/5V_(0)`
D. `5/7V_(0)`

Answer

To solve the problem, we need to analyze the motion of a heavy disc that is initially sliding on a horizontal surface without rolling. The disc will eventually start rolling without slipping due to the frictional force acting on it. We will derive the final speed at which the disc starts rolling without slipping.

### Step-by-step Solution:

1. **Initial Conditions**:
- The disc is thrown with an initial speed \( V_0 \) and has no initial angular velocity (i.e., it is not rolling).
- The disc has mass \( m \) and radius \( r \).

2. **Frictional Force**:
- Since the surface is rough, a frictional force will act on the disc. This force will oppose the motion of the disc, reducing its linear speed and increasing its angular speed until it starts rolling without slipping.

3. **Angular Momentum Conservation**:
- We will analyze the angular momentum of the disc about its center of mass.
- Initially, the angular momentum \( L_i \) of the disc can be calculated as:
\[ L_i = m \cdot V_0 \cdot r \]
- This is because the entire mass can be considered to be concentrated at a distance \( r \) from the axis of rotation.

4. **Final Angular Momentum**:
- When the disc starts rolling without slipping, it has both translational and rotational motion.
- The final angular momentum \( L_f \) can be expressed as:
\[ L_f = m \cdot V \cdot r + I \cdot \omega \]
- Here, \( I \) is the moment of inertia of the disc, which is given by \( I = \frac{1}{2} m r^2 \), and \( \omega \) (angular velocity) can be related to the linear velocity \( V \) by:
\[ \omega = \frac{V}{r} \]
- Substituting these into the equation for \( L_f \):
\[ L_f = m \cdot V \cdot r + \left(\frac{1}{2} m r^2\right) \cdot \left(\frac{V}{r}\right) = m \cdot V \cdot r + \frac{1}{2} m V r \]
- Simplifying this gives:
\[ L_f = m \cdot V \cdot r + \frac{1}{2} m V r = \frac{3}{2} m V r \]

5. **Setting Initial and Final Angular Momentum Equal**:
- By the conservation of angular momentum, we set \( L_i = L_f \):
\[ m \cdot V_0 \cdot r = \frac{3}{2} m V \cdot r \]
- Dividing both sides by \( m \cdot r \) (assuming \( m \) and \( r \) are not zero):
\[ V_0 = \frac{3}{2} V \]

6. **Solving for Final Velocity \( V \)**:
- Rearranging the equation gives:
\[ V = \frac{2}{3} V_0 \]

### Final Answer:
The disc will start rolling without slipping when its speed is reduced to:
\[ V = \frac{2}{3} V_0 \]