A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion ?

Options

A. `(3u)/(5)`
B. `(2u)/(5)`
C. `(5u)/(7)`
D. `(2u)/(7)`

Answer

To find the speed of a heavy solid sphere when it starts pure rolling motion after being thrown on a rough horizontal surface with an initial velocity \( u \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The sphere is thrown with an initial velocity \( u \) and is not rolling initially.

2. **Understand the Forces Acting on the Sphere**:
- When the sphere slides on the rough surface, friction acts in the opposite direction to the motion. This frictional force will slow down the linear motion of the sphere and will also cause it to start rotating.

3. **Define Linear and Angular Motion**:
- Let \( v \) be the linear velocity of the sphere at the moment it starts pure rolling.
- Let \( \omega \) be the angular velocity of the sphere at that moment.

4. **Use the Relationship Between Linear and Angular Velocity**:
- For pure rolling motion, the relationship between linear velocity \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
where \( r \) is the radius of the sphere.

5. **Apply Conservation of Angular Momentum**:
- Since the net torque about the point of contact is zero when the sphere starts rolling, we can apply the conservation of angular momentum.
- The initial angular momentum \( L_i \) about the point of contact (point A) is:
\[ L_i = M u r \]
- The final angular momentum \( L_f \) about point A when it starts rolling is:
\[ L_f = I_A \omega \]
where \( I_A \) is the moment of inertia about point A.

6. **Calculate the Moment of Inertia**:
- The moment of inertia \( I \) of a solid sphere about its center is:
\[ I = \frac{2}{5} M r^2 \]
- Using the parallel axis theorem, the moment of inertia about point A is:
\[ I_A = I + M r^2 = \frac{2}{5} M r^2 + M r^2 = \frac{7}{5} M r^2 \]

7. **Set Up the Equation from Conservation of Angular Momentum**:
- Equating the initial and final angular momentum:
\[ M u r = \frac{7}{5} M r^2 \omega \]
- Substitute \( \omega \) with \( \frac{v}{r} \):
\[ M u r = \frac{7}{5} M r^2 \left(\frac{v}{r}\right) \]

8. **Simplify the Equation**:
- Cancel \( M \) and \( r \) (assuming \( r \neq 0 \)):
\[ u = \frac{7}{5} v \]

9. **Solve for \( v \)**:
- Rearranging gives:
\[ v = \frac{5}{7} u \]

### Final Answer:
The speed of the sphere when it starts pure rolling motion is:
\[ v = \frac{5}{7} u \]