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 solve the problem of a heavy solid sphere thrown on a horizontal rough surface with an initial velocity \( u \) without rolling, we need to determine its speed when it starts pure rolling motion. Here’s the step-by-step solution:

### Step 1: Understand the System
A heavy solid sphere is thrown on a rough surface. Initially, it slides without rolling. As it moves, friction will act on it, causing it to roll.

### Step 2: Identify Key Concepts
- **Conservation of Linear Momentum**: Since there are no external horizontal forces acting on the sphere, we can apply the conservation of momentum.
- **Moment of Inertia**: The moment of inertia of a solid sphere about its center is given by \( I_c = \frac{2}{5} m r^2 \). When considering the point of contact, we use the parallel axis theorem to find \( I_p = I_c + md^2 \), where \( d = r \) (the distance from the center to the point of contact).

### Step 3: Calculate the Moment of Inertia
Using the parallel axis theorem:
\[ I_p = I_c + m r^2 = \frac{2}{5} m r^2 + m r^2 = \frac{2}{5} m r^2 + \frac{5}{5} m r^2 = \frac{7}{5} m r^2 \]

### Step 4: Set Up the Conservation of Momentum Equation
Initially, the linear momentum of the sphere is:
\[ p_{initial} = m u \]
When the sphere starts rolling, its momentum can be expressed as:
\[ p_{final} = I_p \omega \]
Where \( \omega \) is the angular velocity. Since pure rolling motion starts, we have the relationship:
\[ v = r \omega \]
Thus, we can express \( \omega \) as:
\[ \omega = \frac{v}{r} \]

### Step 5: Substitute and Solve
Substituting \( \omega \) into the momentum equation:
\[ m u = I_p \cdot \frac{v}{r} \]
Substituting \( I_p \):
\[ m u = \frac{7}{5} m r^2 \cdot \frac{v}{r} \]
This simplifies to:
\[ m u = \frac{7}{5} m r v \]
Canceling \( m \) and \( r \) (assuming \( r \neq 0 \)):
\[ u = \frac{7}{5} v \]
Rearranging gives:
\[ v = \frac{5}{7} u \]

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