Given:
- A solid sphere of radius **r** is placed on a rough horizontal ground.
- Initial angular speed = **ω°**
- Initial linear velocity = **0**
- Coefficient of friction = **π**
- Time when slipping stops is given by **t = (m r ω°) / (n π g)**
We need to determine the value of **n - m**.
### Step 1: Understanding the Motion
Since the sphere is placed with an initial angular velocity **ω°** and no linear velocity, it is initially slipping. Due to friction, a force acts on it, causing:
1. A linear acceleration to the center of mass.
2. An angular deceleration.
The motion continues until the sphere transitions from slipping to pure rolling, i.e., when the rolling condition **v = rω** is met.
### Step 2: Forces Acting on the Sphere
The only external force affecting the motion is kinetic friction **f_k** acting opposite to the direction of motion. Since friction acts tangentially at the bottom of the sphere, it causes:
1. A linear acceleration **a** given by Newton’s Second Law:
- \( f_k = ma \)
- \( \mu_k mg = ma \)
- \( \pi mg = ma \)
- \( a = \pi g \)
2. An angular deceleration **α** due to torque:
- Torque about the center of mass:
\( f_k r = I \alpha \)
- Moment of inertia of a solid sphere about its center:
\( I = \frac{2}{5} m r^2 \)
- \( (\pi mg) r = \left(\frac{2}{5} m r^2\right) \alpha \)
- \( \pi g = \frac{2}{5} r \alpha \)
- \( \alpha = \frac{5\pi g}{2r} \)
### Step 3: Condition for Pure Rolling
Pure rolling occurs when **v = rω**. The velocity and angular velocity at time **t** are:
1. **Linear Velocity**:
- \( v = u + at \)
- \( v = 0 + (\pi g)t \)
- \( v = \pi g t \)
2. **Angular Velocity**:
- \( \omega = \omega° - \alpha t \)
- \( \omega = \omega° - \left(\frac{5\pi g}{2r}\right) t \)
For rolling to begin:
- \( v = r\omega \)
- \( \pi g t = r \left( \omega° - \frac{5\pi g}{2r} t \right) \)
### Step 4: Solving for t
Expanding the equation:
- \( \pi g t = r \omega° - \frac{5\pi g}{2} t \)
- Bringing terms involving **t** to one side:
\( \pi g t + \frac{5\pi g}{2} t = r \omega° \)
- \( t (\pi g + \frac{5\pi g}{2}) = r \omega° \)
- \( t \left( \frac{2\pi g}{2} + \frac{5\pi g}{2} \right) = r \omega° \)
- \( t \left( \frac{7\pi g}{2} \right) = r \omega° \)
- \( t = \frac{2 r \omega°}{7 \pi g} \)
Comparing with given form:
- \( t = \frac{m r \omega°}{n \pi g} \)
- \( \frac{m}{n} = \frac{2}{7} \)
Thus, **m = 2** and **n = 7**.
### Step 5: Finding n - m
- \( n - m = 7 - 2 = 5 \)
### Final Answer:
**5**
