What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination `theta` ?

Options

A. `(2)/(7) tan theta`
B. `(1)/(3)g tan theta`
C. `(1)/(2)tan theta`
D. `(2)/(5)tan theta`

Answer

To find the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination \( \theta \), we can follow these steps:

### Step 1: Identify the forces acting on the sphere
On an inclined plane, the forces acting on the solid sphere are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( f \) acting up the incline (to prevent slipping).

### Step 2: Resolve the gravitational force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Apply Newton's second law for translational motion
For the sphere to roll down the incline, we can write the equation of motion along the incline:
\[ mg \sin \theta - f = ma \]
where \( a \) is the linear acceleration of the center of mass of the sphere.

### Step 4: Apply Newton's second law for rotational motion
For rolling without slipping, the frictional force provides the torque about the center of mass. The torque \( \tau \) due to friction is given by:
\[ \tau = f \cdot r \]
where \( r \) is the radius of the sphere. The moment of inertia \( I \) of a solid sphere about its center of mass is:
\[ I = \frac{2}{5} m r^2 \]
Using the relation \( \tau = I \alpha \) and the fact that \( \alpha = \frac{a}{r} \) (for rolling without slipping), we can write:
\[ f \cdot r = I \cdot \alpha = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ f = \frac{2}{5} m a \]

### Step 5: Substitute \( f \) back into the translational motion equation
Now, substituting \( f \) into the translational motion equation:
\[ mg \sin \theta - \frac{2}{5} m a = ma \]
Rearranging gives:
\[ mg \sin \theta = ma + \frac{2}{5} m a = \frac{7}{5} m a \]
Thus, we can solve for \( a \):
\[ a = \frac{5}{7} g \sin \theta \]

### Step 6: Find the normal force
The normal force \( N \) can be determined from the vertical forces:
\[ N = mg \cos \theta \]

### Step 7: Relate friction to the normal force
The frictional force can also be expressed in terms of the coefficient of friction \( \mu \):
\[ f = \mu N = \mu mg \cos \theta \]

### Step 8: Set the two expressions for friction equal
From the previous steps, we have:
\[ \frac{2}{5} m a = \mu mg \cos \theta \]
Substituting \( a = \frac{5}{7} g \sin \theta \):
\[ \frac{2}{5} m \left(\frac{5}{7} g \sin \theta\right) = \mu mg \cos \theta \]
This simplifies to:
\[ \frac{2}{7} mg \sin \theta = \mu mg \cos \theta \]

### Step 9: Solve for \( \mu \)
Dividing both sides by \( mg \) (assuming \( m \neq 0 \)):
\[ \frac{2}{7} \sin \theta = \mu \cos \theta \]
Thus, we find:
\[ \mu = \frac{2}{7} \tan \theta \]

### Conclusion
The minimum coefficient of friction \( \mu \) required for a solid sphere to roll without slipping on an inclined plane of inclination \( \theta \) is:
\[ \mu = \frac{2}{7} \tan \theta \]