A sphere of mass M rolls without slipping on the inclined plane of inclination `theta`. What should be the minimum coefficient of friction, so that the sphere rolls down without slipping ?

Options

A. ` 2//5 tan theta`
B. `2//7 tan theta`
C. `5//7 tan theta`
D. `tan theta`

Answer

To solve the problem of finding the minimum coefficient of friction (μ) required for a sphere of mass M to roll down an inclined plane without slipping, we can follow these steps:

### Step 1: Identify the forces acting on the sphere
- The forces acting on the sphere include:
1. Gravitational force (mg) acting downward.
2. Normal force (N) acting perpendicular to the inclined plane.
3. Frictional force (f) acting up the incline (opposing the motion).

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

### Step 3: Write the equations of motion
- For the sphere rolling 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 sphere.

### Step 4: Relate linear acceleration and angular acceleration
- Since the sphere rolls without slipping, we have the relation:
\[ a = \alpha r \]
where \( \alpha \) is the angular acceleration and \( r \) is the radius of the sphere.

### Step 5: Write the torque equation
- The torque (\( \tau \)) about the center of the sphere due to friction is:
\[ \tau = f \cdot r = I \alpha \]
The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} M r^2 \]
Substituting \( \alpha = \frac{a}{r} \) into the torque equation gives:
\[ f \cdot r = \frac{2}{5} M r^2 \cdot \frac{a}{r} \]
Simplifying this, we find:
\[ f = \frac{2}{5} M a \]

### Step 6: Substitute the friction force into the motion equation
- Substitute \( f \) back into the equation of motion:
\[ mg \sin \theta - \frac{2}{5} M a = Ma \]
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 7: Find the friction force
- Now substituting \( a \) back into the expression for friction:
\[ f = \frac{2}{5} M a = \frac{2}{5} M \left( \frac{5}{7} g \sin \theta \right) = \frac{2}{7} M g \sin \theta \]

### Step 8: Relate friction to the normal force
- The normal force is given by:
\[ N = mg \cos \theta \]
The maximum static friction is:
\[ f_{\text{max}} = \mu N = \mu mg \cos \theta \]

### Step 9: Set the friction force equal to the maximum static friction
- For rolling without slipping, we need:
\[ \frac{2}{7} M g \sin \theta = \mu mg \cos \theta \]
Canceling \( mg \) from both sides gives:
\[ \frac{2}{7} g \sin \theta = \mu g \cos \theta \]

### Step 10: Solve for the coefficient of friction
- Finally, we can solve for μ:
\[ \mu = \frac{2}{7} \tan \theta \]

### Final Answer
The minimum coefficient of friction required for the sphere to roll down the incline without slipping is:
\[ \mu = \frac{2}{7} \tan \theta \]