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

### Step 1: Identify Forces Acting on the Sphere
When the sphere is on the inclined plane, the forces acting on it 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 (opposing the motion).

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

### Step 3: Write the Equations of Motion
For the sphere to roll without slipping, we can write the following equations:

1. **Translational Motion**:
\[ mg \sin \theta - f = Ma \quad \text{(1)} \]
where \( a \) is the linear acceleration of the sphere.

2. **Rotational Motion**:
The torque \( \tau \) due to friction about the center of the sphere is given by:
\[ \tau = f \cdot r = I \alpha \]
For a solid sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{5} M r^2 \]
The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ a = \alpha r \quad \Rightarrow \quad \alpha = \frac{a}{r} \]
Substituting this into the torque equation gives:
\[ f \cdot r = \frac{2}{5} M r^2 \cdot \frac{a}{r} \]
Simplifying this, we find:
\[ f = \frac{2}{5} M a \quad \text{(2)} \]

### Step 4: Substitute Equation (2) into Equation (1)
Substituting \( f \) from equation (2) into equation (1):
\[ mg \sin \theta - \frac{2}{5} M a = Ma \]
Rearranging gives:
\[ mg \sin \theta = Ma + \frac{2}{5} M a \]
\[ mg \sin \theta = \frac{7}{5} Ma \]
Thus, we can express \( a \) as:
\[ a = \frac{5}{7} g \sin \theta \quad \text{(3)} \]

### Step 5: Find the Frictional Force
Using equation (2) to find the frictional force:
\[ f = \frac{2}{5} M a \]
Substituting \( a \) from equation (3):
\[ f = \frac{2}{5} M \left(\frac{5}{7} g \sin \theta\right) = \frac{2}{7} Mg \sin \theta \quad \text{(4)} \]

### Step 6: Relate Frictional Force to Coefficient of Friction
The maximum static frictional force is given by:
\[ f_{max} = \mu N = \mu (Mg \cos \theta) \quad \text{(5)} \]
Setting equation (4) equal to equation (5) for the minimum coefficient of friction:
\[ \frac{2}{7} Mg \sin \theta = \mu (Mg \cos \theta) \]
Dividing both sides by \( Mg \):
\[ \frac{2}{7} \sin \theta = \mu \cos \theta \]
Thus, the minimum coefficient of friction \( \mu \) is:
\[ \mu = \frac{2}{7} \tan \theta \]

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

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