A solid cylinder is rolling without slipping on a plane having inclination `theta` and the coefficient of static friction `mu_(s)`. The relation between `theta` and `mu_(s)` is

Options

A. `tan theta gt 3 mu_(s)`
B. `tan theta le 3 mu_(s)`
C. `tan theta lt 3 mu_(s)^(2)`
D. None of these

Answer

To find the relationship between the angle of inclination \( \theta \) and the coefficient of static friction \( \mu_s \) for a solid cylinder rolling without slipping on an inclined plane, we can follow these steps:

### Step 1: Identify Forces Acting on the Cylinder
Consider a solid cylinder of mass \( m \) and radius \( r \) rolling down an inclined plane with an angle \( \theta \). The forces acting on the cylinder include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting perpendicular to the surface.
- The static frictional force \( F_s \) acting up the incline.

### 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: Write the Equations of Motion
Using Newton's second law along the incline:
\[ mg \sin \theta - F_s = ma \tag{1} \]
Where \( a \) is the linear acceleration of the center of mass of the cylinder.

### Step 4: Write the Torque Equation
The torque \( \tau \) about the center of mass due to the static friction is given by:
\[ \tau = F_s \cdot r = I \cdot \alpha \tag{2} \]
For a solid cylinder, the moment of inertia \( I \) about its center of mass is \( \frac{1}{2} m r^2 \), and the angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by:
\[ a = \alpha r \implies \alpha = \frac{a}{r} \]
Substituting this into the torque equation gives:
\[ F_s \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \implies F_s = \frac{1}{2} m a \tag{3} \]

### Step 5: Relate Linear and Angular Acceleration
From the rolling condition, we have:
\[ a = \alpha r \tag{4} \]

### Step 6: Substitute \( F_s \) into the Motion Equation
Substituting equation (3) into equation (1):
\[ mg \sin \theta - \frac{1}{2} ma = ma \]
Rearranging gives:
\[ mg \sin \theta = \frac{3}{2} ma \]
Thus, we can express \( a \):
\[ a = \frac{2}{3} g \sin \theta \tag{5} \]

### Step 7: Find the Static Friction Force
Substituting \( a \) from equation (5) into equation (3):
\[ F_s = \frac{1}{2} m \left(\frac{2}{3} g \sin \theta\right) = \frac{1}{3} mg \sin \theta \tag{6} \]

### Step 8: Relate Static Friction to Coefficient of Friction
The maximum static friction force is given by:
\[ F_{s, \text{max}} = \mu_s N = \mu_s mg \cos \theta \]
For rolling without slipping, we require:
\[ F_s \leq F_{s, \text{max}} \]
Substituting from equation (6):
\[ \frac{1}{3} mg \sin \theta \leq \mu_s mg \cos \theta \]
Dividing through by \( mg \) (assuming \( m \neq 0 \)):
\[ \frac{1}{3} \sin \theta \leq \mu_s \cos \theta \]
Rearranging gives:
\[ \tan \theta \leq 3 \mu_s \]

### Conclusion
Thus, the relationship between the angle of inclination \( \theta \) and the coefficient of static friction \( \mu_s \) is:
\[ \tan \theta \leq 3 \mu_s \]