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 solve the problem, we need to establish the relationship between the angle of inclination \( \theta \) and the coefficient of static friction \( \mu_s \) for a solid cylinder rolling down an inclined plane without slipping. Here’s a step-by-step breakdown of the solution:

### Step 1: Analyze the Forces Acting on the Cylinder
When the solid cylinder rolls down the inclined plane, the following forces act on it:
- Gravitational force \( mg \) acting downwards.
- Normal force \( N \) acting perpendicular to the inclined plane.
- Frictional force \( f \) 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
For the cylinder 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 cylinder.

### Step 4: Relate Linear Acceleration to Angular Acceleration
For a solid cylinder, the moment of inertia \( I \) about its center of mass is given by:
\[ I = \frac{1}{2} m r^2 \]
The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is:
\[ a = r \alpha \]
Since the cylinder rolls without slipping, we can also write:
\[ f = I \alpha / r \]
Substituting \( I \) into this equation gives:
\[ f = \frac{1}{2} m r^2 \cdot \frac{a}{r} = \frac{1}{2} m a r \]

### Step 5: Substitute and Rearrange
Substituting \( f \) into the equation of motion gives:
\[ mg \sin \theta - \frac{1}{2} m a = ma \]
Rearranging this, we get:
\[ mg \sin \theta = ma + \frac{1}{2} m a = \frac{3}{2} ma \]
Thus, we can express \( a \) as:
\[ a = \frac{2}{3} g \sin \theta \]

### Step 6: Analyze the Frictional Force
The maximum static friction force can be expressed as:
\[ f \leq \mu_s N \]
where \( N = mg \cos \theta \). Therefore, the frictional force can be written as:
\[ f \leq \mu_s mg \cos \theta \]

### Step 7: Substitute for \( f \)
From the previous steps, we know:
\[ f = mg \sin \theta - ma \]
Substituting \( a = \frac{2}{3} g \sin \theta \) into this gives:
\[ f = mg \sin \theta - m \left(\frac{2}{3} g \sin \theta\right) = mg \sin \theta \left(1 - \frac{2}{3}\right) = \frac{1}{3} mg \sin \theta \]

### Step 8: Set Up the Inequality
Now, substituting \( f \) into the friction inequality gives:
\[ \frac{1}{3} mg \sin \theta \leq \mu_s mg \cos \theta \]
Dividing through by \( mg \) (assuming \( mg \neq 0 \)):
\[ \frac{1}{3} \sin \theta \leq \mu_s \cos \theta \]

### Step 9: Rearranging to Find the Relationship
Rearranging this gives:
\[ \tan \theta \leq 3 \mu_s \]

### Final Result
Thus, the relationship between \( \theta \) and \( \mu_s \) is:
\[ \tan \theta \leq 3 \mu_s \]