A cylinder of radius `R` and mass `M` rolls without slipping down a plane inclined at an angle `theta`. Coeff. of friction between the cylinder and the plane is `mu`. For what maximum inclination `theta`, the cylinder rolls without slipping ?

Options

A. tan`theta` gt `3 mu_s`
B. tan`theta` `le` `3mu_s`
C. tan`theta` lt `3 mu_s`
D. None of these

Explanation

To solve the problem of determining the maximum inclination angle \(\theta\) at which a cylinder rolls without slipping down an inclined plane, we will follow these steps:

### Step 1: Identify Forces Acting on the Cylinder
When the cylinder rolls down the inclined plane, the forces acting on it are:
- The gravitational force \(Mg\) acting vertically downward.
- The component of gravitational force acting down the incline: \(Mg \sin \theta\).
- The normal force \(N\) acting perpendicular to the incline.
- The frictional force \(f\) acting up the incline, which prevents slipping.

### Step 2: Write the Equations of Motion
For translational motion along the incline, we can write the equation:
\[ Mg \sin \theta - f = Ma \]
where \(a\) is the linear acceleration of the center of mass of the cylinder.

For rotational motion, the torque \(\tau\) about the center of mass due to friction is given by:
\[ \tau = f \cdot R \]
This torque causes an angular acceleration \(\alpha\) given by:
\[ \tau = I \alpha \]
For a solid cylinder, the moment of inertia \(I\) is:
\[ I = \frac{1}{2}MR^2 \]
Using the relationship between linear acceleration \(a\) and angular acceleration \(\alpha\):
\[ a = R \alpha \implies \alpha = \frac{a}{R} \]
We can substitute this into the torque equation:
\[ f \cdot R = \frac{1}{2}MR^2 \cdot \frac{a}{R} \]
This simplifies to:
\[ f = \frac{1}{2}Ma \]

### Step 3: Substitute Friction into the Translational Motion Equation
From the friction equation, we have:
\[ f = \mu N \]
The normal force \(N\) can be expressed as:
\[ N = Mg \cos \theta \]
Thus, we can write:
\[ f = \mu Mg \cos \theta \]
Now substituting this expression for \(f\) into the translational motion equation:
\[ Mg \sin \theta - \mu Mg \cos \theta = Ma \]

### Step 4: Eliminate Mass and Rearrange
Dividing through by \(M\) gives:
\[ g \sin \theta - \mu g \cos \theta = a \]
Now substituting \(a\) from the rotational motion equation:
\[ g \sin \theta - \mu g \cos \theta = \frac{1}{2}a \]
Replacing \(a\) with \(2f\):
\[ g \sin \theta - \mu g \cos \theta = \frac{1}{2} \cdot \frac{1}{2}Ma \]
This leads to:
\[ g \sin \theta - \mu g \cos \theta = \frac{1}{2} \cdot \frac{1}{2} \cdot \mu g \cos \theta \]

### Step 5: Solve for \(\theta\)
Rearranging gives:
\[ g \sin \theta = \mu g \cos \theta + \frac{1}{2} \mu g \cos \theta \]
This simplifies to:
\[ g \sin \theta = \frac{3}{2} \mu g \cos \theta \]
Dividing through by \(g\) (assuming \(g \neq 0\)):
\[ \sin \theta = \frac{3}{2} \mu \cos \theta \]
Dividing both sides by \(\cos \theta\):
\[ \tan \theta = \frac{3}{2} \mu \]
Thus, the maximum angle \(\theta\) for which the cylinder rolls without slipping is:
\[ \theta = \tan^{-1}\left(\frac{3}{2} \mu\right) \]

### Final Answer
The maximum inclination angle \(\theta\) at which the cylinder rolls without slipping is:
\[ \theta = \tan^{-1}\left(\frac{3}{2} \mu\right) \]