A solid uniform sphere having a mass M , radius R , and moment of inertia of `(2)/(5)MR^(2)` rolls down a plane inclined at an angle `theta` to the horizontal starting from rest. The coefficient of static friction between the sphere and the plane is `mu`. Then

Options

A. The sphere will always roll without slipping
B. The sphere will always slide
C. The sphere will roll without slipping only if `theta le sin.(7mu)/(2)`
D. The sphere will roll without slipping only if `theta le tan^(-1).(7mu)/(2)`

Explanation

To solve the problem, we need to analyze the conditions under which a solid uniform sphere rolls down an inclined plane without slipping. We will derive the relationship between the coefficient of static friction (μ) and the angle of inclination (θ).

### Step 1: Identify Forces Acting on the Sphere
When the sphere rolls down the incline, the forces acting on it include:
- Gravitational force (Mg) acting downward.
- Normal force (N) acting perpendicular to the inclined plane.
- Frictional force (f) acting up the incline.

### Step 2: Write Down the Equations of Motion
1. **Translational Motion**: The net force acting down the incline is given by:
\[ Mg \sin(\theta) - f = Ma \]
where \( a \) is the linear acceleration of the sphere.

2. **Rotational Motion**: The frictional force provides the torque (τ) about the center of mass:
\[ fR = I \alpha \]
where \( I \) is the moment of inertia of the sphere, \( R \) is the radius, and \( \alpha \) is the angular acceleration. For a solid sphere, \( I = \frac{2}{5}MR^2 \).

### Step 3: Relate Linear and Angular Acceleration
Since the sphere rolls without slipping, the linear acceleration \( a \) and angular acceleration \( \alpha \) are related by:
\[ a = R \alpha \]
Substituting \( \alpha = \frac{a}{R} \) into the torque equation gives:
\[ fR = \left(\frac{2}{5}MR^2\right) \left(\frac{a}{R}\right) \]
This simplifies to:
\[ f = \frac{2}{5}Ma \]

### Step 4: Substitute \( f \) into the Translational Motion Equation
Now, substituting \( f \) back into the translational motion equation:
\[ Mg \sin(\theta) - \frac{2}{5}Ma = Ma \]
This can be rearranged to:
\[ Mg \sin(\theta) = Ma + \frac{2}{5}Ma \]
\[ Mg \sin(\theta) = \frac{7}{5}Ma \]

### Step 5: Solve for \( a \)
Rearranging gives:
\[ a = \frac{5}{7}g \sin(\theta) \]

### Step 6: Apply the Condition for Rolling Without Slipping
For the sphere to roll without slipping, the frictional force must not exceed the maximum static friction, which is given by:
\[ f \leq \mu N \]
The normal force \( N \) is given by:
\[ N = Mg \cos(\theta) \]
Thus, the maximum static friction is:
\[ f_{\text{max}} = \mu Mg \cos(\theta) \]

### Step 7: Set Up the Inequality
From our earlier expression for \( f \):
\[ \frac{2}{5}Ma \leq \mu Mg \cos(\theta) \]
Substituting \( a = \frac{5}{7}g \sin(\theta) \):
\[ \frac{2}{5}M\left(\frac{5}{7}g \sin(\theta)\right) \leq \mu Mg \cos(\theta) \]
This simplifies to:
\[ \frac{2}{7}Mg \sin(\theta) \leq \mu Mg \cos(\theta) \]
Dividing through by \( Mg \) (assuming \( M \neq 0 \)):
\[ \frac{2}{7} \sin(\theta) \leq \mu \cos(\theta) \]

### Step 8: Rearranging the Inequality
Rearranging gives:
\[ \mu \geq \frac{2}{7} \frac{\sin(\theta)}{\cos(\theta)} = \frac{2}{7} \tan(\theta) \]

### Step 9: Final Relation
Thus, we have:
\[ \tan(\theta) \leq \frac{7}{2} \mu \]
or
\[ \theta \leq \tan^{-1}\left(\frac{7}{2} \mu\right) \]

### Conclusion
The condition for the sphere to roll without slipping is:
\[ \theta \leq \tan^{-1}\left(\frac{7}{2} \mu\right) \]