A hollow sphere of mass m and radius R is rolling downdard on a rough inclined plane of inclination `theta`. If the coefficient of friction between the hollow sphere and incline is `mu`, then

Options

A. Friction opposes its translation
B. Friction supports rotation motion
C. On decreasing `theta`, frictional force decreases
D. All of these

Answer

To solve the problem of a hollow sphere of mass \( m \) and radius \( R \) rolling down a rough inclined plane with an angle of inclination \( \theta \) and a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Analyze the Forces Acting on the Sphere
The forces acting on the hollow sphere are:
- The gravitational force \( mg \) acting vertically downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting up the incline, opposing the motion.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- The component parallel to the incline: \( F_{\text{parallel}} = mg \sin \theta \)
- The component perpendicular to the incline: \( F_{\text{perpendicular}} = mg \cos \theta \)

### Step 3: Write the Equation for Normal Force
Since there is no vertical motion, the normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 4: Write the Equation for Frictional Force
The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N = \mu (mg \cos \theta) \]

### Step 5: Apply Newton's Second Law for Translational Motion
For the translational motion of the sphere down the incline, we can write:
\[ mg \sin \theta - F_f = ma \]
Where \( a \) is the linear acceleration of the center of mass of the sphere.

Substituting \( F_f \):
\[ mg \sin \theta - \mu (mg \cos \theta) = ma \]

### Step 6: Apply Newton's Second Law for Rotational Motion
For the rotational motion, we need to consider the torque due to the frictional force:
\[ \tau = I \alpha \]
Where \( \tau \) is the torque, \( I \) is the moment of inertia of the hollow sphere, and \( \alpha \) is the angular acceleration. The moment of inertia \( I \) for a hollow sphere is given by:
\[ I = \frac{2}{3} m R^2 \]
The torque due to friction is:
\[ \tau = F_f R = \mu (mg \cos \theta) R \]

### Step 7: Relate Angular Acceleration to Linear Acceleration
The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) for rolling without slipping is:
\[ a = R \alpha \]
Thus, we can express \( \alpha \) as:
\[ \alpha = \frac{a}{R} \]

### Step 8: Substitute into the Torque Equation
Substituting \( \alpha \) into the torque equation:
\[ \mu (mg \cos \theta) R = \frac{2}{3} m R^2 \left(\frac{a}{R}\right) \]
Simplifying gives:
\[ \mu mg \cos \theta = \frac{2}{3} ma \]

### Step 9: Solve for Acceleration \( a \)
From the equation:
\[ mg \sin \theta - \mu mg \cos \theta = ma \]
We can substitute \( a \) from the torque equation:
\[ mg \sin \theta - \mu mg \cos \theta = m \left(\frac{3}{2} \mu g \cos \theta\right) \]
This leads to:
\[ mg \sin \theta = m \left(\mu g \cos \theta + \frac{3}{2} \mu g \cos \theta\right) \]
\[ g \sin \theta = g \cos \theta \left(\frac{5}{2} \mu\right) \]
Thus, we can solve for \( \mu \):
\[ \mu = \frac{2 \tan \theta}{5} \]

### Conclusion
The coefficient of friction \( \mu \) required for the hollow sphere to roll down the incline without slipping is given by:
\[ \mu = \frac{2 \tan \theta}{5} \]