A hollow sphere rolls without slipping down a plane inclined at an angle of 30° to the horizontal. Its linear acceleration will be

Options

A. `5/14 g`
B. `(2g)/3`
C. `q/2`
D. `(3g)/10`

Explanation

To find the linear acceleration of a hollow sphere rolling down an inclined plane at an angle of 30°, we can follow these steps:

### Step 1: Identify the forces acting on the hollow sphere
The forces acting on the hollow sphere are:
- The gravitational force (weight) \( Mg \), acting downward.
- The normal force \( N \), acting perpendicular to the inclined plane.
- The frictional force \( F \), acting up the incline (which prevents slipping).

### Step 2: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( Mg \cos(30^\circ) \)
- Parallel to the incline: \( Mg \sin(30^\circ) \)

### Step 3: Apply Newton's second law along the incline
According to Newton's second law, the net force acting on the sphere along the incline is equal to the mass times the linear acceleration (\( a \)):
\[ Mg \sin(30^\circ) - F = Ma \]

### Step 4: Relate linear acceleration to angular acceleration
Since the sphere rolls without slipping, we have the relationship between linear acceleration \( a \) and angular acceleration \( \alpha \):
\[ a = \alpha r \]

### Step 5: Write the torque equation
The torque \( \tau \) due to the frictional force \( F \) about the center of mass is given by:
\[ \tau = F \cdot r = I \alpha \]
For a hollow sphere, the moment of inertia \( I \) is:
\[ I = \frac{2}{3} M r^2 \]
Substituting \( I \) into the torque equation gives:
\[ F \cdot r = \frac{2}{3} M r^2 \alpha \]

### Step 6: Substitute \( \alpha \) in terms of \( a \)
From the no-slip condition, we can substitute \( \alpha \) with \( \frac{a}{r} \):
\[ F \cdot r = \frac{2}{3} M r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ F = \frac{2}{3} Ma \]

### Step 7: Substitute \( F \) back into the force equation
Now, substitute \( F \) into the equation from Step 3:
\[ Mg \sin(30^\circ) - \frac{2}{3} Ma = Ma \]

### Step 8: Solve for \( a \)
Rearranging gives:
\[ Mg \sin(30^\circ) = Ma + \frac{2}{3} Ma \]
\[ Mg \sin(30^\circ) = \frac{5}{3} Ma \]
Now, substituting \( \sin(30^\circ) = \frac{1}{2} \):
\[ Mg \cdot \frac{1}{2} = \frac{5}{3} Ma \]
\[ \frac{Mg}{2} = \frac{5}{3} Ma \]
Now, cancel \( M \) from both sides:
\[ \frac{g}{2} = \frac{5}{3} a \]
Multiplying both sides by \( \frac{3}{5} \):
\[ a = \frac{3g}{10} \]

### Final Answer
The linear acceleration of the hollow sphere rolling down the inclined plane is:
\[ \boxed{\frac{3g}{10}} \]