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 sphere
The forces acting on the hollow sphere are:
- The gravitational force (weight) acting downward: \( mg \)
- The normal force acting perpendicular to the inclined plane: \( N \)
- The frictional force acting up the incline: \( f \)

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

### Step 3: Apply Newton's second law along the incline
Using Newton's second law, we can write the equation for the motion along the incline:
\[ mg \sin(30°) - f = ma \]
where \( a \) is the linear acceleration of the sphere.

### Step 4: Relate linear acceleration to angular acceleration
Since the sphere rolls without slipping, the linear acceleration \( a \) is related to the angular acceleration \( \alpha \) by the equation:
\[ a = \alpha r \]
where \( r \) is the radius of the sphere.

### Step 5: Write the torque equation
The frictional force provides the torque that causes the sphere to rotate. The torque \( \tau \) can be expressed as:
\[ \tau = f \cdot r = I \alpha \]
For a hollow sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{3} m r^2 \]
Substituting \( \alpha = \frac{a}{r} \) into the torque equation gives:
\[ f \cdot r = \frac{2}{3} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ f = \frac{2}{3} m a \]

### Step 6: Substitute \( f \) back into the motion equation
Substituting \( f \) into the equation from Step 3:
\[ mg \sin(30°) - \frac{2}{3} m a = ma \]
This can be rearranged to:
\[ mg \sin(30°) = ma + \frac{2}{3} m a \]
\[ mg \sin(30°) = \frac{5}{3} ma \]

### Step 7: Solve for \( a \)
Now, substituting \( \sin(30°) = \frac{1}{2} \):
\[ mg \cdot \frac{1}{2} = \frac{5}{3} ma \]
This simplifies to:
\[ \frac{mg}{2} = \frac{5}{3} ma \]
Dividing both sides by \( m \) gives:
\[ \frac{g}{2} = \frac{5}{3} a \]
Now, solving for \( a \):
\[ a = \frac{3g}{10} \]

### Conclusion
The linear acceleration of the hollow sphere rolling down the incline is:
\[ \boxed{\frac{3g}{10}} \]