An inclined plane makes an angle 30° with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to

Options

A. `(g)/(3)`
B. `(2g)/(3)`
C. `(5g)/(7)`
D. `(5g)/(14)`

Explanation

To find the linear acceleration of a solid sphere rolling down an inclined plane that makes an angle of 30° with the horizontal, we can follow these steps:

### Step 1: Identify the Forces Acting on the Sphere
The forces acting on the sphere include:
- The gravitational force \( mg \) acting downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The component of gravitational force parallel to the incline, which is \( mg \sin \theta \).
- The component of gravitational force perpendicular to the incline, which is \( mg \cos \theta \).

### Step 2: Break Down the Forces
For an inclined plane at an angle \( \theta = 30° \):
- The parallel component of the gravitational force is:
\[ F_{\text{parallel}} = mg \sin(30°) = mg \cdot \frac{1}{2} = \frac{mg}{2} \]
- The perpendicular component is:
\[ F_{\text{perpendicular}} = mg \cos(30°) = mg \cdot \frac{\sqrt{3}}{2} \]

### Step 3: Apply Newton's Second Law
Since the sphere rolls without slipping, we can apply Newton's second law for the linear motion along the incline:
\[ F_{\text{net}} = ma \]
Where \( F_{\text{net}} \) is the net force acting along the incline. The net force is the component of gravity acting down the incline minus the friction force (which we will account for through rotational motion).

### Step 4: Calculate Torque and Angular Acceleration
The torque \( \tau \) about the point of contact (point O) due to the gravitational force is:
\[ \tau = r \cdot F_{\text{parallel}} = r \cdot \frac{mg}{2} \]
Using the moment of inertia \( I \) for a solid sphere about its center of mass, which is \( I = \frac{2}{5}mr^2 \), the moment of inertia about point O (using the parallel axis theorem) is:
\[ I_O = I + m \cdot r^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2 \]

### Step 5: Relate Torque to Angular Acceleration
Using the relationship \( \tau = I \alpha \):
\[ r \cdot \frac{mg}{2} = \frac{7}{5}mr^2 \cdot \alpha \]
Cancelling \( m \) and one \( r \):
\[ \frac{g}{2} = \frac{7}{5}r \alpha \]
Solving for \( \alpha \):
\[ \alpha = \frac{5g}{14r} \]

### Step 6: Relate Angular Acceleration to Linear Acceleration
The linear acceleration \( a \) of the center of mass is related to the angular acceleration by:
\[ a = \alpha r \]
Substituting for \( \alpha \):
\[ a = \left(\frac{5g}{14r}\right) r = \frac{5g}{14} \]

### Final Answer
Thus, the linear acceleration of the solid sphere rolling down the inclined plane is:
\[ a = \frac{5g}{14} \]

### Conclusion
The correct option is option 4: \( \frac{5g}{14} \).

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