A solid sphere rolls without slipping down a `30^(@)` inclined plane. If `g=10 ms^(-2)` then the acceleration of the rolling sphere is

Options

A. `5ms^(-2)`
B. `(7)/(25)ms^(-2)`
C. `(25)/(7)ms^(-2)`
D. `(15)/(7)ms^(-2)`

Explanation

To find the acceleration of a solid sphere rolling without slipping down a 30-degree inclined plane, we can follow these steps:

### Step 1: Identify the forces acting on the sphere
When the sphere rolls down the incline, the forces acting on it are:
- The gravitational force acting downward, which can be split into two components:
- Parallel to the incline: \( F_{\parallel} = mg \sin(\theta) \)
- Perpendicular to the incline: \( F_{\perpendicular} = mg \cos(\theta) \)

Where:
- \( m \) is the mass of the sphere
- \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \))
- \( \theta \) is the angle of the incline (given as \( 30^\circ \))

### Step 2: Write the equation of motion
For a solid sphere rolling without slipping, we can use Newton's second law. The net force acting along the incline is equal to the mass times the acceleration:
\[ F_{\text{net}} = ma \]
Where \( a \) is the linear acceleration of the sphere.

The net force along the incline can be expressed as:
\[ F_{\text{net}} = mg \sin(\theta) - F_{\text{friction}} \]

### Step 3: Relate linear acceleration to angular acceleration
For rolling motion, the linear acceleration \( a \) is related to the angular acceleration \( \alpha \) by the equation:
\[ a = r \alpha \]
Where \( r \) is the radius of the sphere.

The moment of inertia \( I \) for a solid sphere is:
\[ I = \frac{2}{5} m r^2 \]

Using the relationship between torque and angular acceleration:
\[ \tau = I \alpha \]
The torque due to friction \( F_{\text{friction}} \) is:
\[ \tau = F_{\text{friction}} \cdot r \]

### Step 4: Set up the equations
From the torque equation:
\[ F_{\text{friction}} \cdot r = I \alpha \]
Substituting \( I \) and \( \alpha \):
\[ F_{\text{friction}} \cdot r = \left(\frac{2}{5} m r^2\right) \left(\frac{a}{r}\right) \]
This simplifies to:
\[ F_{\text{friction}} = \frac{2}{5} m a \]

### Step 5: Substitute back into the net force equation
Now substitute \( F_{\text{friction}} \) back into the net force equation:
\[ mg \sin(\theta) - \frac{2}{5} m a = ma \]

### Step 6: Solve for acceleration \( a \)
Rearranging the equation gives:
\[ mg \sin(\theta) = ma + \frac{2}{5} m a \]
\[ mg \sin(\theta) = ma \left(1 + \frac{2}{5}\right) \]
\[ mg \sin(\theta) = ma \left(\frac{7}{5}\right) \]

Now, cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g \sin(\theta) = a \left(\frac{7}{5}\right) \]
Thus,
\[ a = \frac{5}{7} g \sin(\theta) \]

### Step 7: Calculate \( a \) using \( g = 10 \, \text{m/s}^2 \) and \( \theta = 30^\circ \)
Now substituting \( g = 10 \, \text{m/s}^2 \) and \( \sin(30^\circ) = \frac{1}{2} \):
\[ a = \frac{5}{7} \cdot 10 \cdot \frac{1}{2} \]
\[ a = \frac{5 \cdot 10}{14} \]
\[ a = \frac{50}{14} \]
\[ a = \frac{25}{7} \]
\[ a \approx 3.57 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the rolling sphere is approximately \( 3.57 \, \text{m/s}^2 \).

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