A sphere rolls down on an inclied plane of inclination `theta`. What is the acceleration as the sphere reaches bottom ?

Options

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

Answer

To find the acceleration of a sphere rolling down an inclined plane of inclination \( \theta \), we can follow these steps:

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

### Step 2: Write the equation of motion
The net force acting on the sphere along the incline is given by:
\[ F_{\text{net}} = mg \sin \theta - F_{\text{friction}} \]
Since the sphere rolls without slipping, we need to account for the frictional force that provides the torque necessary for rolling.

### Step 3: Relate linear acceleration and angular acceleration
For a rolling object, the linear acceleration \( a \) and angular acceleration \( \alpha \) are related by:
\[ a = r \alpha \]
where \( r \) is the radius of the sphere.

### Step 4: Use the moment of inertia
The moment of inertia \( I \) of a solid sphere about its center is given by:
\[ I = \frac{2}{5} m r^2 \]
The torque \( \tau \) due to the frictional force \( f \) is:
\[ \tau = f \cdot r = I \alpha \]
Substituting \( \alpha = \frac{a}{r} \):
\[ f \cdot r = \frac{2}{5} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ f = \frac{2}{5} m a \]

### Step 5: Substitute the frictional force back into the equation of motion
Now, substituting \( f \) back into the net force equation:
\[ mg \sin \theta - \frac{2}{5} m a = m a \]
This simplifies to:
\[ mg \sin \theta = m a + \frac{2}{5} m a \]
\[ mg \sin \theta = m a \left(1 + \frac{2}{5}\right) \]
\[ mg \sin \theta = m a \cdot \frac{7}{5} \]

### Step 6: Solve for acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{5}{7} g \sin \theta \]

### Final Result
Thus, the acceleration of the sphere as it reaches the bottom of the incline is:
\[ a = \frac{5}{7} g \sin \theta \]

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