A disc of radius R is rolling down an inclined plane whose angle of inclination is `theta` Its acceleration would be

Options

A. `(5)/(7)gsintheta`
B. `(2)/(3)gsintheta`
C. `(1)/(2)gsintheta`
D. `(3)/(5)gsintheta`

Answer

To find the acceleration of a disc rolling down an inclined plane, we can follow these steps:

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

### Step 2: Write the equation of motion
Since the disc is rolling without slipping, we can apply Newton's second law. The net force acting down the incline is equal to the mass times the acceleration:
\[ F_{\parallel} = ma \]
Substituting the expression for \( F_{\parallel} \):
\[ mg \sin \theta = ma \]

### Step 3: Consider the rotational motion
For a rolling object, we also need to consider its rotational motion. The torque \( \tau \) about the center of mass due to the gravitational force is given by:
\[ \tau = r \cdot F_{\parallel} = r \cdot mg \sin \theta \]
This torque causes an angular acceleration \( \alpha \) given by:
\[ \tau = I \alpha \]
where \( I \) is the moment of inertia of the disc. For a solid disc, \( I = \frac{1}{2} m r^2 \).

### Step 4: Relate linear and angular acceleration
Since the disc rolls without slipping, the linear acceleration \( a \) and angular acceleration \( \alpha \) are related by:
\[ a = r \alpha \]
Thus, we can express \( \alpha \) as:
\[ \alpha = \frac{a}{r} \]

### Step 5: Substitute into the torque equation
Substituting \( \alpha \) into the torque equation:
\[ r \cdot mg \sin \theta = I \cdot \frac{a}{r} \]
Substituting \( I = \frac{1}{2} m r^2 \):
\[ r \cdot mg \sin \theta = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ mg \sin \theta = \frac{1}{2} m a \]

### Step 6: Solve for acceleration \( a \)
Now we can solve for \( a \):
\[ a = 2g \sin \theta \]

### Step 7: Combine linear and rotational motion
To find the total acceleration, we need to consider both the translational and rotational aspects. The general formula for the acceleration of a rolling object is:
\[ a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \]
For a disc, substituting \( I = \frac{1}{2} m r^2 \):
\[ a = \frac{g \sin \theta}{1 + \frac{1/2 m r^2}{m r^2}} = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Final Answer:
The acceleration of the disc rolling down the inclined plane is:
\[ a = \frac{2g \sin \theta}{3} \]