A cylinder rolls down an inclined plane of inclination `30^@`, the acceleration of cylinder is

Options

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

Answer

To find the acceleration of a cylinder rolling down an inclined plane with an inclination of \(30^\circ\), we can follow these steps:

### Step 1: Identify the forces acting on the cylinder
When the cylinder rolls down the incline, the forces acting on it include:
- The gravitational force acting downward, which can be resolved into two components:
- Perpendicular to the incline: \(mg \cos \theta\)
- Parallel to the incline: \(mg \sin \theta\)

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

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

### Step 4: Apply Newton's second law for rotation
The torque \(\tau\) due to friction is given by:
\[ \tau = F_{\text{friction}} \cdot r = I \alpha \]
where \(I\) is the moment of inertia of the cylinder. For a solid cylinder, \(I = \frac{1}{2} m r^2\).

### Step 5: Substitute angular acceleration
Substituting \(\alpha\) from the relation \(a = r \alpha\) into the torque equation gives:
\[ F_{\text{friction}} \cdot r = \frac{1}{2} m r^2 \cdot \frac{a}{r} \]
This simplifies to:
\[ F_{\text{friction}} = \frac{1}{2} m a \]

### Step 6: Substitute back into the equation of motion
Now, substituting \(F_{\text{friction}} = \frac{1}{2} m a\) into the net force equation:
\[ mg \sin \theta - \frac{1}{2} m a = m a \]
This simplifies to:
\[ mg \sin \theta = \frac{3}{2} m a \]

### Step 7: Solve for acceleration
Dividing both sides by \(m\) and rearranging gives:
\[ a = \frac{2}{3} g \sin \theta \]

### Step 8: Substitute the angle
Now substituting \(\theta = 30^\circ\):
\[ \sin 30^\circ = \frac{1}{2} \]
Thus,
\[ a = \frac{2}{3} g \cdot \frac{1}{2} = \frac{g}{3} \]

### Final Answer
The acceleration of the cylinder rolling down the inclined plane is:
\[ \boxed{\frac{g}{3}} \]