A thin uniform circular ring is rolling down an inclined plane of inclination `30^(@)` without slipping. Its linear acceleration along the inclined plane will be

Options

A. `g/2`
B. `g/3`
C. `g/4`
D. `(2g)/3`

Answer

To find the linear acceleration of a thin uniform circular ring rolling down an inclined plane of inclination \(30^\circ\) without slipping, we can follow these steps:

### Step 1: Draw the Free Body Diagram
- Start by drawing the inclined plane and the circular ring on it.
- Identify the forces acting on the ring:
- The weight of the ring \(mg\) acting vertically downwards.
- The normal force \(N\) acting perpendicular to the inclined plane.
- The frictional force \(f\) acting up the incline (since the ring rolls down without slipping).

### Step 2: Resolve the Weight into Components
- The weight \(mg\) can be resolved into two components:
- Perpendicular to the incline: \(mg \cos \theta\)
- Parallel to the incline: \(mg \sin \theta\)
- For \(\theta = 30^\circ\):
- \(mg \sin 30^\circ = \frac{mg}{2}\)
- \(mg \cos 30^\circ = \frac{mg \sqrt{3}}{2}\)

### Step 3: Apply Newton's Second Law
- For the linear motion along the incline, we can write:
\[ mg \sin \theta - f = ma \]
- Here, \(f\) is the frictional force and \(a\) is the linear acceleration of the ring.

### Step 4: Relate Linear Acceleration to Angular Acceleration
- Since the ring rolls without slipping, the relationship between linear acceleration \(a\) and angular acceleration \(\alpha\) is given by:
\[ a = r \alpha \]
- For a thin ring, the moment of inertia \(I\) about its center is:
\[ I = m r^2 \]

### Step 5: Apply Torque Equation
- The torque \(\tau\) due to friction about the center of the ring is:
\[ \tau = f r \]
- According to the rotational dynamics, this torque is also equal to:
\[ \tau = I \alpha = m r^2 \alpha \]
- Thus, we can write:
\[ f r = m r^2 \alpha \]
- Rearranging gives:
\[ f = m r \alpha \]

### Step 6: Substitute \(f\) in the Linear Motion Equation
- Substitute \(f\) back into the linear motion equation:
\[ mg \sin \theta - m r \alpha = ma \]
- Since \(a = r \alpha\), we can replace \(\alpha\):
\[ mg \sin \theta - m \frac{a}{r} r = ma \]
- Simplifying gives:
\[ mg \sin \theta = ma + ma \]
\[ mg \sin \theta = 2ma \]

### Step 7: Solve for Linear Acceleration \(a\)
- Rearranging gives:
\[ a = \frac{g \sin \theta}{2} \]
- Substituting \(\theta = 30^\circ\):
\[ a = \frac{g \cdot \frac{1}{2}}{2} = \frac{g}{4} \]

### Final Answer
Thus, the linear acceleration of the ring along the inclined plane is:
\[ \boxed{\frac{g}{4}} \]