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

Options

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

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: Identify the formula for linear acceleration
The linear acceleration \(A\) of a body rolling down an inclined plane can be calculated using the formula:
\[ A = \frac{g \sin \theta}{1 + \frac{k^2}{r^2}} \]
where:
- \(g\) is the acceleration due to gravity,
- \(\theta\) is the angle of inclination,
- \(k\) is the radius of gyration,
- \(r\) is the radius of the ring.

### Step 2: Determine the radius of gyration for a thin circular ring
For a thin circular ring, the moment of inertia \(I\) is given by:
\[ I = m r^2 \]
where \(m\) is the mass of the ring. The radius of gyration \(k\) is defined as:
\[ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{m r^2}{m}} = r \]

### Step 3: Substitute the value of \(k\) into the acceleration formula
Now that we know \(k = r\), we can substitute this value into the acceleration formula:
\[ A = \frac{g \sin \theta}{1 + \frac{r^2}{r^2}} = \frac{g \sin \theta}{1 + 1} = \frac{g \sin \theta}{2} \]

### Step 4: Calculate \(\sin 30^\circ\)
We know that:
\[ \sin 30^\circ = \frac{1}{2} \]

### Step 5: Substitute \(\sin 30^\circ\) into the acceleration formula
Substituting \(\sin 30^\circ\) into the equation gives:
\[ A = \frac{g \cdot \frac{1}{2}}{2} = \frac{g}{4} \]

### Conclusion
Thus, the linear acceleration of the thin uniform circular ring rolling down the inclined plane is:
\[ A = \frac{g}{4} \]

### Final Answer
The correct answer is \(A = \frac{g}{4}\).

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