A cylinder is rolling down on a inclined plane of inclination`60^(@)`. What is iths acceleration?

Options

A. `g//3`
B. `g//sqrt3`
C. `sqrt((2g)/(3))`
D. None of these

Explanation

To find the acceleration of a cylinder rolling down an inclined plane with an inclination of \(60^\circ\), we can use the formula for the acceleration of a rolling object. Here are the steps to derive the acceleration:

### Step 1: Identify the relevant formula
For a solid cylinder rolling down an inclined plane, the acceleration \(a\) 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 (approximately \(9.81 \, \text{m/s}^2\)),
- \(\theta\) is the angle of inclination,
- \(k\) is the radius of gyration,
- \(r\) is the radius of the cylinder.

### Step 2: Determine the moment of inertia
For a solid cylinder, the moment of inertia \(I\) is given by:
\[ I = \frac{1}{2} m r^2 \]
From this, we can express the radius of gyration \(k\) as:
\[ k^2 = \frac{I}{m} = \frac{1}{2} r^2 \]

### Step 3: Substitute \(k^2/r^2\) into the formula
Now, substituting \(k^2\) into the acceleration formula:
\[ \frac{k^2}{r^2} = \frac{1/2 \cdot r^2}{r^2} = \frac{1}{2} \]
Thus, the formula for acceleration becomes:
\[ a = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

### Step 4: Calculate \(\sin 60^\circ\)
Now, we need to find \(\sin 60^\circ\):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 5: Substitute \(\sin 60^\circ\) into the acceleration formula
Substituting \(\sin 60^\circ\) into the acceleration formula:
\[ a = \frac{2g \cdot \frac{\sqrt{3}}{2}}{3} = \frac{g \sqrt{3}}{3} \]

### Step 6: Substitute \(g\) to find numerical value
Using \(g \approx 9.81 \, \text{m/s}^2\):
\[ a = \frac{9.81 \cdot \sqrt{3}}{3} \]
Calculating this gives:
\[ a \approx \frac{9.81 \cdot 1.732}{3} \approx \frac{17.00}{3} \approx 5.67 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the cylinder rolling down the inclined plane is approximately \(5.67 \, \text{m/s}^2\).

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