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, we can use the following steps:

### Step 1: Identify the given values
- Inclination angle, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \)

### Step 2: Write the formula for the acceleration of a rolling object
The formula for the acceleration \( a \) of a rolling object down an incline is given by:

\[ a = \frac{g \sin \theta}{1 + \frac{k^2}{r^2}} \]

Where:
- \( k \) is the radius of gyration
- \( r \) is the radius of the cylinder

### Step 3: Determine the moment of inertia for a solid cylinder
The moment of inertia \( I \) for a solid cylinder about its axis is:

\[ I = \frac{1}{2} m r^2 \]

### Step 4: Calculate the radius of gyration \( k \)
For a solid cylinder, the radius of gyration \( k \) is given by:

\[ k = \sqrt{\frac{I}{m}} = \sqrt{\frac{\frac{1}{2} m r^2}{m}} = \frac{r}{\sqrt{2}} \]

### Step 5: Substitute \( k^2 \) into the acceleration formula
Now, we can substitute \( k^2 \) into the acceleration formula:

\[ k^2 = \left(\frac{r}{\sqrt{2}}\right)^2 = \frac{r^2}{2} \]

Thus, the term \( \frac{k^2}{r^2} \) becomes:

\[ \frac{k^2}{r^2} = \frac{\frac{r^2}{2}}{r^2} = \frac{1}{2} \]

### Step 6: Substitute \( \sin \theta \) and \( \frac{k^2}{r^2} \) into the acceleration formula
Now, substituting \( \theta = 30^\circ \) and \( \sin 30^\circ = \frac{1}{2} \):

\[ a = \frac{g \sin 30^\circ}{1 + \frac{1}{2}} = \frac{g \cdot \frac{1}{2}}{1 + \frac{1}{2}} = \frac{g \cdot \frac{1}{2}}{\frac{3}{2}} = \frac{g}{3} \]

### Step 7: Calculate the numerical value of acceleration
Substituting \( g \approx 9.81 \, \text{m/s}^2 \):

\[ a = \frac{9.81}{3} \approx 3.27 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the cylinder rolling down the inclined plane is approximately \( 3.27 \, \text{m/s}^2 \).

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