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

Options

A. 5 `SqrtX 3 /14 g`
B. `g/ 3`
C. `g/2`
D. 5g/14

Answer

To find the acceleration of a solid cylinder rolling down an inclined plane with an inclination of \(60^\circ\), we can use the principles of energy conservation and the equations of motion.

### Step-by-step Solution:

1. **Identify the Forces Acting on the Cylinder:**
The gravitational force acting on the cylinder can be resolved into two components: one perpendicular to the incline and one parallel to the incline. The parallel component causes the cylinder to roll down the incline.

\[ F_{\text{parallel}} = mg \sin(60^\circ) \]

2. **Apply Conservation of Energy:**
The total mechanical energy at the top of the incline (potential energy) will be equal to the total mechanical energy at the bottom (kinetic energy). The initial potential energy is given by:

\[ PE_{\text{initial}} = mgh \]

The kinetic energy at the bottom consists of translational and rotational components:

\[ KE_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]

For a solid cylinder, the moment of inertia \(I\) is:

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

And for pure rolling, we have:

\[ \omega = \frac{v}{r} \]

Substituting \(I\) and \(\omega\) into the kinetic energy expression gives:

\[ KE_{\text{final}} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

3. **Set Up the Energy Conservation Equation:**
Setting the initial potential energy equal to the final kinetic energy:

\[ mgh = \frac{3}{4} mv^2 \]

Canceling \(m\) from both sides gives:

\[ gh = \frac{3}{4} v^2 \]

4. **Relate Height to Distance Along the Incline:**
The distance \(s\) along the incline can be related to the height \(h\) using the sine of the angle:

\[ h = s \sin(60^\circ) \]

Therefore, we can express \(h\) as:

\[ h = s \cdot \frac{\sqrt{3}}{2} \]

5. **Substituting for \(h\):**
Substitute \(h\) back into the energy equation:

\[ g \left(s \cdot \frac{\sqrt{3}}{2}\right) = \frac{3}{4} v^2 \]

Rearranging gives:

\[ v^2 = \frac{4g s \sqrt{3}}{6} \]

6. **Use the Kinematic Equation:**
Using the kinematic equation \(v^2 = u^2 + 2as\) (where \(u = 0\)), we have:

\[ v^2 = 2as \]

Setting the two expressions for \(v^2\) equal gives:

\[ 2as = \frac{4g s \sqrt{3}}{6} \]

Canceling \(s\) (assuming \(s \neq 0\)) gives:

\[ 2a = \frac{4g \sqrt{3}}{6} \]

Simplifying yields:

\[ a = \frac{2g \sqrt{3}}{6} = \frac{g \sqrt{3}}{3} \]

7. **Final Result:**
The acceleration of the cylinder rolling down the incline is:

\[ a = \frac{g}{\sqrt{3}} \]