Question

A cylinder is rolling down an inclined plane of inclination \( 60^\circ \). Its acceleration during rolling down will be \( \frac{x}{\sqrt{3}} \, \text{m/s}^2 \), where \( x = \) _____. (Use \( g = 10 \, \text{m/s}^2 \)).

Answer 1

Correct Answer: 10

Step 1: Formula for Acceleration in Rolling Motion:

- For a cylinder rolling down an incline, the acceleration \( a \) is given by:

\[ a = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}} \]

- For a solid cylinder, \( I_{cm} = \frac{1}{2} MR^2 \).

Step 2: Substitute Values:

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

- Given \( g = 10 \, \text{m/s}^2 \) and \( \theta = 60^\circ \) (so \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)):

\[ a = \frac{10 \times \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} \]

Step 3: Calculate \( a \):

\[ a = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{10 \sqrt{3}}{3} = \frac{x}{\sqrt{3}} \]

- Therefore, \( x = 10 \).

So, the correct answer is: \( x = 10 \)