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

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: $x = 10$