A cylinder is rolling down on an inclined plane of inclination $$60°$$. Its acceleration during rolling down will be $$\frac{x}{\sqrt{3}}$$ m s$$^{-2}$$, where $$x =$$ _______ (use $$g = 10$$ m s$$^{-2}$$).

For a body rolling down an inclined plane without slipping, the linear acceleration ($$a$$) of the COM is given by:

$$a = \frac{g \sin \theta}{1 + \frac{I}{MR^2}}$$

For a solid cylinder, the moment of inertia about its central axis is $$I = \frac{1}{2}MR^2$$.

$$a = \frac{g \sin 60^\circ}{1 + \frac{1}{2}} = \frac{g \sin 60^\circ}{3/2} = \frac{2}{3} g \sin 60^\circ$$

$$a = \frac{2}{3} \times 10 \times \frac{\sqrt{3}}{2} = \frac{10\sqrt{3}}{3} = \frac{10}{\sqrt{3}} \text{ m s}^{-2}$$

$$x = 10$$