Physics - Rotation Question with Solution | TestHub

For rolling motion, the acceleration of the cylinder along the inclined plane can be expressed as

$a=\frac{gsin\theta }{1+\frac{I_{cm}}{M{R}^2}} ...\left(1\right)$

The moment of inertia of the cylinder can be written as

$I_{cm}=\frac{1}{2}M{R}^2 ...\left(2\right)$

From equations (1) and (2), it follows that

$\begin{array}{rcl}a& =& \frac{gsin\theta }{1+\frac{1}{2}}\\ & =& \frac{2\times 10\times \frac{\sqrt{3}}{2}}{3}\\ & =& \frac{10}{\sqrt{3}} \mathrm{m} {\mathrm{s}}^{-2}\end{array}$$$

Therefore, $x=10$.