Question 249 of 280

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

Your answerCheck answer

Show answer and solution▾

Answer: 10

For rolling motion, the acceleration of the cylinder along the inclined plane can be expressed as
$a=\dfrac{gsin\theta }{1+\dfrac{{I}_{cm}}{M{R}^{2}}}...\left(1\right)$
The moment of inertia of the cylinder can be written as
${I}_{cm}=\dfrac{1}{2}M{R}^{2}...\left(2\right)$
From equations (1) and (2), it follows that
$\begin{matrix}a & = & \dfrac{gsin\theta }{1+\dfrac{1}{2}} \\ & = & \dfrac{2\times 10\times \dfrac{\sqrt{3}}{2}}{3} \\ & = & \dfrac{10}{\sqrt{3}}m{s}^{-2}\end{matrix}$
Therefore, $x=10$.