Question

A cylinder rolls down an inclined plane of inclination ${30}^{\circ }$, the acceleration of cylinder is

• A. $\frac{g}{3}$
• B. $g$
• C. $\frac{g}{2}$
• D. $\frac{2g}{3}$

Answer 1

Answer: (A)

$\frac{g}{3}$

Remember that acceleration of a cylinder down a smooth inclined plane is
$a=\frac{g\sin \theta }{(1+\frac{I}{m{R}^2})}$ where $I=\frac{m{R}^2}{2}$ is the moment of
Inertia for
cylinder $a=\frac{g\sin {30}^{\circ }}{(1+\frac{m{R}^2}{2}\times \frac{1}{m{R}^2})}=\frac{g\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{g}{3}$