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)

Remember that acceleration of a cylinder down a smooth inclined plane is
$a=\frac{gsin\theta }{\left(1+\frac{I}{m{R}^2}\right)}$ where $I=\frac{m{R}^2}{2}$ is the moment of Inertia for cylinder
$a=\frac{gsin30^{\circ }}{\left(1+\frac{m{R}^2}{2}\times \frac{1}{m{R}^2}\right)}=\frac{g\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{g}{3}$