Q. A cylinder rolls down an inclined plane of inclination $30^{\circ}$, the acceleration of cylinder is
Solution:
Remember that acceleration of a cylinder down a smooth inclined plane is
$a=\frac{g\,sin\,\theta}{\left(1+\frac{I}{mR^{2}}\right)}$ where $I = \frac{mR^{2}}{2}$ is the moment of Inertia for cylinder
$a=\frac{g\,sin\,30^{\circ}}{\left(1+\frac{mR^{2}}{2}\times\frac{1}{mR^{2}}\right)}=\frac{g\times\frac{1}{2}}{1+\frac{1}{2}}=\frac{g}{3}$
