A solid cylinder of mass ‘m’ rolls without slippling down an inclined plane making an angle $\mathrm{\theta }$ with the horizontal. The fricitional force between the cylinder and the incline is

According to Newton's second law, a rolling cylinder's linear acceleration can be expressed as 

$\mathrm{mgsin\theta }-{\mathrm{f}}_{\mathrm{k}}=\mathrm{ma}$

 where θ  is the angle of inclination and this frictional force causes the cylinder to rotate at an angle of $\mathrm{\tau }=\mathrm{I\alpha }$ where α is the angular acceleration.

${\mathrm{Rf}}_{\mathrm{k}}=\mathrm{I\alpha }\Rightarrow \mathrm{\alpha }=\frac{{\mathrm{Rf}}_{\mathrm{k}}}{\mathrm{I}}$

We know that $\mathrm{\alpha }=\frac{\mathrm{a}}{\mathrm{R}}$
Now, by replacing in the above equation , we obtain $\frac{\mathrm{a}}{\mathrm{R}}=\frac{{\mathrm{Rf}}_{\mathrm{k}}}{\mathrm{I}}$

​${\mathrm{f}}_{\mathrm{k}}=\frac{\mathrm{aI}}{{\mathrm{R}}^2}$

$\mathrm{mgsin\theta }-\frac{\mathrm{aI}}{{\mathrm{R}}^2}=\mathrm{ma}$

$\mathrm{ma}+\frac{\mathrm{aI}}{{\mathrm{R}}^2}=\mathrm{mgsin\theta }$

$\mathrm{I}=\frac{1}{2}{\mathrm{mR}}^2$

$\mathrm{ma}+\frac{\mathrm{a}\left[\frac{1}{2}{\mathrm{mR}}^2\right]}{{\mathrm{R}}^2}=\mathrm{mgsin\theta }$

$\mathrm{ma}+\frac{1}{2}\mathrm{ma}=\mathrm{mgsin\theta }$

$\frac{3}{2}\mathrm{ma}=\mathrm{mgsin\theta }$

$\mathrm{a}=\frac{2}{3}\mathrm{gsin\theta }$

$\therefore \mathrm{mgsin\theta }-{\mathrm{f}}_{\mathrm{k}}=\mathrm{m}\left(\frac{2}{3}\mathrm{gsin\theta }\right)$

${\mathrm{f}}_{\mathrm{k}}=\mathrm{mgsin\theta }-\frac{2}{3}\mathrm{mgsin\theta }=\frac{\mathrm{mgsin\theta }}{3}$