What should be the minimum coefficient of static friction between the plane and the cylinder for the cylinder not to slip on an inclined plane?

Equation of motion

$\mathrm{Mg} \sin \mathrm{\theta }-\mathrm{f} = \mathrm{Ma}------\left(\mathrm{i}\right)$

Also fR = $\mathrm{\tau } = \mathrm{l\alpha } = {\mathrm{Mk}}^2\frac{\mathrm{a}}{\mathrm{R}}-----\left(\mathrm{ii}\right)$

But $\mathrm{a} = \frac{\mathrm{g} \mathrm{sin\theta }}{1+{\displaystyle \frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}}}----------\left(\mathrm{iii}\right)$

Putting value of a in equation (ii)

$\mathrm{f} = \left(\frac{{\mathrm{Mk}}^2}{{\mathrm{R}}^2}\right)\frac{\mathrm{g} \sin \mathrm{\theta }}{1+{\displaystyle \frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}}}$

For cylinder: ${\mathrm{Mk}}^2 = \mathrm{I} = \frac{1}{2}{\mathrm{MR}}^2$

${\mathrm{k}}^2 = \frac{1}{2}{\mathrm{R}}^2; \mathrm{f} = \frac{\mathrm{M}\left({\displaystyle \frac{1}{2}}\right)\mathrm{g} \sin \mathrm{\theta }}{(1+{\displaystyle \frac{1}{2}})} = \frac{1}{3}\mathrm{Mg} \sin \mathrm{\theta }$

In case of static friction, ${\mathrm{f}}_{\mathrm{s}} = \mathrm{\mu N} = \mathrm{\mu Mg} \cos \mathrm{\theta }$

$\frac{1}{3}\mathrm{Mg} \sin \mathrm{\theta } = \mathrm{\mu Mg} \cos \mathrm{\theta } \Rightarrow \mathrm{\mu } = \frac{1}{3}\mathrm{tan\theta }$