A uniform solid disk rolling down an incline making angle $\mathrm{\theta }$ with the horizontal. The minimum coefficient of friction required to maintain pure rolling motion for the disk is

Torque about the CM is caused by friction because the lever arm of the weight force is zero:

$\mathrm{\tau } = \mathrm{fR} = \mathrm{I\alpha }$

$\mathrm{f} = \mathrm{\mu N} = \mathrm{\mu mgcos} \mathrm{\theta }$

$\mathrm{\mu } = \frac{\mathrm{f}}{\mathrm{mgcos} \mathrm{\theta }} = \frac{{\displaystyle \frac{\mathrm{I\alpha }}{\mathrm{R}}}}{\mathrm{mg} \cos \mathrm{\theta }}$

    = $\frac{({\displaystyle \frac{2}{3}}\mathrm{g} \sin \mathrm{\theta })\left({\displaystyle \frac{1}{2}}{\mathrm{mR}}^2\right)}{{\mathrm{R}}^2\mathrm{mgcos\theta }} = (\frac{1}{3}\tan \mathrm{\theta })$