 Solution

Let $a$ be the center of mass acceleration downhill and $F$ the uphill static friction. Translation and rotation about the center of mass obeyThe rolling without slipping constraint is $a=r\alpha$. For the coin's moment of inertia $I=m{r}^2/2$, this gives $F=ma/2$, soThe normal force is $N=mg\cos \theta$. Static friction can enforce rolling without slipping only if $F\le \mu N$, givingThe static friction acts uphill to provide the torque; it is not automatically equal to $\mu N$, attaining that limiting value only at the threshold. For finite $\mu \ge 0$ the result lies between $0$ and $\pi /2$.