= Solution
{c}
Let $a$ be the <center of mass> <acceleration> downhill and $F$ the uphill <static friction>. Translation and rotation about the <center of mass> obey
$$
ma=mg\sin\theta-F,\qquad I\alpha=Fr.
$$
The <rolling without slipping> constraint is $a=r\alpha$. For the coin's <moment of inertia> $I=mr^2/2$, this gives $F=ma/2$, so
$$
\boxed{a=\frac23g\sin\theta,\qquad F=\frac13mg\sin\theta.}
$$
The normal <force> is $N=mg\cos\theta$. <Static friction> can enforce <rolling without slipping> only if $F\le\mu N$, giving
$$
\frac13mg\sin\theta\le\mu mg\cos\theta,
\qquad
\boxed{\theta_{\max}=\arctan(3\mu).}
$$
The <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 $\mu\ge0$ the result lies between $0$ and $\pi/2$.
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