A uniform rod of length L rests against a smooth roller, as shown...

Let the mass of the rod be $m$ and its length be $L$.

The forces acting on the rod are:
1. Weight $mg$ acting vertically downwards at the center of mass, at a distance $\dfrac{L}{2}$ from the lower end.
2. Normal reaction $N_1$ acting vertically upwards at the lower end.
3. Frictional force $f$ acting horizontally inwards at the lower end.
4. Normal reaction $N_2$ from the smooth roller, acting perpendicular to the rod.

The distance of the point of contact of the roller from the lower end along the rod is $x = \dfrac{h}{\sin\theta}$.

Taking torque about the lower end of the rod:
$N_2 \left( \dfrac{h}{\sin\theta} \right) - mg \left( \dfrac{L}{2} \cos\theta \right) = 0$
$\Rightarrow N_2 = \dfrac{mg L \cos\theta \sin\theta}{2h}$

For horizontal equilibrium:
$f = N_2 \sin\theta$

For vertical equilibrium:
$N_1 + N_2 \cos\theta = mg$
$\Rightarrow N_1 = mg - N_2 \cos\theta$

For the minimum angle $\theta$, the rod is on the verge of slipping, so $f = \mu N_1$.
$\mu = \dfrac{f}{N_1} = \dfrac{N_2 \sin\theta}{mg - N_2 \cos\theta}$

Substituting the value of $N_2$:
$\mu = \dfrac{\left( \dfrac{mg L \cos\theta \sin\theta}{2h} \right) \sin\theta}{mg - \left( \dfrac{mg L \cos\theta \sin\theta}{2h} \right) \cos\theta}$

$\mu = \dfrac{\dfrac{L \cos\theta \sin^2\theta}{2h}}{1 - \dfrac{L \cos^2\theta \sin\theta}{2h}}$

$\mu = \dfrac{L \cos\theta \sin^2\theta}{2h - L \cos^2\theta \sin\theta}$