Slip Less!

Taking Torque about A,

$\displaystyle\tau = mg\dfrac{l}{2}\cos\theta = \dfrac{ml^2}{3}\alpha$

$\displaystyle\begin{aligned}\Rightarrow \alpha &= \dfrac{3g\cos\theta}{2l}\\ &=\dfrac{3g}{4l}\end{aligned}$

$\alpha = \dfrac{3g}{4l} \cdots(1)$

Taking Forces at Centre of Mass, (since forces act on the centre of mass)

Acceleration of centre of mass is perpendicular to the rod because it is performing a pure rotation about the point of contact. $\vec{a} =\vec{\alpha} \times \vec{r}$

$\displaystyle\begin{aligned}m\alpha \dfrac{l}{2}\cos\theta &= mg - \mathbb{N}\\(1)\Rightarrow \dfrac{3mg}{16} &= \dfrac{16mg}{16} - \mathbb{N}\\ \mathbb{N}&=\dfrac{13mg}{16}\end{aligned}$

Applying $F=ma$ for the horizontal forces:

$\displaystyle\begin{aligned}\mu \mathbb{N} &= \dfrac{m\alpha l\sin\theta}{2}\\ \text{i.e }\mu\dfrac{13}{16}mg &= \dfrac{3\sqrt{3}}{16}mg\end{aligned}$

$\displaystyle\Rightarrow \boxed{\mu_s = \dfrac{3\sqrt{3}}{13}} \approx 0.3997\\$

For a general theta i found the solution as u=3sinx.cosx/(1+3sin^2(x)) where x=theta.