If it doesn't slip, which means the base of the rod is fixed on the plane.
Let the mass of rod be $m$, the potential energy would be $E_p=mg\frac{b}{2}\mathrm{sin}\theta$, when the angle between rod and plane is $\theta$. Let the angular velocity of the rod be $\omega$, then the kinetic energy is $E_k=\frac{1}{2}I\omega^2$, which $I=\frac{1}{3}mb^2$. Then as the conservation of energy, we have$$E_p+E_k=E_{p0}=mg\frac{b}{2}$$
thus,$$\frac{1}{6}mb^2\omega^2=mg\frac{b}{2}(1-\mathrm{sin}\theta)$$ then,
$$\omega=\sqrt{\frac{3g(1-\mathrm{sin}\theta)}{b}}$$
this is how $\omega$ changes by $\theta$.
If it could slip on the plane, we assume the plane is absolutely smooth.
Let the velocity of the center of mass be $v_c$, we could easily know the horizontal component of $v_c$ must be $0$, because the conservation of the horizontal component of momentum.
We have the velocity of the base of the rod is $\vec{v_b}=\vec{v_c}+\vec{\omega}\times\vec{b}/2$. We noticed that $v_b$ must be horizontal, so the vertical component of $v_b$ is $v_{by}=v_c-\frac{1}{2}\omega b\cos \theta=0$, so,$$v_c=\frac{1}{2}\omega b\cos\theta$$
then the conservation of energy would be,$$\frac{1}{2}mv_c^2+\frac{1}{2}I_c \omega^2=mg\frac{b}{2}(1-\mathrm{sin}\theta)$$where $I_c=\frac{1}{12}mb^2$.Let's Put $v_c$ and $I_c$ in, we get,
$$\omega=\sqrt{\frac{12g(1-\sin\theta)}{b(3\cos^2\theta+1)}}$$
There is an assumption that the rod wouldn't leave the plane, which is true. But I'm not gonna prove it for brevity.
I guess this might be the answer you want. If you have any more questions, let me know.
