homework and exercises - Condition that the Lagrangian energy function $hequivsum_ifrac{partial L}{partialdot q_i}dot q_i-L$ would be same as the mechanical energy $E$

I'm studying Classical Mechanics by Goldstein. I solved a problem but I have a question.

Pro 2.18
A point mass is constrained to move on a massless hoop of radius a fixed in a vertical plane that rotates about its vertical symmetry axis with constant angular speed ω. Obtain the Lagrange equations of motion assuming the only external forces arise from gravity. What are the constants of motion? Show that if ω is greater than a critical value ω0, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if ω < ω0, the only stationary point for the particle is at the bottom of the hoop. What is the value of ω0?

So I proceeded like this solution here.

So here, when we choose only one generalized coordinate $\theta$(polar angle), energy function $h$ is not same as the energy. But in the text (chapter about Lagrangian) it says that if potential $V=V(q)$, $h=E$. For this problem $V=mga \cos \theta$, (or negative, according to how define $\theta$ or axis) so it satisfies the condition that potential only depends on generalized coordinate, not on generalized velocity. So $h$ should be $E$, but apparently not. What is wrong here?

I know that if I set the azimuthal angle as an independent variable, such a contradiction doesn't appear. But I cannot see why I should do that ( the problem says that azimuthal angle is not independent variable, and derivation of $h=E$ says nothing about that.)

Surely something must be wrong with my reasoning, because if the Lagrangian of a system is $L=\frac{1}{2}my'^2+mgy$, we can insert the constant horizontal kinetic energy $\frac{1}{2}mx'^2$ (x is not generalized coordinate here), but that would destroy $h=E$. Can someone explain this to me?