Lagrangian of an inverted pendulum on a moving cart

So I have been trying to derive the equations of motion of the inverted physical pendulum in a cart, but I seem to be confused about the derivation of its Kinetic Energy. I know this physical system is very popular and while I have searched and searched I couldn’t find an answer to my question anywhere.

So I divided the kinetic energy into the cart’s and pendulum’s:
$$ T = T_C + T_P $$

The cart’s one is pretty straight forward $T_C = 1/2 M dot{x}^2$, where I am denoting $x$ the horizontal coordinate of the cart’s point mass.

My trouble is now with the pendulum’s Kinetic Energy. I would assume I would have to sum the translational energy of the pivot point $T_{pivot}=1/2 m dot{x}^2$ to the rotational energy of the pendulum $T_{rot} = 1/2 I dot{theta}^2$, where $I$ is the moment of inertia of the pendulum with respect to the pivot point (Note: the angle $theta$ i chose is with respect to the upper vertical, unlike in the image up there).

With this I got:
$$ mathcal{L} = frac{1}{2}(M+m) dot{x}^2 + frac{1}{2} I dot{theta}^2 – mglcostheta $$

And therefore the equations of motion:
$$ (M+m) ddot{x} = F(t) $$
$$ I ddot{theta} – mgl sintheta = 0 $$

These equations, though, seem too simple compared to the equations I have seen out there for this problem.
I would really appreciate if someone could point out my mistakes.