Rod's Kinetic Energy in the pendulum problem On the MIT OCW for engineer dynamics the cart pendulum problem is solved by using the Lagrange Method here. This is a 2D problem, so rotation only occurs on the Z axis. While obtaining the rotational Kinetic Energy that is defined as:
$V_{r} = \frac{1}{2}I_{zz}\dot{\theta}^{2}$
Where $I_{zz}$ is the zz element of the inertia matrix/tensor and $\dot{\theta}$ is the angular velocity.
The only doubt I have is related to which Inertia Matrix (of the rod) must be used for this problem, this is solved on the MIT2_003SCF11_rec8notes1.pdf from previous link by defining the rod's moment of inertia as:
$^{G}I_{zz}=\frac{1}{12}m_{2}L^{2}$
Instead of:
$^{A}I_{zz}=\frac{1}{3}m_{2}L^{2}$
Where $m_{2}$ is the mass of the rod
For calculating the kinetic energy of the rod due to rotation why we need to use $^{G}I_{zz}$ instead of $^{A}I_{zz}$?? the rod is rotating around the fixed point A, it is not rotating around its own center of mass?? rigth?
Note: the course professor gives an explanation in this video at 10:07 but it is still no clear for me. Look at the last term of T equation from the video if you can:
$T = \frac{1}{2}m_{1}V_{A/o}^{2} + \frac{1}{2}m_{2}(V_{G/o}V_{G/o}) +\frac{1}{2}\omega ^{G}H$

 A: Why moment of inertia cannot be described by the rotation around A
A is moving. You cannot have a separate rotation term if it depends explicitly on the velocity of point A. This becomes more intuitive if you consider the kinetic energy for a point particle on the rod. It is important to keep in mind that the kinetic energy is quadratic: Intuitively you can expect that if the velocity of A and the rotational velocity of the rod is coupled, then the expressions of the kinetic energy of A and the rotation of the rod will become coupled as well.
Why you can consider the rotation around G
If the rotation term is not allowed to depend on the velocity of A, then there is only one option. The only component of rotation independent of the velocity of A is the rotation of the rod around itself (around G).
Is it the only way?
I suppose you could write an expression of the kinetic energy expressed in terms of your desired rotation around A, but this would result in some comprehensive expression where the rotation is somehow coupled to the velocity of A (not recommended).
I actually have this system simulated on my page
https://zymplectic.com/ with a pendulum bob (press "(3) pendulum cart"). The Lagrangian for the cart with a pendulum bob is simpler to derive, because you would express the kinetic energy in terms of the combined Cartesian velocity of the bob.
A: In a previous answer of mine,  I tried to show how the inertia matrix, containing the various moments of inertia of a solid body, can be derived from "first principles". In the second half of that post, I have also included the derivation of the equations of motion for a rotating bar, attached at one end. I use derivations from Newtonian equations, rather than Lagrangian interpretation, but maybe it can still explain why when a rod is rotating around one of its end, one uses the moment of inertia relative to that rod's end. Maybe this will help you get another perspective.
A: Use parallel axis theorem to calculate the K.E of the rod about point A. The additional term will be 1/12ml^2.
