I am modeling the inverted pendulum on a moving cart using Lagrangian methods. I see most examples model the pendulum's kinetic energy as a sum of translational and rotational components (using a $I\dot\theta^2$ term), but I also encounter examples that only use a sum the translational energies (see diagram and text snippet).
Is one method preferred over another? What assumptions would factor into the choice? Comparing both, I arrived at EOM for each with what seemed like a non-trivial difference - namely the $mL^2\dot\theta^2$ being smaller by a factor of two in the translational-only approach.
Update - I have added details of my model's kinetic energy for feedback:
The pendulum bob's position vector and resulting squared-velocity: $$ \vec p= \begin{bmatrix} x+l\sin(\theta) \\ l \cos(\theta) \end{bmatrix} $$
$$ v^2 = (\dot x + l\dot\theta \cos(\theta))^2 + \dot\theta^2 l^2 \sin^2(\theta) $$
KE: $$ KE=\frac{1}{2}M\dot x + \frac{1}{2}m v^2 + \frac{1}{2}I\dot\theta^2 $$
Which becomes: $$ KE=\frac{1}{2}M\dot x + \frac{1}{2}m(\dot x^2+2l\dot x\dot\theta \cos(\theta)+l^2\dot\theta^2) + \frac{1}{2}m l^2 \dot\theta^2 $$
I suspect I have too many terms here.