An inverted physical pendulum is deviated by a small angle $\varphi$ and connected to an oscillating base with oscillation function $a(t)$. The pendulum's mass is $m$ and its center of mass is $l$ units away from the oscillating base (see picture below).
The kinetic energy of the system is given by: \begin{align} K &= E_{\mathrm{tran}} + E_{\mathrm{rot}} &\\ &= \frac{1}{2} \left( m v^2 + \mathcal{I}_0 \dot{\varphi}^2 \right) \end{align} My question is if $\mathcal{I}_0$ is the moment of inertia around the pivot or the center of mass. I'm stuck on deriving the correct form of the Lagrangian, when I handle $\mathcal{I}_0$ as the moment of inertia around the pivot only if I assume that its the moment of inertia around the center of mass the equation is correct. Can someone please explain this?