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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).

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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?

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You do not need to write rotational kinetic energy separately,

$$T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$$

Should suffice, and just write x, y coordinates as a fn of time.

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We can derive the expression for the rotational kinetic energy and find it out together. Imagine we have a particle, rotating about a point $O$ with rotational speed $\omega_o$ and $r_o$ units away from $O$. Its linear velocity will be $V = r_o\omega_o$, and therefore its kinetic energy will be $$K = \tfrac{1}{2}m V^2 = \tfrac{1}{2}m (r_o\omega_o)^2$$

Now, instead of one particle, imagine a body composed by multiple particles, all with the same angular speed $\omega$. Each particle's kinetic energy will be

$$dK = \tfrac{1}{2}r_o^2 \omega_o^2 dm$$

And the kinetic energy of the whole body

$$K = \omega_o^2 \int\limits_{\text{body}} \tfrac{1}{2}r_o^2 dm$$

But notice that the integral part only depends on the geometry and mass distribution of the body, therefore we can assign a property $\mathcal{I}_o$ called mass moment of inertia of the rotation around the pivot point $O$.

So the final answer is that you must use the pivot point to calculate the moment of inertia.

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