# What is the equivalent moment of inertia of a two-pivot pendulum?

I am working on a system that consists of a hanging-pendulum thrust balance with two pivot points, one in the front and one in the back There is a heavy weight applied in the front. I can change the position of the center of gravity by applying counterweights on the rear part. See attached image.. The position of the center of gravity is known as well as the moments of inertia with respect the pivot points

I would like to simplify this system to an equivalent physical pendulum with one pivot point. However, because I have two pivot points, I have two moments of inertia, one with respect to each one of the pivots.

In the beginning I thought about taking the mean average of both, and then calculate the period of a physical pendulum or compound pendulum, just as Hyperphysics suggests: $T=2\pi\sqrt{\frac{I_{y}}{mgLcm}}$, where

• $I_{y}$ is the average moment of inertia,
• $m$ is the mass of the system,
• $L_{cm}$ is the distance pivot-center of mass.

but this just does not seem right. Can somebody give me a hint on how to do this? Better, is it even reasonable to do such simplification? Shall I use a matrix notation? My purpose is to study how the thrust balance will respond when a force is applied upon it.

I have analyzed the oscillations of the system for several counterweights. From the study of those oscillations I can obtain the damping coefficient $\zeta$ and the natural frequency of the undamped system $\omega_{n}$, but I still need $I_{y}$ to complete the equation $\omega_{n}=\sqrt{\frac{k}{I_{y}}}$.

You have two moments which means you need to find the neutral axis which usually coincides with the system centroid along $$I_{xx}$$.
The difficulty is overcome by applying simple rules of statics and equilibrium. $$\Sigma M_A + \Sigma M_B=\Sigma M_{W}$$