Torque and inverted pendulum

Question

I'm supposed to derive a relationship in which the change in gravity $\delta g$ is linked to the change in length of the spring $\delta s$. When the beam is tilted due to increased weight or gravity, the extra weight will tilt, and this will enhance the rotation. I understand this so far, but I can't figure out how to connect these two system. Without the extra mass one could easily used the Hooke's law $\Delta F=m\delta g=\kappa \delta s$. I can't figure out how to fit the extra weight in this. For an inverted pendulum $\tau=mgl\sin\theta$. How do I relate this with the Hooke's law?

Or am I wrong from the beginning? In a stable system without the extra weight, should I still consider not only Hooke's law but also the torque?

Answer 1

I assume that by "extra weight", you mean the mass $m$. If we assume the displacement $\theta$ is small, $M$ will exert a torque of $Mgl \cos(\theta) \sim Mgl$ (with $g$ being either the original $g$ or $g + \delta g$), and $m$ will exert a torque of $mgh sin(\theta) \sim mgh \alpha$. The spring will exert a torque of $Fl cos(\theta) \sim Fl$ with F being the Hooke's law force. It should be easy to create the equation from here, as in equilibrium you want zero torque, not zero force.