# Torque and inverted pendulum

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?

• Is the system in equilibrium? If it's in equilibrium then shouldn't the net torque be zero with respect to any point? Jul 7 '20 at 5:47

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.