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I wonder how to properly write the motion equations for the inverted pendulum on a cart in case of overdamped dynamics. Imagine the system illustrated in Wikipedia placed in a liquid with high viscosity $\beta$. I completely understand how the system with the fixed pendulum base behaves. Let's start from the equation for the classic inverted pendulum: $$ml^2 \ddot \theta - \beta \dot \theta + mgl\sin\theta=0.$$ High values of $\beta$ allow one to neglect the inertia term, so that $$- \beta \dot \theta + mgl\sin\theta = 0.$$

However, as a person who is intrinsically bad at physics, I am greatly confused about what effect the motion of the cart has on the dynamics. The derivations here are for the classic, non-overdamped case, I guess that here everything should be much simpler. Any help is appreciated.

Update: Probably the equations I ask about should take the simplest possible form in the frame of reference centered at the moving base of the pendulum.

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  • $\begingroup$ I think you are forgetting the input $x(t)$ of the cart position. In the end you can find the desired $\ddot{x}(t)$ for given motion $\theta(t)$, $\dot{\theta}(t)$, and $\ddot{\theta}(t)$. $\endgroup$
    – John Alexiou
    Jun 28, 2013 at 13:12
  • $\begingroup$ Evidently, the input $x(t)$ affects the motion of the pendulum, the question is how. At the same time $\ddot x$ (as well as $\ddot \theta$) has no sense due to high viscosity. So the equation for the system motion should comprise at most $\theta$, $\dot \theta$, $x$ and $\dot x$. If we consider the frame of reference centered at the moving cart, we eliminate $x$. But this is where I am stuck. I have no idea how to describe the effect of the torque applied to the cart. $\endgroup$
    – Chester
    Jun 29, 2013 at 15:12
  • $\begingroup$ Remember that any assertion you make as far as the inertia being not important etc, is just an assertion. Start from the fundamentals and then see what happens when $\beta \dot \theta/L > m L \dot{\theta}^2 $ (in force units). $\endgroup$
    – John Alexiou
    Jun 30, 2013 at 2:17

1 Answer 1

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Using $\sum F=m a$ and $\sum M = I \ddot{\theta}$ I arrive at

$$ F = \left( \frac{M+m}{m} \frac{I+m L^2}{L \cos\theta} - m L \cos \theta \right) \ddot{\theta} + \left( \frac{M+m}{m} \frac{\beta \dot\theta}{L \cos\theta} + m L \sin\theta \dot{\theta}^2 \right) $$

now you can play with the values to get what you need.

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  • $\begingroup$ Thank you for the reply. What is $F$ in the left-hand side of your answer? Is it the force applied to the moving cart? $\endgroup$
    – Chester
    Jul 1, 2013 at 5:32
  • $\begingroup$ Yes, $F$ is the horizontal force (control) acting on the cart of mass $M$. $\endgroup$
    – John Alexiou
    Jul 1, 2013 at 14:10

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