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The planar spring-pendulum is modeled by the set of equations $$m\ddot{r}=mr\dotθ^2 + mg\cos\theta − k(r − L)$$ and, $$r^2\ddot{\theta}= −2r\dot r\dotθ − gr\sin\theta$$ First we should define the angular momentum by $p_\theta = mr^2\dot\theta$ and the radial momentum by $p_r = m\dot r$ and rewrite the spring-pendulum system as a set of four first-order ODEs for $x = (r, \theta , p_r, p_\theta )$. So I did like this, but I don't know if it's correct: $$\dot r = p_r/m, ~\dot\theta=p_\theta/mr^2$$ and, $$\dot p_r = mr\dot\theta^2 + mg\cos\theta - k(r-L)$$ and, $$\dot p_\theta = m(-2r\dot r\dot\theta - gr\sin\theta)$$ Now I have to find the equilibrium solution(s), $x_{eq}$ , of the equations, i.e., those solutions for which $x$ is constant.

How do I find these solutions?

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  • $\begingroup$ Check my edit please $\endgroup$ – Nosrati Feb 2 '17 at 10:41
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By the product rule, $$ \dot p_θ=\frac{d}{dt}(mr^2\dotθ)=2mr\dot r\dotθ+mr^2\ddotθ=-mgr\sinθ $$ With your definitions you should also eliminate all of $\dot r, \dotθ$ for $p_r, p_θ$ from the final form of the equations.

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  • $\begingroup$ When I eliminate θ˙ from p˙_r, I get an expression with p_θ: p˙_r = p_θ^2/mr^3 + mgcosθ - k(r-L). Is that correct? $\endgroup$ – Siri Feb 2 '17 at 13:03
  • $\begingroup$ Yes. For the equilibrium solutions it would also be sufficient to set $\dot r=\ddot r=\dotθ=\ddotθ=0$ in the original equations. $\endgroup$ – Lutz Lehmann Feb 2 '17 at 13:43
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Actually, we should have $\dot{p}=-mgr\sin\theta$ and then, put all the first order ODE to $0$, and then the equilibrium result should be $$ r=\frac{mg}{K}+L; $$ $$ \theta=n\pi;$$ and finally $$ p_r=p_\theta=0$$

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