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I'm trying to find the equilibrium points of a given system using Lagrangian mechanics (the system is still not rotating at the beginning). should I find the diagonal matrix for the characteristic matrix even though it is a complex one? or is there more elegant way to do that?

so this is what i got from the Euler Lagrange questions : enter image description here

enter image description here

The Lagrangian I found is:

$$ \mathcal L = T-V= \frac12 R^2(m\dot \theta_1 ^2+M\dot \theta_2^2+m\dot \theta_3^2)-(\frac12 k(\theta_1-\theta_2)^2+\frac12(\theta_2-\theta_3)^2+\frac12(\theta_1-\theta_3)^2+mgR\sin(\theta_1)+MgR\sin(\theta_2)+mgR\sin(\theta_3))$$

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  • $\begingroup$ Can you write your kinetic and potential energy? $\endgroup$
    – Eli
    Commented Dec 19, 2018 at 13:40
  • $\begingroup$ @Eli I added the Lagrangian. $\endgroup$
    – physics major
    Commented Dec 19, 2018 at 14:04
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    $\begingroup$ @physics major I think you should be careful with this Lagrangian because of angles are defined modulo $2\pi$. If $M\gg m$ you also can use physical intuition and suppose that equilibrium corresponds to the M being in the lowest position and two m positioned symmetrically around vertical axis. $\endgroup$
    – Gec
    Commented Dec 19, 2018 at 15:48
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    $\begingroup$ check your spring energy the spring force is: $\left[ \begin {array}{c} kR \left( \vartheta _1- \vartheta _2 \right) +kR \left( \vartheta _1-\vartheta _3 \right) \\ kR \left( \vartheta _3-\vartheta _2 \right) -kR \left( \vartheta _1-\vartheta _3 \right) \\ -kR \left( \vartheta _3-\vartheta _2 \right) -kR \left( \vartheta_1 -\vartheta _2 \right) \end {array} \right] $ Point "3" is mass M $\endgroup$
    – Eli
    Commented Dec 19, 2018 at 20:51

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