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Double pendulum with springs

Nasser M. Abbasi

January 22, 2020   Compiled on May 18, 2020 at 6:22pm
pict
Figure 1:Geometry of the problem

Assuming both springs have the same relaxed length of \(L\). Starting by finding the Lagrangian \(\mathcal{L}=T-V\). For \(m_{1}\)\begin{align*} T_{1} & =\frac{1}{2}m_{1}\left ( \dot{x}_{1}^{2}+\left ( \left ( L+x_{1}\right ) \dot{\theta }_{1}\right ) ^{2}\right ) \\ V_{1} & =-m_{1}g\left ( L+x_{1}\right ) \cos \theta _{1}+\frac{1}{2}k_{1}x_{1}^{2} \end{align*}

And for \(m_{2}\)\begin{align*} T_{2} & =\frac{1}{2}m_{2}\left ( \left ( \dot{x}_{2}+\dot{x}_{1}\cos \left ( \theta _{1}-\theta _{2}\right ) \right ) ^{2}+\left ( \dot{x}_{1}\sin \left ( \theta _{1}-\theta _{2}\right ) \right ) ^{2}\right ) \\ & +\frac{1}{2}m_{2}\left ( \left ( \left ( L+x_{2}\right ) \dot{\theta }_{2}+\left ( L+x_{1}\right ) \dot{\theta }_{1}\cos \left ( \theta _{1}-\theta _{2}\right ) \right ) ^{2}+\left ( \left ( L+x_{1}\right ) \dot{\theta }_{1}\sin \left ( \theta _{1}-\theta _{2}\right ) \right ) ^{2}\right ) \\ V_{2} & =-m_{2}g\left ( \left ( L+x_{1}\right ) \cos \theta _{1}+\left ( L+x_{2}\right ) \cos \theta _{2}\right ) +\frac{1}{2}k_{2}x_{2}^{2} \end{align*}

Hence\[\mathcal{L}=\left ( T_{1}+T_{2}\right ) -\left ( V_{1}+V_{2}\right ) \] There are 4 generalized coordinates, \(x_{1},x_{2},\theta _{1},\theta _{2}\). Now Mathematica is used to obtain the four equations of motion to help with the algebra. Once \(x_{1},x_{2},\theta _{1},\theta _{2}\) are solved for, the position of each mass \(m_{1},m_{2}\) is fully known at each time instance, and each mass motion can be animated. The four equations of motion are

\begin{align*} \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot{x}_{1}}\right ) -\frac{\partial \mathcal{L}}{\partial x_{1}} & =0\\ \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot{x}_{2}}\right ) -\frac{\partial \mathcal{L}}{\partial x_{2}} & =0\\ \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot{\theta }_{1}}\right ) -\frac{\partial \mathcal{L}}{\partial \theta _{1}} & =0\\ \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot{\theta }_{2}}\right ) -\frac{\partial \mathcal{L}}{\partial \theta _{2}} & =0 \end{align*}

The rest is done using Mathematica to help with the algebra

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