HOME

PDF (letter size)

Assuming both springs have the same relaxed length of $L$. Starting by finding the Lagrangian $\mathcal{L}=T-V$. For $m_1$$$\begin{array}{rl}{T}_1& =\frac{1}{2}{m}_1({\dot{x}}_1^2+{\left((L+x_1){\dot{\theta }}_1\right)}^2)\\ V_1& =-m_{1}g(L+x_1)\cos {\theta }_1+\frac{1}{2}{k}_1{x}_1^2\end{array}$$

And for $m_2$$$\begin{array}{rl}{T}_2& =\frac{1}{2}{m}_2({({\dot{x}}_2+{\dot{x}}_1\cos ({\theta }_1-{\theta }_2))}^2+{({\dot{x}}_1\sin ({\theta }_1-{\theta }_2))}^2)\\ & +\frac{1}{2}{m}_2({((L+x_2){\dot{\theta }}_2+(L+x_1){\dot{\theta }}_1\cos ({\theta }_1-{\theta }_2))}^2+{((L+x_1){\dot{\theta }}_1\sin ({\theta }_1-{\theta }_2))}^2)\\ V_2& =-m_{2}g((L+x_1)\cos {\theta }_1+(L+x_2)\cos {\theta }_2)+\frac{1}{2}{k}_2{x}_2^2\end{array}$$

Hence$$\mathcal{L}=(T_1+T_2)-(V_1+V_2)$$ 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

$$\begin{array}{rl}\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial {\dot{x}}_1}\right)-\frac{\partial \mathcal{L}}{\partial x_1}& =0\\ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial {\dot{x}}_2}\right)-\frac{\partial \mathcal{L}}{\partial x_2}& =0\\ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial {\dot{\theta }}_1}\right)-\frac{\partial \mathcal{L}}{\partial {\theta }_1}& =0\\ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial {\dot{\theta }}_2}\right)-\frac{\partial \mathcal{L}}{\partial {\theta }_2}& =0\end{array}$$

The rest is done using Mathematica to help with the algebra

PDF