Tuesday, 3 September 2019

classical mechanics - Lagrangian of two particles connected with a spring, free to rotate


Two particles of different masses $m_1$ and $m_2$ are connected by a massless spring of spring constant $k$ and equilibrium length $d$. The system rests on a frictionless table and may both oscillate and rotate.


I need to find the Lagrangian for this system. I'm not sure if I'm interpreting it correctly, but I think there are 4 degrees of freedom in this problem, $x_1, y_1, x_2, y_2$ or $r_1,\theta_1,r_2,\theta_2$. If I use the former choice I get my Lagrangian to be



$L = \frac{1}{2}m_1(\dot{x_1}^2 + \dot{y_1}^2) + \frac{1}{2}m_2(\dot{x_2}^2 + \dot{y_2}^2) - \frac{1}{2}k(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} -d)^2$.


Does this make any sense? It seems like the EOM would be a mess in this case.




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