Lagrangian of two particles connected with a spring, free to rotate - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-18T09:10:11Z https://physics.stackexchange.com/feeds/question/32609 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/32609 6 Lagrangian of two particles connected with a spring, free to rotate childofsaturn https://physics.stackexchange.com/users/10762 2012-07-22T20:54:20Z 2017-02-07T07:12:08Z <p>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.</p> <p>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</p> <p>$L = \frac{1}{2}m_1(\dot{x_1}^2 + \dot{y_1}^2) + \frac{1}{2}m_2(\dot{x_2}^2 + \dot{y_2}^2) - \frac{1}{2}k(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} -d)^2$.</p> <p>Does this make any sense? It seems like the EOM would be a mess in this case.</p> https://physics.stackexchange.com/questions/32609/-/32638#32638 6 Answer by Ron Maimon for Lagrangian of two particles connected with a spring, free to rotate Ron Maimon https://physics.stackexchange.com/users/4864 2012-07-23T07:25:52Z 2017-02-07T07:12:08Z <p>This is correct, and you should use the rectangular coordinates until later. The equations of motion aren't a mess, because the system has a center of mass conservation law, so you can linearly mix up the variables:</p> <p>$$X = m_1 x_1 + m_2 x_2$$ $$Y= m_1 y_1 + m_2 y_2$$</p> <p>for the center of mass and </p> <p>$$x = x_1 - x_2$$ $$y = y_1 - y_2$$</p> <p>which are the relative coordinates. In terms of this transformation (which is something you should just know), the Lagrangian for the CM becomes that of a free particle, while the Lagrangian for the relative coordinate becomes that of a 2d particle on a spring of finite length</p> <p>$${m (\dot{x}^2 + \dot{y}^2)\over 2} + {k (\sqrt{x^2 + y^2} - d)^2\over 2},$$</p> <p>where m is the reduced mass. Now you can transform the relative coordinates x,y into polar form $r,\theta$ and the $\theta$ equation is expressing conservation of angular momentum. This reduces to a 1d problem for r with a potential.</p> <p>$$V(r) = {k\over 2}(r-d)^2 + {A\over r^2},$$</p> <p>where A is a constant for constant angular momentum an effective centrifugal repulsion plus the attractive potential. </p> https://physics.stackexchange.com/questions/32609/-/32642#32642 4 Answer by Emilio Pisanty for Lagrangian of two particles connected with a spring, free to rotate Emilio Pisanty https://physics.stackexchange.com/users/8563 2012-07-23T08:53:45Z 2012-07-23T08:53:45Z <p>Your lagrangian is right, but it is needlessly complicated. For two isolated masses, it is always best to move to centre-of-mass and relative coordinates, $$X=\frac{m_1x_1+m_2x_2}{m_1+m_2},$$ $$x=x_2-x_1,$$ and similarly for the $y$s. The kinetic energy is then expressed using the total mass $M=m_1+m_2$ and the reduced mass $m$ such that $\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}$, and your lagrangian becomes quite a bit simpler: $$L=\frac{1}{2}M\left(\dot{X}^2+\dot{Y}^2\right)+\frac{1}{2}m\left(\dot{x}^2+ \dot{y}^2 \right)-\frac{1}{2}k\left(r-d\right)^2$$ for $r=\sqrt{x^2+y^2}$.</p>