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Homework Statement
Find the Lagrangian for the double pendulum system given below, where the length of the massless, frictionless and non-extendable wire attaching m_1 is l. m_2 is attached to m_1 through a massless spring of constant k and length r. The spring may only stretch in the m_1-m_2 direction and its unstretched length is l_0. \theta and \phi are the angles of m_1 and m_2 with respect to the red y-axis.
Homework Equations
Lagrangian:
L = T - U
Kinetic Energy of i-th mass:
T_i = \frac{1}{2}m_i \left(\dot{x}_i^2 + \dot{y}_i^2 \right)
Potential Energy of i-th mass:
U_i = m_i g y_i
Potential Energy of the spring:
U_{spr} = \frac{1}{2}kr^2
The Attempt at a Solution
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Since I'm still learning this topic, I need someone to help me by verifying if my work is correct, as there's no solutions manual.
L = T - U = \left \{ \frac{1}{2}m_1 \left(\dot{x}_1^2 + \dot{y}_1^2 \right) - m_1gy_1 \right \} + \left \{ \frac{1}{2}m_2 \left(\dot{x}_2^2 + \dot{y}_2^2 \right) - m_2gy_2 \right \} - \frac{1}{2}k r^2
I'm considering \left(r, \theta, \phi \right) as my generalized coordinates
\left\{\begin{matrix}<br /> x_1& =& l \sin \theta &\\<br /> x_2& =& l \sin \theta + r \sin \phi &\\<br /> y_1 &=& l \cos \theta &\\<br /> y_2 &=& l \cos \theta + r \cos \phi&<br /> \end{matrix}\right.
Plugging those in the original lagrangian yields
\begin{align*} L = \left \{ \frac{1}{2}m_1 \left(l^2\dot{\theta}^2 \cos^2 \theta + l^2\dot{\theta}^2 \sin^2 \theta \right) - m_1gl \cos \theta \right \}\\<br /> + \left \{ \frac{1}{2}m_2 \left(l^2\dot{\theta}^2 \cos^2 \theta + r^2\dot{\phi}^2 \cos^2 \phi + 2lr \dot{\theta} \dot{\phi} \cos \theta \cos \phi + l^2 \dot{\theta}^2 \sin^2 \theta + r^2 \dot{\phi}^2 \sin^2 \phi + 2lr \dot{\theta} \dot{\phi} \sin \theta \sin \phi \right)\\<br /> - m_2g \left( l \cos \theta + r \cos \theta \right) \right \} - \frac{1}{2}k r^2 \end{align*}
which summarizes to
L = \left \{ \frac{1}{2}m_1 \left(l^2\dot{\theta}^2 \right) - m_1gl \cos \theta \right \} + \left \{ \frac{1}{2}m_2 \left(l^2\dot{\theta}^2 + r^2\dot{\phi}^2 + 2lr \dot{\theta} \dot{\phi} \cos \left(\theta-\phi \right)\right) - m_2g \left( l \cos \theta + r \cos \phi\right) \right \} - \frac{1}{2}k r^2
Is this correct? Something tells me it isn't, is there another way of representing r? Also, I haven't used l_0 which was given in the problem statement, and I feel like it should be there in the lagrangian somewhere.
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