Lagrangian of a double pendulum system (with a spring)

[Je m'appelle]

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

[/B]
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

$$\left\{\begin{matrix} x_1& =& l \sin \theta &\\ x_2& =& l \sin \theta + r \sin \phi &\\ y_1 &=& l \cos \theta &\\ y_2 &=& l \cos \theta + r \cos \phi& \end{matrix}\right.$$

Plugging those in the original lagrangian yields

$$\begin{align*} L = \left \{ \frac{1}{2}m_1 \left(l^2\dot{\theta}^2 \cos^2 \theta + l^2\dot{\theta}^2 \sin^2 \theta \right) - m_1gl \cos \theta \right \}\\ + \left \{ \frac{1}{2}m_2 \left(l^2\dot{\theta}^2 \cos^2 \theta + r^2\dot{\phi}^2 \cos^2 \phi + 2lr \dot{\theta} \dot{\phi} \cos \theta \cos \phi + l^2 \dot{\theta}^2 \sin^2 \theta + r^2 \dot{\phi}^2 \sin^2 \phi + 2lr \dot{\theta} \dot{\phi} \sin \theta \sin \phi \right)\\ - m_2g \left( l \cos \theta + r \cos \theta \right) \right \} - \frac{1}{2}k r^2 \end{align*}$$

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.

 

[kuruman]

... through a massless spring of constant k and length r.

... and its unstretched length is $l_0.$

What is the difference between $l_0$ and $r$? I would assume that $r$ is the stretched or compressed length of the spring in which case the elastic potential energy should be ... ?

 

[Fred Wright]

Sorry, the comment above answers your question.