consider the following situation of a double pendulum.

doppel pendulum

We found the moving equations as $$ \ddot{\theta_1}=-L_1\sin\theta_1 + \frac{m_2}{m_1}\cos\theta_2\sin(\theta_2-\theta_1),\\ \ddot{\theta}_2=-L_2\sin\theta_2. $$ Respectively as a system (*) $$ \dot{\theta}_1=w_1\\ \dot{\theta}_2=w_2\\ \dot{w}_1=-L_1\sin\theta_1+\frac{m_2}{m_1}\cos\theta_2\sin(\theta_2-\theta_1)\\ \dot{w}_2=-L_2\sin\theta_2. $$

Now the task is to find a first integral of the system (*). As a hint it is said to find formula for the kinetic and potential energy of the system by working in cartesian coordinates.

I am stuydying mathematics and not physics... so I do not know how to determine kinetic and potential energy... can you help me?


1 Answer 1


The kinetic energy is, in Cartesians,

$$\frac12 m_1 (\dot{x_1}^2 + \dot{y_1}^2) + \frac12 m_2 (\dot{x_2}^2 + \dot{y_2}^2) $$

where $(x_1,y_1)$ is the cartesian coordinates of the first bob, etc. We find these velocities by differentiating the cartesian coordinates of the bob with respect to time.

I express these coordinates with respect to the rest positions of each bob. The first bob is easy:

$$x_1 = L_1 \sin{\theta_1}$$ $$y_1 = L_1 (1-\cos{\theta_1}) $$

The position of the second bob, however, depends on the first (which is why the equations are coupled). Specifically, I consider its rest position to be that along the line extended through the first pendulum. Thus

$$x_2 = x_1 + L_2 \sin{(\theta_2-\theta_1)} $$ $$y_2 = y_1 + L_2 (1-\cos{(\theta_2-\theta_1)}) $$

Using these coordinates, the potential energy is simply $m_1 g y_1+ m_2 g y_2$.

  • $\begingroup$ What is the potential energy when using the coordinates $x_1=L_1\sin\theta_1, y_1=L_1\cos\theta_1, x_2=L_1\sin\theta_1+L_2\sin\theta_2, y_2=L_1\cos\theta_1+L_2\cos\theta_2$? What is the general formula for the potential energy? $\endgroup$
    – user34632
    Jun 8, 2014 at 16:29
  • $\begingroup$ It looks like you're still being targeted. I'll help you. +1. $\endgroup$
    – Tunk-Fey
    Jun 18, 2014 at 6:35
  • 1
    $\begingroup$ @Tunk-Fey: much appreciated. $\endgroup$
    – Ron Gordon
    Jun 18, 2014 at 9:09

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