Exercise 6
A famous task in Lagrangian mechanics: the double pendulum.
The following figure defines our coordinates:
The angles \(\Theta\) and \(\alpha\) are the generalized coordinates in this systems with two degrees of freedom. As usual in math, we define the angles to be positive, if the rotation takes place counterclockwise. Unfortunately, neither the cartesian coordinates nor the generalized coordinates are defined in the textbook. The solution depends on this definition. Keep this in mind when comparing with other solutions.
Let \((x_1, y_1)\) be the cartesian coordinates of the 1st mass, and \((x_2, y_2)\) those for the 2nd one. To simplify things, let \(r\) be 1 meter and both masses be 1 kg.
At first we calculate the Lagrangian: \[ x_1 = \sin \Theta \] \[ y_1 = - \cos \Theta \] \[ x_2 = x_1 + \sin (\Theta + \alpha) = \sin \Theta + \sin (\Theta + \alpha) \] \[ y_2 = y_1 - \cos (\Theta + \alpha) = - \cos \Theta - \cos (\Theta + \alpha) \]
We need the velocities for the kinetic energy: \[ \dot{x}_1 = \dot{\Theta} \cos \Theta \] \[ \dot{y}_1 = \dot{\Theta} \sin \Theta \] \[ \dot{x}_2 = \dot{\Theta} \cos \Theta + ( \dot{\Theta} + \dot{\alpha}) \cos ( \Theta + \alpha) \] \[ \dot{y}_2 = \dot{\Theta} \sin \Theta + ( \dot{\Theta} + \dot{\alpha}) \sin ( \Theta + \alpha) \]
We obtain the kinetic energy \(T_1\) of the 1st mass: \[ T_1 = \frac{1}{2}(\dot{x}^2_1 + \dot{y}^2_1 ) = \frac{1}{2} \dot{\Theta}^2\]
The calculation of \[ T_2 = \frac{1}{2}(\dot{x}^2_2 + \dot{y}^2_2 )\] for the kinetic energy of the 2nd mass is not difficult, but somewhat lengthy. Taking the square, you get an expression of the form \[ \cos \Theta \cos(\Theta + \alpha ) + \sin \Theta \sin (\Theta + \alpha) = \cos \alpha \] as you can verify easily by using the addition theorems.
Hence, we obtain \(T_2\): \[ T_2 = \frac{\dot{\Theta}^2 + (\dot{\Theta} + \dot{\alpha})^2}{2} + \dot{\Theta} ( \dot{\Theta} + \dot{\alpha} ) \cos \alpha \]
We define the potential energy of the 1st mass to be \[ V_1 = g y_1 + g = -g \cos \Theta + g\]
The zero reference of the potential energy can be chosen freely. Our definition is based on the assumption, that the potential energy is zero if the pendulum is hanging straight down, e.g. \(V_1 = 0\) if \(\Theta = 0\).
In the same way we obtain \[ V_2 = gy_2 + 2g = -g (\cos \Theta + \cos (\Theta + \alpha)) + 2g \]
The sum of the potential energies of both masses is:
\[ V = V_1 + V_2 = -g (2 \cos \Theta + \cos (\Theta + \alpha)) + 3g \]
Hence, the Lagrangian is: \[ L = T - V = \frac{1}{2} \dot{\Theta}^2 + \frac{\dot{\Theta}^2 + (\dot{\Theta} + \dot{\alpha})^2}{2} + \dot{\Theta} ( \dot{\Theta} + \dot{\alpha} ) \cos \alpha + g (2 \cos \Theta + \cos (\Theta + \alpha)) \]
where we ommited the constant \(3g\).
Now to the actual task.
We obtain the Euler-Lagrange equation for \(\Theta\) \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L }{\partial \dot{\Theta}} - \frac{\partial L}{\partial \Theta} = 0 \]
\[ \frac{\partial L }{\partial \dot{\Theta}} = \dot{\Theta} + \frac{2\dot{\Theta} + 2 ( \dot{\Theta} + \dot{\alpha})}{2} + [(\dot{\Theta} + \dot{\alpha}) + \dot{\Theta}] \cos \alpha \]
\[ = 3 \dot{\Theta} + \dot{\alpha} + ( 2 \dot{\Theta} + \dot{\alpha}) \cos \alpha \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L }{\partial \dot{\Theta}} = 3 \ddot{\Theta} + \ddot{\alpha} + (2 \ddot{\Theta} + \ddot{\alpha}) \cos \alpha - (2 \dot{\Theta} + \dot{\alpha}) \dot{\alpha} \sin \alpha \]
\[ \frac{\partial L}{\partial \Theta} = -g(2 \sin \Theta + \sin (\Theta + \alpha)) \]
The Euler-Lagrange equation for \(\alpha\): \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L }{\partial \dot{\alpha}} - \frac{\partial L}{\partial \alpha} = 0 \] \[ \frac{\partial L }{\partial \dot{\alpha}} = \dot{\Theta} + \dot{\alpha} + \dot{\Theta} \cos \alpha \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L }{\partial \dot{\alpha}} = \ddot{\Theta} + \ddot{\alpha} + \ddot{\Theta} \cos \alpha - \dot{\Theta} \dot{\alpha} \sin \alpha \] \[ \frac{\partial L}{\partial \alpha} = - \dot{\Theta} (\dot{\Theta} + \dot{\alpha}) \sin \alpha - g \sin (\Theta + \alpha) \]
The equations of motion are a system of non-linear coupled differential equations of 2nd order. There is only a numerical solution. The double pendulum is an example of a chaotic system: after a long time even the smallest changes of the initial conditions result in complete different states of the system (“butterfly effect”).