Total energy in double pendulum system

Question

Given the following double pendulum system as I outline in the picture attached, how can I use the total energy of the system to derive the equations of motion (assuming angles are small of course)?

I know how to do this using Newtons 2nd law, and Lagrangian mechanics ( which I know is probably the better way to do it), but how can I use the fact that $\frac{dE_{total}}{dT} =0$ as one can for a single pendulum? Say if I rewrite my expression for the potential energy, $U$ and kinetic energy, $K$ as $U=\frac{1}{2} \theta^T V \theta$ and $\frac{1}{2} \dot{\theta^T} T\dot{\theta}$ where $T$ and $V$ are 2x2 matrices in our case (essentially using quadratic forms) how can I derive the equations of motion from this using $\frac{dE_{total}}{dt} = 0$. I've tried calculating this but all I get is one line of equations.

Answer 1

This would just be Hamiltonian mechanics. It is very similar to Lagrangian mechanics except instead of getting a few second order differential equations you get twice as many first order differential equations. Essentially you write $H(q,p,t)=E$ where $q$ is your generalized position (your angles) and $p$ is the conjugate momenta. Then $$\frac{dp}{dt}=-\frac{\partial H}{\partial q}$$ and $$\frac{dq}{dt}=\frac{\partial H}{\partial p}$$

Just writing the conjugate momenta can be tricky, so it is usually better to start from the Lagrangian. Then the Legendre transform of the Lagrangian gives you the Hamiltonian and the conjugate momenta are the dual variables.

Answer 2

To obtain die Equations of motion (@ Dale) you need those steps

Step I

obtain the kinetic energy $~T$ and the potential energy $~U$

The Lagrange is $~\mathcal L=T-U$

Step II

The Hamiltonian is the total energy $\mathcal H=T+U~,\mathcal H=\mathcal H(\boldsymbol q~,\boldsymbol{\dot{q}})$

where

$$\boldsymbol q=\begin{bmatrix} \theta_1\\ \theta_2 \end{bmatrix}\qquad, \boldsymbol{ \dot{q}}=\begin{bmatrix} \dot\theta_1\\ \dot\theta_2 \end{bmatrix} $$

Step III

solve this equation $~\boldsymbol p=\frac{\partial \mathcal L}{\partial \boldsymbol{\dot{q}}}~~$ for $~\boldsymbol{\dot{q}}~$ and substitute the result to the Hamiltonian , you obtain $~\mathcal H=\mathcal H(\boldsymbol p~,\boldsymbol q)$

$$\boxed{~\boldsymbol{\dot{q}}=-\frac{\partial \mathcal H}{\partial \boldsymbol p}~}$$

$$\boxed{~\boldsymbol{\dot{p}}=-\frac{\partial \mathcal H}{\partial \boldsymbol q}~}$$