 Double pendulum equations of motion

Write $\Delta ={\theta }_1-{\theta }_2$. Applying the Euler-Lagrange equations to the double pendulum Lagrangian, then dividing by $m{\ell }^2$, givesFor example, the upper generalized momentum is $\partial L/\partial {\dot{\theta }}_1=m{\ell }^2(2{\dot{\theta }}_1+\cos \Delta {\dot{\theta }}_2)$. Its time derivative contains $-m{\ell }^2\sin \Delta ({\dot{\theta }}_1-{\dot{\theta }}_2){\dot{\theta }}_2$. Subtracting $\partial L/\partial {\theta }_1=-m{\ell }^2\sin \Delta {\dot{\theta }}_1{\dot{\theta }}_2-2mg\ell \sin {\theta }_1$ cancels the mixed product and leaves the positive $\sin \Delta {\dot{\theta }}_2^2$ term in the first equation. This cancellation is a useful check on the sign convention.

The matrix multiplying the two angular accelerations isIt is invertible for every configuration, so specified angles and angular velocities determine both angular accelerations. For the same reason $T>0$ whenever the angular velocities are not both zero. The mass cancels from the motion equations, while the timescale is set by $\sqrt{\ell /g}$.

As an exact constructed release, put ${\theta }_1(0)=\alpha$, ${\theta }_2(0)=0$, and both angular velocities zero. Solving the two equations at release yieldsFor $0<\alpha <\pi /2$, the upper rod initially moves back toward vertical, while the lower rod initially moves away from it. The lower pivot's acceleration makes this possible even though the lower rod starts vertical. The small-angle approximation gives ${\ddot{\theta }}_1(0)\simeq -2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)\simeq 2g\alpha /\ell$.

The conserved mechanical energy is $E=T+V$, because the Lagrangian has no explicit time dependence. Directly, $dE/dt$ is $m{\ell }^2$ times the first equation multiplied by ${\dot{\theta }}_1$ plus the second multiplied by ${\dot{\theta }}_2$, so the equations imply $dE/dt=0$. This also supplies a check when computing a trajectory.