If we look at a simple cannonsball that gets shot out we quickly see the cyclic coordinate in the Lagrangian:
$$L=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\dot{y}}^2-mgy$$
Since the coordinate $x$ isn't in the Lagrangian it is a cyclic coordinate. We can check this with checking the following equality:
$$\frac{\partial L}{\partial x}=0$$
This means that the conjugated momentum of $x$ is conserved. As we can see.
$$\frac{\partial L}{\partial\dot{x}}=\frac{d}{d\dot{x}}\left(\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\dot{y}}^2-mgy\right)$$
$$\frac{\partial L}{\partial\dot{x}}=m\dot{x}=\mathrm{constant}$$
However if we consider the same problem with polar coordinates (which would be very clumsy to do) we have a chance of missing the cyclic coordinate. The lagrangian would be:
$$L=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta}}^2\right)-mgr\sin{\theta} \ \ \ \ (1)$$
As we see the Lagrangian is a function of all the coordinates: $L(r,\theta, \dot{r}, \dot{\theta})$
However still the quantity (linear momentum) has to be conserved. Since the coordinate system is just an arbitrary choice.
$$x=r\cos\theta$$
$$ \dot x = \dot r\ \cos\theta-r \dot \theta\ \sin \theta$$
$$\rightarrow p_x=m\dot x\ =m(\dot r\ \cos \theta-r\dot \theta\ \sin \theta\ )$$
My question:
What can you do to equation (1) to check if there is a cyclic coordinate? Is there something you can do to automatically churn out a cyclic coordinate?
