What does this Euler-Lagrange equation represent? I have the following Lagrangian:
$$L=\frac{\mu}{2}\left(\dot{r}^2+r^2\dot\phi^2\right)-U(r),$$
The Euler-Lagrange equations are thus:
$$\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$$
$$\frac{d}{dt}(\mu \dot r)=\mu r\dot{\phi}^2-\frac{\partial}{\partial r}U(r).$$
I am trouble understanding what each Euler-Lagrange equation represent:
For example the first one: $\frac{d}{dt}\left(\mu r^2\dot\phi\right)=0$
What does $\mu r^2\dot\phi\ $ means?
Would it be the angular momentum?
 A: Yes, the term $\mu r^2\dot{\phi}$ represents the angular momentum, and the first Euler equation telling you that the angular momentum is the constant of motion for the given problem. The second Euler equation is just the radial force equation. If you can  try to write newton's equation of motion like
$$F_r=m(\ddot{r}-r\dot{\theta}^2)\ \ \text{and}\ \ \ F_\theta=m(r\ddot{\theta}+2\dot{r}\dot{\theta})$$
Then you can compare it to the Euler equation to get the meaning of the terms.
A: It is the generalized momentum conjugate to $\phi$, defined as
$$p_\phi=\frac{\partial L}{\partial \dot{\phi}}.$$
With your choice of generalized coordinates, $r$ and $\phi$, it is the angular momentum.
As $\phi$ itself does not appear in the Lagrangian, it is said to be an $\textit{ignorable}$ coordinate, so its conjugate momentum is a conserved quantity. This comes from Euler-Lagrange equations, $$\frac{d}{dt}\Big(\frac{\partial L}{\partial \dot{q}_i}\Big)=\frac{\partial L}{\partial q_i},$$
which can be rewritten as $$\frac{dp_i}{dt}=\frac{\partial L}{\partial q_i}$$ with the definition above. You can see that if the Lagrangian does not explicitey depend on a certain generalized coordinate $q_i$, its conjugate momentum $p_i$ is conserved.
