Show that the angular momentum is conserved (Noether) It is the first time I get in touch with classical mechanics and this stuff...
Suppose a mechanical system has $n$ degrees of freedom described by coordinated $q\in V\subset\mathbb{R}^n$, set $v=\dot{q}$. In the situation of particles under the influence of some forces we have the Lagrange function
$$
L(v,q)=\frac{1}{2}vMv-U(q).~~~(*)
$$
where $M$ is a positive diagonal matrix with the masses of the particles as entries and $U$ is the potential corresponding to the forces. Let
$$
p(v,q)=\frac{\partial L}{\partial v}(v,q)
$$
denote the momentum.


Theorem of Noether Let $\Phi(t,q)$ be the flow generated by $f(q)$. If $\Phi$ leaves the Lagrangian invariant, then
    $$
I(v,q)=p(v,q)\cdot f(q)
$$
    is a constant of motion.


Now there is the following task:


Consider $L(v,q)$ from $(*)$ in $\mathbb{R}^3$ with $M=m\mathbb{I}_3$ and suppose $U(q)=U(\lvert q\rvert)$ is rotation invariant. Show that the angular momentum $l= x\wedge p$ is conserved in this case. Here $\wedge$ denotes the cross product in $\mathbb{R}^3$.


I think I have to apply the theorem of Noether, but I do not know how to do so. The first problem for me already is to calculate $p(v,q)$.
 A: If we have a Lagrangain $L(\vec{v},\vec{q})$ and the (infinitesimal) transformation $\vec{q}\to \vec{q} + \epsilon \vec{\sigma}(t)$ leaves it invariant then Noether's theorem states that
$$I = \frac{dL}{d\vec{v}}\cdot \vec{\sigma}$$
is a conserved quantity. For rotations about a direction $\vec{n}$ we have $\vec{\sigma} = \vec{n}\times \vec{q}$ so
$$I = \vec{p} \cdot (\vec{n}\times \vec{q}) = \vec{n}\cdot (\vec{q}\times \vec{p}) = \vec{n}\cdot \vec{l}$$
Since $\vec{n}$ is arbitrary we get that $\vec{l}$ is conserved. For more information see for example this note.

One can also solve this problem explicitly. For the case $M = m I_n$ and $U(\vec{q})  = U(q)$ where $q=|\vec{q}|$ we find
$$\frac{dL}{d\vec{v}} = m\vec{v} \equiv \vec{p},~~~\frac{dL}{d\vec{q}} = -\frac{dU(q)}{dq}\frac{\vec{q}}{q}$$
so from the Euler-Lagrange equations the equation of motion is 
$$\frac{d\vec{p}}{dt} = \frac{dU}{dq}\frac{\vec{q}}{q}$$
Using the equation of motion we can calculate the time derivative of $\vec{l} = \vec{q}\times \vec{p}$ to find
$$\frac{d\vec{l}}{dt} = \frac{d\vec{q}}{dt}\times \vec{p} +  \vec{q}\times \frac{d\vec{p}}{dt} = \frac{\vec{p}\times \vec{p}}{m} + \frac{dU}{dq}\frac{\vec{q}\times\vec{q}}{q} = 0$$
and angular momentum is conserved.
