I'm a bit rusty in this stuff. I'm trying to design an energy function that would allow me to derive a motion equation.
I'm considering the vector field
$$ F(x,y,z) = k\left( \frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},0 \right)^T $$
($k$ positive constant). I've checked the condition
$$ curl(F) = 0 $$ for all $x,y,z$, and indeed I get 0. However the components aren't differentiable in $(x,y) = 0$ which basically confuses me a little bit.
Is there a slightly modification I could do maybe to get something more sensible? But I'm not sure if this matters though.
The other question is I have the potential function
$$ U(x,y,z) = k \sqrt{x^2 + y^2} $$
and I clearly have
$$ \nabla U = F $$
How do I derive the motion equation? My intuition is that since
$$ F = ma = m \frac{d^2x}{dt^2} $$
I have to relate both $U$ and the motion equation, is there some Euler lagrange equation (in general not for this specific case) I should use?
You should use the cylindric coordinates $\phi$, $r$, $z$, where $$[x,y,z]= [r\cos(\phi),r\sin(\phi),z].$$ We then have for the kinetic energy $$ T = \frac{m}{2}\left(r^2\dot{\phi}^2 + \dot{r}^2 + \dot{z}^2\right) $$ and for the potential simply $$ U = k r. $$ Now apply Euler-Lagrange: $$ L = T - V $$ and $$ d_t\partial_{\dot{q}}L = \partial_q L$$ where $q\in \lbrace\phi,r,z\rbrace$. This will give you your equations of motion: $$ \ddot{z} = 0, $$ conservation of momentum in $z$-direction, $$ r\ddot{\phi} + 2\dot{r}\dot{\phi} = 0 = \dot{J}, $$ conservation of angluar momentum $J = mr^2\dot{\phi}$, and finally $$ m\ddot{r} = mr\dot{\phi}^2-k = \frac{J^2}{mr^3} - k. $$ $J$ in the last equation is constant because of the conservation of angluar momentum. So now you are left with a decoupled equation for $r$.