Hi almost every student knows the rotating bucket with a fluid problem as described here:
I wanted to do the same for a rotating fluid in free fall (like in the ISS) to show that it will take the shape of a sphere.
Then we also consider by a symmetry of the final result that we can guess the analysis along the axis $Oz$ of the final shape. In this case, the velocity components will be $$v_x=-\omega y, v_y=\omega x, v_z=0.$$ Taking Euler's equation:
$$\frac{\partial \vec{v}}{\partial t}+(\vec{v} \cdot \nabla) \vec{v}=-\frac{1}{\rho} \operatorname{grad} p+\vec g.$$
Considering that $\partial \vec{v} / \partial t=0$, the projections on the three axis on Euler's equation are (we take into account the radial self-gravity of the fluid with the Shell theorem $g(r)=g_0(r/R)$):
$\begin{aligned} & x \omega^2=-\frac{1}{\rho} \frac{\partial p}{\partial x}+g_0 \frac{\sqrt{x^2+y^2+z^2}}{R} \\ & y \omega^2=-\frac{1}{\rho} \frac{\partial p}{\partial y}+g_0 \frac{\sqrt{x^2+y^2+z^2}}{R} \\ & 0=-\frac{1}{\rho} \frac{\partial p}{\partial z}+g_0 \frac{\sqrt{x^2+y^2+z^2}}{R}\end{aligned}$
If I integrate that stuff as we do for the rotating cylindrical bucket i don't fall back at the end on the equation of a spherical shape. For example see how ugly is the integration of the first line:
$\dfrac{1}{2}x^2\omega=-\dfrac{p}{\rho}+\frac{1}{2} x \sqrt{x^2+y^2+z^2}+\frac{1}{2} \ln \left(x+\sqrt{x^2+y^2+z^2}\right) y^2+\frac{1}{2} \ln \left(x+\sqrt{x^2+y^2+z^2}\right) z^2$
I can't see any possible way that the final result to give the equation of a sphere?
Any idea?
PS: I will try tomorrow in spherical coordinates to see if it helps.
