Shape of water in rotating bucket I need to show that the surface of water in a bucket rotating with constant angular velocity will have parabolic shape. I'm quite confused by this problem, but here's what I did:
$$\vec{F}_{cf} + \vec{F}_{grav} = -\vec{\nabla} U = m(\vec{\Omega}\times\vec{r})-mg\hat{z}$$ where $F_{cf}$ is the centrifugal force, $F_{grav}$ is the force of gravity, $U$ is potential energy, $\vec{r}:=(x,y,z)$.
So $\nabla U(z) = mg-m\Omega^2 z$, hence $$U(z) = gmz-\frac{1}{2}\Omega^2 z^2+C$$
which is a parabola.
In this approach I was trying to use the fact that the surface is equipotential for $F_{cf}+F_{grav}$. But apparently my approach doesn't quite hold any water. Please give some suggestion on approaching this problem, as I have no other idea.
 A: Your potential energy function $U(z)$ doesn't show at all that the water surface is parabolic. What you need to find is the functional form of the rotating water surface, i.e. the surface height $z$ as a function of $r$ in cylindrical coordinates $r$ and $z$. (Because of rotational symmetry $\phi$ is not necessary.) The centrifugal force is $$F_{cf}=m\omega^2 r$$ and the gravitational force is $$F_{grav}=-mg$$ The water surface is orthogonal to the direction of the resultant force $$\vec F=\vec F_{cf}+\vec F_{grav}$$ Thus the slope of the water surface is $$\frac {dz(r)}{dr}=\frac{|F_{cf}|}{|F_{grav}|}=\frac {m\omega^2 r}{mg}$$ From this we get by integration $$z-z_0=\frac{1}{2g}\omega^2 r^2$$ Thus we get indeed a parabolic surface in the rotating water in the bucket. 
A: This problem has different solutions. One of them is to use the principal of stationary action. It is convenient to use cylindrical coordinates:
$$\vec{R} = \vec{R}(r, \phi, z).$$ 
Let us consider a water volume rotating like a whole body i.e. each small 
volume of water (blob) has the same angular velocity $\Omega$. 
Next step is to reduce this problem to a static problem. In order to 
attain this we should take a look at the water from the reference frame which also rotates with angular velocity $\Omega$ (around the same axis as the water volume). From that point of view the water is at rest. 
Then there are two forces acting on a water blob of mass $m$: gravitational $mg$ and centrifugal $m\Omega^2r$. What are the potentials of these forces?
So, they are
$$V_{gr} = mgz,\quad V_{cf} = -\frac{m\Omega^2r^2}{2}$$
It's up to you to check that the above potentials cause the right forces. 
Therefore the water possesses only potential energy. Now we can write the action:
$$S = -\int_{t_1}^{t_2}dt\int_{vol}d^3r\Bigl\{\rho gz-\frac{\rho\Omega^2r^2}{2}\Bigr\} = \mathrm{const}\cdot \int_{vol}\mathcal{L}d^3r,$$
where $\mathcal{L}$ is Lagrangian or Lagrangian density. But, of course, we must take into account that entire mass of the water $M_w$ is conserved or 
$$\int_{vol}\rho d^3 r = M_{w} = \mathrm{const}.$$
In cylindrical coordinates the above integrals are ($\rho=1$)
$$2\pi\cdot\int_{0}^{R}rdr\int_{0}^{z(r)}\Big\{...\Big\}dz = \pi\cdot\int_{0}^{R}\Big\{gz^2r-\Omega^2zr^3\Big\} dr$$
and
$$2\pi\cdot\int_{0}^{R}rdr\int_{0}^{z(r)}dz=2\pi\int_{0}^{R}rzdr$$
So we have a variation problem with additional condition. It is easy to
solve this using Lagrange multipliers' method.
Necessary condition for $S$ to have extrema with additional
condition (about $M_w$) is the existing of a such multiplier $\lambda$ that
$$\frac{d}{dr}\frac{\partial G}{\partial \frac{\partial z}{\partial r}} - \frac{\partial G}{\partial z} = 0,$$ where 
$$G = gz^2r-\Omega^2zr^3 + \lambda rz.$$
Which gives
$$2gzr - \Omega^2r^3 + \lambda r = 0$$
or
$$z(r) = \frac{\Omega^2r^2}{2g} - \lambda.$$
Voila.
A: Consider the following cylindrical container with liquid to be rotating at uniform $\omega$:

Consider an infinitesimal liquid element $\mathrm{d}m$ at height $h$ above the minimum of the parabola. The forces acting on it are:
1) gravity:
$$g\mathrm{d}m$$
2) the centripetal force:
$$\mathrm{d}F_c=\omega^2r\mathrm{d}m$$
Consider the angle $\alpha$:
$$\tan\alpha=\frac{\mathrm{d}h}{\mathrm{d}r}=\frac{F_c}{g\mathrm{d}m}=\frac{\omega^2r\mathrm{d}m}{g\mathrm{d}m}=\frac{\omega^2r}{g}$$
This means that:
$$\omega^2r\mathrm{d}r-g\mathrm{d}h=0$$
Integrate this differential equation:
$$\int_0^r\omega^2r\mathrm{d}r=\int_0^hg\mathrm{d}h$$
$$\frac{\omega^2r^2}{2}=gh$$
$$\implies h=\frac{\omega^2r^2}{2g}$$
This is a quadratic parabola.
A: I'm pretty late to the question but I think I know a simpler way. You can solve this really easily with conservation of energy, the kinetic energy of a circle of water at a constant radius is $\frac{m r^2 \Omega^2}{2}$, in a idealised system the energy has nowhere else to go but increasing the waters gravitational potential energy, so $mgz = \frac{m r^2 \Omega^2}{2}$ or $z = \frac{r^2 \Omega^2}{2g}$. 
