Shape of water in rotating bucket

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $z = \frac{r^2 \Omega^2}{2g}$.