When a liquid in a cylinder is spun on a turntable, it will experience a vortex or whirlpool effect, as shown below:
I have seen and understood a proof for the value of $h$ by taking the motion of a component of liquid on the surface from the reference frame of the cylinder and then using calculus to show that the curve is a parabola.
However, my teacher told me that this can be shown using Bernoulli's principle (at least the formula for h can be derived). When I tried to use it, assuming that the pressure at all points on the liquid surface is the same, I got the result, but $h$ is negative, i.e. since the liquid near the curved face of the cylinder is moving but the liquid at the center is stationary, the liquid should be higher at the center since the higher potential energy equals the kinetic energy of the liquid at the ends.
Where am I going wrong in my approach?
Best Answer
Let $H\equiv u^2/2+p/\rho+gz=\Omega^2r^2/2+p/\rho+gz$. From Navier-Stokes equation for steady inviscid flow we may derive the following general form of Bernoulli equation (see Fluid Dynamics by Batchelor, Chapter 3): $$\nabla H=\mathbf{u}\times\omega$$ in which $\mathbf{u}$ is the velocity with magnitude $u$, $p$ is pressure, $\rho$ is density, $g$ is gravitational acceleration, $z$ is height from some reference position, $\Omega$ is the rotation speed about the vertical axis, $r$ is the radial distance from the rotation axis, and $\omega\equiv\nabla\times\mathbf{u}$ is the vorticity. What the equation above says is that $H$ varies in a direction perpendicular to both velocity and vorticity. Equivalently, $H$ is constant along streamlines and vortexlines. In the particular case of irrotational flow, i.e. $\omega=0$, we have $\nabla H=0$, and therefore $H$ has the same value throughout the fluid.
But a liquid in solid body rotation is not irrotational. In fact the vorticity has the constant value $\omega=2\Omega\mathbf{e}_z$ everywhere in the fluid, in which $\mathbf{e}_z$ is the unit vector in the vertically upward direction. Therefore $H$ varies from one streamline to another (the streamlines in solid-body rotation are circles concentric with the rotation axis). If $\mathbf{e}_\theta$ be the azimuthal unit vector then $\mathbf{u}=\Omega r\mathbf{e}_\theta$. Therefore: $$\nabla H=2\Omega^2r~\mathbf{e}_\theta\times\mathbf{e}_z=2\Omega^2r~\mathbf{e}_r\\ \frac{dH}{dr}=2\Omega^2r\\ \therefore\quad H=\Omega^2r^2+\textrm{constant}$$
Thus we have the Bernoulli equation for a streamline on the free surface lying at radial distance $r$ from the rotation axis: $$H=\Omega^2r^2+\textrm{constant}=\frac{\Omega^2r^2}{2}+\frac{p}{\rho}+gz\\ \therefore\quad\frac{\Omega^2r^2}{2}-gz=\textrm{constant}$$ in which constant pressure (at the free surface) has been absorbed into the other constant. This is the equation for the free surface of a liquid in solid body rotation.