The relationship between height of water and the rotational speed inside a cup When I stir the water inside a cup with a speed $v$ ( using a stick), I find that a vortex will be formed inside the cup, with the middle part of the water at the lowest point ( with the height $h_0$), and the water level gradually rises until it reaches the wall of the cup, with the height ($h_1$).
I also observe that the faster I stir the water, the steeper the difference ($h_1$-$h_0$) is. 
What is the relationship between the height of the water and the rotational speed? Anyway to derive it?
 A: Because of viscosity, this is actually a very hard problem. A much easier problem is that of a cup of water on a rotating turntable; in that setup, the angular velocity of all the water is the same, and the differential pressure across an infinitesimal cylinder of liquid centered about the axis of rotation is proportional to the distance to the axis. It follows that the surface is a parabola in that case. The shape is given by (see for example this article)
$$y(r) = \frac{\omega^2r^2}{2g}$$
Unfortunately, when you add viscosity (stirring a stationary cup), the velocity profile is no longer so easy to determine. The moment you stop stirring, the angular velocity changes even more - there is no "easy" solution.
A: Consider a cup filled with a liquid whose density is $\rho$ which is rotating about its axis of symmetry with a constant angular velodity $\omega$.
Let the height to the lowest point of the liquid surface from the bottom of the vessel to be $H$ and the height difference is $h$, such that the height of the liquid column near the wall of the cup to be $H+h$.
First we need a coordinate system. As the cup has cylindrical symmetry, It's natural to choose cylindrical coordinate system. Let's take the centre of the base of the cup as the origin.
Let's take a small volume of liquid, $dv$, which is at $(r,\theta,z)$ and the pressure at there to be P. Then,
Using Newton's Second law,
$\vec{F} =  m \vec a$
$$\vec{F_p}- \rho g dv\;\hat k= -\rho r \omega^2 dv\; \hat{r}$$
Where $\vec{F_p}$ is the force due to the pressure difference. But we know, $$\vec{F_p}= -\vec{\nabla}P\cdot dv$$
From above two equations, we can write,
$$\vec{\nabla}P= -\rho g \; \hat k + \rho r \omega^2 \; \hat r $$
Then, $$ \frac{\partial P}{\partial z}=-\rho g \quad \mathrm{and} \quad \frac{\partial P}{\partial r} = \rho r\omega^2$$
As  $\frac{\partial P}{\partial \theta}=0$, The total differential, $dP$ is given by,  $$dP = \frac{\partial P}{\partial z}dz + \frac{\partial P}{\partial r}dr$$
Then $$dP = \rho r\omega^2dr-\rho g dz
$$
$$\Rightarrow \int{dP} = \int{\rho r\omega^2}\; dr-\int{\rho g}\; dz$$
$$\Rightarrow P=\frac{1}{2} \rho r^2 \omega^2 - \rho gz + P_0$$
Where $P_0$ is the pressure at the origin. Consider Two points on the liquid surface where one$(=A)$ is at the lowest place and the other$(=B)$ is at a highest place,
So $A=(0,0,H)$ and $B=(R,\theta,H+h)$
$$P_A=P_0-\rho g H$$
$$P_B=P_0-\rho g (H+h) + \frac{1}{2}\rho R^2 \omega^2$$
But, the total pressure at point A must equal to the total pressure at B as both of them are on the surface of the fluid.
This yeilds to our final result,
$$h=\frac{R^2 \omega^2}{2g}$$
Hope this helps...
