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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?

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    $\begingroup$ Can you derive the equation for the steady state height profile if the entire cup is rotating about its axis? $\endgroup$ – Chet Miller May 31 '16 at 4:24
  • $\begingroup$ @ChesterMiller , not too sure. Can you elaborate in the answer? $\endgroup$ – Graviton May 31 '16 at 4:54
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    $\begingroup$ See Floris' answer. You have centripetal acceleration of the fluid that requires a radial pressure gradient to sustain. The radial pressure gradient is provided hydrostatically by the radial profile of depth. $\endgroup$ – Chet Miller May 31 '16 at 10:27
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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.

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  • $\begingroup$ what about the governing differential equations? can we derive that? $\endgroup$ – Graviton Mar 29 '17 at 10:47

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