On the Surface of a Whirlpool

G.R.Dixon, August 7, 2008

In this article an equation is derived for the surface of a liquid rotating in a cylindrical container. The following assumption (obviously an idealization) is made: all of the molecules in the liquid maintain their positions *relative to one another* at all rotation rates.

If w is the angular rotation rate, then when w=0 the liquid’s surface is horizontal (flat). When the liquid rotates, its surface cups upward.

Choose the xy axes such that their origin is at the lowest point in the cupped surface. And consider a cross section of the surface, lying in the xy plane.

A surface molecule of mass m is acted upon by two forces: (1) a force, F, that acts normal to the surface, and (2) the downward acting gravitational force, mg. Assuming the molecule has no vertical acceleration (or velocity, for that matter) the vertical component of F must be equal and oppositely directed to the gravitational force:

. (1)

The horizontal component of F must account for the molecule’s radial acceleration toward the y axis. If w is the rotational rate, and if the molecule is x distant from the y axis, then wx is the molecule’s speed:

. (2)

The radial acceleration is

. (3)

And Newton’s 2^{nd} law stipulates that

. (4)

Now a tangent to the liquid’s surface is perpendicular to F. The slope of this tangent is accordingly

. (5)

Or, since the slope of the surface cross section is dy/dx,

. (6)

Rearranging:

. (7)

Integrating both sides produces

. (8)

Since the origin of the xy axes is at the bottom of the curve, K=0. The equation for the surface is therefore

. (9)

This is the equation of a *parabola*. Fig. 1 plots y(x) for w=10p/sec (5 revolutions per second) and g=9.8m/sec^{2}.

Figure 1

y vs x, Liquid Surface

Caveats.

In any real situation, the assumption that the molecules all remain at rest relative to each other is of course not true. There will be thermal agitation. And there will be friction where the molecules move over the inner surfaces of the container. In general there will be laminar flow (in cylindrical coordinates), coupled with turbulence. It is safe to say that the exact shape of the cupped surface (or whirlpool) will depend upon many factors, including the coefficient of friction between the liquid and the container walls, the average rate of rotation, the container dimensions and the liquid depth (to name but a few).

In brief, there is no single solution, not even in a given case. (Any real surface will have small bulges and dents in it.)

Aside from such provisos, it is interesting that Eq. 9 can be derived from a fairly straightforward application of Newton’s 2^{nd} law.

**Postscript.**

I have since learned from a colleague on MySpace.com that containers of molten glass are spun as a first step in making parabolic mirrors for telescopes! Of course the entire container is spun in those cases, and there is no significant friction between the container walls and the molten glass. In any case, it would seem that Eq. 9 might have merit.