I have a question regarding the the height of a fluid spinning due to a drain in a sink. We are following Feynmans Lecture of physics. It should be "easy to see" that, that the height is something like $z=\omega^2*r^2/2g$

Now the way we proceded was like this:

We've watched the stream lines defined as $C_r=\oint{\vec{v}*\vec{dl}} = \omega*r^2*2\pi \Rightarrow v(r) = \frac{C_r}{r2\pi}$, Assuming the streamline is constant at r. the tangential velocity goes as 1/r—it’s just from the conservation of angular momentum, like the skater pulling in her arms. Also the radial velocity goes as 1/r. Ignoring the tangential motion, we have water going radially inward toward a hole. Because we don't lose any mass, it follows that the radial velocity is proportional to 1/r. So the total velocity also increases as 1/r, and the water goes in along Archimedean spirals. The air-water surface is all at atmospheric pressure, so it must have the property: $gz+1/2*v^2=$constant.

But v is proportional to 1/r, so the shape of the surface is $(z-z_0)=k/r^2$

I don't fully understand the steps. I can see, that the velocity is proportional to 1/r, but why does that define the shape of the surface? I can see what they did there: "$(z-z_0)=k/r^2\equiv\frac{dz}{dr}=1/r$" but I don't see why this holds. And at the end, how do we obtain our desired equation for the height $z=\omega^2*r^2/2g$?

Feynman can be publicly read on the Caltech homepage. The full text to my problem is on http://www.feynmanlectures.caltech.edu/II_40.html chapter 40-4

Thanks in advance.


1 Answer 1


I think there are a couple of ways you can solve a problem like this.

Method 1:

You can consider the net forces on a "small clump" at the surface of the fluid. Assuming the fluid is incompressible, then the density is uniform throughout. A clump is a arbitrarily small mass on the surface, which is the product of that uniform density with a volume element. The mass is the density times that volume element.

There are a couple of forces operating on the clump at the surface. There's gravity pulling the clump straight down. There's the centripetal force pulling the fluid radially outward.

Apart from rotational motion, the clump is stable, i.e. there has no net acceleration in the r or z direction. So you need to set the z and r components of the centripetal force and gravity to be equal and opposite.

Method 2:

You can set up the integral for total energy and find the function that minimizes the volume integral using the calculus of variations.:

$$\int_0^R[\rho g z(r) +(1/2)\rho\omega^2r^2](2\pi r)\sqrt{1+(dz/dr)^2}dr$$

$\rho$ is the density, assumed to be uniform. $\omega$ is the angular velocity of the clump. $g$ is the gravitational acceleration.

Use the calculus of variations on the integral, essentially the same concept as finding the equations of motion from the Lagrangian.


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