You’ve probably played around with whirlpools in your water bottle. In this problem, you’ll be analyzing an idealized form of this experiment.

You have a cylinder filled with water. Put it on a spinning disc so that the cylinder is spinning around it’s primary axis at an angular speed $\omega$. Gravity points downwards with a magnitude $g$. Assume that each water element rotates at the same rate as the disc. What’s the shape of the whirlpool?

There’s at least two ways to solve this question. Here’s the first, quicker method.

### First solution

First, a close up of the top part of the whirlpool. The $r$ axis is drawn on the diagram, the dark blue curve is the top of the water, and the light blue line is a tangent. The orange forces act on a point $r$ away from the centre.

$F$ points perpendicular to the surface. It is due to the difference between atmospheric pressure and the pressure of the water. Meanwhile, gravity points straight down.

Balancing centripetal force in the horizontal direction,

$\displaystyle F \sin{\theta} = m \omega^2 r$

Balancing forces in the vertical direction,

$\displaystyle F \cos{\theta} = mg$

Dividing the two equations, we get

$\displaystyle \tan{\theta} = \frac{\omega^2 r}{g}$

Let’s say that the shape of the curve at the top is $h(r)$. Then

$\displaystyle \tan{\theta} = dh/dr$

Therefore

$\displaystyle dh/dr = \frac{\omega^2 }{g} r$

Doing a quick integration,

$\displaystyle \int_{0}^{h} \, dh = h = \frac{\omega^2 }{g} \int r \, dr = \frac{\omega^2 r^2}{2g} + C$

And that’s the shape of the curve! It’s a parabola! The constant is there just because we could be measuring $h$ from any initial height. It does not really matter. How about the points not on the surface though? How do we know that they’re also rotating at an angular speed $\omega$? Here’s a more complete solution that might make you feel better.

### Second solution

Let the height of the the surface measured from the bottom of the cylinder be $h(r)$. Let the pressure at a height $z$ above the bottom of the cylinder and a radius $r$ be $p(r, z)$. Consider a small cylindrical element as shown.

If the density of the liquid is $\rho$, then the mass is

$\displaystyle dm = \rho r d \theta \cdot dr \cdot dz$

What’s the net force in the $r$ direction? If the pressure increases by $dp$ as you go from $r$ to $r+dr$, then the force is

$\displaystyle F = - dp \cdot A$

where $A$ is the area the pressure acts on. The negative sign is right because if the pressure increases along the $r$ axis, then the element will feel a force along the negative $r$ axis.

What’s the next observation? Our volume element is pretty much a box! The area is just

$\displaystyle A = r d \theta \cdot dz$

Therefore

$\displaystyle F = -dp r d\theta \cdot dz$

Since the small element is in circular motion with angular velocity $\omega$,

$\displaystyle F = - dm \, \omega^2 r$

Using our formula for $F$ and $dm$, we get

$\displaystyle F = -dm \, \omega^2 r = -( \rho r d \theta \, dr \, dz) \, \omega^2 r = -dp \, r \, d \theta \, dz$

Simplifying,

$\displaystyle dp = \rho \omega^2 dr$

$\displaystyle \Rightarrow \partial{p}/\partial{r} = \rho \omega^2$

Why the partial derivatives? Because we’ve always been measuring $p$ from the same $\theta$ and $z$. Now we need an expression for $p$ in terms of $h$. Observe that even when the water is rotating,

$\displaystyle dp = - \rho g \, dz$

because the $z$ component of force on our mass element has to cancel out. Again, the negative sign is there because as you increase $z$, the pressure becomes smaller.

We can integrate this immediately to get

$\displaystyle p(r, z) = -\rho g z + p(r, 0)$

How do we determine $p(r, 0)$? It’s just $\rho g h(r)$. This gives us

$\displaystyle p(r, z) = \rho g (h(r)-z)$

Therefore

$\displaystyle \partial{p}/\partial{r} = \rho g dh/dr$

Pluggin into

$\displaystyle \partial{p}/ \partial{r} = \rho \omega^2$

We get

$\displaystyle \rho g dh/dr = \rho \omega^2 \Rightarrow dh/dr = \frac{\omega^2}{g}$

This is the same equation as before and the solution goes on the same way.

Hopefully you now have a better understanding of how whirlpools work. If you liked this post, please subscribe. Like always, leave me a comment if you have any questions!