# What are the equations of motion of a hole in a soap bubble?

Imagine the following situation: I have a thin stationary water film, like a soap bubble, suspended inside a large ring. I throw a small loop of string onto the film and punch a hole inside it. How can I describe the motion of the hole in the water film bounded by the loop of string? The surface tension of the surrounding film will tend to minimize the ratio between the length of the boundary and the circumference, so the hole will disk shaped. Furthermore, if the weight of the string that bounds the hole is smaller than the weight of a water film with the surface given by the inside of two, then the effective mass of the hole will be negative, i.e. if the film is subject to a gravitational field, then the hole will tend to move upwards.

What is the correct way to describe such a system? How can I derive its equations of motion?

• Fantastic question. I have no idea how to begin, though. Shall be eagerly awaiting an answer: my guess is is that one will be forthcoming - I do know that the maths dept at Australian national uni was active in this field when I studied there and, it seems, still is Sep 10 '13 at 9:16
• More on soap bubbles: physics.stackexchange.com/q/69124/2451 and physics.stackexchange.com/q/40802/2451 Sep 10 '13 at 9:16

Let the large ring be placed vertically. For simplicity, consider a weightless string, which thickness is equal to the film thickness. Also, assume that the hole is quite far away from the edge of the ring so that we can ignore the surface phenomena of the film. Then the speed of the rising hole can be estimated as follows:

The buoyant force acting on the hole:

$$F_b=\rho gV=\rho g\pi R^2h$$ where $\rho$ is density of water, $g$ is acceleration of gravity, $R$ is radius of the hole and $h$ is the thickness of the film.

Next, we need a formula for the drag on the moving hole . We can use a formula for an infinite cylinder, which moves slowly in a fluid, perpendicular to its axis:

$$F_d=\frac{4\pi\eta v}{\ln\frac{3.7\nu}{Rv}}$$ where $\eta$ is dynamic viscosity and $\nu$ is kinematic viscosity of water.

This is the drag per unit length of the cylinder. The derivation of the formula is given for example in: H.Lamb, Hydrodynamics.

Now, equating the buoyant force and the drag force we get an equation for the rising speed $v$ of the hole:

$$gR^2=\frac{4\nu v}{\ln\frac{3.7\nu}{Rv}}$$ Here we used the formula $\nu=\frac{\eta}{\rho}$. This is a transcendental equation.

To get some estimation we use $\nu=0.01\frac{cm^2}{s}$ at $20^\circ C$ and $g=1000\frac{cm}{s^2}$. Let's introduce a new variable $x=\frac{3.7\nu}{Rv}$. Then the equation can be written as follows:

$$x\ln x=\frac{14.8\nu^2}{gR^3}$$ Now we have seen that $x\approx 1$ holds for all real values ​​for the radius of the hole $R$. That means we get a simple estimation for $v$:

$$v=\frac{3.7\nu}{R}$$ For example a hole with radius $R=1cm$ moves up with speed $v=0.037\frac{cm}{s}\approx 0.4\frac{mm}{s}$

• @UwF This is only an approximation to get an order of magnitude. That means that the estimate is roughly within a factor of 10. Also, $x>1$ not $x=1$, but very close to $1$. Sep 12 '13 at 15:53
• @UwF The drag formula is valid only at small Reynolds numbers i.e. at laminar flow regime. By definition, Reynolds number: $Re=\frac{Rv}{\nu}=\frac{3.7}{x}$ so $x=\frac{3.7}{Re}$. The flow is laminar when $Re<2300$. At given case $x\approx 1$ or $Re\approx 3.7$ so the flow is laminar and the drag formula applies. You can not choose the value of $x$ arbitrarily, it must satisfy the transcendental equation. Note that $x\leqslant 1$ does not satisfy this equation at all. Sep 13 '13 at 16:51