Why doesn't a tied balloon behave like a pendulum?

It is well known that a tied weight will oscilate under the effect of gravity if left from aside, like a pendulum. However, if we tie a helium balloon to the ground from and left it form the floor (not exactly from where it is tied, but from a side) it will go upwards until the string is not loose without oscilation.

Why is this like that? How is the force of the helium pulling up different from the gravity pulling down in the pendulum example?

• My initial guess: The balloon has a very small mass and friction is large (large surface area), so the oscillation is very damped. Sep 29, 2014 at 9:42
• @Jasper: that should be an answer. Sep 29, 2014 at 10:47
• @John: Thx for the advice! Just added it as an answer. Next time I'll put such an answer there, just wasn't sure if such a short statement would qualify. Sep 29, 2014 at 10:55
• It's not precisely the same. The net force on the balloon is upward, but the weight of the tether is still directed downward. Assuming a fairly chunky piece of string or similar, this force is probably not negligible compared to the buoyancy of the balloon. Sep 29, 2014 at 11:58
• Gravity acts on mass. Buoyancy is a function of lack of mass compared to the substrate. Mass is required for inertia. A pendulum works because the gravity acting on the mass of the object is balanced by the object's inertia. Sep 30, 2014 at 20:48

The balloon has a very small mass and friction is large (large surface area), so the oscillation is very damped.

• But if you put the helium balloon in a vacuum then... er... it would burst and the string would fall to the floor. Oh well. Sep 29, 2014 at 13:04
• Not necessarily.. A strong enough material can easily hold back the pressure of the helium inside, but the problem is that in vacuum the balloon would not "rise". I would expect that you could do the experiment in air with a weird-shaped "balloon" with very small surface area in the direction of motion, which would oscillate for a short period. Sep 29, 2014 at 13:11
• You could do it in a superfluid - which behaves as if friction is null. But one of the few physical superfluids we know is liquid helium 4. Which is cold. And not much denser than helium. And will shrivel the helium balloon. Oh well. Sep 29, 2014 at 14:01
• @Richerby: Actually thought you posed an interesting question of how to it would be possible to see the oscillatory motion. ;) Sep 29, 2014 at 14:14
• Rather than saying "the oscillation is very damped", it's better to say that there is no oscillation. Oscillations only exist if the system is underdamped. In an overdamped system (which almost certainly is the case here), the oscillatory behavior of an underdamped system becomes a sum of two exponential decays. Sep 29, 2014 at 15:48

Tied balloons do behave like a pendulum, you only need really massive ones:

You can see it live in a video.

Hot air balloons have such a big amount of..erm...hot air that during the start you can expect oscillations because while the surface area is big, the mass inside is so big that the dampening is low enough. They are not exactly like a pendulum because the hot air can move freely (and mixes with colder air inside), but you can see that it is definitely oscillating.

To answer counterarguments: Hot air balloons are mostly started with very low wind speed which is nearly calm at the ground, so it is not the wind. And the oscillating does also happen if the burner is shut off.

Actually, it does behave exactly like a pendulum. The equations of motion are exactly the same. The issue, as Jasper pointed out, is damping. When you think of a "normal" pendulum, you are considering a lightly damped oscillator. The balloon, as I will prove below, is heavily damped.

For a damped harmonic oscillator (which a pendulum approaches for small deflections), the general equation is

$$m\ddot x + \mu \dot x + k\cdot x = 0$$

Where $x$ is displacement, $m$ is the mass, $\mu$ is the drag coefficient, and $k$ is the "spring" coefficient.

Sometimes this equation is rewritten as

$$\ddot x + 2 \zeta \omega_0 \dot x + \omega_0^2 x = 0$$

Where $\zeta$ is called the damping ratio, a dimensionless number. - more about that in a minute.

The solution to this equation depends on the degree of damping (the magnitude of $\zeta$). It is easy to see that

$$\omega_0 = \sqrt{\frac{k}{m}}$$ and $$\zeta = \frac{\mu}{2m\omega_0} = \frac{\mu}{2\sqrt{m k}}$$

We solve this by using a trial solution:

$$x = A e^{\gamma t}\\ \gamma^2 + 2 \zeta \omega_0 \gamma + \omega_0^2=0$$

which is an equation with two (complex) roots:

$$\gamma = \frac{-2\zeta \omega_0\pm \sqrt{4 \zeta^2 \omega_0^2 - 4 \omega_0^2}}{2}\\ =\omega_0(-\zeta \pm \sqrt{\zeta^2-1})$$

When $\zeta \gt 1$ the roots are real, and the equation is that of an overdamped oscillator - meaning that it never oscillates, just slowly returns to the equilibrium position:

$$x(t) = A e^{\gamma_+ t} + B e^{\gamma_- t}$$

Where $\gamma_+$ and $\gamma_-$ are the two roots of the above equation.

All that remains is to prove that for a balloon, $\zeta \gt 1$.

Since a helium filled ballon is "lighter than air" we can put an upper boundary on the estimate of the mass from the mass of an equivalent body of air. For a 25 cm diameter sphere, the volume is approximately 8 liters, so the mass is < 8 gram (1 kg / cubic meter is a nice approximation of the density of air: it's a little higher, but we are estimating here).

Next, we note that a sphere moving through a medium has an apparent inertia that is its own inertia plus half the inertia of the displaced medium, so we add 4 grams for $m=12 \text{g}$ - except that since the balloon is lighter than air we will say it is $10\text{ g}$ and allow for a tension in the string of $2\text{ g} = 0.02\ N$ (we will assume massless string…)

The drag coefficient of a sphere in air is a function of the Reynolds number. If the sphere moves at 10 cm/s, we compute

$$R = \frac{D V \rho}{\mu}~2000$$

This means the coefficient of drag is 0.43, and the drag force given by

$$F = \frac12\rho v^2 A c_d$$

this is not linear with velocity, so we need to make a further approximation that the average velocity of the sphere during its motion is 5 cm/sec - then we can compute $\zeta$ as

$$\zeta = \frac12 \frac{\rho v A c_d}{2\sqrt{mk}}$$

Now we need to convert the effective buoyancy to a "spring constant", using the small angle approximation.

For a pendulum of length $l$ and (small) deflection $x$,

$$F = \frac{mgx}{l}\\ k = \frac{mg}{l}$$

In this case for a balloon on a 40 cm string with $0.02\ N$ of buoyancy, $k = \frac{0.02}{0.4} = 0.05 N/m$

This gives us

$$\zeta = \frac{1\cdot 0.05 \cdot 2000 \cdot 0.43}{4\cdot \sqrt{0.012 \cdot 0.05}}\approx400$$

So yes - $\zeta \gt 1$ which we set out to prove (quod erat demonstrandum). We conclude this is a heavily damped system, and the solution is a slow return to the equilibrium position without oscillation.