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?
Answer
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.
And that is your explanation.
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