Friday, November 13, 2009

Balloon Boy and the Spherical Cow (FPOTW)

Sorry, no balloon boy here, at least in the popular media sense... sorry.
Rather than my usual Fermi problem, today I would like to present a physics thought experiment.





Real life physics is not a textbook exercise. This is because reality rarely gives you all the variables. It is up to you, as a scientist, a physicist and as a critical thinker to determine what is germane to the solution of a real life problem.

A common joke among physicists is as follows:

One day a physicist was asked by a friend of his, a dairy farmer, to help design a better milking machine for her business. The physicist readily agreed, and asked for a couple of days to think about it, after which he would present his results and design to his friend.
Sure enough, a couple of days later the physicist come over to the farm with his laptop and a beautiful power-point presentation titled: "Explorations towards a better milking machine."

After setting up the projector and gathering his notes, the physicist presents his findings:

"After much though and calculation, I have come up with a workable solution to the cow milking problem. First, let us assume that the cow is a sphere of radius r, homogeneously filled with milk..."

8)

Although the following problem is not meant to be solved exactly, it can be explored to any depth you wish, by any method you choose! You can solve a simple rule of thumb Fermi type problem, and have plenty left to explore by adding an udder to a spherical cow. ;)


The situation is thus:

A helium balloon as shown above is at neutral buoyancy a distance above the ground. Attached a given distance below the basket is a bucket. In the basket is an identical bucket filled to the brim with b-b's. In the bottom of the basket is a hole.
At time t=0 the b-b's are poured through the hole at a given rate, where they fall straight down and into the bucket hung below.

Describe the subsequent motion of the balloon.

That's it.

Hint:
Ask yourself if the color of the spots on the "spherical cow" have an effect on the solution?
Have fun!

8 comments:

  1. The balloon will experience infinitesimal oscillations with a period roughly sqrt(2L/g), and a slow, steady drop, assuming no thermal, humidity or wind fluctuations.

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  2. Where L is the distance of the bucket below the basket.

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  3. How did you come up with that solution Jer?

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  4. Jer pulled it out of his ass, you know... the same place where you pulled the problem?

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  5. @ Jer,

    Well, I would think that since there has to be some air resistance since the balloon has to be neutrally buoyant in something, there is a terminal velocity associated with the bb. Therefore energy is not conserved.

    If a single bb was released and not caught, the balloon would start rising at a certain rate. Since the bb is in fact caught, and the energy is not conserved, I would assume the balloon would rise a certain distance during the drop time, but not be able to be pushed back down to its prior position. So, neglecting any oscillations, the balloon would increase in altitude, not decrease.
    I don't want to do the math right now, but will later.

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  6. Oh I didn't bother to sanity-check it, so I half pulled it out of my ass. And you're correct, Nowitzky, I neglected air resistance. I came up with the oscillation period because that's the length of time of perturbation of the system from the bb's weight being lost. I agree the air resistance would siphon off some of the mechanical energy, as would the balloon's & bucket's rise during the intervening time. How much, I haven't calculated. I don't feel like solving that DEQ. But essentially I was thinking it'd create a low-amplitude, decaying oscillation in the balloon for each drop.
    However, the reason for the balloon's loss of altitude: Loss of helium. Helium is really hard to contain outside of a solid metal tank, as it can diffuse through many materials due to its tiny molecular size. It'll diffuse out of the balloon, causing it to gradually sink.

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  7. How long is the diffusion time compared to the drop of the b-b? Also, say the balloon rises a little bit, now the air is a bit less dense. This would be an unstable equilibrium and the balloon would run away if the diffusion time is large.
    Is a DEQ really needed in a back of the envelope problem?
    Finally, would there really be a small oscillation? The potential is not correct for this is it?

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  8. For simplify, let's ignore air resistance and assume no loss of helium at first.

    In this case we can treat the balloon system as an harmonic oscillator with mass M.

    At t=0 the oscillator is at rest.(initial state)
    Next consider a time moment at which the last b-b fall into the bucket hung below.
    Mass of the oscillator remains unchanged but the center of mass is moved downwards by a length.
    That means some amount of potential energy is released.
    Part of this dissipates to the heat(collisions between b-b's and the bucket are inelastic)and the remaining is absorbed by oscillator.
    So the final state is oscillating balloon.
    Problem now is how much energy is dissipated and how much is absorbed.
    My guess is that this is a random variable which depends on the b-b's mass, on a rate they are poured.
    The subsequent motion of the balloon between the initial state and the final state is a some kind of random walk around the equilibrium point, i think.

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