Physics of the Bouncy-Bounce

SteelyPalooza came off very well, despite high disaster potential. We were, after all, inviting a dozen five-year-olds plus assorted siblings to our house, on a day when Kate and The Pip were out of commission due to coxsackie virus. Everything went smoothly, though: the kids loved the bouncy-bounce, SteelyKid's playset and playhouse, and the six-person tent we set up out back. They're getting close to an age where they can entertain themselves, so I was even able to enjoy snatches of adult conversation during brief lulls in my hosting duties.

This was, however, thoroughly exhausting, so really heavy blogging is out. But let's try to ease back into the physics-blogging thing by talking about one of the big hits of the party, the bouncy-bounce. We rented a bounce castle from a local company specializing in inflatable party games, and the kids careened around inside it to the general delight of all. And once they had all left, but before the delivery guys showed up to pick it up, I joined SteelyKid inside for a bit, as you can see in the "featured image" above. Hey, I paid for the thing, and I figured that any damage could be attributed to the kids...

Anyway, bouncing up and down was enjoyable enough, and I did a couple of belly-flops for SteelyKid's amusement. Despite the impression you get from the outside, though, hitting that surface with your head hurts. SteelyKid and her friends thought it was great fun to bounce as high as they could manage, and basically sit/fall from the highest point onto their butts. They'd bang their heads in the ensuing collision, but popped right back up giggling. The one time I tried it, though, I got my bell rung pretty good. Which got me to wondering, while I waited for the cheeping cartoon birds to dissipate, why it seemed to be harder on me than on them.

The obvious physics here is Hooke's Law for the force due to a stretched or compressed spring, most famously stated as "ceiiinosssttuv." That's a Latin cryptogram that unscrambles to "Ut tensio, sic vis," or "As the extension, so the force." Hooke did it that way so he could establish priority without giving up his advantage over other proto-scientists investigating the same thing. You think Open Access publishing has problems now, try 1660...

Anyway, in modern mathematical notation, this is generally written:

$latex F_{spring} = -ks $

F is the force, k is the "spring constant" characteristic of a particular spring, and s is the stretch away from the equilibrium length of the spring (if the spring is compressed, s is a negative number). The minus sign is in there to ensure that the force is in the correct direction-- for most of this, we'll only worry about the magnitude, so I'll blithely ignore it.

Now, you might object that there aren't any actual springs here, but to a physicist, basically everything behaves like a spring, to a first approximation, because the behavior of a spring is an easy problem to solve. And there's definitely some spring-like nature to the bouncy-bounce: if nobody is inside, the surface is at a particular level, but a person stepping on it pushes the surface down by an amount that depends on their mass-- you can see this in the picture above, as SteelyKid is definitely tilted down into the depression created by my fat ass, which is depressing the red band I'm sitting on well below the equilibrium line seen in the blue band behind us. this is a Hooke's Law sort of behavior-- the more the surface is compressed, the greater the force, so any person inside compresses it until the force from the surface is equal to the gravitational force pulling them down.

We can find the compression due to a given weight by setting forces equal to each other, or find the spring constant given the weight and comrpession, but what we really care about is more of a dynamic behavior-- that is, how much does the balloon compress when a person of a given mass jumps on it, rather than just standing? This is a slightly trickier problem, that I love to give to students in our intro mechanics class, who alas tend not to share my enthusiasm for it.

For the case where you stand still, the solution method is just to set the force of the compressed balloon equal to the force of gravity, and solve for whatever you want to find. In the case of a falling person, though, that's not enough, because the person is initially in motion. The temptation of balance-of-forces is hard to resist, though, leading to lots of invented fictitious forces, or just flat-out ignoring the common-sense observation that the compression is greater for a bouncing person than a standing one.

The simplest way to get the right answer is to use energy methods. A compressed spring has a potential energy given by:

$latex U_{spring}=\frac{1}{2}ks^2 $

where k and s are defined as above, and U is the symbol for potential energy because reasons. Maybe "potential energy" starts with "u" in German, or something, I don't know. Anyway, to find the maximum compression of the balloon, you would set this energy equal to the energy of the moving person when they hit the bag. That, in turn, can be determined from the maximum height h from which they fall, where all their energy is gravitational potential energy:

$latex U_{grav} = m g h$

where m is their mass, and g = 9.8 N/kg is th strength of gravity near the surface of the Earth.

Setting these energies equal gives:

$latex \frac{1}{2}ks^2 = mgh$

which we solve to find s:

$latex s_{max} = \sqrt{\frac{2mgh}{k}} $

(The compression for the static case is just $latex s=mg/k $, which you can show easily. For homework, calculate the ratio of dynamic to static compression, and explain why that makes sense. Send your answers to Rhett for grading.)

That's the maximum compression, at the point where the bouncing person comes to a stop for an instant before reversing direction. This, in turn, gives the maximum force experienced by the bouncing person, from Hooke's Law:

$latex F_{max} = k s_{max} = \sqrt{2 m g h k} $

For the purposes of comparing the experience of bouncing for me versus SteelyKid, g and k are constants, so the difference is in h and m. My mass is roughly seven times hers, and she's a little more than half my height, so if we each just dropped to the ground, my center of mass would fall just about twice as far as hers (there's a subtlety here having to do with the fact that h is implicitly measured from the point of maximum compression, not the equilibrium point of the surface but that's not going to make enough difference to change this substantially). That means the stuff in the square root is about 14 times larger for me than her, so I experience a force that's about 3.7 times larger than she does. Which might be why it hurt my head to hit the balloon surface, but doesn't really affect her.

Then again, it's not clear that force is really the right thing to measure, here. When people talk about impacts, after all, they usually cast it in terms of the acceleration that the person experiences-- g-forces, and that kind of thing. But if I do that, things get a little odd-- force is, after all, related to acceleration through Newton's second law (since neither SteelyKid nor I are moving at relativistic velocities or changing mass during the fall, I can get away with this), $latex F = ma $. But then, the maximum acceleration becomes:

$latex a_{max} = \sqrt{\frac{2ghk}{m}}$

That actually goes down as the mass increases-- according to this analysis, the acceleration I experience would be the square root of 2/7ths times that she feels, or a bit more than half. Which would suggest that the collision ought to be easier on me than on her. So, maybe I'm just old and stiff while she's young and flexible, and even though the actual jarring is less for me than her, the resilience of youth is more than enough to overcome it. Which is kind of depressing.

Of course, this could also be a case where the spherical-daddy approximation breaks down. That is, the analysis above implicitly assumes that both of us are point masses, which is manifestly not true. And it's much more not-true for me than for her-- it could be that a proper analysis of this needs to include the rotational motion-- maybe the brain-rattling impact I felt was a whiplash kind of effect, due to my head moving much faster than my center of mass because it's way out at the end of a long lever arm, as it were. Maybe when you include that, the actual acceleration of my skull is greater than hers.

That, however, is not an analysis I'm going to go through at 10pm on the Sunday night of a very long weekend. Especially when I'm drinking a few of the leftover beers that we don't have room for in the fridge. So, if you want to test that hypothesis, do it on your own paper and post it in the comments. Or submit it to Rhett to be graded for extra credit points...

If you, like my students, prefer numbers with your math, we can attempt an order-of-magnitude estiamte: from the picture above, I'd guess that I'm compressing the balloon by on the order of 10cm in that. My mass is on the high side of 120kg, so let's call it 12 cm, because then we can call g 10 N/kg, and end up with a spring constant of 10,000 N/m. My center of mass is normally a bit over 1m off the ground, so in a straight drop, I would experience a maximum force of about 4900 N, or an acceleration of about 41 m/s/s. That's a hair over 4 times the acceleration of gravity, so more or less what you'd feel on a good roller coaster. by the analysis above, that would have SteelyKid pulling around 7.5 g's, so I'm probably overestimating the spring constant-- doubling the estimated static compression would bring those down to a bit less than 3 and a bit over 5, which seems more reasonable. And this suggests that lifting the spherical daddy approximation is essential, because the collision felt worse than any roller coaster I've been on...