Playground Physics

The playground outside SteedlyKid’s day care, amazingly in this litigious age, has a merry-go-round, a rotating disc with a really good bearing. The kids can really get the thing flying, which is kind of terrifying at times. But on the bright side, it’s an excellent venue for the physics of angular momentum:

In the embedded video, SteelyKid runs in and starts the merry-go-round spinning, then jumps aboard and goes around one full rotation before moving to the center for a few rotations, and then starting to move back out.

This isn’t exactly what I was hoping for– amazingly, a four-and-two-thirds-year-old doesn’t take direction all that well– but let’s dip into Rhett Allain territory and see what physics we can extract from this.

This is, fundamentally, an angular momentum conservation problem– having ridden the thing, I can attest that the bearing is good enough that it doesn’t slow all that much during the 30-second span of the video, so we expect that the rate of rotation should increase as she moves to the center. If I were Rhett, I’d go full bore with Tracker Video and map the rotation in detail, but I’m lazy, so let’s just count frames.

At frame 258 she has both feet on the outer rim, and at frame 330, she has come back to basically the same position, so 72 frames to complete one full rotation. Ideally, she would’ve hung on the outside for a little longer, but alas, she moves to the center after just over one rotation. Once there, she stays for about three full rotations, from frame 481 to frame 649, for 56 frames per rotation. So the rate is, indeed, higher, by about 29%.

What can we learn from this? Well, the angular momentum of the system is equal to the moment of inertia of the system multiplied by the rotation rate. When SteelyKid is on the outer edge, she contributes a fairly significant moment of inertia, but when she moves to the center, her moment of inertia is basically zero. The moment of inertia depends on the mass and the size, so let’s see if we can work out the mass of the merry-go-round.

The merry-go-round is basically a metal disk, and I know the formula for the moment of inertia of a disk is:

$latex I_{disk} = \frac{1}{2}MR^2$

Making the gross simplification that SteelyKid is a point mass located at the outer radius of the disk (which isn’t a great approximation, but makes the math easy), the total moment of inertia of the system once she’s on board is:

$latex I_{initial} = \frac{1}{2}MR^2 + mR^2$

(where m is SteelyKid’s mass). This simplifies to:

$latex I_{initial} = \frac{1}{2}MR^2 (1+2\frac{m}{M})$

so, it’s the moment of inertia of the disk plus a bit that depends on the mass ratio. The final moment of inertia is just the moment of inertia of the disk by itself. We can then invoke conservation of angular momentum to say that:

$latex I_{initial} \omega_{initial} = I_{final} \omega_{final} $

$latex \frac{\omega_{final}}{\omega_{initial}} = 1+2\frac{m}{M} $

The nice thing about this is that we don’t need any details about the sizes or masses to work this out– just knowing the frames per rotation will give us SteelyKid’s mass as a fraction of the mass of the disk, which for the numbers above turns out to be exactly 1/7th.

So, what’s that suggest for the mass of the disk? Well, SteelyKid hasn’t been on a scale in a while, but I’d guess her mass is around 50lbs, so the disk would be 350lbs (do your own metric conversions). That… seems kind of high, but this is very much a back-of-the-envelope upper limit. In fact, I worked it out on the literal back of a literal envelope, so there. It’s not ridiculously unreasonable, I don’t think.

A more detailed analysis would need to treat the bars separately, and probably take into account SteelyKid’s non-point-mass nature. A really rough attempt to account for her non-zero angular momentum at the center suggests this isn’t a terribly large correction, but I suspect the bars are a non-trivial contribution to the mass, and thus the moment of inertia of the merry-go-round. If you want extra credit, feel free to work all that stuff out (for scaling purposes, SteelyKid is around 42″/ 110cm tall), and send it to Rhett for grading.

What I had initially hoped to get was video of her running and jumping directly onto the disk, because that’s an exam and homework problem we use a lot, but she wasn’t into making that leap. Another way to refine this measurement would be to have somebody else– me, say– do the same move-to-the-center thing, and check the two masses for consistency. As a bonus, you’d get the comedy value of seeing me get really dizzy and seasick, but I’m not sure SteelyKid can hold the camera steady enough. Maybe I’ll take a tripod over there some weekend and see what we can get. I can do the run-and-jump myself, for that matter.

A video of the merry-go-round spinning by itself for a long-ish time would also allow an estimate of the frictional torque (it does eventually slow and stop), which is another correction factor. My impression from riding it, though, is that this isn’t a big deal.

Anyway, there’s another fun-with-everyday-physics exercise for you. I’ll try to get a better video one of these days (this was shot yesterday because for once we were the only ones on the playground and The Pip was home sick) with more rotations on the outside edge and maybe it’ll make its way into next year’s mechanics classes.