I’m teaching relativity in a course with an astronomy prefix, which means I’m obliged to talk about stars and stuff. Yesterday’s lecture was about neutron stars, and how their existence was confirmed by the discovery of pulsars (with the story of Jocelyn Bell Burnell included). This requires some discussion of angular momentum to explain how something that big ends up spinning that fast (cribbing a bit from these online notes), so I needed a good demonstration of angular momentum. Which is when I remembered this 2013 post with SteelyKid on the playground, where I estimated the mass of the merry-go-round at the JCC from how the rotation changed as she moved.
Of course, that effect isn’t as dramatic as it could be, because her mass is pretty small. It would work a lot better with my fat ass on the merry-go-round; and, indeed, it turns out I have video of that:
My mass is considerably larger than SteelyKid’s, so when I move to the center, you get a much more pronounced increase in rotation rate. In fact, there’s an outtake of this where she spun it up too fast, and I was unable to drag myself all the way to the center against the centrifugal force. Also, I got really dizzy, and had to sit down for a bit…
Anyway, if I were using this in a majors course, I’d go through the calculation of the monments of inertia and all that. This is a Gen Ed class, though, and I’m crazy busy. But that doesn’t mean you can’t do it for homework. So, your assignment, should you choose to accept it, is to determine the mass of the merry-go-round from the above video. You need some numbers for scale, so I am almost exactly 2m tall (definitely, in sneakers), and my mass is around 127 kg.
Is the mass you estimate from this video and those numbers consistent with my earlier estimate? What systematic sources of uncertainty are there in this calculation? Type your calculation up in no more than 5 single-spaced pages, and send it to Rhett for grading.
(This video was shot back in April, when we also looked at the Coriolis effect on the merry-go-round, but I didn’t upload it until yesterday when I wanted it for class…)
Ballpark w1 = 1/4 rot/s with you on the rim, and w2 = 1 rot/s with you at the center. Looks to be a 1.5 m diameter merry go round, so R = 3/4 m, and we approximate it as a uniform disc. You probably won’t be too upset at being approximated as a point mass.
Angular momentum is L = Iw, and conservation of angular momentum is I1*w1 = I2*w2. The moment of inertia of a disk about its center is MR^2 / 2, where M is the mass of the disk. A point mass has no moment of inertia about its own centerline, so you don’t contribute to the moment of inertia when you’re at the center, and I2 is simply the moment of inertia of the disk. By the parallel axis theorem, the moment of inertia of a point mass about an axis a distance r from the point is mr^2, and since you’re at the rim of the merry go round, r = R. So I1 = MR^2 / 2 + mR^2.
Now, solve for M.
I1w1 = I2w2
MR^2 w1 / 2 + mR^2 w1 = MR^2 w2 / 2
MR^2 (w2 – w1) / 2 = mR^2 w1
M = m * 2 w1 / (w2 – w1)
Dimensions work out, mass = mass * length / length. Plugging in values,
M = m * 2 * 1/4 / 3/4
M = m * 2/3
So, if you’re m = 170 kg, the merry go round is M = 57 kg, which is about 120 lb for Americans like me. Seems reasonable.
If you were truly a point mass, the merry go round would be spinning a bit faster with you at the center, and about the same with you at the edge. So if w2 were bigger, that means the factor would actually be a bit lower than 2/3, which means the mass of the merry go round is probably a little bit overestimated.
At the same time, the mass of the merry go round is probably biased a little toward the center (assuming the hub is sturdy, and the bars are spaced more closely at the center than the outside) versus uniformly distributed. That means w1 is bigger than it would be under the uniform distribution assumption, and w2 wouldn’t change much, because you’re more massive. That would drive the mass factor up, which means the mass of the merry go round is probably a little underestimated.
So the net effect of those assumptions is an approximate wash.
Wait, 170 kg? Oops. I would have recognized that as a problem if I knew metric better. Sorry, Chad. At least I approximated you as a point mass.
Also, I apparently don’t know the difference between 1/3 and 2/3. 170 kg * 2/3 would have been 114 kg. But 127 kg * 2/3 is 85 kg, or about 187 lb. Still a reasonable number.