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Bob, a budding physicist, is sitting on a swivel chair at rest, holding out a large motorized flywheel in front of him. The wheel is oriented parallel to the ground.

Bob's swivel chair is a magical, 100% completely frictionless swivel chair, and Bob's weight is entirely supported by it (i.e. his feet aren't touching the ground). When Bob turns on the motorized flywheel, the wheel starts to spin counterclockwise. Naturally, as a physicist would expect, Bob in his chair immediately begins to rotate clockwise, since total angular momentum must be conserved.

After a fun bout of spinning, Bob deactivates the motor and the flywheel abruptly stops turning. Naturally, because angular momentum is conserved, Bob's chair also abruptly stops spinning so that Bob is once again at rest.

Sandra, another physicist, sees the fun Bob is having and decides to try the same thing with her own swivel chair. However, Sandra's chair isn't a magic frictionless chair like Bob's. It is very well oiled and has a low coefficient of friction, but the coefficient isn't exactly zero.

Undeterred, Sandra starts up her counterclockwise-spinning flywheel in the same way as Bob did, and rejoices as she begins to spin clockwise in her chair. But then, when she kills the motor and the flywheel abruptly stops spinning, she discovers to her surprise that she is not at rest. Instead, something very peculiar—some might even say "remarkable"—happens.

She repeats the experiment several times and notes with amazement that the peculiar phenomenon is reproducible no matter what the starting orientation of her chair and no matter how long she runs the flywheel before stopping it.

What is the phenomenon? What happens when Sandra stops the flywheel?

COTO
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    ...Momentum? No? – Engineer Toast Jul 10 '15 at 17:55
  • How do we interpret the flywheel effect on their hands? Since they aren't exploding, I assume we can consider their hands and bodies being indestructible also? – Quark Jul 10 '15 at 18:04
  • It seems she must not be a very wise physicist if she thinks spinning around is quite peculiar.... – Mark N Jul 10 '15 at 18:11
  • @Quark: When I say "abruptly", suppose that means "at the greatest rate that doesn't break their hands". It actually doesn't matter; it just makes the problem easier to visualize. – COTO Jul 10 '15 at 18:12
  • @EngineerToast maybe of the wheel? – Mark N Jul 10 '15 at 18:14
  • @MarkN: I may not be wise, but I'm a physicist, and at least I found the phenomenon in question peculiar when I first observed it. ;) – COTO Jul 10 '15 at 18:14
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    @COTO I would think Bob's chair would be MUCH more peculiar instead ;-P – Mark N Jul 10 '15 at 18:17
  • @MarkN: He got it from "Ernie" of puzzling.SE fame, so they were kind of expecting it. – COTO Jul 10 '15 at 18:20
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    How is this a puzzle? Looks like it'd be better off on Physics.SE. But I haven't VTCed, in case you manage to convince me of its validity :-) – Rand al'Thor Jul 12 '15 at 22:31
  • could you verify my solution, OP? – hjhjhj57 Jul 18 '15 at 22:40
  • @randal'thor: It asks a puzzling question that can be solved by modeling the system and drawing conclusions from the model, the same way as countless other puzzles here. I obviously misjudged the audience with this one. There seem to be a lot of computer science degrees among the readership and not so many physics degrees. ;) – COTO Jul 19 '15 at 11:57

3 Answers3

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I believe what happens is that

Once the flywheel stops abruptly, Sandra slows down with the flywheel and as it comes to rest, Sandra also momentarily comes to rest very slightly before the flywheel, but then starts spinning in the opposite direction, in the same direction as the flywheel when it was spinning.

I am not very sure, but I feel the cause for this might be

When the flywheel slows down, Sandra slows down faster as some of her angular momentum is lost due to the frictional force. So Sandra comes to a stop but the flywheel still has some momentum, and as it grinds to a halt, to conserve momentum, Sandra gains some in the direction of the flywheel's rotation.

CodeNewbie
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  • Yup, if only I typed it faster... :'''( – Going hamateur Jul 10 '15 at 18:33
  • You're getting warm. It's true that she'll be spinning counterclockwise. But something even more specific can be said about the rate at which she'll be spinning, and the remarkable thing is that this fact doesn't depend on specific value of the coefficient of friction, or how fast she was spinning just before the flywheel stopped. – COTO Jul 10 '15 at 18:46
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Solution:

First of all, if we assumed instantaneous signal transmission (the whole chair knows exactly when the wheel stops) and the wheel stopped instantly friction wouldn't come into play, since the momentum would instantly disappear. From this we see the phenomena you're talking about arises from the time it takes the whole body to notice the wheel stopped.

Now, to exacerbate this quality we may imagine that Sandra is floating around in space holding the wheel with some extra large tweezers, in such a way that when the wheel stops it takes a noticeable time, say $t$ seconds, for the signal to travel to Sandra.

Now it's not that hard to figure out what will happen! As the wheel stops, Sandra will keep spinning for $t$ seconds, time during which friction will continue to act as an opposing force and Sandra will continue spinning!

So when Sandra notices that the wheel stopped and her chair stops spinning, the wheel will continue 'seeing' the friction opposing Sandra's rotation for $t$ seconds, experiencing said force during this time and making it move in the opposite direction as Sandra did. In fact, the wheel's acceleration will be proportional to the friction.

Ultimately the reverse angular momentum that Sandra gained during the arresting of the flywheel momentum will restore her orientation at rest to precisely her starting orientation. This phenomenon is known as damping-induced self-recovery.

COTO
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hjhjhj57
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  • You're correct that the arrest induces a rotational momentum in the opposite direction. This particular phenomenon doesn't depend on the finite $t$ during which momentum is arrested. I actually posted the correct answer in the comments to hamateur's answer (see my comments on damping-induced self-recovery). Since he isn't around to write up the explanation, I extend the same offer to you: include a brief explanation in your solution and I'll award you the bounty. ;) – COTO Jul 19 '15 at 00:54
  • I know the only important fact is that $0<t$, not its value. What do you want me to explain? – hjhjhj57 Jul 19 '15 at 00:59
  • I've added the explanation since the bounty is set to expire. The key observation is that $\theta {\rm final} = \theta _{\rm initial} ;\forall ; \theta _{\rm initial}, t{\rm flywheel\ on}, \omega {\rm flywheel}, t{\rm flywheel\ arrest}$. – COTO Jul 19 '15 at 08:16
  • I don't see how the final position will be equal to the initial position... – hjhjhj57 Jul 19 '15 at 08:37
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    Set up the Lagrangian for the entire system, and include a damping term $-\mu\dot{\theta}$ with constant $\mu > 0$ in the equation of motion for the chair, which is the standard (and most accurate) way to model friction for swivel chairs. The solution for the entire system can be analytically derived, and you'll find that all terms in the post-switch equation decay to zero, with the dominant term being of the form $\exp\left( c_0 + c_1\mu t\right)$ with $c_1 < 0$. The same phenomenon is observable in viscous fluids. Check out the video. ;) – COTO Jul 19 '15 at 11:50
  • cool! I'll work out the math and see.. – hjhjhj57 Jul 19 '15 at 21:17
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Okay here we go freshman physics!

She is spinning the opposite way that she was while the wheel was spinning!
Angular momentum is conserved in the universe, not in here system due to friction.
So the angular momentum lost to friction (that effectively goes to the earth, not that the earth notices much)
But in the system the motorized wheel still had its full angular momentum.
So the net angular momentum is in the way the wheel was spinning, but now it is not spinning so something (or someone) else has to spin that way.
So our spinning physicist is going around the other way.

Update to address COTO's Comment:

SHE SPINS BACK AS LONG AS SHE SPUN FORWARD!

How much angular momentum was lost? Well Torque = Force*distance (for perpendicular)
Force is from friction so $\mu$*NormalForcedistance
So torque is constant while the physicist is spinning
So Net angular momentum shift is Time$\mu$*Weight*distance
Well now the wheel stops and we spin the other way
So we are getting torqued, in the other direction
well the net angular momentum shift divided by Time*$\mu$*weight*distance
wait, what is time? time is how long the physicist went in the initial direction
so how long will it take them to stop?
well AM/Torque = Time*$\mu$*weight*distance/($\mu$*weight*distance) =time
wait what?
SHE SPINS BACK AS LONG AS SHE SPUN FORWARD!
even if she waits till she stops the first way, since she is no longer going she is no longer getting torqued!

Going hamateur
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  • You are correct, but see my comment to CodeNewbie. – COTO Jul 10 '15 at 18:47
  • @COTO I think I got it – Going hamateur Jul 10 '15 at 19:35
  • You might want to check out the MathJax tutorial. It will make everything look much nicer. And just so you know, when you do something like r*t*u, it will show up with italics, like rtu. – mmking Jul 10 '15 at 20:22
  • I might have used better characters and stuff but I just now had to use google to remember angular momentum is j... – Going hamateur Jul 10 '15 at 20:29
  • You're getting even warmer, but I still think the solution can be more specific. In particular: suppose we denote the angle of her chair as $\theta$, we define clockwise rotation as positive $d\theta$, we let her initial orientation (before starting the wheel) be $\theta = \theta _0 = 0$, we let her orientation at the moment the flywheel stops be $\theta _s$, and we let her final orientation (when she comes to rest) be $\theta _f$. In a mathematical statement using these symbols, what can be said about the relationship between them? – COTO Jul 10 '15 at 20:42
  • Uggh I was thinking it might have been something like this because of my experience (traumatic) with precession... – Going hamateur Jul 10 '15 at 20:45
  • @COTO So I did a very brief thought experiment, if you start it up instantly, wait 1 time unit then shut it down, you have travelled roughly 1$\omega$, then you stop and go backwards at velocity 1$\mu$N and you go about half that distance by average velocity math. and quite frankly I keep expecting for final $\theta$ to equal initial $\theta$ but it doesnt seem in that case to work out... as $\omega > \muN/2$ can be the case – Going hamateur Jul 14 '15 at 17:22
  • Actually, $\theta$ does decay to the initial $\theta$ in a damped second-order response, regardless of $\omega$ or $\theta _s$. The phenomenon is called "damping-induced self-recovery". A demonstration is viewable here. If you post a brief explanation in your solution, I'll give you the bounty. Your answer is closest anyway. ;) – COTO Jul 14 '15 at 22:25
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    @COTO so I briefly looked into it. It blows my mind, didnt go in depth enough to write it up (yet), but apparently the whole thing where cats reorient themselves 180 degrees while falling and while having 0 angular momentum is related to this so that is extremely cool. – Going hamateur Jul 15 '15 at 14:19