# “Why does gyroscope levitate?” stumps 3 physics profs

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1. Dec 7, 2014

### Farina

More specifically... question stumps 3 physics profs, PF, textbooks, and entire Internet.

It’s easy to find explanations of why a gyroscope precesses. What’s not so easy to find – even after spending, literally, hours thinking about it, surfing the web (including PhysicsForums), reading textbooks, and asking physics instructors with PhDs, is this: in terms of torque, torque arms, angular momentum, right-hand rules, cross products, and – especially – simple force equilibrium analysis, WHY DOES A SPINNING GYROSCOPE LEVITATE?

Here’s, specifically, how I’m stumped:

Take a toy gyroscope or equivalently a bicycle wheel gyroscope (I’m using both). Orientate the gyroscope so its axis of rotation is horizontal. Suspend one end of the axis by a rope, hold the other end of the axis with your hand – keeping it horizontal. Remove your hand, and of course the gyroscope falls and is left dangling on the rope with its axis now vertical. Now repeat with a spinning gyroscope. This time when you let go the gyroscope remains horizontal as it precesses. As framed by the following, WHY DOES A SPINNING GYROSCOPE LEVITATE?

For a non-spinning gyroscope…

Tension force from the string acts on one end of the wheel’s axis of rotation in the +y direction. A gravitational force acts on the COM in the –y direction. The negative y-forces are greater than the +y forces, as evidenced by the COM moving downward – thus the gyroscope flops over.

For a spinning gyroscope…

Same +y tension force, same –y gravitational force, but now there must be an additional +y force since the gyroscope exhibits no net vertical displacement, like it did when it was not spinning (its COM now remains in the same horizontal plane). Here’s the big question: what is the genesis of this new +y force? What gives rise to it in terms of underlying torques, torque arms, angular momentum, right-hand rules, cross products, etc., etc.

Thanks for any insight.

2. Dec 7, 2014

### Staff: Mentor

It doesn't levitate. The rope is holding it up. If you let go of the rope or cut it then the gyroscope will fall at 9.8 m/s^2.

Gyroscopes do not levitate. Unless you want to say "being held up by string" = "levitation", in which case everything levitates.

I suspect the people you mentioned were not stumped, but just trying to find a polite way to tell you that your question is counter factual.

Last edited: Dec 7, 2014
3. Dec 7, 2014

### SteamKing

Staff Emeritus
Wow! I guess the 'entire' internet wasn't stumped after all.

4. Dec 7, 2014

### Farina

Nope. Thanks for the response, though. This doesn't answer the question as framed. This is pretty much like all the other the other qualitative explanations I've seen, which all side-step explaining in terms of resultant vertical forces (and how they arise).

Once again...

non-spinning gyroscope + string on one end of axis = falls (sum of net vertical forces is downward)
spinning gyroscope + string on one end of axis = levitates (sum of net vertical forces is zero)

Why?

Turns out maybe this does stump the entire Internet.

5. Dec 7, 2014

### Farina

I didn't mean that literally. Thanks for the sarcasm though. Maybe it's time to tune-up your "tongue-in-cheek" nuance detector.

6. Dec 7, 2014

### Farina

I'm thinking it's something akin to how the moon is "continually falling" towards the Earth: for each little bit it falls towards the Earth due to the Earth's gravity, it's inertia moves it a little bit tangentially -- thus circular motion. I'm still not, though, sure of this and need an analytical explanation.

At this moment in time the entire Internet remains stumped ;)

7. Dec 7, 2014

### Staff: Mentor

Or maybe you should avoid "tounge-in-cheek" since it doesn't come through in text very well.

Btw, there was no sarcasm in my response whatsoever. Your question is counterfactual. Gyroscopes do not levitate by the usual meaning of the word. Or (since it seems that you are using a non-standard meaning of the word) everything levitates so you shouldn't be surprised that a gyroscope levitates also.

8. Dec 7, 2014

### Staff: Mentor

Ignoring your absurd claims of "levitation" and even sillier claims that you have "stumped" anyone besides yourself.

The difference between the two scenarios is simply that the tension in the string is different in the two cases. In the non-spinning case the tension is less than the weight, so it accelerates downward. In the spinning case the tension is equal to the weight so it does not accelerate.

Last edited: Dec 7, 2014
9. Dec 7, 2014

### Farina

Most mainstream reader get this kind of tongue-in-cheekiness. Responding with a "...but just trying to find a polite way to tell you that your question is counter factual" isn't abrasive? Next time try something like "perhaps there's a better way to frame you question, e.g. ..." and you'll be less off-putting.

Anyway, the question is not counter-factual. You simply might not agree with its terminology. Big difference.

Gyroscopes absolutely "levitate" according to either the colloquial meaning of the term, or the literal meaning of the term. Google "gyroscope levitation" and you'll learn this. I don't want to go down a rat-trap debate of the meaning of the word levitate; I'm looking for an analytical explanation in the terms I framed.

You have not done this; the entire Internet remains stumped.

10. Dec 7, 2014

### Staff: Mentor

This is certainly not the colloquial meaning of the term. I have never heard anyone but you claim that "levitate" means "sum of net vertical forces = 0". By your usage everything levitates. As I sit here in my chair I am levitating since the sum of net vertical forces is 0. So is my desk, computer, house, and pretty much everything that I see around me. Everything is "levitating" by your usage, so why shouldn't a gyroscope also?

Yes, I did. The tension in the rope is equal to the weight in the spinning case, so $m a = \Sigma F = 0$. In the non-spinning case the tension is less than the weight so $m a = \Sigma F < 0$.

There is no "new +y" force. There is simply a difference in the "+y tension force" between the two cases.

Last edited: Dec 7, 2014
11. Dec 7, 2014

### phinds

Nonsense.

12. Dec 7, 2014

### cj

I'm in the same Physics & Math Club at my school as Farina, and have been part of these conversations along with 3 physics professors (who have PhDs from University of Michigan, UC-Berkley, and somewhere else). Farina threw up her arms in frustration and went outside to feed the ducks, so I'm giving it a try...

This is not a dumb question, it has not been answered in this forum - as far as I can tell - and has not been answered anywhere else either, as far as we can tell, according the vertical force analysis. Maybe I'm not asking the question correctly?

We well understand that the word levitate means: "rise or cause to rise and hover in the air, especially by means of supernatural or magical power." Let's forget about using this word, and instead focus on this:

unless a gyroscope (with one end of its axis tethered to a string) is spinning, it'll flop down.

Someone mentioned "... the tension in the rope is equal to the weight in the spinning case, so ma=ΣF=0. In the non-spinning case the tension is less than the weight so ma=ΣF<0."

This response begs the exact question we are asking: why "In the the non-spinning case the tension is less than the weight so ma=ΣF<0."

Am I correct in assuming something creates an additional vertical force that, when combined with the tension force, keeps the axis horizontal? What, specifically (in terms of forces, torques, angular momentum, right hand rules, etc.) creates this force?

Why does a spinning gyroscope's axis (the other one being attached to a suspended string) remain horizontal, while one that isn't spinning flops over?

13. Dec 7, 2014

### rcgldr

Since the gyroscope is supported at one end, it experiences a downwards torque due to gravity which effectively acts at the center of mass of the gyroscope. The reaction to the downward torque is a rotation about a near vertical axis, which generates an opposing torque to keep the gyroscope nearly horizontal. If the rotation is too slow, the gyroscope will tilt downwards. If the rotation is too fast, the gyroscope will tilt upwards. If the gyroscope is released at just the right rate of rotation, it will remain nearly horizontal. The direction of rotation follows right hand rule. If the gyroscope is supported at the "left" end as viewed from the observer, and the top of the gyro is spinning towards the observer, then the gyro will rotate counter clockwise as viewed from above.

If the gyroscope is suspended from a string, then the entire system will move in a spiral like cycle, with the center of mass moving about a horizontal plane, with the radius of the entire system cycling inwards and outwards, and the rate of rotation of the gyro changing so that it's center of mass remains on a horizontal plane. There was a good video in video #9 from

http://www.gyroscopes.org/1974lecture.asp

but trying to view or download the videos which are stored at gyro.biz result in a DNS error
Fortunately, it was archived and can be viewed from this link, the pattern starts at 1:30 into the video:

http://web.archive.org/web/20140219053656/http://www.gyros.biz/lecture/wmv/9.wmv

Last edited: Dec 7, 2014
14. Dec 7, 2014

### Staff: Mentor

15. Dec 7, 2014

### cj

We pored over this and found lot's about precession, but nothing about why the axis remains horizontal. This was mentioned in the very first post in this thread.

16. Dec 7, 2014

### Staff: Mentor

Consider a top leaning at a 45 degree angle while spinning on the ground. What gives rise to the +y force necessary to keep it upright against gravity? The only difference between this top and your spinning wheel is the angle of the axis from vertical, 45 degrees vs 90.

17. Dec 7, 2014

### voko

Spinning or not, there is torque due to tension and gravity. Spinning or not, this torque changes angular momentum. A stationary wheel has zero angular momentum, so the torque simply increases the angular momentum in the direction the torque ("flopping down"). When the wheel is spinning, it has non-zero angular momentum in a direction perpendicular to the torque. Perpendicular is important; that does not increase the angular momentum in the direction of the torque, but makes it rotate.

This is similar to the motion in a central force field (gravity). If the initial velocity is zero, the motion is in a straight line toward the gravitation centre. If there is any velocity not aligned with that line, the orbit is a conic section.

18. Dec 7, 2014

### A.T.

You would at least need a levitating cat for that.

19. Dec 7, 2014

### VantagePoint72

I think this analogy is very useful, and I want to expand it a little. Farina and cj, terminological issues aside, I think the issue here isn't necessarily that you're not understanding the explanations you're being given. It's that you're not seeing why your original question isn't actually well-formed. There's a very particular way of explaining the problem that you want to be given. The problem is that such an explanation does not exist because it's not the reason for the phenomenon in question. I'm going to use voko's analogy to clarify what I mean.

I gather from your question that what you want is an explanation in terms kind of some force that acts directly on the free end of gyroscope and holds it up, i.e. an upward torque to counter the downward torque of gravity. There is no such counter-torque, and to insist on being shown one is to misunderstand what is going on during gyroscopic precession. Consider a planet orbiting the sun. Gravity tends to pull it inwards, and yet a circular orbit allows it to stay a fixed distance from the sun. Now imagine you came to the forum and asked:

"Why does an orbiting planet stay a fixed distance from the sun? What is the force that keeps it from falling inward? In the case of no angular velocity, the planet simply falls towards the sun. In the orbiting case, there is no radial acceleration so some force must be opposing gravity to keep it from falling. Three physics professors and the entire internet have not been able to give me an explanation in terms of force equilibrium of how orbiting planets don't fall toward the sun."

Do you see how this question is based on a fundamental misunderstanding? Of course there isn't a force opposing gravity, that's the whole point—there's net centripetal acceleration. Forces are not balanced along the radial axis! Sure, if you really wanted to, you could think in terms of the planet's non-inertial reference frame and conclude the the centrifugal force exactly opposes the gravitational force according to the planet. But that is a lot of unnecessary work for what is not really a good, fundamental explanation based on inertial reference frames. The question has wrong assumptions built right into it about how orbits work, and no amount of correct explanations would convince the asker until they were able to see that.

The same is the case here, with really the only difference being that the vectorial nature of torques and angular momentum is harder to picture than the vectorial nature of linear forces and momentum. There is no torque acting on the other side of the gyroscope to keep it from flopping over. We would not see precession if there were! The torque about a horizontal axis due to gravity is manifestly not balanced by another torque because precession occurs, just as the inward force on an orbiting planet is manifestly not balanced some outward force because circular orbital motion occurs!

Spend some time reflecting on this and hopefully you will see that you did not stump the internet; you only stumped yourselves.

Last edited: Dec 7, 2014
20. Dec 7, 2014

### rcgldr

It might help to consider how helicopters have to deal with gyroscopic precession effects on pitch and roll. The cyclic inputs are "advanced" by 90° to compensate for the gyroscopic reaction of the helicopter. A roll or pitch torque on the rotor results in a pitch or roll response 90° "behind" the actual torque applied to the main rotor (via an angle of attack on the main rotor blades that cycle up and down on each revolution of the main rotor via the cyclic portion of the swash plate), and in turn that response exerts a pitch or roll torque onto the rest of the helicopter, so what could be called a "reaction" torque is a real torque.

Last edited: Dec 7, 2014