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

• Farina
In summary, the conversation discusses the question of why a spinning gyroscope remains horizontal and appears to levitate when suspended on a string, while a non-spinning gyroscope falls. The individual asking the question has searched for an explanation, but has not found a satisfactory one. They have provided a detailed explanation of their thought process and asked for insight into the underlying forces at play. However, the responses suggest that the gyroscope does not truly levitate and that the individual's question may be based on a misunderstanding.
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.

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.

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Doug Huffman
Wow! I guess the 'entire' internet wasn't stumped after all.

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.

DaleSpam said:
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.

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

SteamKing said:
Wow! I guess the 'entire' internet wasn't stumped after all.

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 ;)

Farina said:
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.

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

Farina said:
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?
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.

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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.

DaleSpam said:
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.

Farina said:
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.
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?

Farina said:
I'm looking for an analytical explanation in the terms I framed.

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

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Doc Al and Greg Bernhardt
Farina said:
Gyroscopes absolutely "levitate" according to either the colloquial meaning of the term ...
Nonsense.

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?

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

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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.

russ_watters said:

Farina said:
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.

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.

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

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.

VantagePoint72 and Bandersnatch
SteamKing said:
Wow! I guess the 'entire' internet wasn't stumped after all.
You would at least need a levitating cat for that.

voko said:
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.

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.

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Astronuc, davenn, DrClaude and 2 others
LastOneStanding said:
an upward torque to counter the downward torque of gravity.
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.

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rcgldr said:
There is an upward torque. 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.

I don't know anything about aeronautics and particularly not helicopters so this is mostly Greek to me. But it sounds, as best as I can follow, like you're comparing apples to oranges. There is a real, reactive centrifugal force in circular motion as well but it does not prevent things from falling inward; it's a reaction force to the centripetal force and acts on the central body. There cannot possibly be an upward torque in this problem: all external forces on the gyroscope are fully specified (both of them, gravity and tension) and neither is capable of exerting upward torque about the hanging end. Forces do not appear out of no where, unless you transform to a non-inertial frame. It looks you are describing internal forces, just like the case of a centrifuge, and they have no bearing on the overall rigid body motion. Obviously they are important in their own right for ensuring things like helicopters do not fall apart, but they are not relevant here.

cj said:
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
No, there is no additional vertical force. If there were then the tension would be equal in the two cases.

The math gets very complicated, so I will keep it conceptual. In the initial setup, for both the spinning and non spinning gyros, the weight (acting at the center of mass) makes a torque about an axis which passes through the end of the string and is perpendicular to both the string and the axle of the gyro. The tension makes no torque about this axis. These are the ONLY two forces acting on the gyro, so the change in angular momentum about this axis is horizontal in both cases. For the non spinning gyro, this change in angular momentum increases the magnitude of the angular velocity about the axis thereby causing it to "fall". For the spinning gyro, this same change in angular momentum merely changes the direction of the angular momentum causing it to precess instead. So, from considerations of the weight alone we can determine the motion.

The tension in the rope then simply acts as a constraint force, providing whatever force is necessary to constrain the end of the axle so that it remains fixed. This can be determined by Newton's laws given the acceleration of the center of mass determined above.

The reaction to the torque produced by gravity and the support at the non-moving end of the gyroscope is a precession of the gyroscope, and the gyroscope doesn't tilt downwards. So there's no "upwards" torque, just no downwards reaction to the downwards torque. Say you add a weight to apply more downwards external force to the end of the gyroscope, the gyroscope responds by precessing at a faster rate. If you consider the weight as a separate component, then the end of the gyroscope exerts an upwards force onto the weight equal and opposing to the downwards force that the weight exerts onto the end of the gyroscope, but that's a reaction force exerted by the gyroscope onto the weight (along with gravity), and not the net force / torque exerted onto the gyroscope.

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rcgldr said:
If you consider the weight (or the source of the downwards force) as a separate component, then the end of the gyroscope exerts an upwards force onto the weight equal and opposing to the downwards force that the weight exerts onto the end of the gyroscope.

I really don't see that you're saying anything different than I did. In both the case of the ring and the hanging weight, the upward forces you're talking about are forces the gyroscope applies to other things. That is, as I said, not the same question as whether there is an upwards force holding the gyroscope up—to which the answer is no.

You're muddling up what I think is a very straightforward discussion with irrelevant alternative experiments, meanwhile not saying anything I didn't. OP asked how to explain the gyroscope not tipping over in terms of balanced torques. The answer is that torque is not balanced. That's all there is to it.

cj said:
We pored over this and found lot's about precession, but nothing about why the axis remains horizontal.
It remains horizontal because it precesses.

The question is based on the naive notion that torques make stuff rotate around the torque vector, so the torque from gravity should flip the gyro. But torque is actually the change of the current angular momentum vector per time. So how the object rotates under torque depends not only on the torque itself, but also the current angular momentum.

One must admit that the usual treatments in beginners textbooks are quite superficial and sometimes oversimplified. The correct analysis, on the other hand, is standard textbook stuff. You introduce the Euler angles, use the Lagrange formalism (since forces and torques are very complicated, particularly in this case; for me they are pretty unintuitive on a quantitative level in almost all cases). You find an excellent treatment on gyroscopes, using the force-torque approach as well as the (for me much simpler) analytical approach in

A. Sommerfeld, Lectures on Theoretical Physics, vol. 1 (Mechanics)

The Sommerfeld lecture series is among the best theoretical-physics texts ever written.

Of course, the spinning top was one of Sommerfelds specialties, because he wrote an exhaustive monograph on it together with the famous mathematician Felix Klein. I only know the German edition, but I'm pretty sure that it's available in English too. There you can even find a treatment of rotations, using quaternions, but that's pretty out of fashion nowadays (and it's not used in Sommerfeld's lectures either).

Astronuc
LastOneStanding said:
... torque is not balanced ...
From wiki: spinning top ... The result is that the torque exerted by gravity – via the pitching motion – elicits gyroscopic precession (which in turn yields a counter torque against the gravity torque) rather than causing the spinning top to fall to its side. near the bottom of this section:

http://en.wikipedia.org/wiki/Precession#Torque-induced

update / correction - A net torque is applied to the gyroscope, which reacts by precessing instead of tilting downwards (assuming no restriction to precession). The precession coexists with an internal reaction torque within the gyroscope, but that is different than the net external torque that is exerted onto the gyroscope.

Prior to this correction, the original version of this post had incorrectly stated that there was no net torque on the gyroscope since it didn't tilt downwards, and some later posts have quoted the original post. Rather than mis-lead anyone reading this thread, I've corrected this post, but left this note so that the later posts quoting the original post aren't confusing.

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VantagePoint72
rcgldr said:
if there was a net torque on the gyroscope, then the gyroscope would experience angular acceleration in the direction of the torque
There is a net torque on the gyroscope, that’s why the direction of the angular momentum vector changes. If there was no net torque on the gyroscope, the angular momentum vector would be constant, and the gyro would not precess.

mheslep and vanhees71
rcgldr said:
so there is no net torque on the gyroscope, just a steady rate of precession
This is self contradictory. If there were no net torque then the angular momentum would be constant: ##\Sigma \tau = I dL/dt##. Precession means that the angular momentum is changing, therefore there must be some net torque.

The key to understanding why gyroscopes don't levitate is that they don't produce any linear forces - only torques. If you try to make them into a levitating machine the gyro simply tries to bend the machine rather than lift it.

rcgldr said:
no net torque on the gyroscope ...
I corrected my previous post. A net torque on a gyroscope results in precession coexistant with an internal torque that opposes the net torque but doesn't change the fact that there is still a net torque exerted onto the gyroscope.

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rcgldr said:
an internal torque that opposes the net torque
The net torque is the sum of all external torques and fully determines how the angular momentum changes. So whatever that " internal torque" is supposed to be, it's not relevant here.

VantagePoint72
@rcgldr, I always appreciate it when someone acknowledges having made a mistake, but I think it would be best if you just did so in a post of its own rather than going back and editing all your past posts. This whole conversation is now very confusing to anyone just joining because you've removed all the incorrect statements from your previous posts. This is not good forum practice.

This is very helpful - thank you.

From nearly the beginning we said we realized we might not be asking a valid question. We asked, in an early post, for someone to answer the question as asked, or politely suggest how the question itself might be better phrased. Aside: we did indeed also post about a possible analogy to the gravitational/centripetal metaphor you mention.

It is disconcerting to have moderators not read the entire post(s). Also, simply posting links to sources isn't exactly a value-added service (especially since we clearly said we visited all the usual suspect sites). Do moderators actually get paid for this kind of service? Lastly, getting hung up on arcane definitions ("levitate") that are not germane to the real issue is off-putting, as is terse/sarcastic/rude responses. This is true for anyone, and it's especially true for moderators.

Lastly, for the umpteenth time we jokingly used the term "...stumped the internet." If you can't sense this, then either get out and socialize more, and simply ignore it. I mean, really?

Sorry for the rant, but a few of us in my group are very cognizant of polite customer service. I think PF needs a jolt of it.
LastOneStanding said:
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.

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