Precession at LOW angular velocity

[Opus_723]

So I've been working on understanding precession better. I had a whole other thread where it was explained to me how a torque applied to an object with angular momentum can result in precession. But this is sort of a different question, so I'd like to start a new thread for it.

In my physics book, a formula is derived for precession:

\Omega=\frac{T}{L}

I get that tops and gyroscopes and such rotate around the vertical faster as they slow down. I can even see, looking at the acceleration vectors along the rim of a wheel under torque, that a point on the wheel with a small ω would spend more time under acceleration, and thus end up with a greater maximum perpendicular velocity compared to a wheel with greater ω and the same torque. But what bugs me is that, using this formula and that vector approach, as the angular momentum approaches zero, the precession velocity would go to infinity. Everyday experience seems to contradict this. If I hold a bicycle wheel that is just barely spinning, and I apply a torque to it, it doesn't shoot off to the side at near infinite speed!

My book seems to hint at an answer, pointing out that the above equation is only valid if Ω is much smaller than ω. But that doesn't seem to be enough to get me started. Can anyone help me understand this?

 

[ardie]

If the gyroscope's spin slows down (for example, due to friction), its angular momentum decreases and so the rate of precession increases. This continues until the device is unable to rotate fast enough to support its own weight, then it stops precessing and falls off its support. I can't quite put my finger on what assumption exactly is violated unless I see the full text of the proof, it depends largely on notation etc...

 

[ardie]

I just had an epiphany about this while reading Goldstein. its such a great book.
So if you imagine the top's centre of mass to be somewhere in the middle of it; while it is standing at an angle, the centre of mass is not exactly in the middle, but pulls the top down towards the outside, causing precession. At relatively large angular velocity, the Centre of Mass is essentially constantly varying in a circular fashion and its average position becomes the middle of the top again, hence the top can balance itself so as to stay up rather than fall down to the sides. At low angular velocities the centre of mass is shifted again towards the outsides and the argument above fails; hence the top eventually falls off.

 

[Opus_723]

I don't think that answers my question. What I want to know is: Why don't wheels spinning at low speeds seem to precess dramatically? Not specifically for tops, but wheels in general. If you look at torque vectors or linear acceleration vectors, it seems to me that the rate of precession should increase as angular speed decreases (like you said). But if I have a slowly spinning bicycle wheel, and I try to tilt it, it is easy to make it go where I want. It doesn't move sideways very much.

 

[ardie]

that's because of something called torque, more information here:
http://en.wikipedia.org/wiki/Torque

 

[Opus_723]

...I'm afraid that's not very helpful.

I know you are applying torque to the wheel. I just want to know why the movement of the wheel isn't out of phase with the torque, as in precession. Particularly since the math seems to show that precession velocity increases the slower a wheel is spinning.

 

[ardie]

Since you insist, I will try to explain this more clearly. First to try and clear up some ambiguities in your very own description. here is a diagram that will help clarify what I talk about:
http://www.maths.surrey.ac.uk/explore/michaelspages/Spin.htm
It is CRUCIAL, that you specify which rotation you talk about when you say they rotate. Your bicycle, and just about any other everyday rotating object, have only one axis of rotation, which is why precession is not very commonly observed in these objects.

I get that tops and gyroscopes and such rotate around the vertical faster as they slow down.

so this is an example that makes absolutely no sense when you consider the diagram. What rotation around vertical from which set of coordinates? and what slows down as a result?

I can even see, looking at the acceleration vectors along the rim of a wheel under torque, that a point on the wheel with a small ω would spend more time under acceleration, and thus end up with a greater maximum perpendicular velocity compared to a wheel with greater ω and the same torque.

That's not completely true, or perhaps I misunderstand the way you write this. The angular velocity at any point, on a solid wheel, is related to linear velocity by:
ω = v x r. Let's just ignore the cross product and say its just v.r (i.e. circularly symmetric case). the angular velocity at each point on the wheel is the same; that is to say if you draw a radial line, from the centre of the circle to the outer edge, then this line travels around the circle at the same rate (as one entity). so at a point where r is small, v must be large so that ω is constant along this line. similarly at large r, v is smaller to compensate.

But what bugs me is that, using this formula and that vector approach, as the angular momentum approaches zero, the precession velocity would go to infinity. Everyday experience seems to contradict this. If I hold a bicycle wheel that is just barely spinning, and I apply a torque to it, it doesn't shoot off to the side at near infinite speed!

OK so let's first establish that precession doesn't happen in a wheel, unless you get very creative with it and exert an angular momentum along another axis other than it's own centre.
lets go back to the diagram I linked for you. Now imagine you are standing at the top of the z' (PRIMED!) axis, on the very top of the spinning top. Imagine now that some very good person has set up a flag above the z-axis (unprimed!) so that you can look around and see this flag. You will now notice that the flag appears to be rotating around you at some rate \varphi. That is to say if you have a stopwatch, and press start when you see it right in front of you (while standing still) then some time later, you will see it again, and so you can count how fast it goes around you.
Now let's imagine that gravity forces an acceleration in the negative z-axis (UNPRIMED!). And now you have finished your juice, and are happy to go for a walk. Let's also imagine you have a hovering skateboard from "back to the future" and the disk's rotation doesn't take you with it as you move about. If you now go the outermost point on the disk, and hold up a pendulum, you will notice the pendulum will stay completely parallel to the z-axis (unprimed) it is a bit hard to imagine this I guess. Nonetheless the next step is the bottom line, and I really hope you will get it. Imagine the time stops now; the force of acceleration that you observe where you stand, points in such a way as to make the angle \vartheta, that the top makes with the z-axis, large. In other words it wants the top to drop perpendicularly to the ground. However if you let time pass now, and arrive at the opposite side of where you were at just before when the time stopped, the force that want's to ground the top points in exactly the opposite direction. So long as the rotation about the z-axis is fast, you can say that on average, the force that wants to make the top drop to the ground is zero (cancels in all directions due to symmetry).
This is a very simple analysis and that is why one cannot explain what the role of ψ is here. So what do we actually mean by precession? We mean that over time this angle θ starts to get bigger as the top cannot continue its motion indefinitely, perhaps due to friction or other sources. The downward forces eventually dominate and the top drops down.
It may be possible to enforce this type of motion on a wheel for a very short time. put the wheel on its side on the ground (flat ground like wood or so). then hold up the wheel by just one side at say 70 degrees to the horizontal, so that one side is on the ground and the other in your hand. now push the side that you are holding the wheel in your hand strongly such as to make the wheel roll, as you would usually(but at 70 degrees). You will notice that the wheel will not immediately drop, but will start spinning around its centre, as well as around a big imaginary circle like shape on the flour, but eventually it hits the flour. we say the wheel precessed around its axis of rotation before coming to rest.

 

[ardie]

i can't edit the bit that i wrongly talked about angular velocity. v = r x ω. so at larger r, v will be larger since ω is the same in circular rotation. v is the linear velocity of each point pointing straight like a tangent on the circle

 

[Opus_723]

I'm afraid this isn't going anywhere.

 

[Wood Croft]

That's because your physics book probably ommit the important consideration that the precession must be negligible to the torque for the formula \Omega = \frac{T}{L} to hold.
Check the derivation again. The precession creates an additional component to the angular velocity. If your textbook ignores this effect, than that's your answer.

 

[AlephZero]

It is theoretically possible for a wheel of a spinning top to precess in a "perfect circle" around the vertical axis, but in practice there is always some amount of "nodding oscillation" about the circle (the technical term for this motion is "nutation"). When the spin is fast and the precession is slow, this is stable. If you spin a top fast and then try to "knock it over" by tapping it sideways, it doesn't fall over but continues to precess, plus a bit of "wobbling" motion.

As the top slows down, the precession eventually becomes unstable, and the top suddenly falls over.

Your formula doesn't say anything about whether the precession is stable or unstable. Theoretically, a wheel spinnng slowly counld precess very fast as the formula predicts, but since this is an unstable situation you will never see it happen in practice.