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Gyroscopic procession relationships

  1. Apr 12, 2004 #1
    good day to you, good sir or madam!
    i am in a bit of a predicament with a physics assessment of mine; i need to know the mathematical relationships between the precessional frequency, the couple (or torque, as i have read it called) and the angular momentum of a gyroscope, and (here comes the tricky bit) i would like to understand it...
    i just need someone to point me in the right direction, i mean i hardly know which questions to ask or where to start looking (sob sob). :confused:

    i come with the rather unsound basis of a-level physics, which is why it will be an especially challenging task for you get me to understand it. good luck (and thanks).

  2. jcsd
  3. Apr 12, 2004 #2
    Hey when you figure it out let me know, I found a website that might be of help but I am still going through it trying to figure it out.

    Also, I still don't understand the precession of a projectile, as in my post in the mechanical/aerospace forum that nobody seems to know the answer to.

    I also asked an engineering forum (I got a userid just to post there) but it turns out they don't like students and my userid was revoked for asking a student question even though I knew the answer and just wanted to know the why of it.

    here is the website I found, good luck, because it still doesn't make sense to me.

    http://www.euclideanspace.com/physics/dynamics/rotation/rotationconcepts/index.htm [Broken]
    Last edited by a moderator: May 1, 2017
  4. Apr 12, 2004 #3


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    If you understand vectors, then this is just an application of vector algebra. An understanding of calculus may help. You should also probably draw a picture or two.

    τ be the vector quantity torque
    L be the vector quantity angular momentum.

    τ = dL/dt

    This is the basic equation. Gravity will act on the center of mass, and some constraint force will act on the pivot point (which holds the gyroscope up while at the same time constrains the precession to a plane perpendicular to the gravitational force). Together, these two forces induce τ. If you imagine L as an arrow, this induced τ pushes this arrow to the side at a rate given by the above equation.

    The tilt must also be considered in order to relate the precessional frequency. Consider a tilt angle θ (the axis of the gyroscope is titled θ from vertical). τ acts on the radial component of L (that is, eρ.L = L sinθ) in the transverse (eφ) direction. Looking at it from the top, the L arrow makes a projection L sinθ in this view. The torque pushes this projection around in a circle with a tangential rate of τ.

    You can think of the precession as little steps in this direction that each take an amount of time dt. Let Ω be the precessional frequency. In analogy to the relationship

    ω r = v

    you can see that

    Ω L sinθ = τ

    In other words, the gyroscope completes one precession every ( L sinθ ) / ( 2π τ ).

    - Most treatments deal with the torque in terms of the force and moment arm, so it may seem quite different.
    - In the above treatment in this post, L indicates the total angular momentum of the system, to which the precession can contribute.
    Last edited: Apr 12, 2004
  5. Apr 13, 2004 #4

    just a couple of things;
    1. where does the 2pi come from in that equation?
    2. if you use force and moment arm, would it only seem different, or would it be different?
    oh, and 3. how relevant is the angular momentum which the precession contributes? say, my bicycle wheel on my string is spinning at 300 rpm, and its precessional frequency is only 0.2Hz, is it big enough to make a difference?

    also, i would like to know why the gyroscope/bicycle wheel stays completely horizontal when spinning quickly, but starts drooping (for lack of a better word) when spinning more slowly. does it only seem to stay horizontal when spinning quickly, because the drooping is so slow, or is it some other factor...
    maybe even the angular momentum contributed by the precession itself (as this contribution gets bigger as precessional frequency increases, if i'm not mistaken)?
    basically, why does the force and moment of arm not make a rapidly spinning wheel droop, while it does make a slow wheel droop? i know this is stupid but shouldn't, according to the maths, a wheel which is stationary, have an infintite precessional frequency??

    thanks a lot.
  6. Apr 13, 2004 #5


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    I am only assuming that the formula I derived is correct. I can't find it anywhere, so I haven't verified it.
    I myself am not perfectly comfortable with how well I understand this stuff; i.e. I'm not an expert.

    It comes from ω = 2 π f. I am not answering in this manner as a dodge, but out of faith that you will know what I'm talking about.

    When I first decided to do a search for the expression to verify the formula that I gave, I thought I had screwed something up. But, upon further inspection, it seems to me that it only appears to be different. I think the difference is primarily due to the explicit use of torque, which doesn't explicitly introduce a sinθ to cancel the one that appears in the formula.

    You have two things working for you. 1) The contribution is negligible as you have noticed. 2) The very nature of the precession induces a vertical angular momentum vector, which, as you can see from the previous derivation, is not acted on by the torque if a perfect constraint is assumed at the pivot point. Is it big enough to make a difference? Well, that's a judgement call.

    When you first start the experiment, let's say you start out with an angular momentum L(t = 0). There are two "bad" things that influence the wheel: 1) axle friction, and 2) imperfect constraint at the pivot. The axle friction causes L(t > 0) < L(t = 0) internally. But angular momentum is conserved, so it doesn't just disappear. It has an effect on the pivot point, which actually can move a little. It starts to move "out of the way" so to speak, and "drops" the gyroscope. You also get a little bit of the effect from the added angular momentum due to the precession, but this doesn't really become noticeable until the gyroscope starts to wobble. That wobbling is the precessional angular momentum "getting all tangled up" with the spinning angular momentum.

    Sort of. For a frictionless axle and a perfectly rigid pivot point, it would stay horizontal for a very long time. But it would eventually droop even in this case. Can you guess why?

    Yes, this is part of it, as I understand.

    Conservation of angular momentum. A slow wheel has less angular momentum to redirect. This gives the pivot point a better chance to "move out of the way" and "drop" the gyroscope.

    According to the formula that I gave, it sure seems that way. I can't think of a real satisfactory answer for this one, so this answer is kind of a dodge. I'll do my best.

    The torque comes from the force of gravity on the center of gravity and the constraint force on the pivot point. This gives a torque vector that is horizontal. The torque vector tells you how fast the angular momentum changes in that direction. If there is already angular momentum in a different direction, then the angular momentum from the torque adds to it, and effects the gyroscope to precess according to the sum of the vectors from t -> t + dt (integration). If there is no initial angular momentum, then the angular momentum changes from 0 to some value dL in the time dt (according to the basic formula that I gave in the previous post). So, after a time dt, the angular momentum of the gyroscope is dL. This is directed horizonally, which corresponds to a rotation about a horizontal axis, or a rotation from upright to not so upright. As the torque continues to act, it does so in the same direction for every dt. Thus, all of the dL's are in the same direction, and the gyroscope continues to rotate from upright to drooping, without precession.

    As a final note, consider that the formula in my previous post for the precession period was derived under some artificial assumptions: |L| = constant, perfect constraint force on pivot point, and I think there should be some integrability condition that τ tchar < L or something like that, where tchar is some characteristic time. I appologize for not emphasizing them. In other words, I don't think that the derivation that I gave even allows the gyroscope the possibility to droop.
    Last edited: Apr 13, 2004
  7. Apr 13, 2004 #6
    thank you, this is really helping me a lot!

    one thing i don't understand yet: that stuff about the pivot "moving out of the way" and "dropping" the gyroscope... what makes the pivot move out of the way, and how does this cause the gyroscope to droop?
    is what you mean the opposite of the "rigid" pivot you mentioned, or is that a different thing alltogether? if it is i'm not sure what you mean by that either...

    oh, and i can't imagine why a frictionless gyroscope would ever droop. please explain.

  8. Apr 13, 2004 #7


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    I wouldn't say "opposite." I was distinguishing from a pivot with perfect constraint. Ideally, you can assume that the pivot point is infinitesimally small and that it does not accelerate wrt any inertial frame. Then, whatever torquing takes place can be considered about the pivot point. If the pivot point "gives," then the torque is not about that point, and the precessional angular momentum can come into effect, as well as an angular momentum transfer into the pivot by the axle friction.

    My wording was probably confusing. I was talking specifically about an absence of axle friction, not friction in general (the biggest hint). Furthermore, I intended the gyroscope to be here on earth, with no "special" environmental accomodations (i.e. vacuum chamber is not allowed, hint, hint). It isn't all that fascinating, but there is one slightly interesting consequence: the angular momentum is transfered out of the system by this mechanism.
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