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Torque-free precession

  1. Apr 17, 2005 #1
    I am trying to figure out which way a spinning dinner plate (idealized as a disk) precesses when you toss it in the air. I can figure out the rate of precession in terms of the spin rate and the angle the spin axis makes with the precession axis. I believe it is [tex]\Omega_{pr} \cos{\theta} = 2 \Omega_{sp}[/tex]. The question is, if you toss a clockwise-spinning plate at a shallow angle, does it precess clockwise or counterclockwise? I tried adding up the torques on each mass element of the disk, and the requirement that the net torque vanishes indicates that the precession is counterclockwise. However, when I tried it with a real plate, it looked like the precession was clockwise. Of course, it precesses pretty fast and the wobbles are small, so it's hard to see. I taped a plastic drinking straw to the center of the plate (sticking up) to make the precession easier to see, and got the same result. Any ideas?
  2. jcsd
  3. Apr 18, 2005 #2
    To have a precession you need to have an external torque. If torque is zero, no precession occur- the angular momentum must be conserved. The visible "precession" could be if the plate has an initial momentum not along its axis of symmetry
    In that case, the motion in the air can be more complex and unstable because of the turbulence.

    CORRECTION: if the initial angular momentum is along the axis which does not pass from the center of mass, then you will have a precession. Its sign depend on the initial conditions. If initial angular momentum is along the axis, which pass through
    the center of mass, then the gravitation force will provide no torque for that momentum, even it is not along the normal of the disk.
    Last edited: Apr 18, 2005
  4. Apr 18, 2005 #3
    The plate is in free fall. The net force of gravity acts through the center of mass and produces no torque (with respect to the center of mass). Even if you calculated the torque w.r.t some other reference point outside the plate, it would not cause the kind of precession you mention. The plate just falls down. Therefore, gravity does not affect the spinning/precessing motion.

    But the plate does precess! Even with no external torque. Try it. BTW, I recommend doing it outside in a grassy area for safety.

    edit: also notice that the wobble motion (precession) is regular--it's periodic. So it is not the result of air turbulence.
    Last edited: Apr 18, 2005
  5. Apr 18, 2005 #4


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    I don't understand why you think the plate should precess in any specific direction. In the absence of external torques, the plate should obey Euler's equations.

    For reference these are

    I1 dw1/dt = w1w2(I2-I3)
    I2 dw2/dt = w2w3(I3-I1)
    I3 dw3/dt = w1w2(I1-I2)

    I1, I2, and I3 are the moments of inertia of the plate around its three primary axis, and w1,w2,w3 are the instantaneous angular velocity of the plate in a body-centered coordinate system.

    If we let I1 and w1 represent the spin axis of the plate, it's obvious that if w2=w3=0 then no precession occurs.

    In a more realistic case, w2 and w3 might have small values, and the motion will be more complicated, but it will depend on the exact inital conditions, there is nothing in the physics I can see to give it a preferred direction of precession.

    Note that for your plate, if we call I1 the main axis, I2 and I3 will be equal, or almost equal.


    This is the traditional analysis, but there's an easier (IMO) way to cary out the problem. This is to simply note that L (the total angular momentum) must remain constant.
  6. Apr 18, 2005 #5
    I agree, it is not clear from Euler's equations that there should be a preferred precession direction. In the two books I've looked at (Landau's and Goldstein's), they draw the picture so that the total angular momentum makes an acute angle with the spin angular velocity. But they do not say that it must be this way, so I suspect that is just a drawing convention. It makes it easier to visualize the various projections. But this acute angle is in fact what I observe every time I toss the spinning plate. I don't know how the initial conditions could be made different so that it would precess the other way. I toss it slightly different each time, but it does the same thing.
    Yes. That is what I was using when I added up the torques of all the mass elements and said that the total must be zero. Actually, I did the calculation for a thin ring. That is sufficient since the disk is made up of concentric rings. This calculation gives the correct precession rate, but apparently the wrong direction. Here is how I am going about it:

    The total angular velocity is [tex]\vec{\Omega}_{tot} = \vec{\Omega}_{pr} + \vec{\Omega}_{sp}[/tex]. The velocity of a particular mass element is
    [tex]\dot{\vec{r}} = \vec{\Omega}_{tot} \times \vec{r}[/tex].

    The acceleration is (omitting gravity)
    [tex]\ddot{\vec{r}} = \dot{\vec{\Omega}}_{tot} \times \vec{r} + \vec{\Omega}_{tot} \times \dot{\vec{r}}[/tex].

    Using [tex]\dot{\vec{\Omega}}_{pr} = 0[/tex] and [tex]\dot{\vec{\Omega}}_{sp} = \vec{\Omega}_{pr} \times \vec{\Omega}_{sp}[/tex] and substituting for [tex]\dot{\vec{r}}[/tex],

    [tex]\ddot{\vec{r}} = (\vec{\Omega}_{pr} \times \vec{\Omega}_{sp}) \times \vec{r} + \vec{\Omega}_{tot} \times (\vec{\Omega}_{tot} \times \vec{r})[/tex].

    Apply BAC-CAB a couple of times and note that [tex]\vec{\Omega}_{sp} \cdot \vec{r} = 0[/tex] to get

    [tex]\ddot{\vec{r}} \ = \ (\vec{\Omega}_{pr} \cdot \vec{r})(\vec{\Omega}_{pr} + 2\vec{\Omega}_{sp}) - \Omega_{tot}^2 \vec{r}[/tex].

    then form the cross product of this with [tex]\vec{r} dm[/tex] to get the torque on the mass element

    [tex]d\vec{N} = (\vec{\Omega}_{pr} \cdot \vec{r})\vec{r} \times (\vec{\Omega}_{pr} + 2\vec{\Omega}_{sp}) dm[/tex], and integrate around the ring. To do the integral, I took the ring to lie in the xy-plane (principal axes) and precession angular velocity to make an angle theta with the z-body axis, so that it has a component [tex]\Omega_{pr} \sin \theta[/tex] along the positive x-axis. In other words, I chose the x- and y-axes this way, since you are free to do that for a symmetrical top. It turned out that in order for the integral to vanish, [tex]\vec{\Omega}_{pr} + 2\vec{\Omega}_{sp}[/tex] must be proportional to [tex]\hat{e}_x[/tex], which means that the z-component of the precession is [tex]{\Omega}_{pr} \cos \theta = -2\Omega_{sp}[/tex]. So it looks like the precession should be in the direction opposite what I observe.

    edit: actually, I'm pretty sure my choice of axes to perform the integral has nothing to do with principal axes. They were just chosen to make the integration convenient, and happen to coincide with the principal axes. And by the way, my calculations of the velocity and acceleration are in the non-rotating "lab" frame. So the components of the angular velocity differ from those used in Euler's equations.
    Last edited: Apr 18, 2005
  7. Apr 18, 2005 #6
    Ok, I mistook precession for a change of the angular momentum.
    Still, to have a precession, you need to have a torque. By the way, the plate is not exactly in a free fall, it experiences a lift because of the rotation. That is especially true for freesbee.
    To check this I propose to measure the time for a nonrotating plate to fall with the time for rotating plate with zero vertical velocity to fall. If air supports the plate then you have the situation similar to the inclined spinner on a solid surface.
  8. Apr 18, 2005 #7
    I do not quite understand how the precession of the ring laying in xy-axis look like?
  9. Apr 18, 2005 #8
    . Go to google and type in "torque-free precession" or "regular precession".

    I didn't mean the plate's motion stays in one plane. Those coordinates were chosen for convenience of evaluating the integral, and are valid for an instant in time.
  10. Apr 18, 2005 #9

    Andrew Mason

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    I know that precession can be complicated. But I must be missing something here: if you have a spinning plate and it is precessing, the vector normal to its plane of spin keeps changing direction. This means that the angular momentum due to plate spin is not constant.

    In order for there to be 0 net torque (ie. an absence of aerodynamic torques and gravitational torque due to mechanical support), the angular momentum due to precession must be non-constant as well but complementary to the plate spin so that they sum to a constant total angular momentum vector at all times.

    I am not sure how this can happen. Are you sure that the plate precession is not due to some torque from somewhere?

  11. Apr 18, 2005 #10
    I see...
    what you have here is a rotation of the symmetry axis around the angular momentum, which is constant. It can have any sign. The sign of the [tex]cos\theta[/tex] will depend on the sign of that precession frequency.
    look here http://www.du.edu/~jcalvert/phys/ellipso.htm#Iner
    Neveretheless I think that air efect will be important for some dinner plates, exept for very heavy, or slow moving tableware.
  12. Apr 18, 2005 #11
    Right, the axis of the plate is rotating. And this means the total angular velocity vector (spin plus precession) is also rotating. But here's the catch: the total angular momentum is not generally in the same direction as the total angular velocity.

    [tex]\vec{L} = \mathbf{I} \vec{\Omega}[/tex], where I is the inertia tensor.

    So the total angular momentum can still remain constant in the non-rotating frame while the total angular velocity changes. This means that, in the lab frame, the inertia tensor is time-dependent. This is because the orientation of the body changes with time, relative to the fixed axes of the lab.

    edit: this was in response to AM's post.
  13. Apr 18, 2005 #12
    Hey that's a good site. Thanks. It will take me a while to get thru all of it, but it looks to be very helpful.

    The particular plate I am using is quite heavy (it's an "Oneida Classic" cafeteria-style plate, designed to take a beating) so I am pretty sure the effects of air can be neglected.
  14. Apr 19, 2005 #13
    Here is some more evidence in favor of the direction of precession opposite what I've observed:

    Let the 3-axis (or z-axis) be the axis of spin, which is a principal axis of the plate. The other two principal axes through the CM can be chosen somewhat arbitrarily (they have to be orthogonal, of course). Then the principal moment of inertia I3 about the axis of spin is twice the moments of inertia about the other two axes. That is, I1 = I2 and I3 = 2 * I1. In the rotating principal axis coordinate system, the relationship between total angular momentum and total angular velocity is:

    (L1, L2, L3) = (I1*w1, I2*w2, I3*w3) = I1*(w1, w2, 2*w3). Here the w's are the components of the total angular velocity.

    Thus we can see that the angular momentum vector makes a smaller angle with the 3-axis than does the angular velocity vector. Also, the components of the angular momentum are the same sign as those of the total angular velocity. The spin axis of the plate precesses about the angular momentum vector, which is constant in the non-rotating lab frame. From these considerations, it follows that the total angular momentum (and thus the precession angular velocity) makes an obtuse angle with the spin angular velocity vector. At least it looks that way from the picture I drew.

    So it appears that for a given spin rate and tilt angle, the shape of the object determines both the rate and direction of precession. If it were an oblong body instead of a flat one, I think it would precess in the opposite direction with a lower frequency.

    Maybe my eyes were just playing tricks on me when I was observing the thrown plate? Has anyone else tried it? :confused:
  15. Apr 19, 2005 #14
    look, the components along the symmetry axis are independent, so you can make them positive or negative and they will remain the same. Do you assume that w1, w2, and w3 can be only positive?
  16. Apr 19, 2005 #15
    As far as I can tell, I did not assume anything about w1, w2, and w3.

    edit: However, I should point out that when I said the angular momentum makes a smaller angle with the 3-axis than the angular velocity, I did not necessarily mean the positive 3-axis. I should have been more clear about that. For instance, if w3 is negative, then the angular momentum makes a smaller angle with the negative 3-axis than the angular velocity does.
    Last edited: Apr 19, 2005
  17. Apr 20, 2005 #16
    I read over the section on Euler angles in Landau's book more carefully, and realized that he actually solves this problem for a (force-free) general symmetric top. The conclusion is that, in the case of a flat disk, the spin angular velocity vector makes an obtuse angle with the precession angular velocity vector. This agrees with what I calculated.

    So now I'm wondering what could cause me to see it precess the other way. I've tried to think of things that could give the illusion that it is precessing the wrong way. For one thing, I'm looking at the plate from below once I let go of it. But I'm certain I've been taking that into account. Another possibility is that the plastic straw I taped to the center is not perfectly perpendicular to the plate's surface, or that it isn't exactly in the center, and that what I'm seeing is just the spin of the plate. But the precession rate is about twice the spin rate, so it seems like the precession would overcome the wobbling due to spin. It would be cool if someone else tried the experiment. Maybe I just need glasses :smile: .
    Last edited: Apr 20, 2005
  18. Apr 22, 2005 #17
    Yah, maybe u do :rofl: U need a torq to have persession!!! I learn this in high school! The wobling plate has NOTHING TO DO WITH PRESESSION!! U are throwing it with a wobble, thats all.
  19. Apr 23, 2005 #18
    Hi, I am a Physicist (3rd year physics student in the UK), so I know all about gyroscopes. What the last poster said is in fact true. If gravity did not produce a torque on the top/gyroscope, it would not precess. It would just spin about its axis. There is no torque on the spinning plate because it is not supported at one end. Therefore its angular momentum must be constant. How could it precess if its angular momentum is constant?? What you see is just instability about the axis of rotation. Also, your math is wrong. Have you studied any calculus? I think you need it for this problem. And what is "BAC-CAB"?

    Let me know if you have any questions :smile:
  20. Feb 3, 2010 #19

    Take a look at this thread of mine (which went nowhere):


    When you toss the plate with a clockwise twist, you also are giving it an off-axis impulse that starts the plate moving upward and wobbling. What is surprising is that for a disc shape, the resulting spin direction is counterclockwise. Most of the twisting motion you gave it went into the precession, which is indeed clockwise.
  21. Feb 5, 2010 #20


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    I can recommend the following article by Eugene Butikov:
    free rotation of an axially symmetrical body

    The article is illustrated with a Java applet. This interactive animation allows exploring both the case of a spinning prolate body and a spinning oblate body. (Spinning dinner plate = spinning oblate body).
    Last edited: Feb 6, 2010
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