1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Puzzling precession of the torque-free symmetric top

  1. Aug 6, 2014 #1
    Hello everyone,

    as I know the regular precession of torque-free symmetric top is such a cliche, I'll try to keep derivations short. The goal is, as I pointed out, to inspect a behaviour of the torque-free symmetric top in terms of precession rate, rotation rate and nutation angle. One way is to attack this problem via Euler dynamic and kinematic equations, which can by quite time-consuming, so I'll give only an outline of derivation. We'll start with Euler dynamic equations, while we keep in mind, that for symmetric top the first two moments of inertia are equal, so ## J_x = J_y \equiv j ##, the third one is usually different ## J_y \equiv J \neq j ##. Net torque exerted on the top is zero, so the left sides of kinematic equations are zeroes:

    $$ 0 = j \dot{\omega}_1 + (J - j) \omega_2 \omega_3 \\
    0 = j \dot{\omega}_2 + (j - J) \omega_1 \omega_3 \\
    0 = J \dot{\omega}_3 + (j - j) \omega_1 \omega_2 $$

    Third equation implies ##\omega_3 = const.##, first two ##\omega_1^2 + \omega_2^2 = const.## We plug this into Euler kinematic equations, shake them a little (it can be proved without ansatz, by using integrals of motion (conservation laws), but the shorter way is to use ansatz immediately) and we get what we were looking for:

    $$ \Omega_{precession} = \frac{\Omega_{rotation}}{\left( \frac{j}{J} - 1 \right) cos \left( \theta \right)} $$

    where, as the subscript suggests, the ## \omega_{precession} ## stands for precession angular speed, ## \omega_{rotation} ## stands for angular velocity of the rotation of the top and ## \theta ## is constant nutation angle.

    I found other and much simpler way to derive this equation; first we take kinetic energy of the free symmetric top and express it in terms of Euler angles, ## \varphi, \psi, \theta ## (precession, rotation and nutation angle, respectively):

    $$ T = \frac{1}{2} \left( j \omega_1^2 + j \omega_2^2 + J \omega_3^2 \right) = \frac{j}{2} \left[ sin^2 \left( \theta \right) \dot{\varphi}^2 + \dot{\theta}^2 \right] + \frac{J}{2} \left[ cos \left( \theta \right) \dot{\varphi} + \dot{\psi} \right]^2 $$

    The potential energy ##U## is zero, as the body is torque-free, so it could be as well force-free (and any force causing free net torque is irrelevant to Lagrange equations for Euler angles). So the total Lagrangian ##L## of the top is equal to kinetic energy ##T##. Now we use Lagrange equation for the generalized coordinate ##\theta## (we can think of Euler angles as of generalized coordinates):

    $$ \frac{\mathrm{d} }{\mathrm{d} t} \frac{\partial L}{\partial \dot{\theta}} - \frac{\partial L}{\partial \theta} = 0 \\
    j \ddot{\theta} - j sin \left( \theta \right) cos \left( \theta \right) \dot{\varphi}^2 - J \left[ cos \left( \theta \right) \dot{\varphi} + \dot{\psi} \right] \left[ - sin \left( \theta \right) \right] \dot{\varphi} = 0 $$

    Now we plug in the ansatz of regular precession: ##\dot{\varphi} = \Omega_{precession} = const., \dot{\psi} = \Omega_{rotation} = const., \theta = const.## we easily obtain:

    $$ - j sin \left( \theta \right) cos \left( \theta \right) \Omega_{precession}^2 - J \left[ cos \left( \theta \right) \Omega_{precession} + \Omega_{rotation} \right] \left[ - sin \left( \theta \right) \right] \Omega_{precession} = 0 \\
    j cos \left( \theta \right) \Omega_{precession} - J \left[ cos \left( \theta \right) \Omega_{precession} + \Omega_{rotation} \right] = 0 \\
    \left( j - J \right) cos \left( \theta \right) \Omega_{precession} = J \Omega_{rotation} \\
    \Omega_{precession} = \frac{\Omega_{rotation}}{\left( \frac{j}{J} - 1 \right) cos \left( \theta \right)} $$

    I wanted to provide derivation of the formula to make clear that I didn't make it up.

    Now whit is puzzling me is the fact, that the usual way to make top spin with precession is to exert an initial torque on it and then leave it to its own motion. We can give it initial rotational angular speed, ##\Omega_{rotation}## and it will gain some precession according to nutation angle, ##\theta## and ratio of moments of inertia. The strange thing is, that as the ratio ##j/J## approaches 1 (so the top is becoming a spherically symmetric), the precession angular velocity is diverging to infinity, when rotational angular velocity is nonzero, which is, I assume, wrong. I found this very result (how the rate of the precession depends on the rotational speed of the top) in my old textbook of theoretical mechanics, derived via kinematic angles and similar result expressed via other constants in Ландау-Лифшиц (Landau-Lifšic). Now I assume, it's very unlikely that so many physicists would derive this formula incorrectly, so I think it's just me not being able to understand this issue.

    The second problem is, that the formula I found on wiki here seems to neglect -1 (holds true for very wide top J >> j), but I don't understand why. The formula derived above is not that complicated, so that such neglections would be necessary.

    So where the truth lies? Is this formula right? But how come, that precession goes to infinity, when the object is becoming spherical? We know, that spherically symmetric torque-free top spins just about one axis of symmetry and precession does not occur (it is obvious from Euler dynamic equations). How come, that in limit ##j/J \to 1## precession rate diverges to infinity and for ##j = J## precession rate becomes identically zero?

    I'd appreciate any help on this topic and I hope someone will finally put end to this misery of mine of not understanding of the regular precession of the torque-free symmetric top. Thanks :)
  2. jcsd
  3. Aug 16, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
  4. Aug 16, 2014 #3
    I am not sure how you obtained your equation with the -1 term. In the Landau-Lifschits book referenced, they have (33.4): $$ \Omega_3 = {M \over I_3} \cos \theta $$ and (33.5) $$ \Omega_{\text{pr}} = {M \over I_1} ,$$ where ## \Omega_3 ## and ## \Omega_{\text{pr}} ## are the velocities of rotation and precession, correspondingly. So $$ \Omega_{\text{pr}} = { I_3 \Omega_3 \over I_1 \cos \theta }, $$ which is similar to the equation in Wikipedia.

    So there must be a mistake in your derivation.

    Physically, a spherically symmetric top cannot have any precession. That is explained in the same book, in that same section 33, just a few paragraphs above.
  5. Aug 18, 2014 #4
    Thanks for your answer, but now I think I absolutely don't understand the Eulerian angles. It's said that Eulerian angles gives position in space relative to some inertial coordinate system (usually laboratory z-axis). But result for spherically symmetric torque-free body gives [itex] \varphi = \Omega_{\varphi}t + \varphi_0, \theta = const., \psi = 0 [/itex] (can be solved via euler dynamic and kinematic equations or lagrange equations). So okay, the body rotates in eulerian angle [itex] \varphi [/itex], but it's clear that it rotates around vector [itex] \vec{L} [/itex] (angular momentum). This leads me to more difficult cases - are Eulerian angles always defined relative to angular momentum vector, even if there's torque and momentum vector is changing in laboratory coordinate system? The frame of moving vector of angular momentum clearly cannot be considered an 'inertial'.

    Mostly, the problem is, that we do not have relationship [itex] \vec{M} (t) [/itex] in laboratory coordinate system, because there can be forces (and therefore possibly torque) attached to a body (for example a thrust on the side of the cube - when the cube rotates itself, so does the torque and therefore momentum vector, another example a wheel, which can rotate about axis of symmetry and also about some global axis, like front wheels of car). So we need to know relationship [itex] \vec{M} \left( \omega_1, \omega_2, \omega_3, \varphi, \theta, \psi \right) [/itex], where \omega_i are components of the angular velocity about the principal axes and [itex] \varphi, \theta, \psi [/itex] are Eulerian angles. Another problem is, that even if we know how to write proper equations, it's something like running in circles - the Eulerian angles are defined relative to angular momentum, but it's position in inertial frame is changing due to torque ([itex] \dot{\vec{L}} = \vec{M} [/itex]), but torque is defined via Eulerian angles. A bug?

    Another thing that causes my confusion is the fact, that, in general, the axis of rotation might change, move in the body. Consider a ball and a thrust parallel to it's surface, fixed to it. The initial axis of rotation is unknown and it's not a axis of symmetry, there's no argument for the axes to cross center of mass of the ball. As the motion develops (via yet unknown equations, because we don't know the torque, if we don't know exact position of axis of rotation), this axis might move, so the torque changes in time in the inertial frame.

    Example of the cube with fixed thrust parallel to it's side: we know, that thrust is a constant force (constant in cube reference frame), consider it's always an [itex] \xi [/itex] direction (so in body's x-axis), [itex] \vec{F} = \left( F, 0, 0 \right) [/itex] The torque (in body axes) is [itex] \vec{M} = \vec{r} \times \vec{F} [/itex] where [itex] \vec{r} [/itex] is a vector from the point from which torque is measured to the point where force is applied. Now this is a problem, because we don't know, where the torque is measured (we need a value, a number in proper units to plug it into Euler dynamic equations), because we don't know where the axis of rotation will be in the time the torque will be applied!

    All this is very confusing for me :(
  6. Aug 18, 2014 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not sure where your error is. You didn't define [itex]\theta[/itex] or show your derivation in the first part.

    Here's yet another way to look at this. I'm going to use [itex]I_{12}[/itex] for the moments of inertia about the body x and y axes, [itex]I_3[/itex] for the moment of inertia about the body z axis, and [itex]\omega[/itex] for the magnitude of the angular velocity vector. With this, the Euler equation become
    0 &= I_{12}\dot \omega_1 + (I_3 - I_{12})\omega_3 \omega_2 \\
    0 &= I_{12}\dot \omega_2 - (I_3 - I_{12})\omega_3 \omega_1 \\
    0 &= I_3 \dot \omega_3

    So obviously [itex]\omega_3[/itex] is constant. The first two equations can be re-written in a simpler form if we define [itex]\Omega_b = \frac {I_3 - I_{12}}{I_{12}} \omega_3[/itex]. With this, the first two equations become
    \dot \omega_1 + \Omega_b \omega_2 &= 0 \\
    \dot\omega_2 - \Omega_b \omega_1 &= 0
    That's the equation for a coupled harmonic oscillator. Ignoring an arbitrary phase angle, the solution to this is
    \omega_1 &= \omega_{12}\cos \Omega_b t \\
    \omega_2 &= \omega_{12}\sin \Omega_b t
    With the constant third component, [itex]\omega_3[/itex], the magnitude of the angular velocity is constant [itex]\omega^2 = \omega_{12}^2 + \omega_3^2[/itex]. Defining angle $\beta$ as the angle between the angular velocity vector and the body z axis, the angular velocity can be expressed as
    \omega_1 &= \omega \sin \beta \cos \Omega_b t \\
    \omega_2 &= \omega \sin \beta \sin \Omega_b t \\
    \omega_3 &= \omega \cos \beta
    So now we can see what [itex]\Omega_b[/itex] physically represents. From the perspective of the body frame, the angular velocity vector is precessing about the symmetry axis at a rate of [itex]\Omega_b[/itex].

    What do things look like from the perspective of the fixed frame? After far too much work (not shown) for such a simple result, the answer is that the angular velocity vector precesses about the angular momentum axis at a rate [itex]\Omega_s = ||\vec L||\,/\,I_{12}[/itex], which is voko's result from Landau & Lifshitz.

    What about a spherically symmetric body? In this case, the angle between the angular momentum and angular velocity vectors is zero. The body rotates at some constant rate [itex]\Omega_s[/itex] about the z axis, rotates by zero degrees about the x' axis, and rotates at some constant rate [itex]\omega[/itex] about the z'' axis. There's a problem here: That null rotation means the z and z'' axes are one and the same. You could look at angular velocity as being pure precession, with no rotation; as pure rotation, with no precession; or as anything in between. All we can say is that [itex]\Omega_s+\omega=||\vec \omega||[/itex]. The best way to resolve the ambiguity is to say that [itex]\Omega_s=0[/itex]. BTW, there's a name for this situation: Gimbal lock. Gimbal lock is a demon that plagues robotic arms.
  7. Aug 19, 2014 #6
    It is true that the Euler angles are defined with respect to a fixed frame (*). However, the components of dynamical variables in Euler's equations for rigid body are taken with respect to the co-rotaing frame of reference and expressed via the Euler angles and their derivatives. The referenced book shows the details in section 35.

    (*) But there is more than one way to do so. Tricky, I know.
  8. Aug 19, 2014 #7
    D H: The Eulerian angles I used are defined: take the principal axes aligned with laboratory axes, which are x y z. Now rotate principal axes (with whole body) along z axis couterclockwise by angle [itex] \varphi [/itex]. You've just created new coordinate system, x', y', z'. It's clear, that z = z'. Now rotate body and it's principal axes about x' axis counterclockwise by angle [itex] \theta [/itex]. You've defined new coordinate system, x'', y'', z''. The last rotation is about z'' counterclockwise by angle [itex] \psi [/itex]. Principal axes become [itex] \xi, \eta, \zeta [/itex]. The convention is sometimes different, the first and third angle stays the same, but [itex] \theta [/itex] is rotation about y' axis. Of course, the kinematic equations depend on this convention. Now the whole derivation of kinematic equations depends on this convention, but it's clearly said, that these are defined with respect to some inertial (ideally fixed laboratory) frame (the initial coordinate system, x, y, z).

    But if we solve motions for symmetric body, the result is: [itex] \varphi = \Omega_{\varphi} t + \varphi_0, \theta = const, \psi = 0 [/itex]. But it can't be true, that these angles were defined by z-x'-z'' rotation, because now the vector of angular momentum can lie e.g. at the x-axis! At the beginning I wrote, that "rotate principal axes (with the whole body) along z axis couterclockwise by angle [itex] \varphi [/itex]..." - if this was true, then the body could rotate only about z axis. Now it's true, that the frame, where angular momentum vector is static, is fixed, but what about more difficult problems? With nonzero torque, the angular momentum can move relative to a fixed frame and also relative to a body axes. Then it cannot be true, that "the Euler angles are defined with respect to a fixed frame", because they rely on the vector of angular momentum.

    I'm sorry I'm bothering you with this, but I want to really understand that. I can't solve problems with nonzero torque, because I don't understand how the body would move in fixed frame.
  9. Aug 19, 2014 #8
    I'm not sure what the problem really is, but your equations could be suffering from "gimbal lock".


    It happens when you rotate around more than one axis, depending on which one you move the first you get different results. They call it "loss of freedom", to me it looks just undefined. Anyway, to avoid "gimbal lock" instead of Euler angles use quaternions.
  10. Aug 19, 2014 #9
    Now everyone is refering to this gimbal lock thing, but that's not what I have problem with. [itex] \varphi = \Omega_{\varphi}t + \varphi_0, \theta = const., \psi = 0 [/itex] is not an example of loss of the degree of freedom, when [itex] \theta \neq 0 [/itex].

    Okay, so if there's no apparent problem, try this: you have a cube, it has 6 sides. You were given a moment of inertia of the cube. There's force in the center of the one side (say, the upper side, but that's irrelevant) and this force is rotating around (shown in the picture). The force is perpendicular to the same side all the time and it's initial point is always in the center of that side. Now write kinematic and dynamic equations for this problem (so you'll have to figure out torque to plug into dynamic equations). Suppose you have the answer, the functions [itex] \varphi (t), \theta (t), \psi (t) [/itex]. How would move the vertices in the fixed frame (laboratory)? How would move the center of gravity of the cube in the fixed frame? We'll have to know the time evolution of euler angles, so we can know which direction the force F would accelerate the cube in the next moment.

    This example shows if there's is or isn't problem. Do we know torque caused by the force F(t)? Do we have to express it in the principal coordinate system? If yes, what point would be referential, so where'll be the axis of rotation? (we certainly know, that center of gravity of the cube will move - that's how the forces work, even if the force is not located at the center of gravity, the center will move due to newton's equation. Now we can look at this two ways - either the cube don't move, but the axis of rotation is not at the center of mass, or there's present movement and rotation around axis of rotation crossing the center of gravity at the same time.)

    For me, the problem is even to write proper equations and figure out the resulting movement and that's the cause I think, I don't understand it.

    Attached Files:

  11. Aug 19, 2014 #10

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That's definitely not the right way to look at this particular problem. You are not rotating the object. You are trying to describe how the object is oriented at any point in time, and how that orientation changes over time.

    Imagine that the object's principal axes, its angular momentum vector, and it's angular velocity vector are shining pillars. These pillars move around as the object rotates -- except not the angular momentum vector. It's frozen in one direction. You can take advantage of this. Call that the inertial frame's z axis. Now let's freeze time. If you look at the angular momentum vector, angular velocity vector, and principal axis, you'll see the all lie in the same plane. Let loose your hold on time for a bit, and then freeze it again. The axes have moved, but those three axes are still coplanar. You can take advantage of this, too. We don't have a complete lab frame yet. We need an x and a y axis. Let's define the x axis so that it's perpendicular to the plane on which all three of those coplanar vectors lie. Now you have specified two of the three axes of your inertial reference frame. That's all you need to specify. To have a right-handed coordinate system, there's only one direction that third axis can be.

    Now let's make a clone of our newly minted inertial reference frame so we can rotate it. We're just rotating axes here. The object isn't be rotated, nor is our inertial frame. Our reference frame is already fairly nicely aligned. Rotate our duplicate frame by an angle $\beta$ about the x axis so that the z axis points along the body z axis. With one more rotation about the z axis by some angle $\gamma$ and we have aligned that intermediate frame with the body frame.

    Now it's time to unfreeze time again. Let the body rotate a bit and then stop time again. Now our inertial x axis isn't aligned so nicely. So let's make another clone of our inertial frame so we can rotate it to align with the body frame. Last time we used those three coplanar items to pick our x axis. We can do the same thing again with our cloned frame. Rotate it about the inertial z axis by an angle $\alpha(t)$ to make the x axis normal to the angular momentum / angular velocity / symmetry axis plane. Now you rotate the cloned frame about its rotated x axis to make the z axis align with the body z axis. This will be the same angle $\beta$ as the previous time. Finally, rotate about the body z to align our frame with the body axes. This won't be the same angle $\gamma$ as the first time around. Let's call it $\gamma(t)$ instead to denote that this too needs a time-varying rotation.

    This sequence of rotations starting with a rotation by an angle $\alpha(t)$ about the inertial z axis, followed by an angle $\beta$ about the rotated x axis, which in turn is followed by a rotation about an angle $\gamma(t)$ about the body z axis is an Euler sequence. In this case, the second rotation is of a constant angle. In general, all three will be functions of time.

    In this case, the behaviors of $\alpha(t)$ and $\gamma(t)$ are very simple: They're just linear functions of time. The time derivative of $\alpha(t)$ is the inertial frame precession rate. The time derivative of $\gamma(t)$ -- that's what I called $\omega_3$ in my first post (and possibly what you called $\Omega_{rotation}$).
  12. Aug 19, 2014 #11
    I do not entirely understand how the force is applied. Is the direction of the force fixed with respect to the cube, or is it independent? In the former case the solution is straight-forward. Take the principal axes of inertia and let them be your mobile coordinate system. The force is constant in that system, so you obtain its components, as well as the components of torque, easily. Euler's equations then follow just as easily, and I think you can solve them and obtain angular velocity without having to deal with Euler's angles. Try it.
  13. Aug 19, 2014 #12
    Now I don't understand, how this can be true. Equation [itex] \dot{\vec{L}} = \vec{M} [/itex] where [itex] \vec{M} [/itex] is torque, holds true. So when torque is exerted on the body, the angular momentum vector will certainly move. It cannot be frozen in one direction.

    The image shows, that the force is rotating with respect to cube (imagine massless rotating wheel attached to a side of the cube with the thrust - like jet - the effect would be force rotating around)
  14. Aug 19, 2014 #13
    I suggest you first solve the simpler problem with the fixed force as I explained above.
  15. Aug 19, 2014 #14

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    The subject of this thread is the torque-free symmetric top, isn't it? There is not external torque. The angular momentum vector does not move from the perspective of an inertial observer.

    You can look at that precession from either the perspective of an inertial frame or from that of a body fixed frame. From the perspective of an inertial frame, there is no torque, or [itex]\dot {\vec L} = 0[/itex]. But [itex]\vec L = I\vec\omega[/itex], so this means [itex]\dot L = 0 = \dot I \vec \omega + I \dot{\vec\omega}[/itex]. From the perspective of an inertial frame, the inertia tensor of a rotating object with a non-spherical mass distribution is *not* constant. The inertia tensor rotates with the body. Lumpiness here becomes lumpiness over there as the body rotates. Since [itex]\dot I[/itex] is not zero, the only way we can have [itex]\dot I \vec \omega + I \dot{\vec\omega} = 0[/itex] is [itex]\vec \omega[/itex] also is not constant, and changes exactly enough to counter changes in the inertial frame inertia tensor.

    Notice that I didn't do any math. You do not want to do that math in the inertial frame. There's a much easier way to do it, which is to look at things from the perspective of a body-fixed frame. The body-fixed frame is a rotating frame. Since you're covering this topic, I expect you've already covered rotating frames. The time derivative of a vector as observed by an inertial observer is not equal to the time derivative as observed by a rotating observer. They are however related via the transport theorem. (It's not always called that, but I suspect you did learn it). Given some vector quantity [itex]\vec q[/itex], the time derivative from the perspective of an inertial and rotating observer are related via
    [tex]\left(\frac {d\vec q}{dt}\right)_{\text{inertial}} =
    \left(\frac {d\vec q}{dt}\right)_{\text{rotating}} + \vec \omega \times \vec q[/tex]
    Letting [itex]\vec q = \vec L[/itex], this becomes
    [tex]\left(\frac {d\vec L}{dt}\right)_{\text{inertial}} =
    \left(\frac {d\vec L}{dt}\right)_{\text{rotating}} + \vec \omega \times \vec L[/tex]
    The left-hand side is zero. From the perspective of the rotating frame,
    [tex]\frac{d\vec L}{dt} = -\vec \omega \times \vec L = \vec\omega \times (I \vec \omega)[/tex].
    When you learned about rotating frames you learned about fictitious forces. This is a fictitious torque, sometimes called the Euler torque. If you play with the right hand side, you should see that it reduces to Euler's equations in the case of a diagonal inertia tensor.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Puzzling precession of the torque-free symmetric top
  1. Torque-free precession (Replies: 20)

  2. Torque Free Precession (Replies: 6)