I wanted to ask the following question:

Can a body that experiences no forces whatsoever precess? Let's say I have a body in space - no gravity or anything - can I make it precess without applying any forces or torques? If so, how? Under what conditions? What would its movement look like? Could you give an example of something like this?

  • $\begingroup$ Hmmm ... the "tennis racket theorem" (which also goes by other names) may be of interest here. $\endgroup$ Feb 27 '18 at 22:36
  • $\begingroup$ Check en.wikipedia.org/wiki/Precession for torque free precession. $\endgroup$
    – npojo
    Feb 27 '18 at 23:05
  • $\begingroup$ @dmckee That's a new name on me. It makes me think immediately of John McEnroe hurling his racket across the court. To be pedantic, the TR theorem is about stability, so you would still need microtorques to set the instability and wild precession off. $\endgroup$ Feb 27 '18 at 23:21
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    $\begingroup$ @WetSavannaAnimalakaRodVance Being an experimental type, I never worry that their won't be &%#@ perturbations to upset an unstable equilibrium. Y0u can't get rid of them if you try. $\endgroup$ Feb 28 '18 at 2:03

Any spinning body with a non scalar inertia tensor (i.e. the inertia tensor is not proportional to the identity matrix) will have its angular momentum and angular velocity vectors in different directions unless the angular velocity is along a principal axis (one of the three orthogonal eigenvectors of the symmetric inertia tensor). Moreover, from the standpoint of an inertial observer, the inertia tensor will change with time as the body spins relative to the observer.

Thus, in torque free conditions, the angular momentum must stay constant; therefore, from the above considerations, the body's angular velocity is constantly changing.

This is very hard to visualize!

There is another, distinct effect that does not quite qualify as torque free known as the Tennis Racket Effect. This arises when all three principal moments of inertia (i.e. eigenvalues of the inertia tensor) differ. In this case, rotation about the eigenvectors corresponding to the highest two inertias is stable, but rotation about the third is not. That is, if a torque no matter how small changes the angular velocity when the body rotates about this third axis, the perturbation to the motion grows with time, and the body can flap about wildly. See the video in the linked Wikipedia article.

  • $\begingroup$ Aka The Polhode Rolls without Slipping on the Herpolhode Lying in the Invariable Plane. $\endgroup$ Feb 27 '18 at 23:57
  • $\begingroup$ @DavidHammen Google translate can't handle that one! $\endgroup$ Feb 28 '18 at 0:03
  • $\begingroup$ Translate to what language? That's English, from Goldstein, Classical Mechanics. $\endgroup$ Feb 28 '18 at 0:30
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    $\begingroup$ @DavidHammen Sorry, I was commenting on the sheer number of words that I did not know living together in a single sentence! $\endgroup$ Feb 28 '18 at 0:34

The axis of rotation changing even when there is no applied torque is an effect which is possible in 3-dimensions but not in 2-dimentions.

The angular momentum about an axis $\vec L$ is given by $\vec L = I \vec \omega$ where $I$ is the moment of inertia about the axis and $\vec \omega$ is the angular velocity about the axis.

Assume that you are observing an asymmetric body with total angular momentum $\vec L$ then if there are no external forces acting on the body the total angular momentum of the body is constant.

Assume a frame of reference fixed relative to the centre of mass of the body (which does not move) then the x, y and z components of the angular momentum must be constant - $\vec L = \vec L_x + \vec L_y +\vec L_z$.

Because the body is asymmetric as it rotates relative to the frame of reference its moment of inertia about an axis will change eg from $I_x$ to $I_x^\prime$ and this in turn will change the angular velocity about that axis from $\omega_x$ to $\omega_x^\prime$ subject to the constraint that the angular momentum in the x-direction, $L_x = I_x \omega_x$ to $I_x^ \prime \omega_x^\prime$, stays constant.

This means that what you observe as the rotation of the object is the initial angular velocity, $\vec \omega = \vec \omega_x + \vec \omega_y +\vec \omega_z$, changing to a new angular velocity, $\vec \omega^\prime = \vec \omega_x^\prime + \vec \omega_y^\prime +\vec \omega_z^\prime$.
So you observe a change in the direction of the axis of rotation although the total angular momentum has not changed with $\vec L = I_x \vec \omega_x + I_y \vec \omega_y+ I_z \vec \omega_z=I_x^ \prime \vec \omega_x^\prime+I_y^ \prime \vec \omega_y^\prime+I_z^ \prime \vec \omega_z^\prime$.

A full analysis requires one to keep the total kinetic energy of the object, $\frac{1}{2}I_{\rm x} \omega_{\rm x}^2+\frac{1}{2}I_{\rm y} \omega_{\rm y}^2+\frac{1}{2}I_{\rm z} \omega^2_{\rm z}$, constant and hence also equal to $\frac{1}{2}I'_{\rm x} {\omega'}_{\rm x}^2+\frac{1}{2}I'_{\rm y} {\omega'}_{\rm y}^2+\frac{1}{2}I'_{\rm z} {\omega'}^2_{\rm z}$.

Here are some links which may be of interest?

Wikipedia - Precession - Torque Free

Wikipedia - Poinsot's Construction

A wonderful 3D rigid body simulation and start by looking at the angular momentum vector and the angular velocity vector to observe the precession.

Article - Rigid body motion in stereo 3D simulation


Allow me rephrase your question, it seems to me that the following formulation is closer to the case you are thinking about:

In the absence of any external force, can a rigid, axially symmetric body move in a way so that its motion is not axially symmetric?

(Of course I phrased the question that way specifically for the answer to be 'yes'.)

A spinning axially symmetric object can have a sustained wobble. The nature of this wobble is that the symmetry axis sweeps out a cone. This wobble can just as well occur on top of a precessing motion, in which case the wobble is called 'nutation'. When a qyroscope wheel is in a combined precession and nutation motion the amplitude of the nutation motion is smaller than the amplitude of the precession motion, and the frequency of the nutation is higher (in most cases far higher) than the frequency of the precession motion.

For the wobble in the no-external-force case:
One might have the expectation that when you throw some object, giving it a spin, then at the instant that you let go the object settles to a non-wobbling spin. But in fact an object, when thrown with spin, can and will have a sustained wobble if that is how it happened to be thrown.

There is a story known as 'Feynmans wobbling plate'. Feynman was sitting in a cafeteria of the University where he had a teaching position, and as Feynman recounted:"some guy, fooling around, throws a plate in the air. As the plate went up in the air I saw it wobble, and I noticed the red medallion of Cornell on the plate going around. It was pretty obvious to me that the medallion went around faster than the wobbling."

There's a youtube video, titled Feynmans wobbling plate. It shows some footage of an actually thrown plate, followed by a mathematical discussion of the rotational dynamics.

In my rephrased version of the question I added the condition 'axially symmetric body' because that is a case where one might perhaps expect that it cannot wobble - whereas in fact it can. By contrast, the two earlier answers to this question discuss less symmetric shapes, where you do expect complicated tumbling motion.


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