I recall from a course on classical dynamics that angular velocity is a 3-dimensional vector, and angular velocity can be added and subtracted. From this, my understanding is that an object can "rotate about two axes at once" only in the sense that we can add up the two angular velocity vectors and this defines a perfectly sensible rotation. But the final rotation can also be described as a single vector (the sum of the original two) and is thus just rotation about a single axis. Basically, rotation that cannot be expressed as rotation about a single axis cannot exist.

But in this video of a sphere rotating around two axes, it doesn't seem to do this: I can't see any point that holds still relative to the camera (which would also be still relative to the center of the sphere). I don't think I'm missing such a point, either: The gears sweep over all points on the sphere, and none of the gear teeth are stationary.

Further, the hairy ball theorem seems to imply that there must always be a point at zero velocity, at least instantaneously. But maybe that point of zero velocity is moving around constantly, thanks to some acceleration. Without external forces, parts of the sphere can accelerate thanks to internal forces (that's how it holds its shape while it rotates!), so is it possible that there is some internal force causing such an acceleration?

Or is my understanding of angular velocity incorrect, or is there some hidden force that is implicitly acting on the sphere in the animation?

  • $\begingroup$ With my current physics understanding, answer is yes - an object can rotate around couple of axis simultaneously. It's called gyroscopic precession. Check for example an Earth axial precession- Earth rotation axis spins in a loop with $26~000$ years period. $\endgroup$ Dec 22 '20 at 18:16
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – R. Emery
    Dec 22 '20 at 18:19
  • $\begingroup$ The video you link to shows the image of a sphere. Towards the end of your question you shift to writing about 'sphere' and 'ball', but at the start of your question you use the more general term 'object'. This makes a lot of difference. In physics it is convention to treat a sphere as a perfect sphere. Also, you do not mention explicitly whether you are asking about an object in free motion, or a driven object $\endgroup$
    – Cleonis
    Dec 22 '20 at 18:51
  • $\begingroup$ Possible duplicate: Two axes for rotational motion $\endgroup$
    – Qmechanic
    Dec 22 '20 at 20:02

But maybe that point of zero velocity is moving around constantly, thanks to some acceleration.

There is always an instantaneous axis of rotation for any rigid body, but that doesn't mean that that rotation axis is fixed. This means that (as you suspect) the "point of zero velocity" is moving around constantly.

We can actually write out $\vec{\omega}$ as a function of time by recalling that angular velocity vectors are additive between rotating reference frames. We can imagine going to an intermediate frame related to the space frame by an angular velocity $\omega_1 \hat{z}$. In this frame, the angular velocity of the body is something like $\omega_2 \hat{x}'$, where $\hat{x}'$ is the $x$-axis of the intermediate frame. So the overall angular velocity vector is something like $$ \vec{\omega} = \omega_1 \hat{z} + \omega_2 \hat{x}' = \omega_1 \hat{z} + \omega_2 (\cos (\omega_1 t) \hat{x} + \sin(\omega_1 t) \hat{y}), $$ since $\hat{x}' = \cos (\omega_1 t) \hat{x} + \sin(\omega_1 t) \hat{y}$. At any time, the point on the surface of the sphere along this direction is instantaneously at rest; but that "instantaneous rest point" is constantly shifting, both in space and relative to the body.

In fact, it is well-known that a rigid body will generally not have a constant angular velocity $\vec{\omega}$, even in the absence of torques. In the body frame, the angular velocity vector is governed by Euler's equations, and unless $\vec{\omega}$ is aligned with one of the principal axes of the body, we will generally have $\dot{\vec{\omega}} \neq 0$. In the space frame, $\vec{L}$ is fixed in the absence of torques (not $\vec{\omega}$), so we generally have $\dot{\vec{\omega}} \neq 0$ in this frame as well. This is most readily seen in the case of a symmetric top (with two principal moments equal). In the body frame, the symmetry axis is fixed and $\vec{\omega}$ and $\vec{L}$ precess around it at a uniform rate. In the space frame, $\vec{L}$ is fixed, and $\vec{\omega}$ and the symmetry axis precess around it at a uniform rate. In neither frame is $\vec{\omega}$ fixed.

That said, you also ask

[I]s there some hidden force that is implicitly acting on the sphere in the animation?

and the answer to that is "yes, probably". If the body is a perfect sphere, then all the principal moments are the same, and so $\vec{L} = I \vec{\omega}$ for some scalar $I$. This means that $\dot{\vec{\omega}} \neq 0$ (as in the example in the animation) implies that $\dot{\vec{L}} \neq 0$, which would imply that some non-zero torque would be necessary to maintain this rotation. More generally, if one knows the angular velocity in the body frame, one can use Euler's equations to calculate the torque necessary to maintain that rotation (as seen in the body frame).

  • $\begingroup$ Does your assertion extend to a perfect sphere? I mean a sphere with markings, so you can track the surface through time. Other than that: perfect symmetry. $\endgroup$
    – Cleonis
    Dec 22 '20 at 18:46
  • $\begingroup$ As a comment says, what about the various precessions of the Earth? Are they only possible because Earth is not a perfect sphere? And Euler's equations give a non-constant angular velocity in the body frame; does that also hold in an inertial reference frame? $\endgroup$
    – Sam Jaques
    Dec 22 '20 at 19:51
  • $\begingroup$ There is an ambiguity in the expression 'rotation axis'. There is the Dzhanibekov effect We observe there is no external torque, and the rotating object is rigid, hence we know the angular momentum of that object is conserved. But our perception wants to interpret the motion of that object as rotation around some axis that aligns with features of the shape of that object. How to define the expression 'rotation axis'? The definition should be: the orientation of the angular momentum. $\endgroup$
    – Cleonis
    Dec 22 '20 at 19:57
  • $\begingroup$ @SamJaques: Yes, for any body with three equal principal moments, a changing $\vec{\omega}$ requires an external torque. There's some subtlety in the "various precessions of the Earth" — the precession of equinoxes, for example, is due to the Sun's torque on the Earth's equatorial bulge, so it's not a "torque-free" effect. But I think that a spherically symmetric Earth would have no net torque on it due to the Sun's gravitational field and so wouldn't precess in any event. As far as the motion of $\vec{\omega}$ in the space frame, see the new example at the end of the third paragraph. $\endgroup$ Dec 22 '20 at 20:11
  • $\begingroup$ @Cleonis: You seem to be saying that the notion of the vector $\vec{\omega}$ is ambiguous, but I don't think it is. At any time $t$ there is an orthogonal matrix $\mathbf{O}(t)$ such that the body coordinates $\vec{x}_b$ are related to the spatial coordinates $\vec{x}_s$ by $\vec{x}_s = \mathbf{O} \vec{x}_b$. It can then be shown that there is a unique vector $\vec{\omega}$ such that $d \vec{x}_s /dt = \vec{\omega} \times \vec{x}_b$ (assuming $\vec{x}_b$ is constant in the body frame.) That's what I mean by "rotation axis". ... $\endgroup$ Dec 22 '20 at 20:20

I think you're confusing angular velocity with angular acceleration. At any moment, the angular velocity of a rigid body is a single vector. That means the velocity of all points of the body is $\vec{v} = \vec{r} \times \vec{\omega}$ where $\vec{r}$ is defined relative to some instantaneous center of rotation, and thus that all points on the body lying along $\vec{\omega}$ are instantaneously at rest.

This doesn't mean that those points can never move. If the angular velocity is changing over time, as it is in this example, then the set of points instantaneously at rest is also constantly changing. That's why it's hard to see in the video. The change in angular velocity is due to the torques exerted on the sphere from the rest of the setup. (More generally, angular velocity can change even in the absence of torques, because it need not be parallel to angular momentum. But for a sphere the two are proportional, so this subtlety doesn't appear.)

More generally, in $d$ spatial dimensions the angular velocity is a rank $2$ antisymmetric tensor. Therefore, in $4$ spatial dimensions the angular velocity has $6$ independent components and therefore can't be thought of as a vector, so there is no meaningful notion of an axis of rotation.


Your question is a geometric question.
I will address your question in a purely geometric way.

Take a trackball device and for vivid visualization you can take a very large trackball.

The pickups inside the device sense the motion of the ball, and the trackball device sends signals to the computer, and the cursor moves accordingly.

To describe axes of rotation I will use the same terms as in aviation:

  • Roll
  • Pitch
  • Yaw

I assume trackballs are designed to disregard yawing motion of the ball, so as not to cause confusion.

That leaves pitching of the trackball and rolling of the trackball.

Visualize putting two fingertips on the trackball, one fingertip applying purely pitching motion, and the other finger applying purely rolling motion.

So now you are simultaneously applying two rotations to the trackball: pitching and rolling.

(Actually, since you are applying two rotations simultaneously the trackball will swivel underneath each fingertip. The trackball simply has to swivel to accommodate that you are applying two rotations simultaneously. Each fingertip should continue to move so as to apply purely pitching and purely rolling respectively.)

Now comes the crucial bit. As you are applying two rotations simultaneously, is there something you can do to make the cursor on the screen move in a way that is impossible to get with only one finger? That one finger is then of course free to push the trackball in any direction.


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