I want to clear up an ambiguity that seems to exist in my textbook. I am told that a rotating body will have an angular momentum vector parallel to the rotation axis only if the rotation axis is along a symmetry axis through the centre of mass.

However, consider a thin rod rotating about an axis that is fixed at one of its ends. By thin I mean essentially one-dimensional, so that all particles lie along a line that is perpendicular to the rotational axis, and hence the angular momentum vectors of all particles will be parallel to the rotation axis.

This axis is clearly not a symmetry axis, as it is neither symmetrical nor passes through the centre of mass. Yet the net angular momentum of the system is definitely parallel to the axis.

Is this an exception to the rule? And if so, are there any other non-trivial exceptions that can be identified?


That's a sloppy explanation.

It's more simple than that. Angular momentum and angular velocity are parallel, if your axis lies along one of the eigen-axes of the object: of course, symmetry axes are usually eigen-axes (defined by the eigenvectors of the moment of inertia tensor). That's obvious from the definition $\vec{\Gamma}=J\vec{\omega}$ which is an eigenvalue problem.

Here, we said nothing of the axis going through the center of mass, it's not required to be mounted through the center. Consider a point mass at some radius. It's obviously $\Gamma = \vec{r}\times m\vec{v}$, which is perpendicular to the plane of $\vec{r}$ and $\vec{v}$ -- the same as the axis of rotation.

Physically, two things can make your setup excentric and vibrating... if $\Gamma$ is not parallel to $\omega$, then you have oscillating TORQUE acting on the axis, which will cause vibrations and ruin your bearings. But... if the center of mass is not on the axis, then FORCES will act on the axis, as the linear momentum is not conserved (you have to spin the velocity vector of the center of mass, which requires a force). This will likewise want to ruin your bearings, just in a different way (no net torque, but plenty of forces).

In that sense... balanced rotation requires both conditions to be met, but no... center of mass position has nothing to do with the angular momentum and axis being parallel.


Linear algebra tells you that there are always 3 mutually perpendicular axes, each with possibly its own moment of inertia, for which the angular momentum is parallel to the angular velocity vector. This means that even for objects with no symmetry at all, these special axes exist: that's why the symmetry definition is incomplete. If you find a symmetry axis, then it is an eigen-axis, but if there is no symmetry, you must unfortunately compute the entire thing and use linear algebra to find these directions.

If inertia is equal for any pair (or all three) axes, any direction between those axes are also good. This way, a cylinder has one axis aligned with its symmetry axis, but instead of having only two other special axes (like a general blob would have), any axis, perpendicular to the long symmetry axis also has this property. Even more special case, for a sphere, cube, or a cylinder with very special aspect ratio (or cone with special aspect ratio, or a lot of other objects too, including some with no symmetry), all (not just 3 perpendicular) axes are equivalent: they all have the same moment of inertia, and they all keep the angular momentum parallel to the rotation axis.

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  • $\begingroup$ Your explanation is beyond my level of knowledge, as I am studying introductory physics. Eigenvalues and tensors are beyond my current knowledge. But at least you have demonstrated to me that the definition of axis of symmetry is more complicated than what can be understood in an introductory course, and hence why the description of such is vague and described intuitively through examples (all of which have axes passing through the centre of mass). Knowing that the axis does not need to pass through the cm (unlike what is explicitly stated in the book) answers my question. Thanks! $\endgroup$ – esotechnica Jan 12 '16 at 11:09
  • $\begingroup$ The problem with the "symmetry "definition is that the shape can be very weird, without any symmetry axis at all, and there are still 3 mutually perpendicular axes, along which the angular momentum is parallel to the axis (and in case the moment of inertia is the same along two of these axes, all direction in that plane are okay as well - in the case of a cylinder, for instance). That's why I pointed out the mathematical definition. If you "guess" these axes from the symmetry, then of course it's ok. $\endgroup$ – orion Jan 12 '16 at 15:07
  • $\begingroup$ $\Gamma$ can be not parallel to $\omega$ while there is NO torque. See for example, en.wikipedia.org/wiki/Poinsot%27s_ellipsoid $\endgroup$ – Tarek Apr 2 '18 at 14:35

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