# Can the direction of angular momentum and angular velocity differ?

While studying rotational mechanics, I came across a section where it mentioned that angular momentum may not necessarily be parallel to angular velocity. My thoughts were as follows:

Angular momentum ($L$) has the relation $L=I\omega$ where $\omega$ is angular velocity and $I$ is the moment of inertia, so following this relation, it seems they should be in the same direction. Why are they not?

Consider a thin rectangular block with width $w$, height $h$ resting along the xy plane as shown below. The mass of the block is $m$. The mass moment of inertia (tensor) of the block about point A is

$${\bf I}_A = m \begin{vmatrix} \frac{h^2}{3} & -\frac{w h}{4} & 0 \\ -\frac{w h}{4} & \frac{w^2}{3} & 0 \\ 0 & 0 & \frac{w^2+h^2}{3} \end{vmatrix}$$

This was derived from the definition (as seen on https://physics.stackexchange.com/a/244969/392)

If this block is rotating along the x axis with a rotational velocity $$\boldsymbol{\omega} = \begin{pmatrix} \Omega \\ 0 \\ 0 \end{pmatrix}$$ then the angular momentum about point A is

$${\bf L}_A = m \Omega\,\begin{pmatrix} \frac{h^2}{3} \\ -\frac{w h}{4} \\ 0 \end{pmatrix}$$

As you can see, there is a component of angular momentum in the y direction. The angular momentum vector forms an angle $\psi = -\tan^{-1} \left( \frac{3 w}{4 h} \right)$

In the figure below you see the direction of angular momentum, and the circle about which the center of mass is going to orbit due to precession. • Great example, +1. – Pirx Dec 29 '16 at 14:56
• I don't think I understand the example. What force is being applied to the brick to allow the center of mass to rotate? – Harry Johnston Dec 30 '16 at 1:06
• It is a little misleading. There have to be reaction forces at A but no reaction torques to show this behavior. @HarryJohnston – John Alexiou Dec 30 '16 at 22:09
• So the idea is that point A is being held in place, but allowed to rotate freely? That makes sense. Thanks. Oh, and the height of the brick is h rather than w as your first sentence says? I assume that's just a typo. – Harry Johnston Jan 1 '17 at 1:14
• @HarryJohnston - thanks for the fix. $h$ is the height. Yes point A is fixed in space, like a top. – John Alexiou Jan 1 '17 at 4:41

Physicsapproval, you already apparently know that the moment of inertia $I$ tensor (inertia tensor for short) is indeed a tensor rather than a scalar. If it was a scalar, then by definition angular momentum and angular velocity would always be parallel. This is not necessarily the case due to the tensorial nature of moment of inertia is tensorial.

The inertia tensor of an arbitrary three dimensional rigid body as expressed in an arbitrary set of orthogonal Cartesian axes can be expressed in terms of a 3x3 matrix that is (a) symmetric and (b) positive semidefinite. These two facts mean that one can always choose a set of orthogonal axes in which the inertia tensor is diagonal. There are three distinct cases for a 3x3 diagonal matrix:

• All three diagonal elements are equal to one another,
• Two of the three diagonal elements are equal to one another, but the third is a distinct quantity, and
• The three diagonal elements are different quantities.

In the first case, $\mathrm I \vec \omega$ will always be parallel to $\vec \omega$. In the second case, $\mathrm I \vec \omega$ is parallel to $\vec \omega$ if $\omega$ is directed along the axis of symmetry or has zero component along that axis. In the third case, $\mathrm I \vec \omega$ is parallel to $\vec \omega$ if and only if $\vec \omega$ is parallel to one of the eigenaxes of the inertia tensor.

Suppose that the inertia tensor (when orthogonalized) has three distinct elements and that angular velocity has at least two non-zero elements when expressed in terms of the coordinate system that makes the inertia tensor orthogonal. In this case, \begin{aligned} \mathrm I &= \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \\ \vec \omega &= \phantom{\,\,\,0}\begin{bmatrix} \omega_a \\ \omega_b \\ \omega_c \end{bmatrix} \end{aligned} where $a$, $b$, $c$ are distinct and at least two of $\omega_a$, $\omega_b$, and $\omega_c$ are non-zero. This means that $$\mathrm I \vec \omega = \begin{bmatrix} a\, \omega_a \\ b\, \omega_b \\ c\, \omega_c \end{bmatrix}$$ cannot be parallel to $\vec \omega$.

Proof: $\vec \omega$ and $\mathrm I \vec \omega$ are parallel (or antiparallel) only if $\omega \times (\mathrm I \vec \omega)$ is the zero vector. From the above, this is $$\omega \times (\mathrm I \vec \omega) = \begin{bmatrix} (b-c) \omega_b \omega_c \\ (c-a) \omega_c \omega_a \\ (a-b) \omega_a \omega_b \end{bmatrix}$$ Since $a$, $b$, $c$ are distinct, each of $b-c$, $c-a$, and $a-b$ is non-zero. Since at least two of $\omega_a$, $\omega_b$, and $\omega_c$ are non-zero, there exists some combination $\omega_i \omega_j$ that is non-zero. Thus there is at least one element of this vector that is non-zero.

• When are you going to add the example that is better than the example in the other answer? – GreenAsJade Dec 29 '16 at 23:31
• "This is not necessarily the case due to the tensorial nature of moment of inertia is tensorial" has some funky grammar going on. – Emilio Pisanty Dec 30 '16 at 2:30

This only holds when the product of inertia is 0.

From matrix algebra, multiplying a (n x 1) vector(x) by an (n x n) matrix (A) will scale the vector components in eigenvector direction by respective eigenvalues. $$Ax = b;$$ If eigenvectors are not aligned with coordinates of the defined vector then resulting vector (b) will not have the same direction as x. Eigenvectors will be aligned to coordinates only when non-diagonal components of A are zero.