angular momentumangular velocityclassical-mechanics
I know that angular momentum $\vec{L}$ and angular velocity $\vec{\omega}$ of a rigid body doesn't point into the same direction in general. However if your body spins around a principal axis, $\vec{L}$ and $\vec{\omega}$ point into the same direction.
In which other situations is this is exactly or at least approximately true? Please give reasons or a mathematical derivations why it is true in a given situation.
As it happens, there are indeed limits to how different the directions of $\vec L$ and $\vec \omega$ can be. This is because the moment of inertia tensor is something called positive semidefinite, which requires the inner product of any vector $\vec\omega$ with its transformation under the tensor, $I\vec\omega$ to be non-negative: $$ \vec\omega \cdot I\vec\omega \geq 0. $$ In particular, this means that the angle between $\vec L$ and $\vec \omega$ cannot be greater than 90°.
The moment of inertia, as a matrix, is normally defined with the components $$ I_{ij} = \sum_k \left(r_{(k)}^2\delta_{ij} -x_i^{(k)}x_j^{(k)}\right)m_k \tag 1 $$ for a cloud of particles with masses $m_k$, or as the integral $$ I_{ij} = \int\left(r^2\delta_{ij} -x_ix_j\right) \rho(\mathbf r)\mathrm d^3\mathbf r \tag 2 $$ for a continuous mass distribution. Normally you calculate the components about the principal axes of the moment of inertia, which is a frame in which the off-diagonal components like $$ I_{xy} = -\int \,xy\,\rho(\mathbf r)\mathrm d^3\mathbf r $$ vanish, and you're only left with the more usual diagonal components of the form $$ I_{xx} = \int \left(y^2+z^2\right)\rho(\mathbf r)\mathrm d^3\mathbf r. $$ In the general case, however, one needs to stick with the full matrix expressions $(1)$ and $(2)$, and it is not immediately obvious how one can prove from those expressions that the moment of inertia is positive semidefinite, so it is worth exploring that in a bit of detail.
(In a frame aligned with the principal axes this is easy, since in that frame you have $$ \vec\omega \cdot I\omega = \sum_j I_{jj}\omega_j^2 \geq 0 $$ as all the $I_{jj}$ are non-negative. However, the existence of such a frame is not a trivial fact - it essentially requires us to know that the matrix $I_{ij}$ is diagonalizable and to find such a diagonalization. As such, it is better to work in the general case.)
The main point about the positive semidefiniteness of $I$ is that it encodes a much simpler relationship than its components lead you to believe. To see this relationship, you start with the fundamental definition of the angular momentum of a single particle, $$ \vec L= \vec r \times \vec p = \vec r\times m\vec v, \tag 3 $$ and then you use the fundamental definition of the angular velocity vector - that any particle rotating about the origin with (vectorial) angular velocity $\vec \omega$ will have velocity $$ \vec v = \vec \omega \times\vec r, \tag 4 $$ and you put this into the expression for the angular momentum: $$ \vec L = \vec r\times m(\vec \omega\times\vec r). \tag 5 $$ This looks complicated, but the key thing to realize is that it is linear in $\vec\omega$: if you put in a linear combination like $a_1\vec\omega_1 + a_2\vec\omega_2$ in for $\vec\omega$, you will get out $a_1\vec L_1 + a_2\vec L_2$. The equation above is a linear relationship between two vectors, and all such relationships can be expressed in matrix form: $$ \vec L = I \vec \omega. $$ The moment of inertia tensor is exactly that matrix.
However, the matrix form of the relationship is not necessarily the most useful way to phrase it. As an example, let me show the proof that $I$ is positive semidefinite, directly from the relationship $(5)$ above. I want to show that $\vec\omega\cdot\vec L\geq 0$, so I start by calculating that inner product: \begin{align} \vec\omega\cdot\vec L & = \vec\omega\cdot\left(\vec r\times m\,\vec v \right) \\ & = m\vec r\cdot\left(\vec v\times \vec\omega \right) , \end{align} where I've used the cyclical property of the scalar triple product, because the cross product $\vec v\times \vec\omega$ is easy to calculate since $\vec v$ and $\vec\omega$ are orthogonal by construction. Putting in the definition $(4)$ for the velocity, we get a vector triple product, which is again easy to resolve: \begin{align} \vec\omega\cdot\vec L & = m\vec r\cdot\left((\vec \omega \times\vec r)\times \vec\omega \right) \\ & = m\vec r\cdot\left(\omega^2\vec r - (\vec\omega\cdot\vec r)\vec \omega \right) \\ & = m\left(\omega^2 r^2 - (\vec\omega\cdot\vec r)^2 \right) . \end{align} Here we're essentially done, because we know that $|\vec\omega\cdot\vec r|$ must be no larger than $r\omega$, so we get the desired $\vec\omega\cdot I\vec \omega = \vec\omega\cdot\vec L \geq 0$. Finally, if we have a cloud of particles or a continuous mass distribution, we simply sum or integrate over this inequality.
OK, so the above is a (long-winded) explanation of why this $I$ matrix thing obeys this positive-semidefinite property. What does this mean geometrically? Well, expanding the inner product into an angle form, it gives us $$ \vec\omega\cdot\vec L = \omega L\cos(\theta)\geq 0, $$ where $\theta$ is the angle between the angular momentum and the angular velocity; this inequality translates into $\theta \leq \pi/2=90^\circ$.
This bound is pretty much the best you can do, and it is easy to find examples where $\vec \omega$ and $\vec L$ are arbitrarily close to orthogonal. The simplest is a long, thin rod of length $L$ and radius $w$ that's rotating rapidly about an axis that is almost (but not quite) aligned with the rod:
Putting the rod along the $x$ axis, you get that the moment of inertia and angular velocity are given by $$ I= \begin{pmatrix} \frac12 Mw^2 & 0 & 0 \\ 0 & \frac13 ML^2 & 0\\ 0 & 0 & \frac13 ML^2 \end{pmatrix} \quad\text{and}\quad \vec \omega= \begin{pmatrix} \Omega \\ \delta \\ 0 \end{pmatrix}, $$ respectively, and that gives you $$ \vec L= \begin{pmatrix} \frac12 Mw^2\Omega \\ \frac13 ML^2\delta \\ 0 \end{pmatrix} $$ for the angular momentum, $$ \vec\omega\cdot \vec L = \frac12 Mw^2\Omega^2 + \frac13 ML^2\delta^2 $$ for the inner product, and $$ \cos(\theta) = \frac{\vec\omega\cdot \vec L}{L\omega} = \frac{ \frac12 w^2\Omega^2 + \frac13 L^2\delta^2 }{ \sqrt{\frac14 w^4\Omega^2 +\frac19 L^4\delta^2} \sqrt{\Omega^2+\delta^2} } \approx \frac{ 3 w^2/L^2 + 2 \delta^2/\Omega^2 }{ \sqrt{9 w^4/L^4 +4 \delta^2/\Omega^2} } $$ for the angle. Fixing the rod parameters at $w\ll L$, and then choosing an angular velocity with $\Omega\gg \delta$ and such that $$ \frac wL \ll \frac \delta\Omega $$ (i.e. the rod is much thinner than the spin axis is close to the rod axis), the cosine approximates to $$ \cos(\theta) \approx \frac{\delta}{\Omega} +\frac{3\,w^2/L^2}{2\,\delta/\Omega} , $$ which can be made arbitrarily small, with the angle $$ \theta \approx \frac \pi2 -\left(\frac{\delta}{\Omega} + \frac{3\,w^2/L^2}{2\,\delta/\Omega}\right) $$ arbitrarily close to 90°.
More geometrically, spinning a thin rod about its axis gives an angular momentum component along this axis which is very small, and any misalignment will give angular momentum describing rotation about the axis of misalignment which, if the rod is thin and long enough, can dwarf the angular momentum from the spinning itself.
Finally, it's important to note that the angular momentum and the angular velocity can never be fully orthogonal. This can be seen by expressing the orthogonality relation $\vec\omega \cdot \vec L=0$ in the principal frame of the body, where it takes the form $$ \vec\omega \cdot \vec L = I_{xx}\omega_x^2 + I_{yy}\omega_y^2 + I_{zz}\omega_z^2 =0, $$ i.e. the sum of three non-negative terms, all of which must be zero for the sum to vanish. This is possible by having e.g. $I_{xx}=0$ and $\omega_y=\omega_z=0$, but in that case (and all such cases) you will obtain $\vec L=0$, which doesn't really count as "orthogonal" to $\vec\omega$.
Why does not the precession rate of the angular velocity vector give exactly the frequency of wobbling? In other words, how come the precession of $\vec\omega$ is different to the line of nodes rotation?
Short answer: Because the precession rate $\Omega$ corresponds to the precession of $\vec\omega$ in the body frame, not in the Earth frame. The variables in the Euler equations, $I_i$, $\omega_i$ and $\tau_i$, are in the body frame.
The angular velocity of the disk has two contributions, the component $\vec\omega'$ due to the spin and the component $\vec\omega''$ due to a rotation of the inclined disc about a vertical axis. The latter corresponds to the wobbling as viewed by someone in the Earth frame.
As we can see from the figure, the resultant angular velocity $\vec\omega$ is always off the symmetry axis of the disk, which means it has a non-vanishing projection in the plane of the disk (dashed line). At same time, the axes fixed in the plane of the disk, $\rho_1$ and $\rho_2$ rotates with spin $s$ (viewed by someone in the Earth frame). Hence, someone in the disk frame will see the axes fixed and the projection of $\omega$ in this plane rotating with rate $s$. That is why the precession rate equals the spin.
On the other hand, the wobbling effect is due to the non-vanishing $\vec\omega''$. A $2\pi$ rotation about the vertical line corresponds to a complete oscillation (wobbling) therefore the frequency of the wobbling is actually equal to the magnitude of $\vec\omega''$.
Best Answer
There are no "other" examples. The condition that $\vec \omega$ and $$ \vec L = I_{\rm tensor} \cdot \vec \omega $$ point to the same direction i.e. $$ (\vec L=) I_{\rm tensor} \cdot \vec \omega = k \vec \omega $$ where $k$ is a real number (and no longer a tensor) is a definition of an eigenvector of $I_{\rm tensor}$: both $\vec \omega$ and $\vec L$ are eigenvectors of the moment of inertia in such a case.
Eigenvectors of the moment of inertia tensor are called the principal axes. They may always be chosen to be orthogonal to each other and the tensor has the form $$ I_{\rm tensor} = {\rm diag} (I_1,I_2,I_3) $$ in the coordinate system given by these principal axes. In a generic case, the three principal axes are uniquely determined by the tensor (and the axes may be shown to be orthogonal to each other, as eigenvectors of any Hermitian operator).
The only subtlety appears if some of the entries $I_1,I_2,I_3$ coincide. If two of them are equal, any combination of the two vectors is an eigenvector and there is a freedom in choosing these two principal axes (the third axis corresponding to a different eigenvalue is still unambiguous).
If $I_1=I_2=I_3$, then the tensor is proportional to the unit matrix and any vector is an eigenvector of it with the same eigenvalue $I_1=I_2=I_3$. In that case, the choice of principal axes is totally arbitrary. If 2 or 3 eigenvalues coincide, the orthogonality of the basis isn't guaranteed but we still usually impose the extra conditions that the "principal axes" should be orthogonal to each other.
One could discuss various examples – particular shapes with particular tensors of the moment of inertia. But the important fact here is that for the purposes of the spin and angular momentum, only the tensor of the moment of inertia matters. It may be calculated from any matter distribution but everything else besides the directions of the three axes and the three eigenvalues $I_1,I_2,I_3$ is irrelevant for discussions on the angular momentum.