Car moving on a ball in space Consider a car of mass $m$ moving on the surface of a ball (think of it as earth) with moment of inertia $I$, floating in a vacuum.
Let the car slowly (adiabatically) drive around the circle of colatitude $\phi$. Suppose that, after one circle, the north pole of the ball is pointing at the same direction as before.
What are the possible values of $\phi$ other than $0, \frac 1 2\pi$ and $\pi$?
I think this has some Lie-algebra solution, though I haven't figured it out.
Partial result: by conjugating with a rotation around the north pole, we see that any initial segment of the rotation must commute with the rest of the rotation, thus they must be around the same axis. Then, since the total rotation must both be around the north pole and around the colatitude $\pi/2-\phi$ pole, it must be zero.
 A: I'll assume in the following that the mass of the car $m$ is negligible compared to the mass of the ball $M$, so during the motion, the center of the ball remains fixed. However, in order for the problem to be interesting, I will not neglect $mR^2$ compared to $I$, and write their ratio $I'=\frac{mR^2}{I}$, the relevant physical parameter. This is compatible with the previous assumption when the mass distribution of the ball is spherically symmetric, but more concentrated at its center. The problem is that of rigid bodies, no need for Lie algebras (though implicitly, we are referring to $SO(3)$).
I will also assume that the system (in an inertial reference frame) is initially at rest, so the equation of motion is determined by $\vec L = 0$ for the total angular momentum. It turns out that the rate at which the car moves does not change the total final rotation. In fact if you are familiar with gauge theory, this is actually a Wilson's loop, which is purely geometric, so I'll assume for simplicity that the car executes the circle at constant azimuthal speed $\dot \phi = \omega$ ($\theta$ being the colatitude, physics convention) in the reference of the ball, and $z$ axis is the North-South axis. This is compatible with the initial condition, you'll have to imagine a vanishingly short impulse at the beginning to set it in motion.
For the bookkeeping are three frames: the first inertial frame $\mathcal R_1$, in which the problem is stated, the second one $\mathcal R_2$ is the one of the ball, and $\mathcal R_3$ the one of the car. They are all rotated with respect to one another, writing $R_{ij}$ the rotation to go from $\mathcal R_i$ to $\mathcal R_j$, you want $R_{21}$ and see whether it is a $z-$rotation, and I already know $R_{32} = R_z(\phi)$. I'll write $\vec \Omega$ the angular velocity of $\mathcal R_2$ with respect to $\mathcal R_1$ in the frame $\mathcal R_2$, $\vec \Omega_c = \omega \vec e_z$, the angular velocity of $\mathcal R_3$ with respect to $\mathcal R_2$ in $\mathcal R_2$ (which is constant).
Using the condition $\vec L = 0$, you get
$$
\vec \Omega =  I' \omega \sin \theta \vec e_\theta
$$
with $\vec e_\theta$ is the colatitude basis vector at the car's position. Geometrically, $\vec \Omega$, of length $I'\sin \theta \omega$ precesses about $\vec e_z$ at speed $\omega$, at angle $\theta$. Technically, I'm done at this stage, as the rotation is simply a time ordered exponential of $\vec \Omega$, standard procedure when you go from the Lie algebra to the Lie group. However, it is difficult to extract information from it, and I can actually fully compute it by another method.
Let $\vec X$ (resp. $\vec x$) represent the position vector of a fixed point of $\mathcal R_1$ in the frame $\mathcal R_2$ (resp. in the frame $\mathcal R_1$). Since $\vec x = R_{21}\vec X$, by figuring out $\vec X$, the problem is solved. It obeys the equation:
$$
\frac{d}{dt}\vec X +\vec \Omega \times \vec X = 0
$$
so according to our previous results, this is simply a Bloch equation, and the same method of resolution applies. It is readily solved by going into $\mathcal R_3$, and I'll summarize it here. Let's denote $\vec X = R_{32}\vec Y$, ie the position vector of the same point this time in $\mathcal R_3$. I now have (chose coordinates so that the car stays in the $x-z$ plane):
$$
\frac{d}{dt}\vec Y +\omega \vec u \times \vec Y = 0 \\
\vec u = (1-I'\sin^2\theta) \vec e_z + I'\sin\theta \cos\theta \vec e_x
$$
with now $\vec u$ constant, of direction $\hat u$ and magnitude $u$, which is easily solvable. Therefore $R_{31} = R(\hat u,u\phi)$ (notation $R(\vec n,\alpha)$ of ration about unit vector $\vec n$ of angle $\alpha$), so $R_{21} = R(-\hat u,u\phi)R_z(-\phi)$ (notice that as advertized, no $\omega$ dependence, only geometric dependence via $\phi$), and the final rotation is:
$$
R_{21} = R(\hat u,2\pi u)
$$
For $I’\neq 0$, the North Pole regains its original position iff the rotation is about the $z$ axis, ie $\theta=0$, or is trivial, ie $u = n\in\mathbb N$:
$$
n^2 = (1-I'\sin^2\theta)^2+(I'\sin\theta\cos\theta)^2
$$
For all values of $I’$, you also have the other $2$ trivial solutions $\theta=0,\pi$ ($n=1$). For fixed $I'$, there is always only a finite number of colatitudes possible. You can check that if $I'\leq 3$, you only have the $3$ trivial solutions $\theta=0,\pi/2,\pi$.
Hope this helps, and tell me if something isn't clear (tried my best, but it's pretty dense).
