# Spinning Wheel subjected to Gravitational Force

have a question about the dynamics of a suspended horizontally rotating wheel subjected to gravitational force like in following picture I took from here:

At the beginning the wheel has an angular momentum $$\vec{L}:=L \vec{e_x}$$ in the horizontal plane, which coincides with the axis of rotation (considered as the initial condition before the gravitatitional force start to act). Then we let the gravitation act on the wheel and we observe that the wheel axis performs a precession movement on the horizontal plane.

For sake of completeness there is also a video discussing it.

The theory says that gravity exerts torque $$M \vec{e_y}$$ (directing into the drawn 2d plane) on the wheel around the mounting point as a fixed point, so that according to the angular momentum law, in an infinitely small period of time $$\tau$$ the new angular momentum $$L'$$ has the value $$\vec{L} + \Delta L = \vec{e_x } + \tau \cdot M \vec{e_y}$$ also rotates infinitesimally in the horizontal xy plane. That's clear so far.

Question: What I don't understand is why does the axis of rotation follow presisely this movement, ie tries to align with respect the direction of the angular momentum evolving directed by gravitational force on the horizontal plane to $$\vec{L} \to \vec{L} + \Delta\vec{L}$$ as showed in the picture? In general,the angular momentum and the axis of rotation of the considered rigid body are not necessarily parallel. Only if the rotation is about one of it's major axes.

So the question is not why the angular momentum moves that way (it's clear to me), but why the angular velocity pseudovecor of the wheel "strives" towards this alignment with respect to the direction of the angular momentum? Which principle forces this behaviour?

• user267839: "So the question is [...] why the angular velocity pseudovecor of the wheel "strives" towards this alignment with respect to the direction of the angular momentum?" -- To clarify (at least for myself): Do you therefore mean to ask why (generally) the "nutation wobble amplitude dies down" as sketched there, Fig. 1 b) ? (I found this article in this related answer, btw.) This dampening is briefly mentioned in "Goldstein" ... Jul 6 at 17:48
• @user12262: my question is as you quoted why the angular velocity tries to become parallel to the angular momentum. Having answered this, this consequntly implies that nutation wobble amplitute dies after we again turn off the gravitational force. So I don' t think that this is equivalent: if we leave the force on for all the time, then this wobbling would persist probably in some cases with same amplitude also for all the time , but the angular velocity pseudovector would still try to follow the changing angular momentum. And the question is why the latter effect happens. Jul 6 at 18:39
• so in short: if we turn the gravitation off, then the answer of my problem would provide an answer to nutation amplitude dying phenomena, but not the other way around Jul 6 at 18:45

The angular momentum $$~\vec L~=\mathbf I\,\vec\omega~$$

assume that the moment of inertia $$~\mathbf I= \rm diag[I_x~,I_y~\,I_z]$$

from here

$$\vec L\times\vec \omega= \begin{bmatrix} (I_y-I_z)\omega_y\,\omega_z \\ (I_z-I_x)\omega_z\omega_x \\ (I_x-I_y)\omega_x\omega_y\\ \end{bmatrix}\tag 1$$

$$~I_y=I_z~$$ (because the symmetry)
thus from equation (1) the x components of the result is zero, this means that $$~L_x~$$ is parallel (same direction) to $$~\omega_x~$$ (spin direction)
• well, your reasonings show so far I understand you correctly that at the beginning*(!) - say for $t=0$ for time parameter $t$, ie *before we have 'turned on' the gravitation force -the angular velocity $\vec\omega= \vec{\omega}(0)$ is parallel to the angular momentum $\vec L = \vec{L}(0)$. Yes, but that's already true by assumption(=intial condition). My question why the direction of the angular velocity follow the angular momentum when the system evolves in time, ie after we have turned on the gravitational force, not only at the beginning, see also the linked video. Jul 11 at 10:00
• The point is that when we turn of the gravitational force is turned on und by general torque law the angular momentum turns infinitesimally to $\vec{L} + \Delta L = \vec{e_x } + \tau \cdot M \vec{e_y}$. The moment of intentia $\mathbf I= \mathbf{I}(t)$ in in general time dependent matrix/tensor, so naively thinking there is no reason why after some time has past the angular velucity $\vec{\omega}(t)$ should be still parallel to $\vec{L}(t)$, since for $t \neq 0$ the moment of inertia $\mathbf{I}(t)$ may not be of diagonal shape anymore. That's the point. Jul 11 at 10:09