What you have is an articulated system with two bodies. The math is quite complex to solve the general problem. I have done so for you.
Consider the two bearings A and B with gravity acting on the negative z direction and the disk spinning about the x axis. The bearing reactions are
$$ \begin{align}
A_y & = \frac{I_{flip}}{\ell} \dot{\Omega} \\
B_y & = -\frac{I_{flip}}{\ell} \dot{\Omega} \\
A_z & = \frac{m_{disk}}{2}g - \frac{I_{spin}}{\ell} \Omega \omega \\
B_z & = \frac{m_{disk}}{2}g + \frac{I_{spin}}{\ell} \Omega \omega \\
\end{align} $$
where the spin speed is $\omega$, the pivot speed $\Omega$ (and their derivatives $\dot{\boxed{}}$), $\ell$ is the separation between the two bearings and $m_{disk}$, $I_{spin}$ and $I_{flip}$ are the inertial properties of the spinning disk.
This answer expands on the answer by contributor Farcher.
For an intuitive explanation see my 2012 discussion of gyroscopic precession.
That discussion is not in the abstract. (Abstract is to invoke angular momentum vector properties, and operations such as the vector cross product.) Instead it capitalizes on symmetry to connect the phenomenon of gyroscopic precession back to linear mechanics intuition.
See also the following article by Svilen Kostov and Daniel Hammer:
It has to go down a little, in order to go around
Svilen Kostov and Daniel Hammer report on a tabletop experiment that they performed.
They obtained quantitive results, and they report that the amount of drop that they measure is in accordance with the laws of motion.
When a gyroscope wheel is released the center of mass drops a little. The center of mass must drop. Not dropping (a little) is ruled out by the laws of motion.
The center of mass dropping a little always occurs, but almost always there are factors that mask that drop.
When the gyro wheel is released suddenly the resulting motion is a superposition of nutation and gyroscopic precession. Visually the superposition of nutation and gyroscopic precession presents itself as the gyro wheel precessing and bobbing up and down. The midpoint of that bobbing-up-and-down motion is actully lower than the starting height, but the bobbing motion makes it hard to see that.
Usually in classroom demonstrations the gyro wheel is released gingerly. In effect the demonstrator is selectively preventing the nutation while allowing the gyroscopic precession. Releasing gingerly will hide the drop.
It is common in classroom demonstrations to make the gyro wheel spin very fast. When the wheel spins very fast the nutation is very fast and has small amplitude (and damps out quickly), so the nutation tends to go unnoticed. The faster the spin rate the slower the corresponding gyroscopic precession, which corresponds to only a small drop; again the drop tends to go unnoticed.
There is a wide-spread belief system among physicists that in the case of gyroscopic precession the precession happens instead of dropping. That is, according to that belief system there is no drop.
(In the answer by contributor Farcher that mistake is not made. It is pointed out, correctly, that the wheel does drop)
Derek Muller (the author/presenter of the Veritasium channel) shares the erroneous belief system, by the looks of it.
The Veritasium explanation for the phenomenon of gyroscopic precession boils down to stating: 'Gyroscopic precession occurs because the vector cross product says so.'
Best Answer
No.
You spin the gyroscope (upright) so that it has an angular velocity $\omega$. The kinetic energy associated with this motion is:
$$ \frac{1}{2}I_{\mathrm{axle}}\omega^2,$$ where $I_{\mathrm{axle}}$ is the moment of intertia of the gyro about its axle.
The gyroscope then falls a bit under gravity and then it starts precessing. The angular velocity of the gyroscope about its axle is still $\omega$ -- assuming a point contact with the ground and no air resistance, there was no torque in that direction that could have reduced it.
The gyroscope then starts precessing about the $z$ axis (gravity) with angular velocity $\Omega$. This introduces a kinetic energy $$ \frac{1}{2}I_z \Omega^2,$$ where $I_z$ is the momentum of intertia of the gyro assembly about the $z$ axis, not about its axle.
This kinetic energy, not originally present in the system, comes form the gravitational potential energy $mgL(1-\cos\theta)$ lost when the gyro drops by angle $\theta$ before precessing.
Angular momentum (about an axis and/or a point) is only conserved in a closed system, i.e. a system which no external torques act upon.
For the $\omega$-associated angular momentum, that's still conserved as said before, there are no torques tangential to the gyro that can slow down its rotation.
For the total angular momentum, about (say) the point of contact, gravity provides the external torque.
Hence you expect a change in angular momentum:
$$ \Delta \mathbf{L} = \Gamma \Delta t = (\mathbf{r}\times m\mathbf{g})\Delta t,$$
which is exactly what causes precession.