I How Does Table Rotation Direction Affect Gyro Stability?

[wrobel]

Here is a gyro on a rotating table. The gyro has two degrees of freedom:
its axis can tilt in the vertical plane that is fixed relative the table.
It turns out that stability of the vertical equilibrium
of the axis depends in particular on which direction we rotate the table.

 

[wrobel]

a sketch of the solution

 

[ergospherical]

With reference to @wrobel's post in #2,\begin{align*}
H = \dfrac{\partial L}{\partial \dot{q}^i} \dot{q}^i - L = \dfrac{1}{2} A \dot{\theta}^2 + \frac{1}{2} B \dot{\varphi}^2 - \dfrac{1}{2}\Omega^2 (A\sin^2{\theta} + B\cos^2{\theta})
\end{align*}The variable $\varphi$ being cyclic means $\dfrac{\partial L}{\partial \dot{\varphi}} = B(\Omega \cos{\theta} + \dot{\varphi}) \equiv c$ is a constant, then upon eliminating $\dot{\varphi}$ from the Hamiltonian:\begin{align*}
H = \dfrac{1}{2} A \dot{\theta}^2 - c\Omega \cos{\theta} - \dfrac{1}{2} A \Omega^2 \sin^2{\theta} + \dfrac{c^2}{2B}
\end{align*}the last term is a constant so can be ignored, $H \mapsto H - c^2/2B$. Then we may examine $V_c \equiv - c\Omega \cos{\theta} - \dfrac{1}{2} A \Omega^2 \sin^2{\theta}$.\begin{align*}
V_c' &= \Omega \sin{\theta}(c - A\Omega \cos{\theta}) \\
V_c'' &= \Omega (c \cos{\theta} - A\Omega \cos{2\theta})
\end{align*}The equilibrium position at $\theta = 0$ is stable if $\Omega(c-A\Omega) > 0$.
The equilibrium position at $\theta = \pi$ is stable if $-\Omega(c+A\Omega) > 0$.
Meanwhile the equilibrium positions at $\theta = \mathrm{arccos}(c/A\Omega)$ and $\theta = 2\pi - \mathrm{arccos}(c/A\Omega)$ are stable if $\Omega(-\frac{c^2}{A\Omega} + A\Omega) > 0$.

 

[wrobel]

I feel I have got something different

 

[Swamp Thing]

If the turntable could be stopped dead just when the gyroscope's axis is halfway through the process of flipping and is exactly horizontal, would the axis also freeze immediately in the horizontal position? Or would it continue to precess due to inertia? Or would it oscillate around the horizontal and then settle down somewhere?

 

[wrobel]

I think you can understand that substituting $\Omega(t)=\Omega_0\ne 0$ for $t<t^*$ and $\Omega(t)=0$ for $t\ge t^*$
into the equations.
You will need stitch the solution properly

 

[Cleonis]

About what is demonstrated with the setup in that video:

It is a setup where the rotation rate of the rotating platform is transferred to the spinning gyro wheel by the outer ring of the gimbal mounting; the orientation of the outer ring of the gimbal mounting is fixed relative to the rotating platform.

Under those circumstances: the gyro wheel spinning in the same direction as the rotating platform is a state of lower potential energy than the state of counter-rotating.

I will use the following words to refer to axes of rotation: spinning, pitching, swiveling
spinning - rotation of the gyro wheel
pitching - rotation of the inner gimbal ring
swiveling - rotation of the outer gimbal ring

Consider what would happen if the starting configuration is that the spin axis of the gyro wheel is parallel to the plane of the rotating platform. As you set the rotating platform into motion: the outer gimbal ring imparts that rotating motion to the spinning gyro wheel. In response the gyro wheel proceeds with pitching motion.

(For explanation of that pitching response: see the following discussion on physics.stackexchange, written by me, of gyroscopic precession )

The tendency to impart pitching motion is the strongest when the spin axis is parallel to the plane of the rotating platform. At the point where the spin axis is aligned with the rotation axis of the rotating platform there is no longer a pitching tendency.

The pitching response is quite vigorous, so the gyro wheel tends to overshoot the perpendicular orientation. Once the pitching motion has gone passed the perpendicular: the pitching tendency arises again, acting once again to pitch the gyro wheel to alignment with the rotation axis of the rotating platform.

In the video you see that when the angular velocity of the rotating platform is reversed the gyro wheel pitches 180 degrees. The gyro wheel was very close to being perfectly anti-aligned, and if it would actually be very, very close to being perfectly anti-aligned the anti-alignment would not immediately result in the gyro wheel pitching 180 degrees. But there is enough of an intitial angle to give a bit of leverage, and pitching motion commences.

As the gyro wheel pitches into alignment (with the rotation axis of the rotating platform) you see an oscillation that dampens out after a couple of swings. That gives an indication of the amount of friction in the gimbal bearings. That friction removes the energy that can dissipate, and from then on the system is in the lowest potential energy state that is available. The lowest available energy state is with the spin axis of the gyro wheel is aligned with the rotation axis of the rotating platform.

Every time the angular velocity of the rotating platform is reversed the system is replenished with potential energy, which is then dissipated by way of the gyro wheel pitching to alignment.Gyrocompass
The effect demonstrated in the video is used in a navigational device called 'Gyrocompass'. The gyro wheel of a gyro compass is suspended in such a way that the spin axis of the gyro wheel tends to go into alignment with the Earth's rotation. When the spin axis of the gyro wheel is not aligned with the Earth's rotation that rotation is transferred to the gyro wheel. Calibrated friction of motion of the gimbal suspension is optimized for the gyro wheel proceeding to alignment with the Earth's rotation.

Submarines don't have GPS reception while submerged. A gyrocompass allows the crew of a submarine to see the orientation of the submarine (with respect to the Earth).

A pure gyroscope will have some non-zero drift. By contrast: for a gyrocompass it is intrinsically impossible to have any drift. A gyrocompass in operation will intrinsically align with the Earth's axis.

 

[wrobel]

Under those circumstances: the gyro wheel spinning in the same direction as the rotating platform is a state of lower potential energy

for all the positions the potential energy is the same : the center of mass does not change its altitude

 

[Cleonis]

for all the positions the potential energy is the same : the center of mass does not change its altitude

Of course: the center of mass does not change height, that is not at issue.

Here is what I was referring to: the motion of pitching 180 degrees means a force is doing work, in the sense that things will only accelerate from one location to another when a force is exerted.

So in this particular context the potential I am referring to is the potential difference that corresponds to the integral of the work done in pitching the gyro wheel 180 degrees.

There is the rapid 180 degrees pitching motion, and after the gyro wheel axis has arrived at the new orientation the pitching motion does not suddenly stop, it ends in the form of decaying oscillation. The decaying oscillation finalizes the dissipation of energy.

We have that every time that the gyro wheel pitches 180 degrees energy is dissipated. So: we need to identify what act is replenishing that energy. When the angular velocity of the rotating platform is reversed the angular velocity of the gyro wheel relative to the rotating platform is changed. That is the act of replenishing energy.
Rather: the angular velocity of the rotating platform is maintained manually. During the motion of pitching down there is a corresponding torque that is transferred from the gyro wheel to the rotating platform. If the rotation rate of the rotating platform would not be maintained that torque would slightly decrease the angular velocity of the rotating platform. The act of actively maintaining the angular velocity of the rotating platfrom provides the required influx of energy.