I don't think the "pushing against the wall" analogy is appropriate here. It would be impossible to move the wall, but the forces required to reorient the gyroscope are of the same magnitude required when the gyroscope is not spinning, because the object's moment of inertia is the same in both cases. However, it seems impossible due to its counterintuitive nature.

Think of it in terms of vectors.

When you apply opposing forces to both ends of the axis of the gyroscope, you are applying a torque about the x (or y, but let's say x) axis.

But you also have the angular momentum vector in the system (actually a psuedovector, but nevermind).

Both vectors together produce a movement in a direction that is somewhere between the two contributing vectors' direction. Your force is redirected, in other words.

The sensation you get that you "can't" get the gyroscope to the orientation you want is just the result of your efforts continually being redirected away from your goal by the momentum vector.

In theory, getting to the goal should require the same amount of force for the gyroscope regardless of whether it is spinning or not. However, doing so requires that move it in a very counterintuitive way, and even trying to correct for the other contributing vector just perpetuates this.

two concentric and counterrotating flywheels preclude all precession forces regardless of which plane the axis is rotated in. this is assuming the connection between the two flywheels is sufficiently strong--it make break from tension/compression due to each flywheel experiencing its own forces.
refer to the diagram i just drew up.

the black rectangles are the two flywheels, the connecting line is the physical connection and also the axle which both flywheels are concentric. the red arrows show the direction of angular momentum (along x axis), while the red circles indicate the direction of rotation (around x axis).

the blue arrows indicate the precession forces experienced by both flywheels when the whole system is rotated in the direction indicated by the curved blue arrow. this is what causes tension/compression in the connecting bar, but otherwise zero torque on the system as a whole.

the green arrows indicate the same forces if the system was rotated the other way (counter to the blue curved arrow).

the situation is similar for rotation of the system in any other plane.

## Best Answer

The answer to your question is that there is no minimum force required to produce the required rotation.

What you need is a torque to start the flywheel rotating about the vertical axis and then at an appropriate time when the required angle is reached you need to switch off the torque.

The size of the applied torques and hence forces will depend on how rapidly you wish to change the angle.

The larger the torques and forces the faster will be the change.

In the left hand diagram the torque $\tau$ is provided by a couple of magnitude $Fd$ where $d$ is the perpendicular distance between the lines of action of the two forces.

Now you might think that the directions of the forces (parallel to the vertical axis about which you wish to rotate the flywheel) and hence the torque is incorrect; it is not and for many totally counter-intuitive.

I have omitted the vector notation in my diagrams to make them clearer.

In the right hand diagram the angular velocity of the flywheel is $\vec \omega$ and you want to rotate the flywheel about the vertical axis so that its new angular velocity is $\vec \omega +\Delta \vec \omega$ with the magnitudes of both these angular velocities the same.

The initial angular velocity, the final angular velocity and the change in angular velocity $\Delta \vec \omega$ are all in a plane which is orthogonal to the axis of rotation.

Now the direction of the change of angular velocity represents a rotation which is clockwise (right hand grip rule) when look from thr right of the diagram and this is also the direction of the torque provided by the couple.

So you apply a torque to the system, allow the flywheel to rotate an appropriate amount and then switch off the torque.

Applying a larger force/torque will get the flywheel to the final position faster.

In fact the flywheel is a acting like a gyroscope undergoing precession which Professor Lewin calls the most non-intuitive part of mechanics and I suggest you view this video from about 14:00 for further insight?

In the video one of the forces producing the torque on the bicycle wheel is the downward weight of the wheel and the other is the upward force exerted by the rope on the axle of the wheel.