# Spinning flywheel energy “absorption”?

Consider a flywheel, mounted such that its AXIS can be rotated anywhere on the XY plane (about the Z axis), set spinning, initially oriented along the X-axis. Once spinning, the flywheel resists changes to it's axis of rotation, but if we apply enough force (red arrows), we can rotate the axis over to the Y direction and again for another $90°$, to the -X direction. After another $180°$, the direction is reversed again back to where it started. We have expended energy reversing the vector component of the wheels angular momentum twice.

My question: Where did the energy go (the energy we put into the $360°$ rotation)? What was it transformed into? That is, what evidence is there of work having been done?

None of the torque you applied in doing the rotation was applied in the direction of the change in orientation. $\vec{\tau} \cdot \vec{\omega} = 0$. No work is done. Precession is a weird, counterintuitive thing: in order rotate the flywheel about the $z$ axis, the torque you apply has no component along said axis. For example, the initial torque required, given that the axis of the flywheel is parallel with the $x$ axis, is actually about the $y$ axis.

If "mounted such that its axis can be rotated anywhere on the XY plane" is meant to indicate that the axis is constrained to that plane, say by a frictionless circular groove, then no externally applied torque is required to do the rotation. Once the flywheel axis is rotating at a certain rate about the $z$ axis, it will maintain that rotation forever--it's analogous to sliding a block across a frictionless tabletop. All the necessary force is supplied as normal force by the groove surface.

• The initial axis of rotation is in the X-direction, parallel to the X axis. Torque is applied to swing that axis to the Y direction - exactly like holding a spinning bicycle wheel in your hands and twisting its axis of rotation horizontally ie in the XY plane. Sorry if I didn't describe it clearly enough. – EHT Jan 25 '18 at 15:53
• You did explain it clearly (at least I think so). Is there something that's still bugging you? – Ben51 Jan 25 '18 at 15:56
• I am trying to illustrate the case of a flywheel spinning so it's axis is horizontal, say oriented North South. Now, we force it into an East-West direction - still horizontal. Like it (and its bearing mounts) were mounted on a large turntable. – EHT Jan 25 '18 at 16:07
• I think the setup is clear. What do the orange arrows represent in your diagram? Motion, or force? – Ben51 Jan 25 '18 at 16:08
• The force we put onto the shaft to swing the axis around. I know it should be a circular type arrow, I just dont have one in this drawing program. – EHT Jan 25 '18 at 16:10

Rotational dynamics can be counter-intuitive.

Your diagrams are incorrect in that the orange arrow show the directions of change of the angular momentum not the force acting on the axis of the spinning disc.

Also one force alone which does not pass through the centre of mass of a body will not only cause the disc to change its angular momentum it will also accelerate its centre of mass ie the centre will also undergo transnational motion.

To go from diagram 1 to diagram 2 the angular momentum of the body must change and your diagram 1 shows such an initial change $\Delta \vec L$ with the orange arrow.

$$\vec L_{\rm old} + \Delta \vec L = \vec L_{\rm new}$$

For that change in angular momentum to occur a torque $\vec \tau$ must be applied to the axle of the disc.

This torque $\vec \tau$ could be achieved by applying two forces $\vec F$ on the axle, one going into the screen $\otimes$ and one coming out of the screen $\odot$ , to form a couple.

The axis of rotation of the disc will move $\vec L_{\rm new}$ but note the the displacement of each of the forces is at right angles to the direction of the forces, so the forces do no work on the disc ie its kinetic energy stays the same.

Counter-intuitive?
To me it is in that pushing and puling the axis into and out of the screen causes the axis of the disc to undergo a clockwise rotation.

Please, can we use the word 'steer' when we are turning the axle in one of the axes orthogonal to its rotational axis?
In theory the ideal flywheel has the same energy in any axle orientation, so it shouldn't require work to 'steer' it. It should be possible to steer the axle effortlessly and with zero losses.

[Aside: With no friction or flexing, wouldn't an ideal flywheel exhibit infinite reaction to any applied 'steering' torque? Can there be precession without flexing the flywheel? ]

In reality, energy is 'lost' as heat as materials flex both in the wheel and in the external frame of reference from which the torque is applied; and RPM loss results from bearing friction.

If the flywheel is prevented from precessing (as in a ship's gyrostabiliser) then further RPM loss comes from work done in resisting the steering torque (via the axle bearings), and causing the water to pile up higher against the ship. Note that energy lost in this way is resistive loss and can only be restored by spinning up the wheel again.