Minimal rotational kinetic energy for a gyroscope to precess

[Malwina]

Feynman does a good job on precession. In particular fig. 20-6 and its environs.

Okay, I understand how the equations for new Torques are derived. Does moment of inertia (I) work analogously, because τ = Iα, so
Iz′=Iz
Ix′=Ixcosθ+Iycosθ
Iy'=Iycosθ-Ixsinθ?

 

[hutchphd]

The moment of inertia is fixed (in the frame of reference attached to the body). It is a matrix that is diagonal in the coordinate system that recognizes the symmetry of the body. Those are the body coordinates usually chosen. Your question is rather general.....I do not know what you need here, please be specific.

 

[Malwina]

Generally, I want to calculate the energy loss in the gyroscope overtime. To do that I did the evaluation of the dependency of KE on ωp as haruspex sugested in #14. Now I realized that there might be a problem with the formulas, as I think the spin axis and precession axis are not parallel, but perpendicular at the end point (when the gyroscope tipped to the horizontal). Now I'm looking for a way to incorporate this into the formula for KE associated with precession, but I do not know how.

Basically, I am looking for a formula relating KE with ωp.

 

[Malwina]

Oh! Nevermind! I misunderstood what the parallel axis theorem refered to. It is not the spin and precession axes that are parallel, but spin about the side of the disc and the spin about the centroid of the disc. The formulas work then, thank you.

My new idea is to put the gyroscope at the horizontal initially and since I found a function expressing KE as a function of ωp, I can calculate KE at different times by measuring ωp.