Free-floating top; relation of precession angle and precession and spin rates

[hkyriazi]

Does anyone know of a formula describing the relationship between precession rate, precession angle, and spin rate for a top (or gyroscope, or any hard, non-spherical object) spinning freely, in zero gravity? All the analyses I've been able to find are for tops under constant torque (gravity), or with their tip held in place in some way.

Specifically, I'm trying to analyze what happens to a baton-like object (like the ones that majorettes twirl in marching bands) as it goes from pure twirling (which has, by my definition, an axial spin rate of zero, and 90 deg precession angle), under a series of frictionless taps at opposite ends and sides, to the pure spinning state (zero deg precession angle, zero twirl/precession rate, and much higher spin rate than the initial twirl/precession rate).

I've had this posted for almost a full day, with no takers, so I'll add that the Earth is known to precess once every 26,000 years or so, but the explanations I've found on the web seem to attribute this to the constant force the equatorial bulge feels from the sun and other planets.

In order to prove to myself that an object can indeed precess under weightless and force-free conditions, I spun a smooth surfaced pen off my fingertips into the air (one must spin it very rapidly, with a finger-snap movement) with a slight jerk as I let it go, and could see (with a good spin and flick) that it precessed nicely until it hit the floor. (If one does this with a pencil, which is usually hexagonal in cross-section, the air resistance effect is so great that it causes very bizarre movements through the air, obscuring the precession.)

--Harold Kyriazi

 

[techmologist]

Well, you know what they say. Better late than never.

Objects like batons, footballs (American), some pens and most pencils have one axis of symmetry, and luckily those are the easy ones to analyze. Any solid object has three principal moments of inertia: I₁, I₂, and I₃. For an object with an axis of symmetry, two of these are equal, say I₂ = I₃ = I. You can always label the principal axes so this is the case.

In the attached diagram, there is a potato shaped object representing an ellipsoid of revolution. \bold{\omega} is the total angular velocity of the potato, which is decomposed into the spin angular velocity \bold{\omega}_s and the angular velocity of precession \bold{\omega}_{pr}. The precession angular velocity is directed along the constant angular momentum vector L. Also shown are the unit vectors e₁ and e₂, which are the the directions of the first and second principal axes at that instant.

Since the potato is assumed to have an axis of symmetry, at any instant we can choose the second and third principal axes such that the angular momentum and angular velocity lie in the plane of the first two principal axes. Let \omega_1 and \omega_2 be the components of the total angular velocity along the 1 and 2 axes, respectively. Then the components of the angular momentum are L_1 = I_1 \omega_1 and L_2 = I_2\omega_2. Then from the diagram

L_2 = L\sin\theta = I_2\omega_2
\omega_2 = \omega_{pr}\sin \theta

from which we get

\omega_{pr} = \frac{L}{I_2}. (*)

But we also have

L_1 = L \cos \theta = I_1 \omega_1 = I_1 (\omega_s + \omega_{pr}\cos\theta). (**)

Now from (*) we get

\omega_{pr} \cos \theta= \frac{L\cos\theta}{I_2}.

and from (**) we get

L \cos \theta =I_1 (\omega_s + \omega_{pr}\cos\theta), which combined with the previous equation gives

\omega_{pr} \cos \theta = \frac{I_1 (\omega_s + \omega_{pr}\cos\theta)}{I_2}

This can be solved to get

\frac{I_2 - I_1}{I_1}\omega_{pr} \cos \theta = \omega_s, which relates the spin rate, precession rate, and precession angle.

For a baton, I₂ >> I₁, so the precession is slow compared to the spin rate.

 

[hkyriazi]

Thanks, Techmologist. Better late than never to see the better-late-than-never answer, too!