(adsbygoogle = window.adsbygoogle || []).push({}); 1. Statement of the problem

Imagine a magical, marching band majorette-type baton floating and twirling in front of you, in the horizontal (x-y) plane, with mass M, length L, width L/30, and angular velocity omega. It need not be magical to float -- we could imagine ourselves to be space station astronauts in orbit about the earth, i.e., in free fall. It is "magical" in that it has an internal mechanism that allows it to pull its arms in, and extend that mass outward in the + and - z directions, so that it expends energy doing what skaters do when they pull in their arms/legs for a fast spin. In this way, it gradually transforms itself from a twirling baton motion to a spinning baton motion. The angular momentum is the same, and the axis of rotation is still the z axis, but now the baton lies along the z axis rather than in the x-y plane.

It's easy to calculate the energy added to effect this transformation, based upon knowledge about the mass distribution, and thus moments of inertia in the longitudinal vs. transverse orientations.

If instead of utilizing this internal mechanism, and internal energy, we effect this transformation externally, via small, frictionless taps (characterized by purely elastic collisions) onto opposite ends of the baton, so as to force it to line up with the z axis, I'm sure the amount of added energy will be the same, because the final state would be the same. Alternatively, we can imagine that we have the twirling baton in a vertically-oriented cylinder whose sides are shrinking in, forcing the baton into a vertical alignment (and we only have to manually tap its ends initially, to get the baton off its perfectly horizontal initial alignment, so that it doesn't get wedged into the collapsing walls). (Remember, there's no friction with the walls, so the baton wouldn't lose energy.)

The question is, what would be the spin and precession rates at the various precession angles (varying from 90 to 0 degrees), expressed as multiples or fractions of the initial omega, as we effect this transformation externally?

For simplicity's sake, treat the baton like a long, thin rod.

2. Relevant equations:

For a rod spinning about its axis, I_{s}= M/2 R^{2}where R=radius

For a thin rod twirling perpendicular to its axis, I_{t}= M/12 L^{2}(These formulae are taken from a table in Halliday and Resnick, BTW.)

Angular momentum = I omega

Angular kinetic energy = 1/2 I (omega)^{2}

3. Initial stab at the problem

Since angular momentum is conserved in this system, we immediately know that the final spin rate, Omega_{s}, is I_{t}/I_{s}* Omega_{t}= M/12 L^{2}/ M/2 R^{2}* Omega_{t}. Since R= 1/2 L/30, R=L/60, and this yields a final spin rate of 1/12 * 2 * 3600 * Omega_{t}= 600 Omega_{t}. So, it'd spin 600 times faster than it was twirling.

In terms of energy, the twirling mode had 1/2 M/12 L^{2}omega_{t}^{2}= M/24 L^{2}omega_{t}^{2}

The spinning mode has: 1/2 (7200ML^{2}) (600*omega_{t})^{2}= 25 ML^{2}omega_{t}^{2}.

So, there's 600 times more energy (25*24) in the spinnning mode than in the twirling.

From a paper by Butikov on torque-free rotation ("Inertial rotation of a rigid body," Eur. J. Phys.,27(4): 913-922 (2006)) (see http://faculty.ifmo.ru/butikov/Precession.pdf) [Broken], we learn that at small precession angles, the precession speed stands in the same ratio to spin rate as I_{s}does to I_{t}, in this case 1 to 600. But since the spin rate at small precession angles is 600 times the original twirling rate, this seems to say that the smallest observable, final precession rate is the same as the initial twirling rate. This, in turn, seems to tell us that the precession rate may be constant all along (being the same at the beginning and the end, so why not all angles in between?), and as the precession angle gets smaller and smaller, the spin rate simply gets faster and faster, such as to keep the angular momentum constant, and utilize the added energy.

If this is true, then all we need to do to calculate the various spin rates is calculate the intermediate I values for the twirling baton, see how much angular momentum that accounts for, and put all the rest of the momentum into the spinning.

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# Homework Help: A simple rotational problem

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