Homework Help: A simple rotational problem

1. Jan 5, 2007

hkyriazi

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, Is = M/2 R2 where R=radius

For a thin rod twirling perpendicular to its axis, It = M/12 L2 (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, Omegas, is It/Is * Omegat = M/12 L2 / M/2 R2 * Omegat. Since R= 1/2 L/30, R=L/60, and this yields a final spin rate of 1/12 * 2 * 3600 * Omegat = 600 Omegat. So, it'd spin 600 times faster than it was twirling.

In terms of energy, the twirling mode had 1/2 M/12 L2 omegat2 = M/24 L2 omegat2

The spinning mode has: 1/2 (7200ML2) (600*omegat)2 = 25 ML2 omegat2.

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 Is does to It, 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.

Last edited by a moderator: May 2, 2017
2. Jan 8, 2007

hkyriazi

off-line discussion with Dr. Butikov

Since no one has responded, I'll post some exchanges I've had with Dr. Eugene Butikov, to see if I can get some thoughts from folks here on our differing viewpoints. With him, I used the example of a needle rather than a baton, and in the exchange below, focused on the alternative way (shrinking cylinder) of effecting the baton's re-orientation.

My recent post to Dr. Butikov:

And, instead of tapping it manually, we can simply have the needle initially twirling slightly inclined to the horizontal x-y plane. Certainly if we restrict the twirling needle's motion in this way, by a shrinking, vertical cylinder, it will eventually line up vertically, and I assume that, without friction in the collisions with the walls, it'd maintain its vertically oriented angular momentum, and thus would be spinning about its major axis.

Are you saying that this procedure will not result in any spinning of the needle, just faster twirling? Again, I'm going on intuition here, but I imagine that there'd be both twirling and spinning, and if we removed the restricting cylinder at any time, the needle would continue its same motion, just as if it were a spinning, precessing top (albeit a very long and skinny one).

Assuming we're agreed on that (which I'm not at all sure of), the question is, does the needle precess (twirl) at the same rate during the entire procedure? Certainly if there were friction, and the needle were rolling along the inside of the cylinder, it'd make each orbit faster and faster as the walls close in. Or, maybe I've got it backwards, and only if there *is* friction would there be such rolling, which would be observed as spinning. Now I'm really getting confused!

Getting back to my original example, however, another line of reasoning has occurred to me. Let's consider the ordinary case of a top spinning and precessing on a hard surface, under the influence of gravity. Can we not think of the combined forces of gravity and the compression present on the hard surface underneath the top's bottom point, as a constant series of forceful "taps" on a spinning and precessing top, directed along the z-axis? Could we not reverse the direction of these forces, and begin with the top twirling on its side, and gradually get it to spin upright? That seems to be analogous to the original case, involving the taps I imagined making on our otherwise torque-free, twirling needle.

One more thought. I know that when I spin a smooth-surfaced pen out from between my fingers, so that it spins at high speed about its axis, and give it a slight jerk when I let it go up in the air, it precesses, even though I only gave it a jerk in the downward direction (as I threw it upward). Are not the taps I'm imagining analogous to that? If so, why would the needle not also precess?

Dr. Butikov's response:

Your intuition fails you in your discussion of the gedanken experiment with the twirling baton inside a vertically-oriented cylinder whose sides are shrinking in. Any torque-free motion of a symmetrical top can be represented as a superposition of the axial spinning and the regular precession of the axis in space. (This situation is described in detail in my paper which you have mentioned.) However, the constrained motion of the baton inside a cylinder (whose sides are shrinking) is certainly not a torque-free rotation. In this constrained motion neither the total angular momentum nor the energy of the baton is conserved (even in the absence of friction). Hence my program (which simulates the torque-free rotation) has nothing to do with this situation.

Indeed, the ends of the baton inside a cylinder are subjected to the pair of reaction forces exerted by the cylinder surface. In the absence of friction, these forces are directed normally to the surface (toward the axis of the cylinder). When the baton twirling inside a vertically-oriented cylinder makes an angle with respect to the cylinder axis (that is, the baton long axis is describing a circular cone), the total angular momentum vector, being orthogonal to the baton long axis, also moves in a cone. This variation with time of the total angular momentum vector is provided by the above mentioned rotating pair of external reaction forces exerted on the baton by the cylinder surface.

You write: “If we removed the restricting cylinder at any time, the needle would continue its same motion, just as if it were a spinning, precessing top.” – No, this is not the case. After you remove the cylinder, the angular momentum vector immediately stops to move along a cone. The further torque-free motion of the baton will occur with a conserving total angular momentum. This motion will be just the uniform rotation about an axis fixed in space whose direction coincides with the total angular momentum vector (this vector is the total angular momentum at the moment when the cylinder was removed, and preserves its direction further on). And this fixed in space axis of rotation is perpendicular to the baton – the further motion of the baton is again a “twirling” about this fixed axis. This motion cannot be regarded as a precession of a spinning top.

When the walls of the cylinder are forced to shrink (to move slowly inward), the external reaction forces do some work on the baton, thus increasing the energy of the baton. Simultaneously, the torque of these forces increases the horizontal component of the (rotating along a cone) angular momentum vector (whose vertical component remains unchanged). The angular velocity of the baton (directed along the vertical) also increases. In the idealized frictionless model of a system of extremely thin needle inside a cylinder, the energy and total angular momentum increase indefinitely as the radius of the enveloping cylinder tends to zero. However, this infinite increment is not related to the axial spin of the needle. It is related to the constrained conical motion of the needle long axis. The infinite growth of the total angular momentum (of its horizontal component) occurs when the vertex angle of this cone tends to zero.

My response to Dr. Butikov:

So, I gather that you're saying that the needle would not begin to spin inside the constricting cylinder, but would merely slide along on its inner suface, and that if that surface is removed, the needle would immediately go back to a pure twirling state, just as it exhibited in the beginning. What I don't understand, however, is how, in the extreme case where the cylinder has constricted completely, resulting in no observable translational (cone-type) movement of the needle, the initial angular momentum will have disappeared. If it is to be conserved, I would think that the needle must now be spinning.

Perhaps we're running into difficulty with the fiction of frictionless contact, and infinities relating to an infinitely thin needle (which was not my example, however)-- for example, you say the horizontal component of total angular momentum goes to infinity as the vertex of the cone is made to go to zero. We seem to be saying that it'd take infinite energy to reduce that angle to zero. In any case, even with a finite diameter needle, I believe you're saying that no spinning of the needle would occur at any time during the cylinder's contraction.

In order to help train my intuition, let me ask you what would happen to a top spinning and precessing on a surface, under gravity, if, for example, it were on a 747 plane that went into a power dive, such that the plane's contents experienced zero gravity. Let's say that the surface the top is spinning on is attached to the plane, and the initial dive causes the top to rise up off the surface and spin freely in space. This would then correspond to your simulation, and paper, about torque-free rotation. The question is, would the top continue to precess and spin? My intuition tells me that it would, although it would now precess exactly about its center of mass.