# Can a gyroscope under no external force precess?

1. Dec 8, 2006

### hkyriazi

We know intuitively that a gyroscope in zero gravity (or free fall) can spin almost indefinitely with either a zero or 90 degree precession angle (i.e., spinning perfectly along its axis or along its "side," respectively). Why shouldn't it be able to spin at intermediate states, such as with a precession angle of 45 degrees?

If the answer is yes (which I think it is, based not only on the above intuition, but from observing what a rapidly spinning, smooth-surfaced ballpoint pen does after I flick it from between my fingers and before it hits the ground), does anyone here know of an equation that relates the spinning object's distribution of mass (moment of inertia), spin rate, precession rate, and precession angle?

2. Dec 9, 2006

### daniel_i_l

how can a gyroscope "precess" with no external forces? in order for it to precess something has to try to rotate its axes of rotation.

3. Dec 9, 2006

### vanesch

Staff Emeritus
Yes, the axi-symmetric top precesses unless it is rotating around one of its principal axes. What needs to be constant is the angular momentum (given that no torque is working on it). But the angular momentum is given by L = I.w, where I is the tensor of the moments of inertia and w is the rotation velocity (angular velocity vector). L and w are hence not in the same direction in all generality. But L needs to remain constant, so w needs to change, and w turns out to rotate around the axis of L. That is precession.

EDIT: see http://en.wikipedia.org/wiki/Euler's_equations

Last edited: Dec 9, 2006
4. Dec 9, 2006

### hkyriazi

Thanks, Vanesch, for pointing me in the right direction. The wikipedia site, which I'd visited many times (shame on me for not finding what I needed!), does make a mention on that "Euler's Equations" page of "torque-free precession," which immediately answers the question posed here in my subject line.

I'll see if I can figure out how to apply Euler's equations to the specific object I had in mind, which is somewhat like the batons majorettes twirl in marching bands (broadened at the ends, and thinner in the middle).

It seems there should be only two major axes for something cylindrically symmetric - along the baton, and anywhere in the plane at 90 degrees to it, so I'll only have two principal moments of inertia, and two equations. Or do I need to include both the x and y axes separately, even though they're indistinguishable? Given that my math skills are minimal to nonexistent, the safest course will be to include all three axes so I can use the equations as given, and see what I come up with (I can make omega sub-2 and 3 equal, and I sub-2 and 3 equal.)

Thanks again!

5. Dec 9, 2006

### pervect

Staff Emeritus
For a textbook discussion of Euler's equations, you can always try Goldstein's "Classical mechanics", but it sounds like you're well on your way.

6. Dec 12, 2006

### hkyriazi

help needed with Euler's Equations

Thanks, Pervect, for the Goldstein reference. I checked it out of our library but, sad to say, I'm afraid that, acting on my own, I'd have to read and comprehend everything prior to the discussion of Euler's Equations, as well as brush up on my matrix algebra and calculus, to be able to understand and apply them correctly.

To try to bypass that time-consuming unpleasantness for the time being (it seems that this problem should have a relatively simple solution), let me pose a few questions about the torque-free equations, and their relation to my specific problem of a baton going from perfect twirling mode (90 deg precession angle) to perfect spinning mode (0 deg precession angle).

First of all, what puzzles me a bit is that in both extremes, the angular momentum is defined completely by the z-axis (no movement occurs in the z direction for something spinning perfectly about the z-axis), but for the in-between modes (where there is real precession), the spinning motion (though not the precession) does produce movement of the body's mass in the z direction (equal amounts + and -). Does this come from the series of frictionless taps to the opposite sides of the body's ends that I'm imagining to cause the change in its motion? If so, how does it disappear at the end of the exercise (when we go from one extreme to the next, and where, I assume, all the angular momentum comes from one or the other of spinning vs. twirling)? Is this related to the phenomenon of holding a gyroscope while sitting on a chair that can spin freely, turning the gyroscope through 90 degrees a certain way, and causing the chair to spin?

Secondly, and more importantly, the first of the equations with the right-hand side set to zero (see the wikipedia page on Euler's Equations) contains the terms I(sub-1) Omega dot(sub-1), the latter of which if my memory serves me correctly, is a rotational acceleration (time rate of change of angular velocity). Is this correct? And since, for this particular object under no torque, the angular speeds are unchanging, does this change in velocity simply specify the rate of precession (the time rate of change of the positioning of the baton's main axis)? If so, what the heck are omega-2 and omega-3 referring to? Movements in the x and y axes for a spinning top whose principal axis is not set to z?

Frankly, I'm at a loss as to how even to set the problem up. For example, I don't know whether I'm free to set omega-1 as the spin axis and omega-2 as the (same) twirling axis, or whether I should just work with the angular momentum equation, L=w(1)I(1) + w(2)I(2) + w(3)I(3). (I was trying to do the latter, when the above-mentioned "in-between mode" motion along the z-axis confused me.)

Also, I've seen elsewhere that a top's rate of spin and precession are inversely proportional, but that would necessitate dividing by zero in the extreme cases I'm interested in, which seems wrong. So, I was guessing (intuiting) that the spin and precession rates would vary inversely sinusoidally between zero and 90 degrees.

Sorry for blathering on this way.

Any help would be greatly appreciated. (This isn't for any coursework, BTW.)

7. Dec 13, 2006

### vanesch

Staff Emeritus
Be careful with the use of the Euler equations: the $$\omega_1$$ and similar components are the components of the rotation velocity vector on axes fixed to the rigid body: so the reference frame in which these components are expressed is rotating. The rotation velocity vector itself is the vector (in an inertial frame!) which describes the rotationary motion of the rigid body wrt to this inertial frame. Consider that vector $$\bf{\omega}$$, but as a vector (and don't think of it as 3 components in the inertial frame). Now, consider the three unit vectors, fixed to the rigid body, $$\bf{e_1}$$, $$\bf{e_2}$$ and $$\bf{e_3}$$. The components like $$\omega_1$$ which enter into the Euler equations, are simply the in products $$\omega_1 = \bf{\omega . e_1}$$ etc...
where both the vector $$\bf{\omega}$$ and the vector $$\bf{e_1}$$ are evolving in time.

The vector L is fixed (and has hence fixed components in the inertial frame, but not in the rotating body frame). Given that in the body frame (if we choose the directions of e1, e2 and e3 along principal axes), the components of L are given by, respectively $$L_1 = I_1 \omega_1$$ etc, and given that L1 will evolve in time ($$L_1 = L.e_1$$ where L is constant, and e_1 is rotating), $$\omega_1$$ will also evolve in time.
This evolution is exactly what is prescribed by the (torque-less) Euler equations. It is pretty difficult (at least for me) to devellop some intuition for what exactly happens because there are actually two phenomena at work: first of all, because of the rotation of the frame, the L1, L2 and L3 components will vary (fixed vector, projected onto rotating axes).
Next, these L1, L2, and L3 components will fix, through the L1 = I1.w1 equation, the components of the omega vector in the rotating frame. However, because of the different weights (I1, I2 and I3), these components of omega will NOT be the components of a fixed vector on rotating axes: the deficit (because of the different values of I1,I2, and I3) will genuinly make the omega vector change in the fixed reference frame too. This change of the omega vector in the fixed reference frame is called precession.

Last edited: Dec 13, 2006
8. Dec 13, 2006

### hkyriazi

Thanks, Vanesch. That clue is helpful. (I've been pretty clueless so far.) But, man, you're right, this is confusing! When you say that L is fixed in the inertial frame, I suppose you mean it's not zero, but invariant. This seems to be the natural way to view a spinning top in a stationary world - one gives it some angular momentum in a certain direction, and that doesn't change (absent torque), although the top itself changes position.

Next, we assign the body axes (and unit vectors e1-3) according to the object's geometry (in this case, an axially symmetric, baton-like object). Fine. But what's the advantage of having this chosen coordinate system rotate against the inertial one (I assume in a way that makes L there zero)? It sure doesn't help me to picture what's going on any better! Well, let me think out loud for a moment, and see if anything occurs to me. Doing so for a perfectly spinning baton allows me to picture the baton as stationary. No advantage there. Same for one twirling perfectly. But if it's both spinning and twirling (the latter being synonymous with the various degrees of precession), then for sure I'd see some movement of the object with respect to its reference frame. But if I'm spinning along with it, I'm having difficulty picturing how it would behave. If I guessed correctly that the body reference frame is made to rotate such that L becomes zero, it would seem that I'd be spinning faster than the object, and it'd seem to spin and precess backwards a bit...or something like that.

Should I be focusing on understanding Poinsot's Construction, with its discussion of polhodes and herpelhodes, etc.?

Last edited: Dec 13, 2006
9. Dec 15, 2006

### hkyriazi

solutions to Euler's equations for a top under zero torque?

I found a derivation of a solution to the problem I'd posed, but I'm not quite sure how to interpret the solutions. It's from p. 98 of a book called "Cats, Atoms, Gyrons, Aether, and the Universe," by Frank M. Meno (2000).

From Herbert Goldstein's "Classical Physics" (2nd edition), equation 5-40 gives (where I'm using w for omega), for the zero torque condition:

I1(dw1/dt)=(I2-I3)w2w3

I2(dw2/dt)=(I3-I1)w3w1

I3(dw3/dt)=(I1-I2)w1w2

Since we're working with an object with axial symmetry, we can substitute I1 for I2 to yield:

I1(dw1/dt)=(I1-I3)w2w3

I1(dw2/dt)=(I3-I1)w3w1

I3(dw3/dt)=0

Thus, the spin axis is along w3, and its rate is unchanging as the top precesses.

Here it gets interesting. He differentiates the first equation with respect to time, and combines the result with the second equation, to yield the following 2nd order differential equation:

d2w1/dt2+w2(1-I3/I1)2w1=0

I'm OK till this point. Then he says this leads to the suitable solutions:

w1=+w0cos[(1-I3/I1)w3t]

w2=-w0sin[(1-I3/I1)w3t]

My question is, what are these omega noughts?

10. Dec 15, 2006

### vanesch

Staff Emeritus
An integration constant which depends on the initial conditions.

11. Dec 15, 2006

### vanesch

Staff Emeritus
No,no, not at all ! The vector L is the SAME vector as it was in the inertial frame (in other words, you're not supposed to calculate rxp in the rotating frame, but only in the inertial frame) ; only its components are now written in the axes of the rotating frame.
The same goes for w. What is confusing in this business is that the defining equation for the vectors w and L are to be considered only in the inertial frame, but that we write their components in the rotating frame (but not using their defining equations! Indeed, the body being at rest wrt to the rotating frame, w and L would be 0 if we did that!)

Last edited: Dec 15, 2006
12. Dec 16, 2006

### hkyriazi

relating delta phi to spin and precession speeds

Thanks, Vanesch. But after struggling with these equations a bit, and actually thinking about what they mean (I decided that the Omega noughts mentioned above must be the maximum rates, which are multiplied by the sine and cosine functions, which vary between 0 and 1, and that are a function of the initial conditions, as you mentioned), I've decided that they're really not what I'm after after all.

I need an expression relating changes in the precession angle, phi, to the rates of rotation and precession (given a certain angular momentum). (What I want to know is how fast it'll precess at all angles between 0 and 90, given the same angular momentum.) Since the moments of inertia depend on phi, I was able to differentiate the total angular momentum with respect to phi (which is zero), to yield the following expression (where the subscripts s and p refer to spin and precession, respectively, as the mass and object length terms cancelled out, and C is a constant relating the torques of the two major axes):

CWsd(cos2(phi))/d(phi)=-Wpd(sin2(phi))/d(phi)

I'll have to re-check my math, and then see if I can take it any further.

13. Dec 17, 2006

### hkyriazi

Oops! d(omega)/d(phi) not equal to zero

My immediately previous equation was wrong, as I'd overlooked the fact that the various omegas also change with changes in phi (precession angle). (In deriving that previous expression, I set all their derivatives with respect to phi at zero.)

I find it surprising that there's no general solution to the problem of how a torque-free spinning object (e.g., a majorette's baton), of known mass distribution and shape, behaves as one frictionlessly taps it repeatedly to convert its motion from precessing to spinning perfectly (or vice-versa). Specifically, let's say we tap the ends of a horizontally twirling baton (x-y plane) with impacts solely in the +/- z-axis direction, and get it to precess at an 85o angle instead of its initial 90o. What is the new (smaller) twirling rate, and what is the new (non-zero) spin rate about the baton's main axis?

The angular momentum would seem to be conserved (none of the re-orienting taps occur in the plane of rotation), although I'm pretty sure that one must add energy, i.e., the more rapidly spinning object has much greater kinetic energy, just as spinning skaters must do work to pull their arms and legs in, resulting in greater energy in that state.

With knowledge about how changes in phi affect some of the individual components of angular momentum (moments of inertia, but not yet the omegas), the spin rates of the 90o and 0o precession states, and also knowledge about the kinetic energy in these two states, one would think we have enough info to solve the problem about intermediate states.

I'll keep working on it, but any help from people more experienced in this area would be appreciated.

14. Dec 17, 2006

### vanesch

Staff Emeritus
A torque doesn't need to be in the plane of rotation in order to change the angular momentum ! Only forces that go through the center of mass wouldn't result in a torque (and hence in a change in angular momentum).

15. Dec 18, 2006

### hkyriazi

Hmmmm... Embarassingly, I hadn't even thought of my reorienting taps as being torques, but they certainly are an application of force. I envision them as being frictionless and the impacts as taking zero time. I do understand that a torque applied to the center of mass would result only in translational acceleration (absent constraints).

But let me see if I understand. If a spinning top is under the influence of gravity (which may be thought of as a series of infinitessimal downward taps), if the top is always perfectly straight, it'd never fall over, even when it stops spinning. Of course that's never the case, and any real top will precess a bit (even if imperceptibly), and so gravity will gradually accentuate that tilt, and it'll therefore have some component in the plane of rotation (although most of the force will be applied perpendicular to that precessing plane).

But if there were no friction, the top wouldn't slow down, and in this case, there'd be no change resulting from gravity, right? In other words, however much the top was precessing, it'd continue in that state forever, and the total angular momentum would remain constant despite the constant torque. Am I right in thinking that it is only the friction-induced slowing of the spin rate that allows gravity to produce the gradually increasing angle of precession, and greater precessional speed? Or, conversely, would gravity continue to transform the top's movement even in the absence of surface and air friction, until it ended with the top spinning completely on its side (precessing at 90 deg), with the same total angular momentum, but now in a more stable orientation?

16. Dec 18, 2006

### vanesch

Staff Emeritus
Up to now we were only considering a torque-free gyroscope (meaning: not fixed at any other point than its COG). If the gyroscope is suspended at its COG, then the force of gravity plays no role.

The problem with a top, spinning around a fixed point which is not its COG, and moreover on which the force of gravity acts, is a totally different problem!

Nope, that's not possible: a constant torque will have a constant derivative of the angular momentum: dL/dt = torque (equation valid in an inertial frame).
Moreover, the torque will not be constant, because "bound" to the spinning body, which will be moving around, and hence the torque will also spin around.

The solution of the spinning heavy top is known and treated in any serious book on mechanics. For instance, in Goldstein, there's an entire chapter on it (chapter 5 in the second edition which I have). The solution is quite involved and uses elliptical functions.

BTW, the Euler equations are valid in two different cases:
1) axes fixed to the body, at the COG
2) axes fixed to the body, at a fixed point in an inertial frame

In both cases, one has to fill in the total torque working around the center of the system of axes.

Last edited: Dec 18, 2006
17. Dec 19, 2006

### vanesch

Staff Emeritus
I should maybe add something. The Euler equations are the result of the transformation of the equation:
$$\frac{d L}{dt}_{inertial} = \tau$$
where L and $$\tau$$ are calculated in an inertial frame, wrt to a certain fixed point in that frame, O,into a frame which is rotating around an axis which goes through the point O.

In the above expression, both L and $$\tau$$ are vectors, and not to be thought of as 3-tuples. We can hence write down the components of that equation in any coordinate frame: it doesn't need to be an inertial frame. However, there's one thing that goes wrong: that is the derivative to time. The derivative to time of a vector depends on the coordinate system in which we apply it, because we derive the components. Hence any time-dependence of the coordinate frame in which we write down the components will influence the time derivative.

Now, we want to express everything in components of the rotating frame (fixed to the body). The reason is that it is in this frame that there is a simple relation between the angular momentum and the rotation vector, namely through $$L = I \omega$$ where I can be a constant matrix (preferably even diagonal): the inertia tensor of the body expressed in axes which are fixed to the body. However, if we are going to take time derivatives of the components in this frame, we have to take into account the rotation, because the relation $$\frac{d L}{dt}_{inertial} = \tau$$ is only valid in an inertial frame.
It is to correct for the effect of a rotating frame on the time derivative that we have to add a term:
$$\frac{dA}{dt}_{inertial} = \frac{dA}{dt}_{rotating} + \omega \times A$$

Working this out gives you the Euler equations, because we now have everything (including the time derivative) expressed wrt to the rotating frame. It is also because we have limited our expression for the time derivative to a rotating frame without translation, that we cannot apply the result (the Euler equations) around just any point: we need the origin of the rotating frame to be fixed in an inertial frame.
Note that because there isn't any time derivative in the expression for the torque, that you can just as well determine it as seen from the rotating frame as when you look at it in the inertial frame and then project it on the rotating axes. However, we need to work with the angular momentum and the torque around a point which is fixed in an inertial frame.

For a free body, the COG is a point which is fixed in an inertial frame, so we can use it there (with then of course also the tensor of moments of inertia expressed around the COG).
For a body with a point restricted to a fixed position (like the spinning top), then it is THIS point which is fixed in an inertial frame, and hence we have to write out the Euler equations (with the tensor of moments of inertia) calculated around this point, because now the COG is doing weird movements and is not fixed in an inertial frame anymore. We also have to calculate all torques wrt to this point now. Overall gravity will act upon the COG, and hence have a non-zero torque. The right-hand side of the Euler equations will hence not be 0.

Last edited: Dec 19, 2006
18. Dec 28, 2006

### hkyriazi

Thanks, Vanesch. I still have much work to do to answer my question about the relationship between precession rate, precession angle, and spin rate.

In the meantime, let me try to understand your 12-18 post. I understand that the top spinning under gravity, on a point (and not rotating about its center of mass) is a very different problem from one spinning and precessing freely in zero gravity. But, under gravity, is it still not the case that, absent air and surface contact friction, the top would spin forever? (I understand that the angular momentum and torque constantly change as the top precesses, but after each precession, I believe they will have returned to their previous values.)

Somewhat relatedly, is there anything to be gained by wondering what would happen to a top spinning normally (i.e., under gravity, on its tip), that is carried aloft on a 747, which then plunges to simulate zero gravity? I imagine the top would go from precessing about its tip, to precessing about its center of mass, while continuing to spin about its axis.

19. Dec 29, 2006

### vanesch

Staff Emeritus
The motion of a "symmetrical top in gravity" has been solved in closed form ; the solutions are not necessarily closed orbits. What one can say is that the motion is controlled by two periods, and it is only when these two periods are commensurate that the orbit will be closed (and that there will be an overall periodic motion).
The motion of the "asymmetrical top" in gravity is an more complicated problem, but as you point out, in general, the top will never come to a stand still. According to Arnold (mathematical methods of classical mechanics, second edition), there is no known closed form solution to this problem (but it can be studied numerically, or in certain approximations).

If the top is considered fixed at its tip, then you should continue to consider it fixed at its tip. However, what will happen is that the effective force of gravity will diminish: as such, you would have a time-dependent torque which tends to zero. This can be smoothly going, or abrupt, depending on how the airplane plunges. The final motion will be the one of a free top, spinning around its tip (free Euler equations, using the inertia tensor around the tip, and not around the COG).

20. Dec 29, 2006

### hkyriazi

Thanks again, Vanesch. I must ask, embarrassingly, what it means for the motion to be "solved in closed form." Given your subsequent usage, it seems that it may mean it yields an analytical solution, rather than requiring a numerical/computational one.

Also, I'm not sure what a "closed orbit" is. Does it have to do with something I once read about, called nutation, that happens in addition to the precession? (I just found an article on the web about it. I'll read that and see if it adds any insights.)

I'm glad that at least some of my intuitions hold (e.g., a spinning top with no friction would continue forever).

I was imagining the top in the 747 to not be fixed except by gravity, and that it'd float off the surface it was spinning on when the 747 goes into its plunge. I'm imagining that it'd continue its motion almost as it was before, and that its tip would trace out circles as it precesses. (If there were no air friction, I imagine the circle would stay the same size, i.e., its extent of precession wouldn't change.)