PHY6200 W07

Chapter 10: Rotational Motion of Rigid Bodies


In Taylor, read sections 10.7 to 10.8 for today, and 6.1 to 6.2 for Friday.


Euler's equation

Ldot + ω×L = Γ.

This is Euler's equation in component form.

λ1ω1dot - (λ2 - λ32ω3 = Γ1
λ2ω2dot - (λ3 - λ13ω1 = Γ2
λ3ω3dot - (λ1 - λ21ω2 = Γ3

This is a set of coupled, nonlinear differential equations. We will attempt to solve these for some special instances only. As an example, when the object has an axis of symmetry, two of the principal moments are the same, simplifying one of the equations. If we call the 3-axis the axis of symmetry, then λ1 = λ2, and the third Euler equation becomes

λ3ω3dot = Γ3

If the torque Γ3 is known, then this equation can be solved, and the result substituted into the other equations. This is how to proceed with the homework problem with the space station (#5).

Euler's Equations with Zero Torque

A common occurance, especially for space based problems, is an object is subject to zero torque whose motion we'd like to find. This applies to projectiles as well as satellites, and asteroids. With zero torque, Euler's Equations become

λ1ω1dot - (λ2 - λ32ω3 = 0
λ2ω2dot - (λ3 - λ13ω1 = 0
λ3ω3dot - (λ1 - λ21ω2 = 0

These are still coupled, non-linear equations, but we can make some headway with them.

A Body with Three Different Principal Moments

Start with the case that the principal moments are all different. Because of the definition of principal axes, we know that in the case the torque is zero, if the object is spinning about one of the principal axes, then L is parallel to ω and will remain that way. If the object spins about the 2-axis, ω = ωe2, then L = λ2ωe2. This is true in the rotating (body) frame, but it is also true in the space frame since

(dL/dt)space = (dL/dt)body + ω×L = 0

where the derivative on the right is zero because L is constant in the body frame, and the cross product is zero because L is parallel to ω. Let's see if this result emerges from Euler's Equations.

If ω1 and ω2 are initially zero, then the Euler Equations read

λ1ω1dot = 0
λ2ω2dot = 0
λ3ω3dot = 0

indicating that all three components of ω are constant, and must equal their initial values ω = (0, ω, 0). This agrees with the above argument.

If the object doesn't initially rotate around one of the principal axes, then ω is not constant. This arises out of Euler's Equations because, at least two components of ω must be non-zero, so one of the time derivatives is not zero. This will occur for the third component, so if it was initially zero, it won't remain zero. For instance, if ω2 and ω3 are non-zero, then the first equation will yield ω1dot &neq; 0. So, once two of its components are non-zero, then ω is not constant.

We aren't seeking a general solution to this situation. But it is interesting to inquire into the stability of the rotation when initially, two or more components of ω are non-zero. This applies, for instance, to the simple example of a rectangular block with three different dimensions. If it is tossed in the air, rotating mostly about one axis, but with some unavoidable rotation about the other axes, will it continue to rotate mostly about one axis, with some small wobble, or will it tumble erratically?

To answer this question, let's set up the problem with initial angular velocity mostly in the ω3 direction, and small components in ω1 and ω2. Now the Euler equation for the 3 component says that ω3dot is small, or ω3 is approximately constant (at least initially). Treating ω3 as constant, the other two equations become

λ1ω1dot = [(λ2 - λ33] ω2
λ2ω2dot = [(λ3 - λ13] ω1

To get the result we want, differentiate the first equation with respect to time, treating ω3 as constant, and substitute the second equation for ω2dot:

λ1ω1ddot = -[(λ3 - λ2)(λ3 - λ13² / λ1λ2] ω1

If the quantity in square brackets is positive, then ω1 is a constant times a cosine, that is, it varies, but remains bounded. This is the condition for stable rotation. The quantity in brackets is positive if λ3 is bigger than both λ1 and λ2, or smaller than both.

If λ3 is intermediate between the other principal moments, then the quantity in square brackets is negative, and the expression for ω1 is an exponential function of time. This is the unstable case. The rate of rotation about the 1-axis increases until it is no longer small and our approximation breaks down.

Of course, the indices 1, 2, and 3 can be permuted and the same result holds for each of the axes. The result is that an object with three different principal moments can rotate steadily along the axes with the smallest or largest principal moments, but is unstable when rotated about the axis of intermediate principal moment. That is: rotation about the axes of smallest or largest principal moment is stable against small perturbations about the other axes; rotation about the axis of intermediate principal moment is unstable against small perturbation about the other axes. This is an interesting result that you are probably familiar with, but likely haven't given much thought to.

Motion of a Body with Two Equal Moments: Free Precession

We will seek the general solution for the rotation of a body with two equal moments, in the absence of torque. This is the case for an object with axial symmetry such as a top, although axial symmetry is not required (for example, a system of two objects of mass m located at (±2a, 0, 0), and two objects of mass 4m located at (0, ±a, 0) lacks axial symmetry, but has equal moments Ixx and Iyy).

Let the first two moments be equal, λ1 = λ2 = λ. Now the third Euler equation reduces to ω3dot = 0 resulting in ω3 = ω30 = constant. Treating ω3 as a constant, the first two equations can be rewritten in the form

ω1dot = [ω30(λ - λ3)/λ]ω2 = Ωbω2
ω2dot = [-ω30(λ - λ3)/]ω1 = -Ωbω1


Ωb = ω30(λ - λ3)/λ.

The subscript "b" stands for body. We now have a pair of coupled, linear, first order differential equations. They can be solved with a technique we've used before. Let η = ω1 + iω2, with ηdot = ω1dot + iω2dot. Substituting for ω1dot and ω2dot we find

ηdot = Ωb2 -iω1) = -iΩbη

This is a first order, linear differential equation in a complex variable, and the solution is

η = η0e-iΩbt

In the most general case, the object has an initial angular velocity with a component along the 3-axis and a component perpendicular to the 3-axis. For convenience, let us say that at t=0, the component perpendicular to the 3-axis is in the direction of the 1-axis and call it ω10, that is ω(t=0) = (ω10, 0, ω30). Then at t=0, η0 = ω10, and

ω(t) = (ω10 cos(Ωbt), -ω10 sin(Ωbt), ω30)

Remember, this result is for the body frame. It says that ω is of constant magnitude ω = sqrt(ω10² + ω30²), and rotates around the 3-axis forming a cone shape, with the angular frequency Ωb. This cone shape is called a body cone.

The angular momentum in the body frame is

L(t) = (λω1, λω2, λ3ω3)
L(t) = (λω10 cos(Ωbt), -λω10 sin(Ωbt), λ3ω30).

It is easy to see that the magnitude of L is constant, L = sqrt(λ²ω10² + λ3²ω30²), and rotates around the 3-axis forming a cone shape, with angular frequency Ωb, just like ω. The dot product of L and ω is also constant, Lω = λω10² + λ3ω30², meaning that the angle between L and ω is constant. Finally, the cross products e3×L = λω10(sin(Ωbt)e1 + cos(Ωbt)e2) and e3×ω = ω10(sin(Ωbt)e1 + cos(Ωbt)e2) are parallel. Therefore, the three vectors L, ω and e3 lie in the same plane.

Motion in the space frame

In this case, since there are no external torques, we know that L is constant in an inertial frame. (In general, we use the relation (dL/dt)space = (dL/dt)body + ω×L to solve for the behavior of L in the space frame. In this case, (dL/dt)body = λω10Ωb (-sin(Ωbt), -cos(Ωbt), 0), and ω×L = [ω10ω30(λ - λ3)](sin(Ωbt), cos(Ωbt), 0). Substitution of the definition of Ωb reveals that these are negatives, and we obtain the above result that (dL/dt)space = 0, or that L is constant in the space frame.)

The relative geometry (angles and relative directions) of vectors don't change when moving from the body frame to the space frame. That is, the relative positions of L, ω and e3 don't change. The still lie in a plane and are separated by constant angles. But, while e3 is fixed in the body frame, in the space frame, L is constant. Therefore, ω and e3 precess around L in the space frame. The rate of precession is the topic of problem 10.46, and can be shown to be

Ωs = L/λ1.

This motion is called free precession. Recall that there is no torque in this problem; the precession arises from the complex dynamics of rotational motion.

We will leave rotational motion for now to study an alternative approach to mechanics. Afterward we'll return to finish the last two sections of this chapter using Lagrangian mechanics.

© 2007 Robert Harr