PHY5210 W08

Chapter 10: Rotational Motion of Rigid Bodies


In Taylor, read sections 10.7 to 10.8 for today, and 10.7 to 10.8 for Wednesday.


( I - λ )ω = 0.
det(I - λ1) = 0.

Principal Moments and Axes for a Cube Rotating About a Corner

Determine the principal moments and axes for a cube rotating about a corner.

The inertia tensor for a cube with sides of length a oriented with its sides parallel to the x, y, and z axes is given in Taylor, and is

[ 8 -3 -3 ]
I = (1/12) Ma² [ -3 8 -3 ]
[ -3 -3 8 ]

First, find the principal moments by solving the equation

[ 8 - λ' -3 -3 ]
det(I - λ) = 0 = (1/12) Ma² det [ -3 8 - λ' -3 ]
[ -3 -3 8 - λ' ]

where I've used λ = (1/12)Ma² λ' to simplify the expression somewhat. The constant outside the determinant plays no role in finding the zeros. To take the determinant, I will expand about the entries in the first row:

(8 - λ')((8 - λ')² - 3²) - 9(8 - λ' + 3) - 9(8 - λ' + 3) = 0

This can be simplified like:

(8 - λ')(8 - λ' - 3)(8 - λ' + 3) - 18(11 - λ') = (11 - λ')[(8 - λ)(5 - λ') - 18] = 0

and expanding the terms in square brackets

(11 - λ')[22 -13λ' + λ'²] = (11 - λ')² (2 - λ') = 0.

Therefore, the principal moments are λ1 = λ2 = (11/12) Ma² (a double root), and λ3 = (1/6) Ma².

Determine the principal axes by solving ( I - λi ) ei = 0. Beginning with e3 we have

[ 6 -3 -3 ] [ e3x ]
I = (1/12) Ma² [ -3 6 -3 ] [ e3y ] = 0
[ -3 -3 6 ] [ e3z ]

This has the solution e3 = (1/√3)(1, 1, 1), where I've normalized the vector to unit length. This direction corresponds to a body diagonal, a diagonal from the corner the cube is rotating about, to the opposite corner, passing through the middle of the cube.

For e1 and e2 we have

[ -3 -3 -3 ] [ e1x ]
I = (1/12) Ma² [ -3 -3 -3 ] [ e1y ] = 0
[ -3 -3 -3 ] [ e1z ]

where e1 can also be e2. This has an infinite number of solutions for e1 and e2, which can be any pair of orthogonal directions in a plane perpendicular to e3. For instance, we can take e2 = (1/√2)(1, -1, 0), and then find e1 from e1 = e2 × e3 = (1/√6)(-1, -1, 2).

Precession of a Top due to a Weak Torque

To get a flavor of what's involved in solving rotation problems, consider the case of a rotating top subject to a weak torque. The top has axial symmetry, and calling e3 the symmetry axis, with e1 and e2 perpendicular to it and each other, we know that the inertia tensor is diagonal with principal moments λ1 = λ2 = λ, and λ3.

Suppose the top begins spinning about its symmetry axis. In the absence of gravity, there is no torque, and it will continue to spin that way with L = λ3ω = λ3ωe3. With no torque, the angular momentum will remain constant, equal to this value.

Now turn on gravity, supposing that it causes a weak torque Γ = R×Mg. The magnitude of the torque is Γ = RMg sinθ and is directed perpendicular to the plane containing the z and e3 axes. The torque will cause the angular momentum to change, but since it is directed perpendicular to the angular momentum, the direction will change but not the magnitude. The change in L will cause ω1 and ω2 to be small, but non-zero.

e3dot = (MgR / λ3ω)zhat×e3 = &Omegae3
Ω = (MgR / λ3ω)zhat

Euler's Equations

Now we know that (dL/dt) = Γ, L = Iω, we know how to calculate the inertia tensor, and we know how to find the principal moments and principal axes of the inertia tensor, but how does this help to solve real situations. In principal we know everything, but the framework we have is cumbersome for solving real problems. The basic problem is that ω, L, and I can all change as the object rotates with respect to our fixed, inertial coordinate system. A "simplification" is made by choosing a different, non-inertial coordinate system.

If we choose a coordinate system fixed to the rotating object, then at least I doesn't change. The obvious choice is a coordinate system with axes aligned with the principal axes of the object. Now I is constant and diagonal, but since the object is rotating, so is the coordinate system. The dynamics is contained in the relation between torque and angular momentum, and in an inertial frame (called the space frame in the text), this reads

Γ = (dL/dt)space

In chapter 9 we learned how to transform a derivative to a frame rotating with angular velocity ω (called the body frame in the text, since it is attached to the rotating body), namely

(dL/dt)space = (dL/dt)body + ω×L = Ldot + ω×L

where the convention is that a dot represents a time derivative in the rotating frame. Using the body frame (for Γ as well as L), the relation becomes

Ldot + ω×L = Γ.

This is Euler's equation. We will more commonly deal with it 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.

Euler's Equations with Zero Torque

A common occurance, especially for space based problems, is an object subject to zero torque whose motion we'd like to find. This applies to projectiles (neglecting air drag) 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, 0, ω). 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 occurs with Euler's Equations because, at least two components of ω must be non-zero in order that 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 ≠ 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, small 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 use the second equation to substitute 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.

© 2008 Robert Harr