PHY5210 W08

Rotational Motion of Rigid Bodies


In Taylor, read sections 10.8 to 10.9 for today, 10.9 to 10.10 for Wednesday.


L/∂qi - (d/dt)(∂L/∂qidot) = 0
L = T - U

We derived a set of equations for treating the motion of a rigid body in a rotating frame where the principal axes of the body are fixed. These are the Euler equations:

Ldot + ω×L = Γ

or in component form

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

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.

© 2008 Robert Harr