PHY6200 W07

Chapter 10: Rotational Motion of Rigid Bodies


In Taylor, read sections 10.5 to 10.7 for today, and 10.7 to 10.8 for Wednesday.


Finding the Principal Axes; Eigenvalue Equations

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.

© 2007 Robert Harr