PHY5210 W08

Rotational Motion of Rigid Bodies

Reading:

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

Recall

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:

or in component form

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 I_xx and I_yy).

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

where

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

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

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

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 Ω_b. This cone shape is called a body cone.

The angular momentum in the body frame is

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 Ω_b, just like ω. The dot product of L and ω is also constant, L⋅ω = λω₁₀² + λ₃ω₃₀², meaning that the angle between L and ω is constant. Finally, the cross products e₃×L = λω₁₀(sin(Ω_bt)e₁ + cos(Ω_bt)e₂) and e₃×ω = ω₁₀(sin(Ω_bt)e₁ + cos(Ω_bt)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 = λω₁₀Ω_b (-sin(Ω_bt), -cos(Ω_bt), 0), and ω×L = [ω₁₀ω₃₀(λ - λ₃)](sin(Ω_bt), cos(Ω_bt), 0). Substitution of the definition of Ω_b 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

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.