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

∂L/∂q_{i} - (d/dt)(∂L/∂q_{i}dot) = 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:

or in component form

λ_{1}ω_{1}dot - (λ_{2} - λ_{3})ω_{2}ω_{3} = Γ_{1}

λ_{2}ω_{2}dot - (λ_{3} - λ_{1})ω_{3}ω_{1} = Γ_{2}

λ_{3}ω_{3}dot - (λ_{1} - λ_{2})ω_{1}ω_{2} = Γ_{3}

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, λ_{1} = λ_{2} = λ.
Now the third Euler equation reduces to ω_{3}dot = 0 resulting in ω_{3} = ω_{30} = constant.
Treating ω_{3} as a constant, the first two equations can be rewritten in the form

ω_{1}dot = [ω_{30}(λ - λ_{3})/λ]ω_{2} = Ω_{b}ω_{2}

ω_{2}dot = [-ω_{30}(λ - λ_{3})/]ω_{1} = -Ω_{b}ω_{1}

where

Ω_{b} = ω_{30}(λ - λ_{3})/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 = ω_{1} + iω_{2}, with ηdot = ω_{1}dot + iω_{2}dot.
Substituting for ω_{1}dot and ω_{2}dot we find

ηdot = Ω_{b}(ω_{2} -iω_{1}) = -iΩ_{b}η

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

η = η_{0}e^{-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 ω_{10}, that is **ω**(t=0) = (ω_{10}, 0, ω_{30}).
Then at t=0, η_{0} = ω_{10}, and

Remember, this result is for the body frame.
It says that **ω** is of constant magnitude ω = sqrt(ω_{10}² + ω_{30}²), 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(λ²ω_{10}² + λ_{3}²ω_{30}²), 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** = λω_{10}² + λ_{3}ω_{30}², meaning that the angle between **L** and **ω** is constant.
Finally, the cross products **e**_{3}×**L** = λω_{10}(sin(Ω_{b}t)**e**_{1} + cos(Ω_{b}t)**e**_{2}) and **e**_{3}×**ω** = ω_{10}(sin(Ω_{b}t)**e**_{1} + cos(Ω_{b}t)**e**_{2}) are parallel.
Therefore, the three vectors **L**, **ω** and **e**_{3} lie in the same plane.

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 (d**L**/dt)_{space} = (d**L**/dt)_{body} + **ω**×**L** to solve for the behavior of **L** in the space frame.
In this case, (d**L**/dt)_{body} = λω_{10}Ω_{b} (-sin(Ω_{b}t), -cos(Ω_{b}t), 0), and **ω**×**L** = [ω_{10}ω_{30}(λ - λ_{3})](sin(Ω_{b}t), cos(Ω_{b}t), 0).
Substitution of the definition of Ω_{b} reveals that these are negatives, and we obtain the above result that (d**L**/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 **e**_{3} don't change.
The still lie in a plane and are separated by constant angles.
But, while **e**_{3} is fixed in the body frame, in the space frame, **L** is constant.
Therefore, **ω** and **e**_{3} precess around **L** in the space frame.
The rate of precession is the topic of problem 10.46, and can be shown to be

Ω_{s} = L/λ_{1}.

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