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}

This is a set of coupled, nonlinear differential equations. We solved these for certain special cases only. Now that we've learned the Lagrangian formulation of mechanics, we can learn how to treat them in the Lagrangian formalism. Since the Lagrangian is simply kinetic minus potential energy in an inertial system, we need a way to express the kinetic energy of a rotating body in an inertial system. This is normally accomplished with by defining 3 angular coordinates that specify the orientation of the body relative to fixed spatial coordinates. The 3 angular coordinates are called Euler angles.

Since the principal axes represent a rotating coordinate system, they are non-inertial and a Lagrangian cannot be expressed in terms of them. A fixed, inertial coordinate system is needed, but then we need a way to relate the principal axes to it. The Euler angles relate the principal axes of a rotating object to a fixed coordinate system.

The transformation relates (x,y,z) to (**e**_{1}, **e**_{2}, **e**_{3}).
First, rotate about the z axis by φ until the x axis lies in the z-e_{3} plane.
Call this set of axes (x', y', z') and note that z' = z.
Second, rotate about the y' axis by θ until the z' = z axis is aligned with the e_{3} axis.
Call this set of axes (**e**_{1}', **e**_{2}', **e**_{3}') and note that **e**_{3}' = **e**_{3} and **e**_{2}' = y'.
Finally, rotate about the **e**_{3}' = **e**_{3} axis by ψ until **e**_{1}' and **e**_{2}' are aligned with **e**_{1} and **e**_{2}.

Now we can express the angular velocity of the body in terms of the time derivatives of the Euler angles

For an axially symmetric object, with the symmetry axis in the 3-direction, **e**_{1}' and **e**_{2}' serve as principal axes equally as well as **e**_{1} and **e**_{2}.
Therefore, we can express ω in terms of body axes by transforming **z**hat = **e**_{3}cosθ - **e**_{1}'sinθ yielding

L_{3} = λ_{3}ω_{3} = λ_{3}(ψdot + φdot cosθ)

L_{z} = **L**⋅**z**hat = λ_{1}φdot sin²θ + λ_{3}(ψdot + φdot cosθ) cosθ

L_{z} = λ_{1}φdot sin²θ + L_{3} cosθ

T = ½ **ω**^{T} **I** **ω** = ½λ_{1}(φdot² sin²θ + θdot²) + ½λ_{3}(ψdot + φdot cosθ)²

U = MgR cosθ

L = T - U = ½λ_{1}(φdot² sin²θ + θdot²) + ½λ_{3}(ψdot + φdot cosθ)² - MgR cosθ

∂L/∂θ = (d/dt)(∂L/∂θdot)

λ_{1}φdot² sinθcosθ - λ_{3}(ψdot + φdot cosθ)φdot sinθ + MgR sinθ = (d/dt)(λ_{1}θdot) = λ_{1}θddot

∂L/∂φ = 0 = (d/dt)(∂L/∂φdot)

λ_{1}φdot sin²θ + λ_{3}(ψdot + φdot cosθ)cosθ = p_{φ} = L_{z} = const.

∂L/∂ψ = 0 = (d/dt)(∂L/∂ψdot)

λ_{3}(ψdot + φdot cosθ) = p_{ψ} = L_{3} = const.

© 2008 Robert Harr