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

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.

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 (e1, e2, e3). First, rotate about the z axis by φ until the x axis lies in the z-e3 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 e3 axis. Call this set of axes (e1', e2', e3') and note that e3' = e3 and e2' = y'. Finally, rotate about the e3' = e3 axis by ψ until e1' and e2' are aligned with e1 and e2.

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

ω = φdot zhat + θdot e2' + ψdot e3.

For an axially symmetric object, with the symmetry axis in the 3-direction, e1' and e2' serve as principal axes equally as well as e1 and e2. Therefore, we can express ω in terms of body axes by transforming zhat = e3cosθ - e1'sinθ yielding

ω = -φdot sinθ e1' + θdot e2' +(ψdot + φdot cosθ)e3.
L = -λ1φdot sinθ e1' + λ1θdot e2' + λ3(ψdot + φdot cosθ)e3.
L3 = λ3ω3 = λ3(ψdot + φdot cosθ)
Lz = Lzhat = λ1φdot sin²θ + λ3(ψdot + φdot cosθ) cosθ
Lz = λ1φdot sin²θ + L3 cosθ
T = ½ ωT I ω = ½λ1(φdot² sin²θ + θdot²) + ½λ3(ψdot + φdot cosθ)²

Motion of a Spinning Top

U = MgR cosθ
L = T - U = ½λ1(φdot² sin²θ + θdot²) + ½λ3(ψdot + φdot cosθ)² - MgR cosθ
L/∂θ = (d/dt)(∂L/∂θdot)
λ1φdot² sinθcosθ - λ3(ψdot + φdot cosθ)φdot sinθ + MgR sinθ = (d/dt)(λ1θdot) = λ1θddot
L/∂φ = 0 = (d/dt)(∂L/∂φdot)
λ1φdot sin²θ + λ3(ψdot + φdot cosθ)cosθ = pφ = Lz = const.
L/∂ψ = 0 = (d/dt)(∂L/∂ψdot)
λ3(ψdot + φdot cosθ) = pψ = L3 = const.
© 2008 Robert Harr