Rigid Bodies; Euler's Eq of Motion derivation


by logic smogic
Tags: bodies, derivation, euler, motion, rigid
logic smogic
logic smogic is offline
#1
Mar24-08, 04:31 PM
P: 56
Problem
Derive Euler's equations of motion for a rigid body from the Lagrange formulation (for generalized coordinate [tex]\psi[/tex], the third Euler angle).

Applicable Formulae
Euler's equations of motion (what we are trying to derive) are:

[tex]I_{1} \dot{\omega_{1}} - \omega_{2} \omega_{3} (I_{2}-I_{3}) = N_{1}[/tex]
[tex]I_{2} \dot{\omega_{2}} - \omega_{3} \omega_{1} (I_{3}-I_{1}) = N_{2}[/tex]
[tex]I_{3} \dot{\omega_{3}} - \omega_{1} \omega_{2} (I_{1}-I_{2}) = N_{3}[/tex]

Lagrange Formulation:

[tex]\frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q_{j}}} \right) - \frac{\partial T}{\partial q_{j}} = Q_{j} [/tex]

Attempt at a solution

It seems I need an expression for the kinetic energy T of the rigid body in terms of the Euler angles. I would then plug that into the Lagrange equation above for [tex]\psi[/tex], and reduce it to the desired form.

The kinetic energy is given by:

[tex]T = \frac{\vec{\omega}\cdot\bar{I}\cdot\vec{\omega}}{2}[/tex]

and the angular velocity in terms of Euler angles is:

[tex]\vec{\omega_{x,y,z}}=\left( \begin{array}{c} \dot{\phi} sin\theta sin\psi + \dot{\theta} cos\psi \\ \dot{\phi} sin\theta cos\psi - \dot{\theta} sins\psi \\ \dot{\phi}cos\theta + \dot{\psi} \end{array} \right) \cdot \left( \begin{array}{c} \hat{x} \\ \hat{y} \\ \hat{z} \end{array} \right)[/tex]

I've already started working on it, but it seems like a lot of work. Am I on the right track, or is there something I'm missing here?
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