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The equations of motion for rigid body rotation are:

$I\,\dot{\vec{\omega}}+\vec{\omega}\times I\,\vec{\omega}=\vec{\tau}$

How i can calculate this equations using Lagrangian method ?

If i use $L=\frac{1}{2}\vec{\omega}^T\,I\,\vec{\omega}$

i don't get the right equations.

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Let $$ S[ \omega, {\bf p},{\bf r}]= \int \left( \frac 12 I_1 \omega_1^2+ \frac 12 I_2 \omega_2^2+ \frac 12 I_3 \omega_3^2+{\bf p}\cdot (\dot{\bf r}+\omega \times{\bf r}) \right) $$ and vary all three vector variables ($\omega$, ${\bf p}$, and ${\bf r}$) to get an equation for each one. Then eliminate ${\bf p}$ and ${\bf r}$. You will end up with Euler's equations for the angular velocity $\omega$. Since this action is linear in the time derivative $\dot {\bf r}$ it is a Hamiltonian action principle. The vector ${\bf p}$ is Lagrange multiplier enforcing a Lin constraint. For more on Lin constraints see "Lin constraints, Clebsch potentials and variational principles" By Cendra and Marsden,  Physica D 27(1-2):63-89 (1987) 

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  • $\begingroup$ Check my signs! I may have an error. We want to get $\dot {\bf L}+ \omega\times {\bf L}$=0 $\endgroup$
    – mike stone
    Sep 1, 2018 at 18:14
  • $\begingroup$ Thank you, but can you explain why we have constraints? $\endgroup$
    – Eli
    Sep 1, 2018 at 18:15
  • $\begingroup$ No. Their origin is complicated. You need to read Cendra and Marsden. It is interesting that these things were only understood in the mid 20th century despite action principles dating back to the 1700's. $\endgroup$
    – mike stone
    Sep 1, 2018 at 18:17
  • $\begingroup$ I had forgotten that I answered this last year physics.stackexchange.com/q/321836. My mind is going $\endgroup$
    – mike stone
    Sep 2, 2018 at 14:06

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