# Euler equation and conservation of angular momentum (rigid body)

I am a beginner in this field.

Just ask a simple question which confuses me.

Please consider the following:

1. Conservation of angular momentum about fixed point $o$: $\dot{H}_o = M$.
$M$: the total external torque applied to the body about $o$.
2. Euler equation: $I\dot{\omega}+\omega\times I \omega = M$.
$I$: moment of inertia in matrix form (suppose diagonal $I$ for simplicity.)

My question: If there is no external torque ($M=0$), then from 1., we know $\dot{H}_o=0$ and by $H=I\omega$, we know $\dot{\omega}=0$ (Due to rigid body, $I$ is constant).

However, by 2., if $M=0$, $I\dot{\omega}=-\omega\times I \omega$. So $\dot{\omega}\ne 0$.

It confuses to me. Where am I wrong?

• Euler equation is written in the body frame of reference. – Abhijeet Melkani Feb 18 '17 at 20:40
• @A.Melkani It seems to remind me something. Could you explain it clearly? How does that fact solve my problem? I am very weak in identifying this. – sleeve chen Feb 18 '17 at 20:46

So, in the case of zero torque the (physical ie as expressed in space frame of reference) angular velocity vector of a rigid body vector is indeed constant. But $\omega$ in Euler's equations refer to the angular velocity vector expressed in the (moving) body frame of reference. And because the frame of reference is moving the description of the vector is non-constant.
• So how does this fact influence the results of $\dot{\omega}\neq 0$ and the existence of external torque? – sleeve chen Feb 18 '17 at 20:48