# 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

## 1 Answer

The first equation is valid in the space frame of reference while the second one (Euler's equation) is valid in the body frame of reference.

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
• I have edited my answer. Hope it helps. – Abhijeet Melkani Feb 18 '17 at 20:56