# A Euler's Principal Axis

#### kiuhnm

When we solve Euler's differential equations for rigid bodies we find the angular acceleration $\dot{\boldsymbol\omega}$ and then the angular velocity $\boldsymbol\omega$. Integrating $\boldsymbol\omega$ is less straightforward, so we start from a representation of the attitude, take its derivative and equate it to $\boldsymbol\omega$.

The attitude can be represented as $\boldsymbol\gamma=\phi\hat{\boldsymbol e}$, where $\hat{\boldsymbol e}$ is the principal axis of the rotation. In a course I'm watching online, the professor computes the derivative of $\boldsymbol\gamma$ as follows:$$\dot{\boldsymbol\gamma} = \dot\phi \hat{\boldsymbol e}$$
Wouldn't that be correct only for a fixed $\hat{\boldsymbol e}$? Shouldn't we assume $\hat{\boldsymbol e}$ is changing in time as well?

Related Classical Physics News on Phys.org

#### Hiero

The attitude can be represented as $\boldsymbol\gamma=\phi\hat{\boldsymbol e}$,
That sure bothers me... angles cannot generally be represented by vectors because they do not commute under addition!
(But they do commute differentially, and so angular velocities can be represented by (pseudo-)vectors.)

where $\hat{\boldsymbol e}$ is the principal axis of the rotation. In a course I'm watching online
I’ve never heard of “the” principle axis... only the principle axes (plural). Perhaps you’re thinking about an object with an axis of symmetry, like a top? Maybe you could link the lecture video so people have more context.

#### kiuhnm

I’ve never heard of “the” principle axis... only the principle axes (plural). Perhaps you’re thinking about an object with an axis of symmetry, like a top? Maybe you could link the lecture video so people have more context.
You're probably thinking about the eigendecomposition of the inertia matrix. This is something unrelated to that.

Here's the lecture:
It turns out we're assuming that $\boldsymbol\omega$ is parallel to the principal axis $\hat{\boldsymbol e}$ so, by the transport theorem, the inertial derivative is equal to the derivative wrt the frame $B$. Time to watch the lecture again to see how this assumption is justified.

The lecture is split in short videos. Here's the video with the playlist:
edit: I had it backwards. By making the assumption that $\dot{\boldsymbol\gamma} = \boldsymbol\omega$ either we come to a contradiction or we find a solution which respects that assumption. In this case, everything works out just fine.