This is really basic, I'm sure: For rigid body motion, Euler's equations refer to $L_i$ and $\omega_i$ as measured in the fixed-body frame. But that frame is just that: fixed in the body. So how could such an observer ever measure non-zero $L$ or $\omega$?
[Physics] Motion in the body-fixed frame
angular momentumangular velocityreference framesrotational-kinematics
Related Solutions
Great question; I remember being so confused by this when I first took analytic mechanics.
The components of the angular velocity "in the body frame" aren't zero because when one writes these components, one isn't referring to measurements of the motions of the particles in the body frame (because, of course, the particles are stationary in this frame). Instead, one is referring to angular velocity as measured in an inertial frame but whose components have simply been written with respect to a time-varying basis that is rotating with the body.
In practice, we make measurements of the positions $\mathbf x_i(t)= (x_i(t),y_i(t),z_i(t))$ of the particles in an inertial frame. Then, we note that for a rigid body (let's consider pure rotation for simplicity), the position of each particle $i$ satisfies \begin{align} \mathbf x_i(t) = R(t) \mathbf x_i(0) \end{align} for some time-dependent rotation $R(t)$. Then we compute $\boldsymbol\omega(t) = (\omega^x(t),\omega^y(t),\omega^z(t))$ in the standard way in terms of $R(t)$. To see how this is done in detail, see, for example
https://physics.stackexchange.com/a/74014/19976
Once we have $\boldsymbol\omega$, we can write its components with respect to any basis we like. If we write it in the standard ordered basis $\{\mathbf e_i\}$, then we'll just get $\omega_x(t)$ as its components. If we write it in some basis $\{\mathbf e_{i,B}(t)\}$ that is rotating with the body (like one that points along the principal axes of the body) then we get different components $\omega^i_B(t)$, and these are the body components.
Main Point Reiterated. Angular velocity is being measured with respect to an inertial frame, but its components can be taken with respect to any basis we wish such as one rotating with the body.
Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference:
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$
where $s$ denotes the space fixed frame and $r$ denotes the rotating frame.
That is the relevant equation for the instantaneous co-moving inertial frame. What about a non-comoving space frame? Generalizing this to a space frame in which the origin of the rotating frame is moving results in
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r_0}{dt}\right)_s + \left(\frac{d(\boldsymbol r - \boldsymbol r_0)}{dt}\right)_r +\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_0)$$
where $\boldsymbol r_0$ is the displacement vector from the origin of the space frame to the origin of the rotating frame.
Suppose you use some other point $\boldsymbol r_1$ that is fixed from the perspective of the rotation frame (i.e., $\left(\frac{d(\boldsymbol r_1-\boldsymbol r_0)}{dt}\right)_r \equiv 0$. Go through the math (an exercise I'll leave up to you) and you'll find that
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r_1}{dt}\right)_s + \left(\frac{d(\boldsymbol r - \boldsymbol r_1)}{dt}\right)_r +\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_1)$$
In other words, the angular velocity $\boldsymbol \omega$ is independent of the choice of origin.
Best Answer
I fussed about this as well. My resolution: for these calculations the fixed-body frame is not to be considered as co-moving with the body, but rather a non-rotating frame that instantaneously aligns with the body.
The Euler angles translate between the body and the space frames. The Euler angles are indeed functions of time, and the fixed-body frame is as well, but angular velocity and momentum are measured with respect to a fixed "snapshot" of the body frame at a particular time.