This question is relevant to Euler's angles and Euler's equations for a rigid body.
Why aren't $\omega_1$, $\omega_2$ and $\omega_3 = 0$ in the body frame? How can we measure $\vec\omega$?
Newtonian Mechanics – Why Body Frame Angular Velocity is Nonzero
angular velocitynewtonian-mechanicsreference framesrotational-dynamics
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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.
Does this mean we can somehow add the 3 rotations (which are about different axes) and get an equivalent rotation about some other (single) axis?
Yes. It's one of many consequences of Euler's rotation theorem. This neat little trick of finding the Euler rotation axis works in three dimensional space, and in three dimensional space only. The Lie group SO(3) is a very special space.
A more generic way of looking at rotation is that rotation in any Euclidean space comprises a combinations of rotations parallel to a two dimensional plane rather than about an axis. There's only one plane in two dimensional space, so rotation in 2D space requires but one parameter. There are three pairs of planes of rotation in three dimensional space (the YZ, ZX, and XY planes), so rotation in 3D space requires three parameters. You can think of those planar rotations as being about an axis (with an arbitrary decision as to what qualifies as positive or negative rotation). There are six planes of rotation in four dimensional space, so rotations in 4D space requires six parameters. This can result in something very weird, something you don't see in 2D or 3D space, a Clifford rotation.
Best Answer
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.