If an arbitrary rigid body rotates with angular velocity $\omega_0$ about some axis, can it be said that the body will rotate with an angular velocity $\omega_0 \cos(\theta)$ about an axis which is at an angle of $\theta$ with this axis. If yes, what is the physical significance of such an equation?
Rotational Dynamics – Component of Angular Velocity Along an Axis Inclined at $\theta$
angular velocityrotational-dynamics
Related Solutions
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.
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
You can decompose an angular velocity into components e.g. express it as:
$$ \vec{\omega_0} = \omega_x \vec{i} + \omega_y \vec{j} + \omega_z \vec{k}$$
and this is is essentially what you're doing by calculating the component along some axis at an angle $\theta$, though obviously you will need the value in two other directions as well. However there is no physical significance to the values of the components. They are just the representation of the angular velocity vector in whatever coordinate system you have chosen and will be different in different coordinate systems.