Is the angular velocity of a rigid body about any point the same as that about the axis of rotation. Also, can we even define angular terms (Angular Velocity, Angular Acceleration, etc) about any axis, other than the axis of rotation?
Rotational Dynamics – Understanding The Angular Velocity
angular velocityreference framesrigid-body-dynamicsrotational-dynamicsrotational-kinematics
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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
Given some central point $\vec c_1$, the velocity of this point $\vec v_{c_1}$ with respect to some frame of reference, and an angular velocity $\vec\omega$, the velocity of some other point $\vec x$ on a rigid body is $\vec v_x = \vec v_{c_1} + \vec\omega \times (\vec x - \vec c_1)$.
What if some other point $\vec c_2$ is chosen as the central point? The expression for the velocity of the point $\vec x$ becomes $\vec v_x = \vec v_{c_2} + \vec\omega \times (\vec x - \vec c_2)$, where $\vec v_{c_2} = \vec v_{c_1} + \vec\omega \times (\vec c_2 - \vec c_1)$. The angular velocity does not change. It is the same regardless of which point one chooses as the central point. Angular velocity is a free vector.
Update, because the above apparently isn't satisfactory to some
A rigid body is an object for which a frame of reference exists such that the location of every point on the rigid body is constant from the perspective of this frame. In other words, $(\frac {d\boldsymbol x}{dt})_F = 0$ for every point $x$ in the rigid body, with the derivative taken from the perspective of the fixed frame $F$.
Suppose at some time $t$ you know the location $\boldsymbol X(\boldsymbol c,t)$ of some fixed point $\boldsymbol c$ on the rigid body in some other frame of reference $I$. The location of the point $\boldsymbol x$ in this other frame is related to the location of the point $\boldsymbol c$ via $$\boldsymbol X(\boldsymbol x,t) = \boldsymbol X(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol x - \boldsymbol c) \tag{1}$$ where $\boldsymbol{\mathrm R}_{F\to I}(t)$ is the rotation matrix that transforms coordinates from frame $F$ to frame $I$.
Suppose you know some other point $\boldsymbol c'$, also fixed with respect to the rigid body. The location of the point $\boldsymbol c'$ in the non-fixed frame is $$\boldsymbol X(\boldsymbol c',t) = \boldsymbol X(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol c' - \boldsymbol c)\tag2$$ The location of the point $\boldsymbol x$ in the non-fixed frame can be re-expressed to be in terms of $\boldsymbol c'$: $$\boldsymbol X(\boldsymbol x,t) = \boldsymbol X(\boldsymbol c',t) + \boldsymbol{\mathrm R}_{F\to I}(t)(\boldsymbol x - \boldsymbol c') \tag{3}$$ Differentiating equations (1) and (3) with respect to time yields $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \dot{\boldsymbol{\mathrm R}}_{F\to I}(t)(\boldsymbol x - \boldsymbol c) = \dot{\boldsymbol X}(\boldsymbol c',t) + \dot {\boldsymbol{\mathrm R}}_{F\to I}(t)(\boldsymbol x - \boldsymbol c') \tag4$$ The time derivative of a transformation matrix $\boldsymbol{\mathrm R}(t)$ from one Cartesian frame to another can be written as a product of that matrix and a skew symmetric matrix: $$\dot{\boldsymbol{\mathrm R}}(t) = \boldsymbol{\mathrm R}(t) \boldsymbol{\mathrm S}(t)$$ This is valid for a Euclidean space of any dimensionality. In three dimensional space (and three dimensional space only), such a skew symmetric matrix can be mapped to and from a three dimensional pseudo vector: $$\dot{\boldsymbol{\mathrm R}}(t) = \boldsymbol{\mathrm R}(t) \operatorname{Sk}(\boldsymbol\omega(t))$$ Rewriting equation (4) in terms of this pseudo vector, $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \boldsymbol{\mathrm R}_{F\to I}(t)\left(\boldsymbol\omega\times(\boldsymbol x - \boldsymbol c)\right) = \dot{\boldsymbol X}(\boldsymbol c',t) + \boldsymbol{\mathrm R}_{F\to I}(t)\left(\boldsymbol\omega\times(\boldsymbol x - \boldsymbol c')\right)\tag5$$ Without loss of generality, one can choose a non-fixed frame that is instantaneously co-aligned with the fixed frame at time $t$. With this, equation (5) reduces to $$\dot {\boldsymbol X}(\boldsymbol x,t) = \dot {\boldsymbol X}(\boldsymbol c,t) + \boldsymbol\omega\times(\boldsymbol x - \boldsymbol c) = \dot{\boldsymbol X}(\boldsymbol c',t) + \boldsymbol\omega\times(\boldsymbol x - \boldsymbol c')\tag6$$ This is of course identical to what I originally wrote.
That angular velocity expressed as a skew symmetric matrix is the same regardless of which point one chooses as the origin applies in all Euclidean spaces in which time is the independent parameter of motion. Angular velocity is related to the time derivative of the transformation matrix, and changing origins doesn't change the transformation matrix in an affine transformation (e.g., equations (1) to (3) are affine transformations).
That angular velocity expressed as a pseudo vector is the same regardless of which point one chooses as the origin applies only in three dimensional Euclidean spaces in which time is the independent parameter of motion. That specialization is rather important because we apparently live in a universe that locally appears to be a three dimensional Euclidean space with time as the independent parameter of motion. In other words, we live in a universe where Newtonian mechanics is locally valid. This is the context in which this question was asked and in which I wrote this answer.