Does a spinning object acquire mass due to its rotation? We have settled on a standard way of expressing  mass at relativistic speeds, as per Matt Strassler's  excellent blog.
That is no problem for linear momentum, mass is invariant.  
All I want to know is does the same principle apply to angular momentum, for every object, from golf balls to neutron stars? Or have I pushed the invariance idea into the wrong area?
EDIT I just discovered this post Does rotation increase mass and I really want to avoid semantics about the word mass, but it seems to say mass does increase  END EDIT
 A: Yes, the mass increases according to the energy you've added by $E=mc^2$.
A: I am only considering Special Relativity. When the rotating object is composed of particles with rest mass $m_i$ with rotational speed $v_i=r_i\omega$, where $r_i$ are the distances to the rotational axis and $\omega$ is the angular velocity, the total rest mass of the object is $M= \sum_i m_i$ and the relativistic mass is $$M_\textrm{rel}=\sum_i \frac{m_i}{{\sqrt{1-(v_i/c)^2}}}=\sum_i \frac{m_i}{{\sqrt{1-(r_i\omega/c)^2}}}$$ This doesn't include additional potential energy due to internal tension in the rotating object. The sum can also written as an integral over mass elements dm.
A: The mass of a extended body in rotation is larger than the mass of the same body not rotating, but that increase in mass should not be attributed to any change in the mass of the constituent particles.
Here I am defining the mass of an object (in the usual manner of modern treatments) as the square of that object's energy-momentum four-vector:
$$ m \equiv \frac{1}{c^2}|\mathbf{p}|^2 = \frac{1}{c^2}\sqrt{E^2 - (\vec{p}c)^2} \;.$$ Such masses are Lorentz invariants: they do not change under a Lorentz boost (change in inertial velocity). This is the only meaning of "mass" in modern treatments and in particular the phrase "relativistic mass" is nowhere to be seen.
There is kinetic energy in rotation and that energy contributed to the time-like component of the four momentum (i.e. the $E$ above).
To expand on how the overall body can grow more massive while the particle that make it up do not, we have to examine the four-momentum of a compound system.
Like other vectors, you can add four-vectors component-wise. So trivial compound body consisting of a symmetric rigid1 rotor of total mass $M_\textrm{rot} = 2m$ when not rotating and separation $r$. At a given moment each sub-mass has  three-momentum $\pm\vec{p}$ in the rotor's COM frame and therefore energy $E = \sqrt{(\vec{p}c)^2 + (mc^2)^2}$ in the same frame. That makes their total four-momentum
$$ \mathbf{P}_\textrm{tot} = \left( 2\sqrt{(\vec{p}c)^2 + (mc^2)^2}, \vec{0} \right) \;.$$
The mass of the rotor is 
\begin{align*}
M_\textrm{rot}  
&= \frac{1}{c^2} \left| \mathbf{p}_\textrm{tot} \right| \\
&= \frac{1}{c^2}E^2 \\
&= 2\sqrt{\left(\frac{\vec{p}}{c}\right)^2 + m^2} \;.
\end{align*}
This obviously exceeds $2m$ for any non-zero $\vec{p}$.
The basic lesson here is that the mass of systems is not automatically the sum of the mass of the parts in relativity; a matter often inexplicably left off of lists of things that are different in Einstein's world. One of the most interesting games you can play with this fact is showing that two photons (each of exactly zero mass using the definition above) can none-the-less form a system of non-zero mass.
A word of caution: I've not considered the binding energy of this system which would not be acceptable in a serious treatment as any rotation swift enough to make an appreciable difference to the mass of the system would also necessarily alter the binding energy by an appreciable amount (that is, the assumption of rigidity would be violated). 

1 I know. No truly rigid bodies in relativity. We can assume this thing isn't actually rigid but that we spin it up slowly enough to pretend.
