Let's suppose that we have an ideal homogeneous sphere of mass and finite radius that rotates around a peripheral axis passing through one of its extreme points and that the only point diametrically opposed to the one hinged on the axis reaches at the limit the luminal speed c.
Question: Is the energy to bring this ideal sphere into rotation finite or infinite?
Finite, because the one point on the sphere moving at $c$ contains none of the mass of the sphere. You could try integrating $\rho dV c^2/\sqrt{1-v^2/c^2}$ over the inside of the sphere to see this. I convinced myself by doing the analagous one-dimensional integral for a thin rod whose endpoint is moving at $c$, and the integral was finite.
We'll ignore the usual caveats about the difficulties of rigid bodies in SR, and just assume that the sphere is made up of zillions of tiny particles with magical thrusters that can move them any which-way they want. We can then imagine that all of these particles figure out how to "fly in formation" such that their distances to some origin point that is stationary in some frame is a constant, and such that their density is uniform in that frame.
Let's consider a slightly different case: our magical point particles fly in the formation of a rod pivoted about its end. It's fairly evident that if the answer is finite in this case, it will be finite in the case of the sphere. If we treat the distribution of these particles as continuous with a density $\lambda$ in the desired frame, the energy of the section of rod with length $dr$ at a distance $r$ will be $$ dE = \frac{\lambda dr}{\sqrt{ 1 - v^2/c^2}} = \frac{\lambda dr}{\sqrt{ 1 - r^2/R^2}} $$ where $R$ is the length of the rod. (Note that $v/c = r/R$, since the speed increases linearly with distance.)
Integrating this, we get $$ E = \int_0^R \frac{\lambda dr}{\sqrt{ 1 - r^2/R^2}} = \frac{\pi \lambda R}{2}, $$ which is finite even though the integrand diverges as $r \to R$. So in this very contrived situation, you end up with a finite amount of kinetic energy.
The sphere problem could be done via much the same method, but you'd need to do a triple integral over a sphere whose boundary includes the origin, and I'm not immediately sure how to set that up.
