The storage of kinetic energy in a flywhell? I am reading a book on physics demonstrations and problems, and one of the problems deals with a flywheel which rotates at maximum angular speed. The density of the flywheel is uniform and the question asks for the kinetic energy per kilogram stored in the flywheel. I am confused about the effect of the physical dimensions. Will the thickness $h$ and the radius $r$ affect the storage of kinetic energy? 
 A: There are two things at play: the energy stored in the flywheel, and the stresses due to the rotation that threaten to pull it apart. The question is really asking about their relationship:
The stored energy is given as $$E = \frac12 I \omega^2$$
Where the moment of inertia for a solid disk is given by
$$I = \frac12 m r^2$$
For a given density of material $\rho$ and thickness $t$, the mass is given by $m = \pi r^2 t \rho$ so we can write the energy as
$$E = \frac14 \pi \rho t r^4 \omega^2$$
Now the hoop stress is given by
$$\sigma = \rho r^2 \omega^2$$
When that value is limited by material properties, we see that the energy stored in the flywheel is given by
$$E = \frac14 \pi r^2 t \sigma$$
In other words, for a given stress level, the maximum stored energy goes up with the thickness of the flywheel (no surprise there, since making it thicker is like increasing the number of flywheels - so of course it should scale with that) and also with the radius - bigger flywheels store more energy before they break. That's perhaps not quite so intuitive, but it clearly follows from the equations.
Interestingly, the density of the disk does not appear in the final expression - it contributes to the stored energy and the hoop stress in the same way, so when you express energy in terms of the strength of the material, the density disappears.
Finally - looking at the above expression you can see that most of the energy term now relates to the volume of the disk, so we can get the energy per kilogram by dividing by the mass ($\pi r^2 t \rho$) on both sides:
$$\frac{E}{m} = \frac{\sigma}{4\rho}$$
The energy stored per unit mass does not, in fact, depend on the physical dimensions. Unexpected and interesting result.
A: There is a peculiar result for cylindrical flywheels whereby the centrifugal force exceeds the tensile stress of the material for a certain speed which is measured at the periphery of the wheel, irregardless of the dimensions. So no, the thickness or the radius of the cylinder don't matter (as it turns out).
I can't exactly remember the details, but I think there was also some kind of paradox whereby you got a different answer if you calculated for a solid disc versus a disc with a tiny hole in the center. I think it was something to do with a different fracture mode, but I can't remember.
A: The kinetic energy of a flywheel is given by
$$T = \frac{1}{2} I \omega^2$$
Where $I$ is the moment of inertia and $\omega$ is the angular speed.  The specific kinetic energy (KE per kilogram) is 
$$t = \frac{\frac{1}{2} I \omega^2}{M}$$
The moment of inertia $I$ will be of the form $I = k M R^2$, where $k$ can vary from 0 to 1 depending on the geometry of the flywheel.  Most flywheels will be shaped like a ring, since that makes $k = 1$, and maximizes $t$.  In any case, you may notice that while the thickness $h$ doesn't affect $I$, and therefore doesn't affect $t$, the radius of the flywheel does:
$$t = \frac{\frac{1}{2} (k M R^2) \omega^2}{M}$$
$$t = \frac{1}{2} k R^2 \omega^2$$
If you want to understand what moment of inertia is, and how it is calculated, which is really at the heart of your question, hyperphysics provides a nice overview.  The wikipedia page goes into more detail.  You can also find a list of moments of inertia for different shapes.
