How to derive (the dimensionless coefficient in front of) the moment of inertia for common shapes? Is there a way to derive (the dimensionless coefficient in front of) the moment of inertia for common shapes? I assume it has to do with the density of the shape, but I'm having trouble seeing it.
For example the moment of inertia for a cylinder is $I = \frac12 M R^2$ while the moment of inertia for a hoop is $I = M R^2$.
 A: Briefly, the moment of inertia depends on "how much of the mass is how far away from the axis of rotation". If the radial mass distribution is $m(r)$ then the moment of inertia is given by
$$I=\int m(r) r^2 dr$$
In the case of a hoop, all the material is the same distance $R$ away. But if you take a solid disk, you can think of this disk as being made up of a series of concentric hoops - each hoop bigger, and heavier, than the previous one.
This leads to a simple integral. If we consider a disk of mass M and radius R, then the mass per unit area is $\rho = \frac{M}{\pi R^2}$. Imagine we cut a hoop of radius $r$, thickness $dr$ from this disk. It would have area $2\pi r \ dr$. Its mass would be $2\pi r\ dr\ \rho$, and its moment of inertia $2 \pi r\ dr\ \rho\ r^2 = 2 \pi r\ dr \frac{M}{\pi R^2} r^2 = 2 \frac{M}{R^2} r^3 dr$
Summing the total moment of inertia of all these hoops, we get
$$I = \int_0^R 2 \frac{M}{R^2} r^3 dr = \frac12 M R^2$$
A similar method can be used for any other shape - if you break it up into little segments that have their mass at the same distance from the center of rotation, then the moment of inertia of that segment is easily determined; and then you just add all contributions to give you the total moment of inertia.
