Deriving a formula for the moment of inertia of a pie slice of uniform density Say you have a right cylinder of radius $R$, and you take a pie slice of angle $\theta$ at the origin with mass $M$. How can you determine the moment of inertia?
My teacher says it is impossible to derive its moment of inertia given those two variables, but this problem was in our textbook.
 A: This comes down to a trivial integral, assuming that the relevant axis is the centre of the cylinder:
\begin{align}
I
& = \int_\Omega \rho\:r^2\:\mathrm dV
=\int_0^L\mathrm dz \int_0^\theta\mathrm d\varphi\int_0^Rr\mathrm dr \: \frac{M}{L\theta R^2/2}r^2
\\ & = \frac{2M}{R^2}\int_0^Rr^3\mathrm dr
\\ & = \frac12MR^2.
\end{align}
Note that it is independent of $\theta$ and $L$ (with the only dependence coming if you want to regard $\rho$ instead of $M$ as fixed), as it should be: the relationship between $I$ and $M$ is fixed by the radius of gyration, and this is only a function of the radial density profile.
A: Assuming that the axis of rotation is the axis of symmetry of the cylinder, then the moment of inertia (MI) is the same as that of the cylinder which it came from, ie $\frac12 MR^2$ where $M$ is now the mass of the 'pie slice' rather than the mass of the 'whole pie' (= cylinder).
The explanation is the Stretch Rule, which says that the MI is the same if an object is stretched (or compressed) symmetrically along or around the axis of rotation. If every element of mass is kept at the same distance from the axis during any transformation, then there is no change in the MI.     
