Tensor of inertia The tensor of inertia of a solid sphere is $I_{ii}=\frac{2}{5}MR^2$ about an axis passing through its CM. Why would the tensor of inertia of each hemisphere about that axis be $I_{ii}=\frac{2}{5}mR^2$, where $m=\frac{M}{2}$ is the mass of the hemisphere? Is it because the tensor of inertia of each hemisphere would be expected to be $I_{ii}=\frac{1}{5}MR^2$, but then $M = 2m$, hence $I_{ii}=\frac{2}{5}mR^2$?
 A: Just to avoid confusion. The expression "tensor of inertia" of a body has 9 components and takes care of any direction in which you could rotate the body. However you need a reference point to calculate the components of the tensor, which ist usually the center of mass. (but you could take any point inside or outside the body)
As for your "why"-question: Each component of the tensor is calculated via an Integral
(http://en.wikipedia.org/wiki/Moment_of_inertia) and an integral is a linear operator.
However the argument inside the integral is quadratic, so the tensor component of the hemisphere is only half of the old number, because of the symmetric cut of the solid sphere.
So Ii does matter where you take the mass away! If you would for example create a hollow sphere by taking out half of the mass form the core as a solid sphere, this new hollow sphere would NOT have tensor components with half the value of the old one, but the would be a greater than just half the value of the old one.
Lastly, again just to avoid confusion: The the main diagonal components of the tensor of inertia of a hemisphere with reference to its center of mass are (m is the mass of the hemisphere) :
$ I_{xx} = \frac{83} {320} mr^2 \quad \quad   I_{yy} = \frac{83} {320} mr^2 \quad \quad I_{zz} = \frac{1} {5} mr^2
$
A: Rotation about the center of the sphere and about the center of mass of a hemisphere is different. You have to move the rotation point to the center of the sphere using the parallel axis theorem to arrive at the equivalency equation.
