Parallel axis theorem of stick-ball configuration

Question

I have a system with a rod of mass m and length 2a. Let the origin be in the middle of the rod at x = 0. (Therefore, each end is a distance a away.) A ball of mass m is attached to the far right end. I want to calculate the moment of inertia of the system about the center of mass, which I know is at x = a/2.

I know the answer is $\frac{1}{12}m(2a)^2 + m(a/2)^2 + m(a/2)^2$ but I am not sure where this is coming from. I know to use the parallel axis theorem, but I am unsure of where each term comes from within this expression. An explicit enumeration of this would be very helpful.

Answer 1

The moment of inertia of the rod, about the middle, is $\frac{1}{12}m(2a)^2$.

The center of mass of the rod and mass together is $\frac{a}{2}$ from the middle of the rod. Using the parallel axis theorem, the moment of inertia of the rod about the new center of mass is $\frac{1}{12}m(2a)^2 +m(\frac{a}{2})^2$. That is the first two terms.

The moment of inertia of the mass, $m$, a distance $\frac{a}{2}$ from the combined center of mass, is $m(\frac{a}{2})^2$. That is the last term.