A very light (meaning don't consider mass of the rod) rod is placed along the x axis. It has a mass m1=2.0kg at x=0, a mass m2=1.50kg at x=50cm, and a mass m3=3.0kg at x=100cm.
Find the moment of inertia of the system about a pivot point at x=0.
I=(1/3)M(L^2) -- Parallel Axis Theorem
I= integral of[(r^2)dm]?
The Attempt at a Solution
Because the mass of the rod does not matter, I'm thinking of finding the center of mass between the three masses and treating that as one whole mass at a certain point x. Does that idea make any sense? Any help would be greatly appreciated!