The parallel axis theorem allows us to determine the moment of inertia for an object rotated through an axis that does not pass through its center of mass. Of course the inertia could be derived about any point (let's call it P) by just assuming the coordinate origin is at P, but that would mean re-deriving each geometry's inertia each time. Instead, the common thing is to have tables of inertias for different shapes rotated about an axis that passes through the center of mass C, and then to use the parallel axis theorem to allow one to calculate the inertia the object would have if it rotated about a point P.[br][br]The parallel axis theorem contains the word "parallel" because the new chosen axis must be parallel to the original one. The theorem simply says that we add a term to the inertia about the center of mass C to get the inertia about any other point P. That's noteworthy in itself since it indicates[b] that there is no easier way to rotate an object than about its center of mass[/b]. The result is this:[br][br][center][math]I_P=I_C+mh^2 \\[br]\text{where m is the whole mass and h is the distance from C to P.}[/math][/center]

[color=#1e84cc]EXAMPLE: Consider the inertia calculated for a long rod in the previous section. The result for the inertia of a rod rotated about C along an axis perpendicular to the symmetry axis of the rod is:[br][/color][center][math]I_C=\tfrac{1}{12}mL^2.[/math][/center][color=#1e84cc] Find the inertia of the same object rotated about an axis passing through its end.[br]SOLUTION: Using the parallel axis theorem we get [/color][math]I_P=I_C+mh^2=\tfrac{1}{12}mL^2+m(\tfrac{L}{2})^2=\tfrac{1}{3}mL^2.[/math][color=#1e84cc] This result suggests that it is four times harder to get the rod rotating about its end than about its center, since one third is four times one twelfth, and we saw that already in the last section.[/color]