Theorem of Parallel Axes

The theorem of parallel axes applies to ANY rigid object (of any shape, not necessarily flat), and requires two axes that are PARALLEL to each other, with (at least) one of the two passing through the object's centre of mass. Let axis MOP pass through an arbitrary point O, and let a second, parallel, axis ACB pass through the object's centre of mass C, with the two axes separated by a fixed perpendicular distance $h=CO$. Consider a mass element dm located at point D; drop the perpendicular DN from D onto the line joining O and C (extended if necessary, so that N lies on that line). Then $I_C=\int(DC)^2\,dm$ (about the centre-of-mass axis) and $I_O=\int(DO)^2\,dm$ (about the axis through O).

From the right-angled triangles sharing the foot N, $(DO)^2=(DN)^2+(NO)^2$ and $(DC)^2=(DN)^2+(NC)^2$, and since $NO=NC+CO=NC+h$, $(DO)^2=(DN)^2+(NC+h)^2=(DN)^2+(NC)^2+2(NC)(h)+h^2=(DC)^2+2h(NC)+h^2$ Integrating this over the whole object (multiplying by dm and summing): $I_O=\int(DC)^2\,dm+2h\int(NC)\,dm+h^2\int dm=I_C+2h\int(NC)\,dm+Mh^2$ Now, NC is the (signed) distance of a mass element from the centre of mass, measured along the line joining the two axes -- and by the very DEFINITION of the centre of mass, any mass distribution is symmetric about it in exactly the sense that $\int(NC)\,dm=0$ (positive and negative contributions on either side of C exactly cancel). This eliminates the middle (cross) term entirely, leaving the clean result $I_O=I_C+Mh^2 \qquad \text{(theorem of parallel axes)}$ In words: the moment of inertia of an object about ANY axis equals its moment of inertia about a PARALLEL axis through its own centre of mass, PLUS the product of its total mass and the square of th …