Theorems of Moment of Inertia

Because moment of inertia depends on the axis of rotation and the body's orientation relative to it, the very same body has a different moment of inertia about every different axis one might choose. Two general theorems make it possible to shift between axes without repeating the integration from scratch every time.

1. Parallel axis theorem. Statement: the moment of inertia of a body about any axis equals the sum of (a) its moment of inertia about a parallel axis through its center of mass, and (b) the product of the body's mass and the square of the perpendicular distance between the two axes. Proof. Let $I_C$ be the moment of inertia about an axis $AB$ through the center of mass, and let $DE$ be a parallel axis at perpendicular distance $d$ from it, with moment of inertia $I$ about $DE$ to be found. Consider a point mass $m$ on the body at distance $x$ from the center-of-mass axis $AB$; its distance from $DE$ is then $(x+d)$, so its contribution to $I$ is $m(x+d)^2$. Summing over the whole body:

   $I=\sum m(x+d)^2=\sum mx^2+\sum md^2+2d\sum mx=\sum mx^2+d^2\sum m+2d\sum mx.$

   Here $\sum mx^2=I_C$ and $\sum m=M$ (total mass); and $\sum mx=0$ because $x$ is measured from the center of mass itself, so positive and negative contributions on either side of $AB$ exactly cancel by definition. This leaves

   $\boxed{I=I_C+Md^2},$

   as stated.
2. Perpendicular axis theorem. This theorem holds only for plane laminar objects (flat bodies of negligible thickness). Statement: the moment of inertia of a plane laminar body about an axis perpendicular to its plane equals the sum of its moments of inertia about two mutually perpendicular axes lying in the plane of the body, all three axes intersecting at one common point. Proof. Let the $X$ and $Y$ axes lie in the plane of the lamina and the $Z$ axis be perpendicular to it, all three through a common origin $O$. A representative particle of mass $m$ at coordinates $(x,y)$ is at distance $r=\sqrt{x^2+y^2}$ from $O$ (and hence from the $Z$-axis). Its contribution to $I_Z$ is $mr^2=m(x^2+y^2)=mx^2+my^2$. Summing over the whole lamina: $I_Z=\sum mr^2=\sum mx^2+\sum my^2.$ …