The moment of inertia of a body depends on the chosen rotation axis. When the moment of inertia about an axis through the center of mass is known, the parallel-axis theorem gives the moment of inertia about any parallel axis.

Let $I_{\rm CM}$ be the moment of inertia about an axis through the center of mass, and let a second axis be parallel to it at perpendicular distance $d$. If the body's total mass is $M$, then

$$\boxed{I=I_{\rm CM}+Md^2}.$$

The shifted-axis moment is always at least as large as the center-of-mass value, because $Md^2\ge0$.

Why the theorem works

Choose coordinates in the plane perpendicular to the axes. Let $(x_i',y_i')$ locate each mass element relative to the center of mass, and let the new axis be displaced by $(a,b)$, where

$$d^2=a^2+b^2.$$

Relative to the shifted axis, the squared perpendicular distance is

$$(x_i'+a)^2+(y_i'+b)^2.$$

Therefore

$$I=\sum_i m_i\left[(x_i'+a)^2+(y_i'+b)^2\right].$$

Expanding,

$$I=\sum_i m_i(x_i'^2+y_i'^2) +2a\sum_i m_ix_i' +2b\sum_i m_iy_i' +(a^2+b^2)\sum_i m_i.$$

Because the primed coordinates are measured from the center of mass,

$$\sum_i m_ix_i'=0, \qquad \sum_i m_iy_i'=0.$$

The middle terms vanish, leaving

$$I=I_{\rm CM}+Md^2.$$

The same reasoning extends from discrete masses to continuous bodies.

Worked example

Two equal point masses $m$ lie at $x=\pm a$ in the $xy$-plane. About the $z$-axis through their center of mass,

$$I_{\rm CM}=2ma^2.$$

Now choose a parallel axis displaced a distance $d$ in the $y$ direction. The total mass is $M=2m$, so the theorem gives

$$I=2ma^2+(2m)d^2 =2m(a^2+d^2).$$

Direct calculation confirms the result because each mass is a distance

$$\sqrt{a^2+d^2}$$

from the shifted axis, giving

$$I=2m\left(a^2+d^2\right).$$

The theorem applies only when the two axes are parallel. It cannot be used to change the orientation of the rotation axis.