Moment of Inertia by Integration gave $I$ about axes through the center of mass. Objects often turn about other axes: a door on its hinges, a pendulum on a pin, a disk swinging from its rim. One theorem saves you from integrating again.
Key idea
If $I_{\text{cm}}$ is the moment of inertia about an axis through the center of mass, then about any parallel axis a distance $d$ away, $$I = I_{\text{cm}} + Md^2.$$ The center-of-mass axis gives the smallest $I$ of all parallel axes.
Why it works
Put the center of mass at the origin and the new axis at $x = -d$ (parallel to the old one). A piece at $x$ is now a distance $x + d$ along that direction from the axis: $$I = \int (x + d)^2\,dm = \int x^2\,dm + 2d\int x\,dm + d^2\int dm.$$ The first term is $I_{\text{cm}}$ (the other coordinate's contribution is the same for both axes). The middle term vanishes, because $\int x\,dm = Mx_{\text{cm}} = 0$ at the center of mass. The last term is $Md^2$. The cancellation only happens when you start from the center of mass.
Worked example: A rod on a nail
A uniform $0.5\ \text{kg}$, $1.0\ \text{m}$ stick swings about a nail $0.30\ \text{m}$ from its center. Find $I$ about the nail.
- $I_{\text{cm}} = \tfrac{1}{12}(0.5)(1.0)^2 = 0.0417\ \text{kg·m}^2$.
- $$I = 0.0417 + 0.5(0.30)^2 = 0.0417 + 0.045 = 0.0867\ \text{kg·m}^2$$
Answer: $0.0867\ \text{kg·m}^2$.
Worked example: From one off-center axis to another
A uniform $3\ \text{kg}$, $2\ \text{m}$ rod has $I_{\text{end}} = \tfrac{1}{3}ML^2 = 4\ \text{kg·m}^2$. Find $I$ about a parallel axis $0.5\ \text{m}$ from that end.
- Step back to the center of mass: $I_{\text{cm}} = I_{\text{end}} - M(1)^2 = 4 - 3 = 1\ \text{kg·m}^2$.
- The new axis is $1 - 0.5 = 0.5\ \text{m}$ from the center: $I = 1 + 3(0.5)^2 = 1.75\ \text{kg·m}^2$.
Answer: $1.75\ \text{kg·m}^2$.
Common mistake
Applying the theorem starting from an axis that doesn't pass through the center of mass. In the second example, $I_{\text{end}} + M(0.5)^2 = 4.75\ \text{kg·m}^2$ is wrong: the new axis is closer to the center, so $I$ must be smaller than $I_{\text{end}}$, not bigger. Check: $I$ is smallest at the center of mass and grows as $d^2$ away from it, so always pass through $I_{\text{cm}}$: subtract to get there, then add.
- Disk swinging from a pin at its rim: $\tfrac{1}{2}MR^2 + MR^2 = \tfrac{3}{2}MR^2$.
- Solid ball about a tangent line: $\tfrac{2}{5}MR^2 + MR^2 = \tfrac{7}{5}MR^2$.
- Several parts about one axis: find each part's $I$ about that axis, then add.