Parallel axis theorem

The parallel axis theorem gives the rotational inertia about any axis parallel to one through the center of mass: add the total mass times the square of the distance between the two axes. The reference axis must pass through the center of mass.

I' = I_{\text{cm}} + Md^2

Both the AP Physics 1 and the AP Physics C: Mechanics equation sheets print it, and EK 5.4.B.2 introduces it in AP Physics 1, so it is not a Physics C exclusive. $I_{\text{cm}}$ is the rotational inertia about an axis through the center of mass, $M$ is the total mass of the system, and $d$ is the perpendicular distance between the two parallel axes.

Two conditions, both easy to break.

The reference axis has to pass through the center of mass. The theorem does not hop between two arbitrary axes; going from one off-center axis to another means routing through the center of mass first.
The axes must be parallel. Tilting the axis is a different calculation entirely.

The result can only grow. EK 5.4.B.1 says a rigid system's rotational inertia in a given plane is at a minimum when the rotational axis passes through the system's center of mass. Since $Md^2$ is never negative, shifting the axis away can only increase $I$ .

A worked instance: a uniform rod of mass $M$ and length $L$ has $I_{\text{cm}} = \frac{1}{12}ML^2$ about its center. Swing it about one end and $d = L/2$ , so

I' = \tfrac{1}{12}ML^2 + M\left(\tfrac{L}{2}\right)^2 = \tfrac{1}{3}ML^2

four times the central value. Values of $I_{\text{cm}}$ for extended shapes are supplied on the AP Physics 1 exam rather than memorized. See rotational inertia for what those values depend on.

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