Parallel Axis Theorem

The theorem

The moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass, plus $Md^2$:

$I = I_{cm} + Md^2$

where $d$ is the perpendicular distance between the two parallel axes.

Derivation

Set up coordinates with the CM at the origin. The CM axis is the $z$-axis. The parallel axis passes through point $(a, b, 0)$ at distance $d = \sqrt{a^2+b^2}$ from the CM axis.

For a mass element $m_i$ at $(x_i, y_i, z_i)$:

Distance from CM axis: $r_i^2 = x_i^2 + y_i^2$

Distance from the parallel axis: $r_i'^2 = (x_i-a)^2 + (y_i-b)^2$

$I = \sum m_i r_i'^2 = \sum m_i[(x_i-a)^2 + (y_i-b)^2]$

$= \sum m_i(x_i^2+y_i^2) - 2a\sum m_i x_i - 2b\sum m_i y_i + (a^2+b^2)\sum m_i$

Since origin is at CM: $\sum m_i x_i = M x_{cm} = 0$ and $\sum m_i y_i = 0$

$I = I_{cm} + (a^2+b^2)M = I_{cm} + Md^2$

Applications

Rod about one end:

$I_{cm} = \frac{ML^2}{12}$ (about centre), $d = L/2$:

$I_{end} = \frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2}{3}$

Disc about rim:

$I_{cm} = \frac{MR^2}{2}$ (about centre), $d = R$:

$I_{rim} = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$

Solid sphere about tangent:

$I_{cm} = \frac{2MR^2}{5}$ (about diameter), $d = R$:

$I_{tangent} = \frac{2MR^2}{5} + MR^2 = \frac{7MR^2}{5}$

Important constraint

The parallel axis theorem requires that one of the axes passes through the centre of mass. You cannot apply it between two arbitrary parallel axes — one must be the CM axis.

To go from one non-CM axis to another, first go to the CM axis, then to the target axis:

$I_2 = I_{cm} + Md_2^2 = (I_1 - Md_1^2) + Md_2^2$