Perpendicular Axis Theorem

The theorem

For a planar body (flat lamina) lying in the $xy$-plane:

$I_z = I_x + I_y$

where $I_z$ is the MI about the $z$-axis (perpendicular to the plane), and $I_x$, $I_y$ are the MIs about the $x$ and $y$ axes (in the plane).

The three axes must be mutually perpendicular and pass through the same point.

Derivation

For a planar body in the $xy$-plane, every mass element has $z = 0$.

$I_z = \sum m_i(x_i^2 + y_i^2) = \sum m_i x_i^2 + \sum m_i y_i^2$

But $I_y = \sum m_i x_i^2$ (distance from $y$-axis is $x_i$) and $I_x = \sum m_i y_i^2$ (distance from $x$-axis is $y_i$):

$I_z = I_y + I_x$

$\boxed{I_z = I_x + I_y}$

Critical limitation: planar bodies only

This theorem applies only to flat, planar objects (laminas). It does not apply to 3D objects like spheres or cylinders.

Applications

Disc — diameter vs central axis:

By symmetry, $I_x = I_y$ for a disc (all diameters are equivalent).

We know $I_z = \frac{1}{2}MR^2$ (about central perpendicular axis).

$I_z = I_x + I_y = 2I_x \implies I_x = \frac{I_z}{2} = \frac{MR^2}{4}$

The MI of a disc about any diameter $= \frac{MR^2}{4}$.

Ring — diameter vs central axis:

$I_z = MR^2$ (about central perpendicular axis).

By symmetry $I_x = I_y$:

$MR^2 = 2I_x \implies I_x = \frac{MR^2}{2}$

MI of a ring about any diameter $= \frac{MR^2}{2}$.

Square plate of side $a$:

$I_x = I_y = \frac{Ma^2}{12}$ (about axis through centre parallel to a side)

$I_z = I_x + I_y = \frac{Ma^2}{6}$

Using both theorems together

Often, problems require both the perpendicular and parallel axis theorems in sequence.

Example: MI of a disc about a tangent line in the plane of the disc.

Step 1 (perpendicular axis): $I_{diameter} = \frac{MR^2}{4}$

Step 2 (parallel axis): $I_{tangent} = I_{diameter} + MR^2 = \frac{MR^2}{4} + MR^2 = \frac{5MR^2}{4}$