The Perpendicular Axis Theorem is used when calculating the Moment of Inertia of a rigid lamina. It states that:

The Moment of Inertia about an axis perpendicular to the plane of the lamina and passing through the center of mass of the lamina is equal to the sum of the Moments of Inertia about any two orthogonal axes in the plane of the lamina, also passing through the centre of mass of the lamina.

To make this clearer, I will use an example. Suppose we have a disc in the x-y plane, with its centre at the origin. The disc has a total mass of *m* and a radius of . The Moment of Inertia about (say) the x-axis is relatively hard to work out, but the Moment of Inertia about the z-axis is much easier, and is given by:

I_{z} = 0.5 * ma^{2}

By the Perpendicular Axis theorem:

I_{z} = I_{x} + I_{y}

Because of the symmetry of the disc:

I_{x} = I_{y}

Hence:

I_{z} = 2 * I_{x}

I_{x} = 0.5 * I_{z}

I_{x} = 0.25 * ma^{2}

Tada!

Some people didn't like me pulling I_{z} out of the air...so here is the calculation in more detail:

I_{z} = ∫_{S} r^{2} dm

Where

*S* is the area I am integrating over (in this case the disc) and the mass element

*dm* is given by:

dm = ρ dS

dm = ρ 2πr dr

The elements of area *dS* are thin rings, radius *r*, thickness *dr*. I_{z} now becomes:

I_{z} = ∫_{0}^{a} 2πρ r^{3} dr

I_{z} = 2πρ [ 0.25 r^{4} ]_{0}^{a}

I_{z} = 0.5πρ a^{4}

Substituting back in the value for ρ:

I_{z} = 0.5 * ma^{2}

P.S. Does anyone know how to write a superscript above a subscript for my integral limits?