Perpendicular axis theorem

In physics, the perpendicular axis theorem (or plane figure theorem) can be used to determine the moment of inertia of a rigid object that lies entirely within a plane, about an axis at right angles to the plane, given the moments of inertia of the object about two perpendicular axes lying within the plane. The axes must all pass through a single point in the plane.

Define perpendicular axes "X", "Y", and "Z" (which meet at origin "O") so that the body lies in the "XY" plane, and the "Z" axis is perpendicular to the plane of the body. Let

* "I_X" be the moment of inertia of the body about the "X" axis;
* "I_Y" be the moment of inertia of the body about the "Y" axis; and
* "I_Z" be the moment of inertia of the body about the "Z" axis.

The perpendicular axis theorem states that

: $I_Z = I_X + I_Y$

This rule can be applied with the parallel axis theorem and the stretch rule to find moments of inertia for a variety of shapes.

Proof

Let p be a plane thin uniform lamina. Let $m_i$ be a mass element with perpendicular distance $r_i$ from an axis OZ perpendicular to the plane and passing through O in the plane.

Let OX and OY be two perpendicular axes lying in the plane. Let $a_i$ be the perpendicular distance of $m_i$ from OX and $b_i$ be the perpendicular distance of $m_i$ from OY, both in the plane. Let : $I_x = sum {m_i {a_i}^2}$ be the moment of inertia of p about OX and: $I_y = sum {m_i {b_i}^2}$ be the moment of inertia of p about OY.The moment of inertia of p about OZ is given by: $egin{align}I_z & = sum {m_i {r_i}^2} \& = sum {m_i left( {a_i}^2 + {b_i}^2
ight)} qquad ext{by the Pythagorean theorem} \& = sum {m_i {a_i}^2} + sum {m_i {b_i}^2} \& = I_x + I_yend{align}$

ee also

* Parallel axis theorem
* Stretch rule

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