Huygens-Steiner Theorem

Theorem
If:
 * $I_0$ is the moment of inertia of a body of mass $M$ about some axis through its centre of mass, and
 * $I$ the moment of inertia of that body about another axis parallel to $I_0$

then they are related by:


 * $I=I_0 + Ml^2$

where $l$ is the perpendicular distance between both axes.

Proof
Without loss of generality suppose $I$ is oriented along the z-axis. We have by definition:


 * $I=\Sigma m_j \lambda_j^{2}$


 * $I_0=\Sigma m_j \lambda_j^{'2}$

where:
 * $\lambda_j$ is the position vector to the $j^{\textrm{th}}$ particle from the z-axis;
 * $\lambda_j^'$ is related to $\lambda_j$ by:


 * $\lambda_j=\lambda_j^' + R_\perp$


 * $R_\perp$ is the perpendicular distance from $I$ to the centre of mass of the body.

We therefore have:


 * $I=\Sigma m_j \lambda_j^{2}=\Sigma m_j (\lambda_j^{'2} + 2\lambda_j^' \cdot R_\perp + R_\perp^2)$

The middle term is:


 * $2R_\perp \cdot \Sigma m_j\lambda_j^'=2R_\perp \cdot \Sigma m_j(\lambda_j - R_\perp)=2R_\perp \cdot M(R_\perp - R_\perp)=0$

Thus:

$I=\Sigma m_j \lambda_j^{2}=\Sigma m_j (\lambda_j^{'2} + R_\perp^2)=I_0 + Ml^2$

Note
This theorem is also known as the Huygens-Steiner Theorem, for Christiaan Huygens and Jakob Steiner.