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:


 * $I = I_0 + M l^2$

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

Proof
, suppose $I$ is oriented along the $z$-axis.

By definition of moment of inertia:


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


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

where:
 * $\lambda_j$ is the position vector to the $j$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:


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

The middle term is:


 * $2R_\perp \cdot \Sigma m_j \lambda_j' = 2 R_\perp \cdot \Sigma m_j \left({\lambda_j - R_\perp}\right) = 2 R_\perp \cdot M \left({R_\perp - R_\perp}\right) = 0$

Thus:

$I = \Sigma m_j \lambda_j^2 = \Sigma m_j \left({\lambda_j'^2 + R_\perp^2}\right) = I_0 + M l^2$

Also known as
This theorem is also known as the Huygens-Steiner Theorem, for and.