Huygens-Steiner Theorem

Theorem
Let $B$ be a body of mass $M$.

Let $I_0$ be the moment of inertia of $B$ about some axis $A$ through the centre of mass of $B$.

Let $I$ the moment of inertia of $B$ about another axiss $A'$ parallel to $A$.

Then $I_0$ and $I$ are related by:


 * $I = I_0 + M l^2$

where $l$ is the perpendicular distance between $A$ and $A'$.

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 center of mass of $B$.

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

The middle term is:


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

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 Parallel Axes Theorem.