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 axis $A'$ 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:


 * $\ds I = \sum m_j \lambda_j^2$


 * $\ds I_0 = \sum m_j \lambda_j'^2$

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:
 * $\ds I = \sum m_j \lambda_j^2 = \sum m_j \paren {\lambda_j'^2 + 2 \lambda_j' \cdot R_\perp + R_\perp^2}$

The middle term is:


 * $\ds 2 R_\perp \cdot \sum m_j \lambda_j' = 2 R_\perp \cdot \sum m_j \paren {\lambda_j - R_\perp} = 2 R_\perp \cdot M \paren {R_\perp - R_\perp} = 0$

Thus:


 * $\ds I = \sum m_j \lambda_j^2 = \sum m_j \paren {\lambda_j'^2 + R_\perp^2} = I_0 + M l^2$