# Huygens-Steiner Theorem

(Redirected from Parallel Axes Theorem)

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## 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

Without loss of generality, 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:

- $I = \Sigma m_j \lambda_j^2 = \Sigma m_j \paren {\lambda_j'^2 + 2 \lambda_j' \cdot R_\perp + R_\perp^2}$

The middle term is:

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

Thus:

- $I = \Sigma m_j \lambda_j^2 = \Sigma m_j \paren {\lambda_j'^2 + R_\perp^2} = I_0 + M l^2$

$\blacksquare$

## Also known as

This theorem is also known as the Parallel Axes Theorem.

## Source of Name

This entry was named for Christiaan Huygens and Jakob Steiner.

## Sources

- 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**parallel axes theorem**