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

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.