Huygens-Steiner 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:
- $\ds I = \sum m_j \lambda_j^2$
- $\ds I_0 = \sum m_j \lambda_j'^2$
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:
- $\ds I = \sum m_j \lambda_j^2 = \sum m_j \paren {\lambda_j'^2 + R_\perp^2} = I_0 + M l^2$
$\blacksquare$
Also known as
The Huygens-Steiner Theorem theorem is also known as the Parallel Axes Theorem.
Source of Name
This entry was named for Christiaan Huygens and Jakob Steiner.
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): parallel axes theorem
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): parallel axes theorem