Huygens-Steiner Theorem

Theorem
If:
 * $$I_0$$ is the moment of inertia of a body of mass $$M$$ about some axis through its centre of mass, and
 * $$I$$ the moment of inertia of that body about another axis parallel to $$I_0$$

then they are related by:


 * $$I=I_0 + Ml^2$$

where $$l$$ is the perpendicular distance between both axes.

Proof
Without loss of generality suppose $$I$$ is oriented along the z-axis. We have by definition:


 * $$I=\Sigma m_j \lambda_j^{2}$$


 * $$I_0=\Sigma m_j \lambda_j^{'2}$$

where:
 * $$\lambda_j$$ is the position vector to the $$j^{\textrm{th}}$$ 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 centre of mass of the body.

We therefore have:


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

The middle term is:


 * $$2R_\perp \cdot \Sigma m_j\lambda_j^'=2R_\perp \cdot \Sigma m_j(\lambda_j - R_\perp)=2R_\perp \cdot M(R_\perp - R_\perp)=0$$

Thus:

$$I=\Sigma m_j \lambda_j^{2}=\Sigma m_j (\lambda_j^{'2} + R_\perp^2)=I_0 + Ml^2$$

Note
This theorem is also known as the Huygens-Steiner Theorem, for Christiaan Huygens and Jakob Steiner.