Parallel Axis Theorem

If the inertia tensor for a set of axes with the center of mass at the origin is calculated, the tensor for any set of parallel axes can be easily derived. The translation of the coordinates is given by

\bgroup\color{black}$\displaystyle \vec{r}\;'=\vec{r}+\vec{a} $\egroup

where \bgroup\color{black}$ \vec{a}$\egroup is a constant vector. We now simply compute the inertia tensor for the new set of axes.

  $\displaystyle I_{ij} = \sum\limits_\alpha m_\alpha\left[\delta_{ij}r_\alpha^2-r_{\alpha i} r_{\alpha j}\right]$    
  $\displaystyle I_{ij}' = \sum\limits_\alpha m_\alpha\left[\delta_{ij}\left(\vec{r}_\alpha +\vec{a}\right)^2-(r_{\alpha i}+a_i)(r_{\alpha j}+a_j)\right]$    
  $\displaystyle I_{ij}'= \sum\limits_\alpha m_\alpha\left[\delta_{ij}\left(r_\alp...
...ight) -(r_{\alpha i}r_{\alpha j}+r_{\alpha i}a_j+a_ir_{\alpha j}+a_ia_j)\right]$    
  $\displaystyle I_{ij}'= I_{ij}+\sum\limits_\alpha m_\alpha\left[\delta_{ij}\left...
..._\alpha\cdot\vec{a}+a^2\right) -(r_{\alpha i}a_j+a_ir_{\alpha j}+a_ia_j)\right]$    
  $\displaystyle I_{ij}'= I_{ij}+\sum\limits_\alpha m_\alpha\left[\delta_{ij}\left(a^2\right)-(a_ia_j)\right]$    
  $\displaystyle I_{ij}'= I_{ij}+M\left[\delta_{ij}a^2-a_ia_j\right]$    

This result is called the Parallel Axis Theorem.
\bgroup\color{black}$\displaystyle \vec{r}\;'=\vec{r}+\vec{a} $\egroup
\bgroup\color{black}$ \displaystyle I_{ij}'=I_{ij}+M\left[\delta_{ij}a^2-a_ia_j\right]$\egroup
It can save us a lot of time recalculating the inertia tensor for some object.

Note that the parallel axis theorem shows how the inertia tensor depends on the origin. Angular momentum, torque, and kinetic energy all depend on the origin. This is physically relevant if the origin is a fixed point in the rotation. The origin should be chosen to satisfy the conditions of the physical problem being solved.

Jim Branson 2012-10-21