structural:expansion, structural:reformatting
Parallel axis theorem

The inertia matrix of a body depends on the choice of the reference point. There is a useful relationship between the inertia matrix relative to the center of mass C {\displaystyle \mathbf {C} } and the inertia matrix relative to another point R {\displaystyle \mathbf {R} }. This relationship is called the parallel axis theorem.

Consider the inertia matrix I R {\displaystyle \mathbf {I_{R}} } obtained for a rigid system of particles measured relative to a reference point R {\displaystyle \mathbf {R} }, given by I R = − ∑ i = 1 n m i 2. {\displaystyle \mathbf {I} _{\mathbf {R} }=-\sum _{i=1}^{n}m_{i}\left^{2}.}

Let C {\displaystyle \mathbf {C} } be the center of mass of the rigid system, then R = (R − C) + C = d + C, {\displaystyle \mathbf {R} =(\mathbf {R} -\mathbf {C})+\mathbf {C} =\mathbf {d} +\mathbf {C},} where d {\displaystyle \mathbf {d} } is the vector from the center of mass C {\displaystyle \mathbf {C} } to the reference point R {\displaystyle \mathbf {R} }. Use this equation to compute the inertia matrix, I R = − ∑ i = 1 n m i 2 = − ∑ i = 1 n m i 2. {\displaystyle \mathbf {I} _{\mathbf {R} }=-\sum _{i=1}^{n}m_{i}^{2}=-\sum _{i=1}^{n}m_{i}^{2}.}

Distribute over the cross product to obtain I R = − (∑ i = 1 n m i 2) + (∑ i = 1 n m i) + (∑ i = 1 n m i) − (∑ i = 1 n m i) 2. {\displaystyle \mathbf {I} _{\mathbf {R} }=-\left(\sum _{i=1}^{n}m_{i}^{2}\right)+\left(\sum _{i=1}^{n}m_{i}\right)+\left(\sum _{i=1}^{n}m_{i}\right)-\left(\sum _{i=1}^{n}m_{i}\right)^{2}.}

The first term is the inertia matrix I C {\displaystyle \mathbf {I_{C}} } relative to the center of mass. The second and third terms are zero by definition of the center of mass C {\displaystyle \mathbf {C} }. And the last term is the total mass of the system multiplied by the square of the skew-symmetric matrix {\displaystyle } constructed from d {\displaystyle \mathbf {d} }.

The result is the parallel axis theorem, I R = I C − M 2, {\displaystyle \mathbf {I} _{\mathbf {R} }=\mathbf {I} _{\mathbf {C} }-M^{2},} where d {\displaystyle \mathbf {d} } is the vector from the center of mass C {\displaystyle \mathbf {C} } to the reference point R {\displaystyle \mathbf {R} }.

Note on the minus sign: By using the skew symmetric matrix of position vectors relative to the reference point, the inertia matrix of each particle has the form − m 2 {\displaystyle -m\left^{2}}, which is similar to the m r 2 {\displaystyle mr^{2}} that appears in planar movement. However, to make this to work out correctly a minus sign is needed. This minus sign can be absorbed into the term m T {\displaystyle m\left^{\mathsf {T}}\left}, if desired, by using the skew-symmetry property of {\displaystyle }.