A Mathematical Review

A.6 Spatial inertia

Assume we have a rigid body with mass m and a coordinate frame located at the body’s center of mass. If {\bf v} and \boldsymbol{\omega} give the translational and rotational velocity of the coordinate frame, then the body’s linear and angular momentum {\bf p} and {\bf L} are given by

{\bf p}=m{\bf v}\quad\text{and}\quad{\bf L}={\bf J}\boldsymbol{\omega}, (A.30)

where {\bf J} is the 3\times 3 rotational inertia with respect to the center of mass. These relationships can be combined into a single equation

\hat{\bf p}={\bf M}\hat{\bf v}, (A.31)

where \hat{\bf p} is the spatial momentum and {\bf M} is a 6\times 6 matrix representing the spatial inertia:

\hat{\bf p}\equiv\left(\begin{matrix}{\bf p}\\
{\bf L}\end{matrix}\right),\qquad{\bf M}\equiv\left(\begin{matrix}m{\bf I}&0\\
0&{\bf J}\end{matrix}\right). (A.32)

The spatial momentum satisfies Newton’s second law, so that

\hat{\bf f}=\frac{d\hat{\bf p}}{dt}={\bf M}\frac{d\hat{\bf v}}{dt}+\dot{\bf M}%
\hat{\bf v}, (A.33)

which can be used to find the acceleration of a body in response to a spatial force.

When the body coordinate frame is not located at the center of mass, then the spatial inertia assumes the more complicated form

\left(\begin{matrix}m{\bf I}&-m[{\bf c}]\\
m[{\bf c}]&{\bf J}-m[{\bf c}][{\bf c}]\end{matrix}\right), (A.34)

where {\bf c} is the center of mass and [{\bf c}] is defined by (A.22).

Like the rotational inertia, the spatial inertia is always symmetric positive definite if m>0.