Inertia matrix

An inertia matrix (also called an inertia tensor or inertia operator) will be derived by considering ${\cal A}$ as a collection of particles that are rigidly attached together (all contact forces between them cancel due to Newton's third law). The expression ${\sigma}(r) dr$ in (13.77) represents the mass of an infinitesimal particle of ${\cal A}$. The moment of momentum of the infinitesimal particle is $r \times {\dot r}{\sigma}(r) dr$. This means that the total moment of momentum of ${\cal A}$ is

$${E}= \int_{{\cal A}(q)} (r \times {\dot r}) \;{\sigma}(r) dr .$$ (13.86)

By using the fact that ${\dot r}= \omega \times r$, the expression becomes

$${E}= \int_{{\cal A}(q)} r \times (\omega \times r) \;{\sigma}(r) dr .$$ (13.87)

Observe that $r$ now appears twice in the integrand. By doing some algebraic manipulations, $\omega$ can be removed from the integrand, and a function that is quadratic in the $r$ variables is obtained (since $r$ is a vector, the function is technically a quadratic form). The first step is to apply the identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$ to obtain

$${E}= \int_{{\cal A}(q)} \big((r \cdot r)\omega - (r \cdot \omega)r\big) {\sigma}(r) dr .$$ (13.88)

The angular velocity can be moved to the right to obtain

$${E}= \left( \int_{{\cal A}(q)} \big((r \cdot r)I_3 - r r^T\big) {\sigma}(r) dr \right) \omega ,$$ (13.89)

in which the integral now occurs over a $3 \times 3$ matrix and $I_3$ is the $3 \times 3$ identity matrix.

Let $I$ be called the inertia matrix and be defined as

$$I(q) = \left( \int_{{\cal A}(q)} \big( (r \cdot r)I_3 - r r^T \big) {\sigma}(r) dr \right) .$$ (13.90)

Using the definition,

$${E}= I \omega .$$ (13.91)

This simplification enables a concise expression of (13.82) as

$${N}(u) = {d{E}\over dt} = \frac{d(I\omega)}{dt} = I \frac{d\omega}{dt} + \frac{dI}{dt} \omega ,$$ (13.92)

which makes use of the chain rule.