Lecture 1. Angular momentum of a rigid body
Angular momentum of a rigid body
For a continuous distribution of mass, we can write the expression for the angular momentum as the integral
\begin{equation}
\vec{L}=\int\vec{r}\times(\vec{\omega}\times\vec{r})dm.
\end{equation}
For a rigid body the angular velocity \(\vec{\omega}\) is necessarily the same for each mass element \(dm\).
The angular momentum depends on the point with respect to which it is calculated. A useful decomposition is achieved by taking
\begin{equation}
\vec{r}=\vec{r}_{cm}+\vec{r}^{\ \prime}\ ,
\end{equation}
where \(\vec{r}_{cm}\) is the position of the center of mass and \(\vec{r}^{\ \prime}\) is the position with respect to the center of mass. Thus, we can write:
\begin{equation}
\vec{L}=\int(\vec{r}_{cm}+\vec{r}^{\ \prime})\times[\vec{\omega}\times(\vec{r}_{cm}+\vec{r}^{\ \prime})]dm
\end{equation}
The angular momentum becomes the sum of four terms:
\begin{align}
\vec{L}&=\int\vec{r}_{cm}\times(\vec{\omega}\times\vec{r}_{cm})dm\\
&+\int\vec{r}_{cm}\times(\vec{\omega}\times\vec{r}^{\ \prime})dm\\
&+\int\vec{r}^{\ \prime}\times(\vec{\omega}\times\vec{r}_{cm})dm\\
&+\int\vec{r}^{\ \prime}\times(\vec{\omega}\times\vec{r}^{\ \prime})dm
\end{align}
By using the vector identity \(\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\times\vec{b})\vec{c}\), we can write the angular momentum in the form
\begin{equation}
\vec{L}=\int\{(\vec{r}\cdot\vec{r})\vec{\omega}-(\vec{r}\cdot\vec{\omega})\vec{r}\}dm(\vec{r})
\end{equation}
The mass element \(dm\) varies with the position \(\vec{r}\) according to the mass distribution. The mass can be distributed either in one dimension with density \(\lambda(\vec{r})\) (mass per unit of length), or in two dimensions with density \(\sigma(\vec{r})\) (mass per unit area), or in three dimensions with density \(\rho(\vec{r})\) (mass per unit of volume). Thus, the mass element can take one of the following forms:
\begin{equation}
dm(\vec{r})=\begin{cases}\lambda(\vec{r})dl&\text{unidimensional case}\\
\sigma(\vec{r})da&\text{twodimensional case}\\
\rho(\vec{r})dv&\text{tridimensional case}\end{cases}
\end{equation}
The element of volume \(dv\), of area \(da\), and of length \(dl\), take different forms in different coordinate systems. Here there are some simple examples:
\begin{equation}
dv=\begin{cases}dxdydz&\text{cartesian}\\
rdrd\theta dz&\text{cylindric}\\
r^2\sin\theta d\phi drd\theta&\text{spheric}\ .\end{cases}
\end{equation}
\begin{equation}
da=\begin{cases}dxdy&\text{cartesian}\\
rd\theta dr&\text{polar}\\
R^2\sin\theta d\phi d\theta&\text{spheric}\ .\end{cases}
\end{equation}
In cartesian coordinates, the element of area can be either \(da=dxdz\) or \(da=dydz\) depending on the plane of integration.
\begin{equation}
dl=\begin{cases}dx&\text{cartesian on a straight line};\\
Rd\theta&\text{polar on the plane}\\
|\vec{r}^{\ \prime}(s)|ds&\text{cartesian in space}\ ,\end{cases}
\end{equation}
where \(\vec{r}(s)\) is a parameterization of the unidimensional distribution of mass in space, \(|\vec{r}^{\ \prime}(s)|=\sqrt{(x^{\prime}(s))^2+(y^{\prime}(s))^2+(z^{\prime}(s))^2}\), and \(\vec{r}^{\ \prime}(s)=d\vec{r}(s)/ds\) is the derivative with respect to the parameter \(s\).
The tensor of inertia
The right side of either Eq.(1) or Eq.(2) is a linear transformation \(\mathcal{I}(\vec{\omega})\) of the angular velocity vector \(\vec{\omega}\). The linear transformation \(\mathcal{I}\) is called the tensor of inertia. To find an expression for this tensor we define the identity linear transformation \(id\) that acts on the angular velocity in the form \(id(\vec{\omega})=\vec{\omega}\), and the tensor linear transformation \(\vec{r}\otimes\vec{r}\) that acts on the angular velocity in the form \((\vec{r}\otimes\vec{r})(\vec{\omega})=\vec{r}(\vec{r}\cdot\vec{\omega})\).
With the previous definitions, we find that the tensor of inertia can be written in the form
\begin{equation}
\mathcal{I}=\int\{r^2id-\vec{r}\otimes\vec{r}\}dm(\vec{r})
\end{equation}
Matrix representation of the tensor of inertia
By selecting some basis, one can express such a tensor in matrix form and write it in the form of a square matrix \(I\) times a column vector
\begin{equation}
\vec{l}=I\vec{\omega}.
\end{equation}
The matrix $I$ is the representation of the inertia