Lecture 1. Angular momentum of a rigid body
Angular momentum of a rigid body
The angular momentum is always calculated with respect to some fixed point. The fixed point can be taken as the origin of a reference system. For a continuous distribution of mass (not necessarily a rigid body), we can write the expression for the angular momentum as the integral
\begin{equation}
\vec{L}=\int\vec{r}\times\vec{v}\ dm\ .
\end{equation}
In the case of a rigid body in rotation about an axis, we can take any point on the rotation axis as the origin of the reference system .
Every point of the rotating rigid body preserves its distance to the rotation axis and rotates with the same angular velocity \(\vec{\omega}\). The rotation axis itself can be in motion either translational (as in the case of a yoyo) or rotational (as in the case of a spinning top), or both.
The angular velocity vector is directed along the rotation axis according with the right hand rule.
The linear velocity \(\vec{v}\) of every element of mass comes from the rotational motion around the axis and depends on its distance to it. Thus, for rotational motion around an axis, the linear velocity is given by
\begin{equation}
\vec{v}=\vec{\omega}\times\vec{r}\ ,
\end{equation}
and consequently, the angular momentum of a rotating rigid body takes the special form
\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 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