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\subsubsection*{The Inertia Tensor (non-examinable)}

$I$ is not inherent to the body itself - it also
depends on the axis of rotation. A more refined
quantity, which is a property only of the rigid
body and contains the necessary information to
compute the moment of inertia about any axis is
the \emph{inertia tensor} $\mathcal{I}$, a $3
\times 3$ matrix. \\
It was already implicit in our expression of
kinetic energy of a rotating object:
\begin{align*}
    T
    &+ \sum_i \half m_i (\bf{\omega} \times
    \bf{x}_i) \cdot (\bf{\omega} \times \bf{x}_i)
    \\
    &= \sum_i \half m_I
    \{|\bf{\omega}||\bf{x}_i|^2 - |\bf{\omega}
    \cdot \bf{x}_i|^2\} \\
    &= \half \bf{\omega}^\top \mathcal{I}
    \bf{\omega}
\end{align*}
where the elements of the matrix are expressed in
terms of $m_i$ and the components of $\bf{x}_i$.
\[ \bf{x}_i = ((x_i)_1, (x_i)_2, (x_i)_3) :=
((x_i)_a)_{a = 1, 2, 3} \]
i.e.
\[ \mathcal{I}_{ab} = \sum_i m_i \{|x_i|^2
\delta_{ab} - (\bf{x}_i)a(\bf{x}_i)_b\} \]
Moment of inertia $\mathcal{I}_{\hat{\bf{n}}}$
about an axis $\hat{\bf{n}}$ is
\[ \mathcal{I}_{\hat{\bf{n}}} = \hat{\bf{n}}^\top
\mathcal{I} \hat{\bf{n}} \]
and one can show that $\bf{L} = \mathcal{I}
\bf{\omega}$. Hence angular momentum not
necessarily in the same direction as the angular
velocity.

\subsection{Motion of Rigid Bodies}

The mos general motion of a rigid body can be
described by its centre of mass following a
trajectory $R(t)$, together with a rotation around
the centre of mass.
\[ \bf{r}_i = \bf{R} + \bf{y}_i \implies
\dot{\bf{r}}_i = \dot{\bf{R}} + \dot{\bf{y}}_i \]
If the body rotates around the centre of mass with
a velocity $\bf{\omega}$,
\[ \dot{\bf{y}}_i = \bf{\omega} \times \bf{y}_i
\implies \dot{\bf{r}}_i = \dot{\bf{R}} +
\bf{\omega} \times (\bf{r}_i - \bf{R}) \tag{$*$} \]
The kinetic energy is
\begin{align*}
    T
    &= \half M \dot{\bf{R}}^2 + \half \sum_i m_i
    \dot{\bf{y}}_i \cdot \dot{\bf{y}}_i \\
    &= \half M \dot{\bf{R}}^2 + \half \bf{T}
    \omega^2
\end{align*}

\subsubsection*{Motion with Rotation About a Different Point}

The centre of mass is the most natural point to
choose. But we could describe the motion as the
trajectory of some other point $\bf{Q}$ and
rotation about $\bf{Q}$. \\
Put $\bf{r}_i = \bf{Q}$ in ($*$):
\[ \dot{\bf{Q}} = \dot{\bf{R}} + \bf{\omega}
\times (\bf{Q} - \bf{R}) \]
Now use this to eliminate $\dot{\bf{R}}$ in ($*$)
to get the motion of any particle with respect to
$\bf{Q}$:
\[ \implies \dot{\bf{r}}_i = \dot{\bf{Q}} +
\bf{\omega} + \bf{\omega} \times (\bf{r}_i -
\bf{Q}) \]
Comparing to ($*$), we see that angular velocity
$\bf{\omega}$ is the same.

\subsubsection*{Example: Rolling}

A hoop, radius $a$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/211cf8fc99c711ec.png}
\end{center}
If the hoop moves without slipping, there is a
relationship between rotational and translational
motion. The \emph{no-slip condition} is the
requirement that point $A$ (defined as the point
of contact with the ground) is stationary, i.e.
has no speed with respect to the ground. \myskip
The angular speed about centre is $\dot{\theta}$.
But can also think of the hoop as rotating about
$A$; also with angular speed $\dot{\theta}$ hence
horizontal velocity of origin is $v =
a\dot{\theta}$. Horizontal velocity of top of hoop
is $2a \dot{\theta}$. \myskip
\ul{Question}: What is speed of general point $P$?
\\
Think of it rotating about $A$:
\[ AP =2a \cos \left( \frac{\theta}{2} \right)
\implies \text{speed of $P$ relative to $A$ is} \]
\[ v' = 2a \cos \left( \frac{\theta}{2} \right)
\dot{\theta} \]
(check: $\theta = 0$, $v' = 2a \dot{\theta}$, and
$\theta = \pi$, $v = 0$ as expected.)

\begin{note*}
    Velocity of $P$ is not tangent to the hoop!
    This would only be the case if the hoop were
    rotating about a fixed point. It's actually
    perpendicular to $AP$, reflecting the fact
    that $P$ is rotating \emph{and} moving forward
    as the hoop moves.
\end{note*}

\noindent
Comment: despite the presence of friction, this is
one case where mechanical energy is conserved -
because point of contact with the ground doesn't
move with respect to the ground, friction does no
work.
