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\[ I = \sum_{i = 1}^N m_i d_i^2 \]
moment of inertia. \myskip
The angular momentum of the body is
\[ L = \sum_i m_i \bf{x}_i \times \dot{\bf{x}}_i =
\sum_i m_i \bf{x}_i \times (\bf{\omega} \times
\bf{x}_i) \]
For $\bf{\omega} = \omega \hat{\bf{n}}$
($\hat{\bf{n}}$ is unit vector),
\begin{align*}
    \bf{L} \cdot \hat{\bf{n}}
    &= \omega \sum_i m_I (\bf{x}_i \times
    (\hat{\bf{n}} \times \bf{x}_i)) \cdot
    \hat{\bf{n}} \\
    &= \omega \sum_i m_i |\bf{x}_i \times
    \hat{\bf{n}}|^2
\end{align*}
the identity can be proven using components:
\begin{align*}
    \bf{x} \times (\hat{\bf{n}} \times \bf{x})
    \cdot \hat{\bf{n}}
    &= \eps_{lkm} x_l \eps{ijk} \hat{n}_i x_j n_k
    \\
    &= (\eps_{mlk} \hat{n}_m x_l)(\eps_{ijk}
    \hat{n}_i x_j) \\
    &= (\bf{x} \times \hat{\bf{n}}) \cdot (\bf{x}
    \times \hat{\bf{n}})
\end{align*}
but $d_i = |\bf{x}_i \times \hat{\bf{n}}|$, so
\[ \bf{L} \cdot \hat{\bf{n}} = I \omega \]
Recall that $\dot{\bf{L}} = \bf{\tau}$, the
torque. If $\bf{\tau} = \tau \hat{\bf{n}}$, then
$\tau = I \dot{\omega}$.

\subsubsection*{Calculating the Moment of Inertia}

We often treat rigid bodies as continuous , with
density distribution $\rho(\bf{x})$ instead of
discrete masses $m_i$. Then, the mass is
\[ M = \int \rho(\bf{x}) \dd V \]
and the moment of inertia is
\[ I = \int \rho(\bf{x}) x_\perp^2 \dd V \]
where $x_\perp = x \sin \phi$ is the perpendicular
distance to the axis of rotation.

\subsubsection*{Examples}

\begin{itemize}
    \item Circular hoop, axis of rotation through
        centre and perpendicular to the hoop. Mass
        $M$, radius $a$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/f0b5c8d28fdb11ec.png}
        \end{center}
        \[ I = \int \rho a^2 \dd V = a^2 \int
        \rho(\bf{x}) \dd V = a^2 M \]
    \item Disc radius $a$.
        \begin{enumerate}[(a)]
            \item axis through centre,
                perpendicular to plane of disc
                (spinning plate).
                \begin{center}
                    \includegraphics[width=0.6\linewidth]
                    {images/132d747c8fdd11ec.png}
                \end{center}
                \[ M = \pi a^2 \rho \]
                (from now on, put $\rho(\bf{x}) =
                \rho$, a constant within the
                body).
                \[ I = \int_0^a r \dd r
                \int_0^{2\pi} \dd \theta \rho r^2
                = \half M a^2 \]
            \item axis of rotation through centre,
                in the plane of the disc (coin
                flip).
                \begin{center}
                    \includegraphics[width=0.6\linewidth]
                    {images/237c1b088fdd11ec.png}
                \end{center}
                \[ I = \int_0^a r \dd r
                \int_0^{2\pi} \dd \theta \rho(r
                \sin\theta)^2 = \frac{1}{4} M a^2 \]
        \end{enumerate}

    \begin{definition*}
        2 dimensional objects such as the disc are
        called \emph{laminas}. (a) and (b) illustrate
        a general fact about laminas. If we take the
        $z$-axis perpendicular to the lamina, we
        obtain the\dots
    \end{definition*}
    
    \begin{theorem*}[Perpendicular Axis Theorem]
        Consider a 2 dimensional object lying in the
        $xy$-plane.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/75fbf9488fdd11ec.png}
        \end{center}
        \[ I_x = \int \rho y^2 \dd A \]
        \[ I_y = \int \rho x^2 \dd A \]
        while
        \[ I_z = \int \rho r^2 \dd A = \int \rho (x^2 +
        y^2) \dd A \]
        hence
        \[ \boxed{I_z = I_x + I_y} \]
        so for example for a disc, by symmetry $I_x =
        I_y$, so $I_z = 2I_x$, which is consistent with
        the previous computations.
    \end{theorem*}
    \item A sphere, radius $a$, mass $M =
        \frac{4}{3} \pi \rho a^3$. Axis of
        rotation through centre
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/084978c08fde11ec.png}
        \end{center}
        spherical polar coordinates with $\theta =
        0$ pointing along axis. $\theta \in [0,
        \pi)$ and $\phi \in [0, 2\pi)$.
        \begin{align*}
            I
            &= \int_0^a r^2 \dd r \int_0^\pi \sin
            \theta \dd \theta \int_0^{2\pi} \dd
            \phi \rho
            \frac{(r\sin\theta)^2}{x_\perp^2} \\
            &= \frac{8}{15} \pi \rho a^5 \\
            &= \frac{2}{5} Ma^2
        \end{align*}
        (note that all the examples have been
        consistent with dimensional analysis, $[I]
        = [M][L]^2$)
    \item disc again about an axis perpendicular
        to disc, passing through a point on
        circumference.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/01d75b328fdf11ec.png}
        \end{center}
        \[ d^2 = (\bf{r} - \bf{a})^2 = r^2 + a^2 -
        2ar \cos\theta \]
        \begin{align*}
            I
            &= \rho \int_0^a r \dd r \int_0^{2\pi}
            \dd \theta (r^2 + a^2 - 2ar
            \cos\theta) \\
            &= \frac{3}{2} \pi \rho a^2 \\
            &= \frac{3}{2} Ma^2
        \end{align*}
        We could have done this quickly using
        the\dots
        \begin{theorem*}[Parallel Axis Theorem]
            A rigid body has mass $M$ and moment
            of inertia $I_{\text{CoM}}$ about an
            axis which passes through the centre
            of mass. Let $I$ be the moment of
            inertia about a parallel axis that
            lies a distance $h$ away. Then
            \[ \hat{\bf{n}} I = I_{\text{CoM}} + M
            h^2 .\]
        \end{theorem*}
        \begin{proof}
            Let $\bf{r}_i = \bf{R} + \bf{y}_i$
            where $\sum_i m_i \bf{y}_i = 0$.
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/12cd3ece8fe011ec.png}
            \end{center}
            Then
            \begin{align*}
                I
                &= \sum_i m_I (\hat{\bf{n}} \times
                \bf{r}_i) \cdot (\hat{\bf{n}}
                \times \bf{r}_i) \\
                &= \sum_i m_i (\hat{\bf{n}} \times
                (\bf{R} + \bf{y}_i)) \cdot
                (\hat{\bf{n}} \times (\bf{R} +
                \bf{y}_i)) \\
                &= \sum_i m_i [\ub{(\hat{\bf{n}}
                \times \bf{y}_i) \cdot
                (\hat{\bf{n}} \times \bf{y}_i)}
                _{d_i^2} +
                \ub{2(\hat{\bf{n}} \times \bf{y}_i)
                \cdot (\hat{\bf{n}} \times \bf{R})}
                _{\propto \sum_i m_i y_i = 0}
                + \ub{(\hat{\bf{n}} \times \bf{R})
                \cdot (\hat{\bf{n}} \times
                \bf{R})}_{h^2}] \\
                &= I_{\text{CoM}} + Mh^2
            \end{align*}
        \end{proof}
        \begin{note*}
            \[ I = I_{\text{CoM}} + Ma^2 =
            \frac{3}{2} Ma^2 \]
        \end{note*}
\end{itemize}