% vim: tw=50
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% \begin{example*}[Swinging Rod]
\myskip\textbf{Example }(Swinging Rod). \\
    This is a pendulum made from a heavy rod of
    mass $m$, length $L$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/a6536da899c711ec.png}
    \end{center}
    Viewing the rod as rotating about the pivot,
    the kinetic energy is
    \[ T = \half I \dot{\theta}^2 \qquad
    \text{with} \qquad I = \frac{1}{3} mL^2 \]
    (from before).
    \begin{note*}
        Could also look at this from the point of
        view of the centre of mass. As we saw
        before, the angular speed is still
        $\dot{\theta} = \text{ angular speed about
        the pivot}$. Speed of the centre of mass
        is
        \[ \frac{L}{2} \dot{\theta} \qquad
        \text{and} \qquad T = \half m \left(
        \frac{L\dot{\theta}}{2} \right)^2 + \half
        I_{\text{CoM}} \dot{\theta}^2 \]
        parallel axis theorem implies that
        \[ I = I_{\text{CoM}} + m \left(
        \frac{L}{2} \right)^2 \]
        so $T$ calculated both ways is identical.
    \end{note*}
    Compared to the pivot, the rod is at a height
    \[ -\frac{L}{2} \cos\theta \implies E = \half
    I\dot{\theta}^2 - \frac{mgL}{2} \cos\theta \]
    \[ \implies \dot{E} = \dot{\theta} \left( I
    \ddot{\theta} + \frac{mgL}{2}\sin\theta
    \right) = 0 \]
    \[ \implies I \ddot{\theta} = -\frac{mgL}{2}
    \sin\theta \]
    is equation of motion. Pendulum:
    \[ \ddot{\theta} = -\frac{g}{L}\sin\theta \]
    Rod:
    \[ \ddot{\theta} = -\frac{3g}{2L} \sin\theta \]
% \end{example*}

\begin{example*}[Rolling Disc]
    A disc of mass $M$ rolls down a slope with the
    plane of the disc being vertical.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/7c827fa499c811ec.png}
    \end{center}
    The moment
    of inertia perpendicular to the plane of the
    disc is $I = \half Ma^2$ (from before). Let
    speed be $v$ and angular speed be $\omega$. No
    slip condition implies that $v = a\omega$. The
    kinetic energy is
    \[ T = \half Mv^2 + \half I\omega^2 = \half
    \left( \frac{I}{a^2} + M \right) v^2 \]
    ($\half Mv^2$ is translational energy and
    $\half I\omega^2$ is rotational energy).
    Total energy is ($x = \text{ distance down
    slope}$):
    \[ E = \half \left( \frac{I}{a^2} + M \right)
    \dot{x}^2 - Mgx \sin\alpha \]
    ($\dot{E} = 0$ to get equation of motion):
    \[ \left( \frac{I}{a^2} + M \right) \ddot{x} =
    Mg \sin\alpha .\]
\end{example*}

\newpage
\section{Non-Inertial Frames}

\subsection{Rotating Frames}

Let $S$ be an inertial frame. $S'$ is a
non-inertial frame, rotating with respect to $S$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/02dcec5699c911ec.png}
\end{center}
\ul{Question}: If you sit in $S'$, what do
Newton's Laws look like? \\
Consider a particle stationary in $S'$. Then from
the perspective of $S$, it moves with velocity
\[ \dot{\bf{r}} = \bf{\omega} \times \bf{r} \]
with $\bf{\omega} = \dot{\theta} \hat{\bf{z}}$ in
the diagram. In general, $\bf{\omega}$ points
along the direction of rotation. We can apply this
formula to the axes of $S'$. Let $\bf{e}_i'$ ($i =
1, 2, 3$) be unit vectors pointing in $x'$, $y'$
and $z'$ directions $S'$. Then these rotate, with
\[ \boxed{\dot{\bf{e}}_i' = \bf{\omega} \times
\bf{e}_i'} \]
(this is the main formula that allows us to
understand motion in rotating forces).

\subsubsection*{Velocity and Acceleration in Rotating Frames}

Consider a particle now moving on a trajectory in
$S'$. The position of the particle as measured in
$S$ is
\[ \bf{r} = r_i \bf{e}_i \]
(note summation convention). Measuring the same
point in $S'$, we write
\[ \bf{r} = r_i' \bf{e}_i' \]
The velocity in frame $S$ is
\[ \dot{\bf{r}} = \dot{r}_i \bf{e}_i \]
(because unit vectors in $S$ don't change with
time) while in frame $S'$ the velocity is
\begin{align*}
    \dot{\bf{r}}
    &= \dot{r}_i' \bf{e}_i' + r_i' \dot{\bf{e}}_i'
    \\
    &= \dot{r}_i'\bf{e}_i' + r_i' (\bf{\omega}
    \times \bf{e}_i') \\
    &= \dot{r}_i' \bf{e}_i' + (\bf{\omega} \times
    \bf{r})
\end{align*}
Now we introduce some new notation to highlight
the physical difference between the two frames. \\
The velocity of the particle as seen by an
observer in $S$ is
\[ \left( \dfrac{\bf{r}}{t} \right)_S := \dot{r}_i
\bf{e}_i \]
The velocity of the particle as seen by an
observer in $S'$ is:
\[ \left( \dfrac{\bf{r}}{t} \right)_{S'} =
\dot{r}_i' \bf{e}_i' .\]
From above, we have
\[ \left( \dfrac{\bf{r}}{t} \right)_S = \left(
\dfrac{\bf{r}}{t} \right)_{S'} + \bf{\omega}
\times \bf{r} \]
For acceleration, we just take another time
derivative. In $S$, we get $\ddot{\bf{r}} =
\ddot{r}_i \bf{e}_i$ whereas in $S'$ we get
\[ \ddot{\bf{r}} = \ddot{r}_i \bf{e}_i' +
\dot{r}_i' \bf{e}_i + \dot{r}_i' (\bf{\omega}
\times \bf{e}_i) \]
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