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\[ \left( \dfrac[2]{\bf{r}}{t} \right)_S = \left(
\dfrac{\bf{r}}{t} \right)_{S'} + 2\bf{\omega}
\times \left( \dfrac{\bf{r}}{t} \right)_{S'} +
\dot{\bf{\omega}} \times \bf{r} + \bf{\omega}
\times (\bf{\omega} \times \bf{r}) \tag{$*$} \]

\subsection{Newton's Laws in a Rotating Frame}

In the inertial frame $S$, we have
\[ \bf{F} = m \left( \dfrac[2]{\bf{r}}{t} \right)_S \]
Then, from ($*$) in $S'$ we have
\[ m \left( \dfrac[2]{\bf{r}}{t} \right)_{S'} =
Fbf - 2m \bf{\omega} \times \left(
\dfrac{\bf{r}}{t} \right)_{S'} - m
\dot{\bf{\omega}} \times \bf{r} - m\bf{\omega}
\times (\bf{\omega} \times \bf{r}) \]
To explain the motion of the particle viewed from
$S'$, we must include the extra terms on the RHS.
They are called \emph{fictitious forces}. They are
necessary to explain the seeming departure from
uniform motion of a free particle due to the
rotating frame.
\begin{itemize}
    \item $2m \bf{\omega} \times \left(
        \dfrac{\bf{r}}{t} \right)_{S'}$
        \emph{Coriolis force}
    \item $m \dot{\bf{\omega}} \times \bf{r}$
        \emph{Euler force}
    \item $m \bf{\omega} \times (\bf{\omega}
        \times \bf{r})$ \emph{Centrifugal force}
\end{itemize}
The most familiar non-inertial frame is this room.
Earth rotates once a day around N-S axis and once
a year around the sun, which rotates around the
galaxy. \\
$R_\text{earth} \approx 6000\mathsf{km}$, earth
spins with an angular frequency of
\[ \omega_{\text{rot}} = \frac{2\pi}{1\text{ day}}
\approx 7 \times 10^{-5} \mathsf{s^{-1}} \]
angular frequency of orbit:
\[ \omega_{\text{orb}} = \frac{2\pi}{1\text{
year}} \approx 2 \times 10^{-7} \mathsf{s^{-1}} \]

\begin{note*}
    \[ \frac{\omega_{\text{rot}}}{\omega_{\text{orb}}}
    \approx 365 = \frac{T_{\text{orb}}}{T_{\text{rot}}} \]
\end{note*}

\noindent
Here, we won't consider the Euler force.

\subsection{Centrifugal Force}

\[ \bf{F}_{\text{cent}} = -m\bf{\omega} \times
(\bf{\omega} \times \bf{r}) = -m (\bf{\omega}
\cdot \bf{r}) \bf{\omega} + m\omega^2 \bf{r} \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/3e440f2a99d011ec.png}
\end{center}
$\bf{\omega} \times \bf{r}$ points into the paper
so $-\bf{\omega} \times (\bf{\omega} \times
\bf{r})$ points away from the axis of rotation, as
shown.
\[ |\bf{F}_{\text{cent}}| = m\omega^2 r \cos\theta
= m\omega^2 d \]
$\bf{F}_{\text{cent}}$ doesn't depend on the
velocity of the particle. In fact, it's a
conservative force:
\[ \bf{F}_{\text{cent}} = -\nabla V \qquad
\text{with} \qquad V = -\frac{m}{2} |\bf{\omega}
\times \bf{r}|^2 \]

\subsubsection*{Example: Apparent Gravity}

Suspend a piece of string from the ceiling.
Centrifugal deflects the string (a bit) from
pointing straight down to the centre of the earth.
But by how much?
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/39acfebc99d111ec.png}
\end{center}
\ul{Question}: How bis is $\phi$? \\
The effective acceleration is due to the
combination of gravity and the centrifugal force
\[ \bf{g}_{\text{eff}} = \bf{g} - \bf{\omega}
\times (\bf{\omega} \times \bf{r}) \]
We resolve the central force
\begin{align*}
    \bf{F}_{\text{cent}}
    &= |\bf{F}| \cos\theta \hat{\bf{r}}
    - |\bf{F}| \sin\theta \hat{\bf{\theta}} \\
    &= m\omega^2 r \cos^2\theta \hat{\bf{r}} -
    m\omega^2 r \cos\theta \sin\theta
    \hat{\bf{\theta}}
\end{align*}
(Note at the north pole, $\theta = \frac{\pi}{2}$
and $F_{\text{cent}} = 0$, as expected). Giving
the effective acceleration:
\begin{align*}
    \bf{g}_{\text{eff}}
    &= -g \hat{\bf{r}} - \bf{\omega} \times
    (\bf{\omega} \times \bf{r}) \\
    &= (-g + \omega^2 R \cos^2 \theta)
    \hat{\bf{r}} - \omega^2 R \cos\theta
    \sin\theta
\end{align*}
We can also resolve tension in the string
\[ \bf{T} = T \cos\phi \hat{\bf{r}} + T\sin\phi
\hat{\bf{\theta}} \]
which balances the $m \bf{g}_{\text{eff}}$ force.
\[ \bf{T} + m \bf{g}_{\text{eff}} = 0 \implies
\tan\phi = \frac{\omega^2 R\cos\theta \sin
\theta}{g - \omega^2 R\cos^2\theta} \]
(Note $\omega^2 R = 3 \times 10^{-2}
\mathsf{ms^{-2}}$). $\phi$ is (approximately)
maximised at
\[ \dfrac{}{\theta} (\cos\theta \sin\theta) = 0 \]
i.e. $\theta \approx 45^\circ$.
\[ \tan\phi \approx \frac{\omega^2 R}{q} \approx
o(10^{-3}) \]
so very small. At the equator, $\theta = 0$, $\phi
= 0$. But gravity is weaker there:
\[ \bf{g}_{\text{eff}} \mid_{\text{equator}} = g -
\omega^2 R \implies g - g_{\text{eff}} = 10^{-2}
\mathsf{ms^{-2}} \]

\begin{example*}[Rotating Water Bucket]
    Gravity acts down on water and centrifugal
    force pushes it to the side.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/09f29d2a99d211ec.png}
    \end{center}
    \ul{Question}: What's the shape of the
    surface? \\
    Assume water spins with bucket. Consider a
    water molecule, mass $m$ on the surface. It's
    potential energy is
    \[ V = mgz - \half m \omega^2 r^2 \]
    If the water molecule could change its energy
    by moving along the surface, then it would.
    But we're looking for the equilibrium shape of
    surface i.e. each point on it has equal
    potential energy. Put $V = \text{ constant}$:
    \[ z = \frac{\omega^2}{2g}r^2 + \text{constant} \]
    i.e. a parabola!
\end{example*}
