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\subsubsection*{Variable Mass Problem}

N2 really is $\bf{F} = \dot{\bf{p}}$ where $p = m
\dot{\bf{x}}$ is momentum. This coincides with the
more familiar $\bf{F} = m \ddot{\bf{x}}$ only
when $m$ doesn't change with $t$, for example
where it \emph{does}: \emph{Rockets}. \myskip
A rocket moves in a straight line with velocity
$\bf{v}(t)$. The mass of the rocket $m(t)$ changes
with time because it propels itself forward by
spitting fuel out behind. \\
Suppose the fuel is emitted at speed $u$ relative
to the rocket. \\
\textbf{Question}: what is $v(t)$? \\
analysis
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/a59e4b6c8f4011ec.png}
\end{center}
momentum conserved (later)
\[  p(t + \delta t) = p_{\text{rocket}}(t + \delta
t) + p_{\text{fuel}}(t + \delta t) \]
\begin{align*}
    p_{\text{rocket}} (t + \delta t)
    &= m(t + \delta t) v(t + \delta t) \\
    &= \left( m(T) + \delta t \dfrac{m}{t} \right)
    \left( v(t) + \delta t \dfrac{v}{t} \right) +
    O(\delta t^2) \\
    &= m(t)v(t) + \delta t \left( m \dfrac{v}{t} +
    v \dfrac{m}{t} \right) + O(\delta t^2) \\
    p_{\text{fuel}}
    &= [m(t) - m(t + \delta t)][v(t + \delta t) -
    u] \\
    &= -\delta \dfrac{m}{t}[v(t) - u] + O(\delta
    t^2)
\end{align*}
substitute in
\[ p(t + \delta t) = p(t) + \left( m \dfrac{v}{t}
+ u \dfrac{m}{t} \right) \delta t + O(\delta t^2) \]
N2:
\[ \lim_{\delta t \to 0} \frac{p(t + \delta t) -
p(t)}{\delta t} = \boxed{F = m(t) \dfrac{v}{t}
+ u \dfrac{m}{t}} \]
(the boxed part is Tsiolkovsky rocket equation
(1903)). For example rocket in deep space (i.e. $F
= 0$) Here, the Tsiolkovsky equation becomes
\[ \dfrac{v}{t} = -\frac{u}{m(t)} \dfrac{m}{t} \]
integrate
\[ \implies v(t) = v_0 + u \ln \left(
\frac{m_0}{m(t)} \right) \]
where the rocket (and fuel inside) has mass $m_0$
when its speed is $v_0$.

\begin{note*}
    burning fuel only increases speed
    logarithmically!
\end{note*}

\noindent
We shall assume that the fuel is burnt at a
constant rate:
\[ \dfrac{m}{t} = -\alpha \implies m(t) = m_0 -
\alpha t \]
\begin{note*}
    $\alpha > 0 \implies \dfrac{m}{t} < 0$.
\end{note*}
\noindent
sub in:
\[ v(t) = v_0 - u \ln \left( 1 - \frac{\alpha
t}{m_0} \right) \]

\begin{note*}
    Only makes sense for $t < \frac{m_0}{\alpha}$,
    since at $t = \frac{m_0}{\alpha}$, all of the
    mass in the rocket is burnt, including the
    rocket ($m = 0$). Assuming this, integrate
    once more to yield the distance travelled:
    \[ x = v_0 t + \frac{um_0}{\alpha} \left[
    \left( 1 - \frac{\alpha t}{m_0} \right) \ln
    \left( 1 - \frac{\alpha t}{m_0} \right)  +
    \frac{\alpha t}{m_0} \right] .\]
\end{note*}

\newpage
\section{Rigid Bodies}

A \emph{rigid body} is an extended object
comprised of $N$ particles that are constrained so
that the relative distance between any two,
$|\bf{x}_i - \bf{x}_j|$, is fixed. \\
A rigid body can undergo two types of motion:
\begin{itemize}
    \item its centre of mass can move
    \item it can rotate
\end{itemize}
We'll begin with rotations of rigid bodies.

\subsection{Angular Velocity}

Consider a single particle rotating in a circle
around the $z$-axis.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d0452abe8f4211ec.png}
\end{center}
The position is
\[ \bf{x} = (d \cos \theta, d \sin\theta, z) \]
\[ \implies \dot{\bf{x}} = (-\dot{\theta} d
\sin\theta, \dot{\theta} d \cos\theta, 0) \]
We can write $\dot{\bf{x}}$ by introducing a new
vector
\[ \bf{\omega} = \dot{\theta} \hat{\bf{z}} \qquad
\text{and} \qquad \dot{\bf{x}} = \bf{\omega}
\times \bf{x} \]
we call $\bf{\omega}$ the \emph{angular velocity}.
In general, we write
\[ \bf{\omega} = \omega \hat{\bf{n}} \]
where $\omega$ is the angular speed,
$|\dot{\theta}|$ and $\hat{\bf{n}}$ is the unit
vector in direction of motion in the right-handed
sense.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/ed17415e8f4211ec.png}
\end{center}
(curl fingers of right hand in direction of motion
and the thumb points at $\hat{\bf{n}}$) \\
The speed of the particle is $v = |\dot{\bf{x}}| =
\omega d$ where $d = |\hat{\bf{n}} \times \bf{x}|
= |\bf{x}| \sin\phi$ is the perpendicular distance
to the axis of rotation. The kinetic energy of the
particle is
\begin{align*}
    T
    &= \half m \dot{\bf{x}}^2 \\
    &= \half m (\bf{\omega} \times \bf{x}) \cdot
    (\bf{\omega} \times \bf{x}) \\
    &= \half m d^2 \omega^2
\end{align*}

\subsubsection*{Moment of Inertia}

For a rigid body, all $N$ particles rotate with
the same angular velocity, so
\[ \dot{\bf{x}}_i = \bf{\omega} \times \bf{x}_i
\qquad i = 1, \dots, N \]
This ensures that
\begin{align*}
    \dfrac{}{t} (\bf{x}_i - \bf{x}_j)^2
    &= 2 (\dot{\bf{x}}_i - \dot{\bf{x}}_j) \cdot
    (\bf{x}_i - \bf{x}_j) \\
    &= 2[\bf{\omega} \times (\bf{x}_i - \bf{x}_j)]
    \cdot (\bf{x}_i - \bf{x}_j) \\
    &= 0
\end{align*}
The kinetic energy is
\[ T = \half \sum_i m_i \dot{\bf{x}}_i \cdot
\dot{\bf{x}}_i := \half \bf{T} \omega^2 \]
where
\[ \bf{T} := \sum_{i = 1}^N m_i d_i^2 \]
is the \emph{moment of inertia} of the body.