A frame of reference (i.e. set of coordinates) that is not accelerating.

Any frame moving at a constant relative velocity to the original is also an inertial frame.

Position | Velocity $\left (dx/dt \right)$ | Acceleration $\left (d^2x/dt^2 \right)$ |
---|---|---|

$$ \begin{align*} x &= x' + Vt' \\ y &= y' \\ z &= z' \end{align*} $$ | $$ \begin{align*} v_x &= v'_x + V \\ v_y &= v'_y \\ v_z &= v'_z \end{align*} $$ | $$ \begin{align*} a_x &= a'_x \\ a_y &= a'_y \\ a_z &= a'_z \end{align*} $$ |

Acceleration is the same in both frames: **Galilean Invariant**

What about non-inertial reference frames?

Where did this come from?

But what about when acceleration is not just a constant value?

What's the easiest $a(t)$ you can think of, where $$\frac{da}{dt} \neq 0$$

- No change in velocity if there is no net force acting.
- A force will cause a proportional acceleration inversely proportional to the mass.
- Forces come in equal and opposite pairs

- $\Delta v = 0 $ if $F_\textrm{net} = 0$
- $\mathbf{F} = m\mathbf{a}$
- $F_{AB} = -F_{BA}$

\begin{equation} \mathbf{F} = m\mathbf{a} \label{eq:newtons2ndlaw} \end{equation}

Or perhaps as: \begin{equation} \mathbf{F} = m \frac{d\mathbf{v}}{dt} = \frac{d\mathbf{p}}{dt} \label{eq:newtons2ndlawP} \end{equation} assuming mass doesn't change w.r.t time.

An object falls from a height $h$. How long does it take to hit the ground?

What is $g$ at the height of say, the International Space Station?

$$\Sigma F = F(v) + \ldots = m a$$

This leads to equations of motion that resemble: $$ m \frac{dv}{dt} = -A v^n $$ or (ignoring the constants and stuff) $$ \dot{v} = -v$$

($\dot{v} \equiv \frac{dv}{dt}$ and $\ddot{v} \equiv \frac{d^2v}{dt^2}$)

If $$\dot{v} = v$$ what is $v(t)$?

What is $x(t)$?

Find the $v(t)$ for a bacterium coasting in a viscous fluid. You can assume it starts with a velocity $v_0$ and then is slowed down by a force given by $F = -bv$ where $b$ is a constant that depends on the size and shape of the bacterium and the viscosity of the fluid.

$$f(v) = b v + c v^2 = f_\textrm{lin} + f_\textrm{quad}$$

\begin{equation} F_\textrm{Stokes} = 6 \pi \eta R v \end{equation}

- $\eta$ is the viscosity of the fluid.
- $R$ is the radius of the object (assumed to be a sphere)
- $v$ is the velocity of the object relative to the fluid

\begin{equation} F_\textrm{Newtonian} = \frac{1}{2} \rho v^2 c_d A \end{equation}

- $\rho$ is the mass density of the fluid.
- $A$ is the cross sectional area of the object
- $v$ is the velocity of the object relative to the fluid
- $c_d$ is the drag coefficient

The ratio of these two forces, the quadratic and the linear drags, can be used to describe systems: $$\mathbf{Re}\; \textrm{or}\; R \propto \frac{f_\textrm{quad}}{f_\textrm{lin}}$$

\begin{equation}
\frac{f_\textrm{quad}}{f_\textrm{lin}} = \frac{cv^2}{bv} = \left(3.2 \times 10^{3} \frac{\textrm{s}}{\textrm{m}^2} \right) R v
\end{equation}

We can take this ratio and use it to figure out which terms are important. All the constant are given by the various parameters of air (for example) and can be merged into one number multiplied by the radius and the speed.

What comes to mind when you see this?

Non-Linear Drag

A ball drops from a height $h$ and reaches the ground a time $t$ later. Let the ball have a mass of $m$ and a radius of $r$. What is the height of the building in terms of $t$? Air resistance is proportional the square of the velocity.

```
m = 1;
g = 9.80;
Cdrag = 0.5;
\[Rho] = 1.29;
r = 0.1;
A = \[Pi] r^2;
vterm = Sqrt[(2 m g )/(Cdrag \[Rho] A )];
yDrag[t_] := vterm^2/g Log[Cosh[(g t)/vterm]]
vDrag[t_] := vterm Tanh[(g t)/vterm]
Plot[yDrag[t], {t, 0, 10}]
Plot[vDrag[t], {t, 0, 10}]
```