The Prodigal Physicist

Dimensional Analysis

First, let’s do a super-fast review of dimensional analysis since it will be your best method of spot checking your calculations, not to mention a handy way of guessing answers on obnoxiously long multiple-choice tests.

When solving for a variable in an equation, dimensional analysis will tell you if you’re on the right track or not. For example, an object with initial position, initial velocity and acceleration (in one direction) of x_o, v₀ and a respectively, can be found at a time t by the equation:

$null$

Regardless of the equation’s significance, we need the value of n.

Using dimensional analysis, or by doing “normal math” on the units, we can find out very quickly. So,

$null$

Since x must be in meters, the seconds must cancel, so n must be 2. For absolutely any equation, just jot down the units and cancel everything you can.

Equations of Motion in Cartesian Coordinates

Kinematics is not actually physics in the truest sense since it only answers the “how” of an arbitrary equation, not the “why” of the system’s behavior. But damnit, it’s useful.

The instantaneous velocity of an object as a function of time is given by an infinitesimal change in position over an infinitesimal period of time or:
v(t) = ^dx/_dt

The instantaneous acceleration is just the derivative of the velocity, or an infinitesimal change in velocity over an infinitesimal period of time:
a(t) = ^dv/_dt

And keep in mind that any variable with ₀ (phonetically “nawt”) means the initial value (t = 0).

Here are the kinematic equations of motion:

$\textbf{v}=\textbf{v}_0+\textbf{a}t$
$\textbf{v}^2=\textbf{v}_0^2+2\textbf{a}(\textbf{x}-\textbf{x}_0)$
$\textbf{x}=\textbf{x}_0+\textbf{v}_0t+\frac{1}{2}\textbf{a}t^2$
$\textbf{x}=\textbf{x}_0+\frac{1}{2}(\textbf{v}_0+\textbf{v})t$

Keep in mind that there are only so many well-formed questions you can ask with these equations:

t (v), (What is the time when the velocity is v?)

t (x), (What is the time when the position is x?)

x (t), (What is the position at time t?)

x (v), (What is the position at a velocity v?)

v (x), (What is the velocity at a position x?)

v (t), (What is the velocity at time t?)

projectile motion

Projectile motion is really just a special case of motion where the acceleration is constant. In the y-direction (+ up and - down), the force is a constant g (= -9.8 m/s² = the force of Earth’s gravity).

Here’s one example, but there are several more in Problem Set 2.

Example: Swish!

What initial velocity is needed to make the shot?

Well, the definition of making the shot is: x (y = 3) = 6
Or, x (t = t_swish) = 6 and y (t = t_swish) = 3

First the x component:

$x=x_0+v_{0x}t$

$6=0+v_0cos45(t_{swish})$

$t_{swish}=\frac{6}{v_0 cos45}$

Then the y component:

$y=y_0+v_{0y}t-gt^2$

$3= 1.8+v_0sin45(t_{swish})-4.9t^2$

But we know the value of t_swish from our calculation of the x component.
Just plug it in:

$3=1.8+v_0sin45(\frac{6}{v_0cos45})-4.9({\frac{6}{v_0cos45})^2}$

$-4.8=\frac{352.8}{v_0^2}\hspace{10 mm}so,\hspace{10 mm}v_0=8.573m/s$

uniform circular motion

For uniform circular motion, it’s much easier to speak in terms of radii and angles instead of x and y components. Using the graph below, we’ll get our new variables.

A particle moving in a circle of radius r will sweep an infinitesimal angle dθ and move a distance s. With constant velocities, we get:

$s=rsind\theta=rd\theta$

(since sine of a very small angle is approximately equal to the angle)

$d\theta=\frac{v}{r}dt\qquad|dv|=\frac{v^2}{r}dt$

And we get the centripetal acceleration of an object moving in a circle:

$\frac{|dv|}{dt}=\frac{v^2}{r}=\textbf{a}=\frac{4\pi^2\textbf{r}}{\tau^2}\hspace{10 mm}where,\hspace{10 mm}\tau=\frac{2\pi\textbf{r}}{\textbf{v}}$

Non-linear coordinate systems

Using these radii and angles, we can construct a set of equations for rotational motion using cylindrical coordinates. And then, we can extrapolate to spherical coordinates!

The kinematic equations of rotational motion:

$\omega=\omega_0+\alpha{t}$
$\omega^2=\omega_0^2+2\alpha\theta$
$\theta=\omega_0t+\frac{1}{2}{\alpha}t^2$
$\theta=\frac{\omega_0+\omega}{2}t$

Spherically speaking, the coordinates become:

$\begin{array}{lll} r=\rho\sin{\phi}&x=r\cos{\theta}&x=\rho\sin\phi\cos{\theta}\\ \\ z=\rho\cos{\phi}&y=r\sin\theta&y=\rho\sin\phi\sin\theta\\ \\ \theta=\theta&z=z&z=\rho\cos\phi \end{array}$

Classical Mechanics

Kinematics

Dimensional Analysis

Equations of Motion in Cartesian Coordinates

projectile motion

Example: Swish!

uniform circular motion

Non-linear coordinate systems

Problem Set 2 » Kinematics