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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>Classical Mechanics: Notes</title>
<meta name="generator" content="Org mode">
<meta name="author" content="George Kontsevich">
<meta name="description" content="Projectors"
>
<link rel="stylesheet" type="text/css" href="../web/worg.css" />
<link rel="shortcut icon" href="../web/panda.svg" type="image/x-icon">
</head>
<body>
<div id="content">
<h1 class="title">Classical Mechanics: Notes</h1>
<div id="table-of-contents">
<h2>Table of Contents</h2>
<div id="text-table-of-contents">
<ul>
<li><a href="#org4e11192">Newton's Laws</a>
<ul>
<li><a href="#org3d05b66">First Law</a></li>
<li><a href="#org8e69e61">Second Law</a>
<ul>
<li><a href="#org09d5bba">Solve for velocity and position</a></li>
</ul>
</li>
<li><a href="#org39e76d9">Third Law</a>
<ul>
<li><a href="#org3d86fc0">Conservation of momentum</a></li>
<li><a href="#org07cd5a9">leveraging cartesian coordinates</a></li>
</ul>
</li>
</ul>
</li>
<li><a href="#orge9f688e">Projectiles</a>
<ul>
<li><a href="#org06a8122">Linear Drag</a>
<ul>
<li><a href="#org16bcd41">Dropping a weight</a></li>
</ul>
</li>
</ul>
</li>
</ul>
</div>
</div>
<div id="outline-container-org4e11192" class="outline-2">
<h2 id="org4e11192">Newton's Laws</h2>
<div class="outline-text-2" id="text-org4e11192">
<dl class="org-dl">
<dt>Mass</dt><dd>resistance to being accelerated</dd>
<dt>Weight</dt><dd>gravitation force of an object</dd>
<dt>Momentum</dt><dd>mass * velocity</dd>
<dt>1 Newton</dt><dd>force to accelerate <i>1 kg</i> by <i>1 m/s<sup>2</sup></i></dd>
</dl>
</div>
<div id="outline-container-org3d05b66" class="outline-3">
<h3 id="org3d05b66">First Law</h3>
<div class="outline-text-3" id="text-org3d05b66">
<blockquote>
<p>
In the absense of forces - a point mass moves with constant velocity <b>v</b> (= zero acceleration)
</p>
</blockquote>
</div>
</div>
<div id="outline-container-org8e69e61" class="outline-3">
<h3 id="org8e69e61">Second Law</h3>
<div class="outline-text-3" id="text-org8e69e61">
<blockquote>
<p>
The net force on a particle is <b>F=ma</b> <br>
Where <b>a = dv/dt = d<sup>2</sup>r/dt<sup>2</sup></b> <br>
In other words, force is a change in momentum <b>p</b>. ie. <b>F= dp/dt</b>
</p>
</blockquote>
</div>
<div id="outline-container-org09d5bba" class="outline-4">
<h4 id="org09d5bba">Solve for velocity and position</h4>
<div class="outline-text-4" id="text-org09d5bba">
<p>
Ex: Accelerting a particle with force <b>F<sub>0</sub></b> we can write out the equation for the second law
</p>
<blockquote>
<p>
F<sub>0</sub> = m ⋅ a(t) = m ⋅ d<sup>2</sup>x(t)/dt<sup>2</sup>
</p>
</blockquote>
<p>
Then by integration we can solve
</p>
<blockquote>
<p>
∫F<sub>0</sub> = ∫m⋅d<sup>2</sup>x(t)/dt<sup>2</sup> <br>
∫F<sub>0</sub>/m = ∫d<sup>2</sup>x(t)/dt<sup>2</sup> <br>
[F<sub>0</sub>/m] t = v<sub>0</sub> + dx(t)/dt <br>
</p>
</blockquote>
<p>
With this, given a starting velocity <b>v<sub>0</sub></b> and a time <b>t</b> we can solve for the velocity <b>dx/dt</b>
</p>
<blockquote>
<p>
∫[F<sub>0</sub>/m] t = ∫v<sub>0</sub> + dx(t)/dt <br>
[F<sub>0</sub>/2m] t<sup>2</sup> = v<sub>0</sub>t + x<sub>0</sub> + x(t) <br>
</p>
</blockquote>
<p>
Integrating once more we get an equation for the position. We just need a starting position <b>x<sub>0</sub></b> to solve
</p>
</div>
</div>
</div>
<div id="outline-container-org39e76d9" class="outline-3">
<h3 id="org39e76d9">Third Law</h3>
<div class="outline-text-3" id="text-org39e76d9">
<blockquote>
<p>
If <code>Object1</code> exerts a force <b>F<sub>21</sub></b> on <code>Object2</code><br>
Then <code>Object2</code> always exerts an opposite force <b>F<sub>12</sub></b> <br>
Such that <b>F<sub>12</sub> = -F<sub>21</sub></b>
</p>
</blockquote>
</div>
<div id="outline-container-org3d86fc0" class="outline-4">
<h4 id="org3d86fc0">Conservation of momentum</h4>
<div class="outline-text-4" id="text-org3d86fc0">
<p>
Givens:
</p>
<ul class="org-ul">
<li>If you have a system with many particles <b>α = [1,2 … N]</b></li>
<li>They extert an influence on each other particle <b>F<sub>α β</sub></b></li>
<li>They are under some external force <b>F<sup>ext</sup><sub>α</sub></b></li>
</ul>
<p>
The total force on each particle will be:
</p>
<blockquote>
<p>
F<sub>α</sub> = ∑<sub>β≠α</sub> F<sub>αβ</sub> + F<sup>ext</sup><sub>α</sub>
</p>
</blockquote>
<p>
This will equal the change in momentum for that particle - <b>dp<sub>α</sub>/dt</b> (b/c Force = change in momentum)
</p>
<blockquote>
<p>
dp<sub>α</sub>/dt = ∑<sub>β≠α</sub> F<sub>αβ</sub> + F<sup>ext</sup><sub>α</sub>
</p>
</blockquote>
<p>
The total momentum of the system will be
</p>
<blockquote>
<p>
P = ∑p<sub>α</sub>
</p>
</blockquote>
<p>
The corresponding change in momentum of the system will be
</p>
<blockquote>
<p>
dP/dt = ∑dp<sub>α</sub>/dt
</p>
</blockquote>
<p>
(the derivative of a sum is the sum of derivatives) <br>
And then we just plug in..
</p>
<blockquote>
<p>
dP/dt = ∑<sub>α</sub>∑<sub>β≠α</sub> F<sub>αβ</sub> + ∑<sub>α</sub>F<sup>ext</sup><sub>α</sub>
</p>
</blockquote>
<p>
But for every term F<sub>ab</sub> = -F<sub>ba</sub>, so all the terms cancel out in the first sum
</p>
<blockquote>
<p>
dP/dt = 0 + ∑<sub>α</sub>F<sup>ext</sup><sub>α</sub> <br>
dP/dt = F<sup>ext</sup>
</p>
</blockquote>
<p>
And the total change in momentum is just the sum all the external forces. The forces internal to the system can't change the momentum of the system as a whole!
</p>
</div>
</div>
<div id="outline-container-org07cd5a9" class="outline-4">
<h4 id="org07cd5a9">leveraging cartesian coordinates</h4>
<div class="outline-text-4" id="text-org07cd5a9">
<p>
We can also break up forces along coordinates to make our lives easier by using unit vectors ex: <i>x<sup>^</sup></i>, <i>y<sup>^</sup></i>
</p>
<blockquote>
<p>
F = F<sub>x</sub>x<sup>^</sup> + F<sub>y</sub>y<sup>^</sup>
</p>
</blockquote>
<p>
And these components can be treated separately b/c they differentiate independently
</p>
<blockquote>
<p>
r = x ⋅ x<sup>^</sup> + y ⋅ y<sup>^</sup> <br>
dr/dt = dx/dt ⋅ x<sup>^</sup> + dy/dt ⋅ y<sup>^</sup> <br>
d<sup>2</sup>r/dt<sup>2</sup> = d<sup>2</sup>x/dt<sup>2</sup> ⋅ x<sup>^</sup> + d<sup>2</sup>y/dt<sup>2</sup> ⋅ y<sup>^</sup>
</p>
</blockquote>
<p>
So our <i>F=ma</i> can be rewritten
</p>
<blockquote>
<p>
F = F<sub>x</sub>x<sup>^</sup> + F<sub>y</sub>y<sup>^</sup> = m ⋅ d<sup>2</sup>x/dt<sup>2</sup> ⋅ x<sup>^</sup> + m ⋅ d<sup>2</sup>y/dt<sup>2</sup> ⋅ y<sup>^</sup>
</p>
</blockquote>
<p>
And each force can be looked at explicitely
</p>
<blockquote>
<p>
F<sub>x</sub> = m ⋅ d<sup>2</sup>x/dt<sup>2</sup> <br>
F<sub>y</sub> = m ⋅ d<sup>2</sup>y/dt<sup>2</sup>
</p>
</blockquote>
<p>
So if you have a sliding block on a plane.. you set your coordinate system so that <i>x<sup>^</sup></i> is along the plane and <i>y<sup>^</sup></i> is orthogonal to the plane and then you can look at the forces separately.
</p>
<p>
In the <i>y</i> direction there is no movement so the normal force cancels out with the force of gravity
</p>
<blockquote>
<p>
F<sub>y</sub> = N - mg cos(θ) = 0 <br>
<i>so we can solve for the normal force</i> N <br>
N = mg cos(θ)
</p>
</blockquote>
<p>
In the <i>x</i> direction there is gravity and friction - where friction is a function of the <i>normal force N</i>
</p>
<blockquote>
<p>
f = μN <br>
f = μmg cos(θ)
</p>
</blockquote>
<p>
And gravity we already know how to calculate
</p>
<blockquote>
<p>
<i>gravitation force in the direction of</i> x<sup>^</sup> = mg sin(θ)
</p>
</blockquote>
<p>
So the total force in <i>x<sup>^</sup></i>
</p>
<blockquote>
<p>
F<sub>x</sub> = m ⋅ d<sup>2</sup>x/dt<sup>2</sup> = mg sin(θ) + f <br>
m ⋅ d<sup>2</sup>x/dt<sup>2</sup> = mg sin(θ) + μmg cos(θ)
</p>
</blockquote>
<p>
And we can solve for <i>d<sup>2</sup>x/dt<sup>2</sup></i>, <i>dx/dt</i>, <i>x</i>
</p>
<blockquote>
<p>
d<sup>2</sup>x/dt<sup>2</sup> = g sin(θ) + μg cos(θ) <br>
dx/dt = [g sin(θ) + μg cos(θ)]t <br>
x(t) = (1/2)[g sin(θ) + μg cos(θ)]t<sup>2</sup>
</p>
</blockquote>
<p>
(these assume the initial velocity and position are <i>0</i>)
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-orge9f688e" class="outline-2">
<h2 id="orge9f688e">Projectiles</h2>
<div class="outline-text-2" id="text-orge9f688e">
<p>
A class of problems involve projectiles traveling through air. These undergo two forms of drag - quadratic and linear grad - with the former dominating for large object and the latter dominating for small things (like small oil drops and mist)
</p>
</div>
<div id="outline-container-org06a8122" class="outline-3">
<h3 id="org06a8122">Linear Drag</h3>
<div class="outline-text-3" id="text-org06a8122">
<p>
This kind of drag decreases the speed of an object proportional to the current speed
</p>
<blockquote>
<p>
dv/dt = -kv
</p>
</blockquote>
<p>
When you have a variable that changes as a function of itself then you should think of the exponential <i>e<sup>-kt</sup></i>. It has the property that its derivative is equal to itself
</p>
<blockquote>
<p>
v = Ae<sup>-kt</sup>
</p>
</blockquote>
</div>
<div id="outline-container-org16bcd41" class="outline-4">
<h4 id="org16bcd41">Dropping a weight</h4>
<div class="outline-text-4" id="text-org16bcd41">
<blockquote>
<p>
F<sub>falling</sub> = F<sub>gravity</sub> + F<sub>drag</sub> <br>
m⋅dv(t)/dt = mg - bv(t)
</p>
</blockquote>
<p>
Solving this is a bit more difficult. You proceed by subsitution..
</p>
<blockquote>
<p>
u=(v(t)-mg/b) <br>
du/dt = dv(t)/dt <br>
∴ <br>
m⋅dv(t)/dt = mg - bv(t) <br>
m (dv(t)/dt) = -b ((v(t)-mg/b) <br>
m du/dt = -b u <br>
</p>
</blockquote>
<p>
We have gotten it into the same form as the simpler problem and therefore
</p>
<blockquote>
<p>
u = Ae<sup>-tb/m</sup> = (v(t)-mg/b) <br>
∴ <br>
v(t) = Ae<sup>-tb/m</sup>+mg/b <br>
<i>and</i> <br>
dv(t)/dt = -A(b/m)e<sup>-tb/m</sup>
</p>
</blockquote>
<p>
<i>A</i> is still a free variable, which we constrain by adding a starting velocity so that at <i>t=0</i> -> <i>v(0) = v<sub>0</sub></i>
</p>
<blockquote>
<p>
v<sub>0</sub> = Ae<sup>0</sup>+mg/b<br>
A = v<sub>0</sub> - mg/b
</p>
</blockquote>
<p>
So the final velocity is
</p>
<blockquote>
<p>
v(t) = (v<sub>0</sub> - mg/b)e<sup>-tb/m</sup>+mg/b <br>
</p>
</blockquote>
<p>
..<br>
Pluggin back into the original equation to double check
</p>
<blockquote>
<p>
m⋅dv(t)/dt = mg - bv(t) <br>
-mAe<sup>-tb/m</sup> = mg - b(Ae<sup>-tb/m</sup>+mg/b) <br>
✓
</p>
</blockquote>
<p>
<b>Note:</b> When the net force is zero and acceleration stops (the terminal velocity)
</p>
<blockquote>
<p>
m⋅dv(t)/dt = mg - bv(t) <br>
0 = mg - bv<sub>terminal</sub> <br>
v<sub>terminal</sub> = mg/b
</p>
</blockquote>
<p>
Alternately you can put in <i>t=∞</i> into our equation for <i>v(t)</i>
</p>
<blockquote>
<p>
v(t) = (v<sub>0</sub> - mg/b)e<sup>-tb/m</sup>+mg/b <br>
v(∞) = (v<sub>0</sub> - mg/b)e<sup>-∞b/m</sup>+mg/b <br>
v(∞) = mg/b
</p>
</blockquote>
</div>
</div>
</div>
</div>
</div>
</body>
</html>