Fix handouts

Advanced/COM/main.tex

@@ -1,32 +0,0 @@
-% use [nosolutions] flag to hide solutions.
-% use [solutions] flag to show solutions.
-\documentclass[
-	solutions,
-	singlenumbering,
-	shortwarning
-]{../../resources/ormc_handout}
-\usepackage{../../resources/macros}
-
-\usepackage{tikz}
-\usetikzlibrary{patterns}
-
-\usepackage{graphicx}
-
-
-
-\uptitlel{Advanced 2}
-\uptitler{\smallurl{}}
-\title{Geometry of Masses I}
-\subtitle{Prepared by Sunny \& Mark on \today{}}
-
-
-
-
-\begin{document}
-	\maketitle
-
-	\input{parts/0 balance}
-	\input{parts/1 continuous}
-	\input{parts/2 pappus}
-
-\end{document}

Advanced/COM/parts/0 balance.tex

@@ -1,160 +0,0 @@
-\section{Balance}
-
-\example{}
-Consider a mass $m_1$ on top of a pin in two-dimensional space. \par
-Due to gravity, the mass exerts a force on the pin at the point of contact. \par
-For simplicity, we'll say that the magnitude of this force is equal the mass of the object---
-that is, $m_1$.
-\begin{center}
-	\begin{tikzpicture}[scale=2]
-		\fill[color = black] (0, 0.1) circle[radius=0.1];
-		\node[above] at (0, 0.20) {$m_1$};
-
-		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
-
-
-		\draw[color = black, opacity = 0.5] (1, 0.1) circle[radius=0.1];
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1, 0) -- (0.85, -0.3) -- (1.15, -0.3) -- cycle;
-
-		\draw[->, line width = 0.5mm] (1, 0) -- (1, -0.5) node[below] {$m_1$};
-		%\draw[->, line width = 0.5mm, dashed] (1, 0) -- (1, 0.5) node[above] {$-m_1$};
-
-		\fill[color = red] (1, 0) circle[radius=0.025];
-	\end{tikzpicture}
-\end{center}
-The pin exerts an opposing force on the mass at the same point, and the system thus stays still.
-
-
-\remark{}<fakeunits>
-Forces, distances, and torques in this handout will be provided in arbitrary (though consistent) units. \par
-We have no need for physical units in this handout.
-
-\example{}
-Now attach this mass to a massless rod and try to balance the resulting system. \par
-As you might expect, it is not stable: the rod pivots and falls down.
-\begin{center}
-	\begin{tikzpicture}[scale=2]
-		\fill[color = black] (-0.3, 0.0) circle[radius=0.1];
-		\node[above] at (-0.3, 0.1) {$m_1$};
-		\draw[-, line width = 0.5mm] (-0.8, 0) -- (0.5, 0);
-
-		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
-
-
-		\draw[color = black, opacity = 0.5] (1.2, 0.0) circle[radius=0.1];
-		\draw[-, line width = 0.5mm, opacity = 0.5] (0.7, 0) -- (1.9, 0);
-
-		\draw[line width = 0.25mm, pattern=north west lines, opacity = 0.5] (1.5, 0) -- (1.35, -0.3) -- (1.65, -0.3) -- cycle;
-
-
-		\draw[->, line width = 0.5mm] (1.2, 0) -- (1.2, -0.5) node[below] {$m_1$};
-		%\draw[->, line width = 0.5mm, dashed] (1.5, 0) -- (1.5, 0.5) node[above] {$f_p$};
-	\end{tikzpicture}
-\end{center}
-This is because the force $m_1$ is offset from the pivot (i.e, the tip of the pin). \par
-It therefore exerts a \textit{torque} on the mass-rod system, causing it to rotate and fall.
-
-\pagebreak
-
-
-\definition{Torque}
-Consider a rod on a single pivot point.
-If a force with magnitude $m_1$ is applied at an offset $d$ from the pivot point,
-the system experiences a \textit{torque} with magnitude $m_1 \times d$.
-\begin{center}
-	\begin{tikzpicture}[scale=2]
-		\draw[-, line width = 0.5mm] (-1.2, 0) -- (0.5, 0);
-		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
-
-		\draw[->, line width = 0.5mm, dashed] (-0.8, 0) -- (-0.8, -0.5) node[below] {$m_1$};
-		\fill[color = red] (-0.8, 0.0) circle[radius=0.05];
-
-
-		\draw[-, line width = 0.3mm, double] (-0.8, 0.1) -- (-0.8, 0.2) -- (0, 0.2) node [midway, above] {$d$} -- (0, 0.1);
-	\end{tikzpicture}
-\end{center}
-
-We'll say that a \textit{positive torque} results in \textit{clockwise} rotation,
-and a \textit{negative torque} results in a \textit{counterclockwise rotation}.
-As stated in \ref{fakeunits}, torque is given in arbitrary \say{torque units}
-consistent with our units of distance and force.
-
-\vspace{2mm}
-
-% I believe the convention used in physics is opposite ours, but that's fine.
-% Positive = clockwise is more intuitive given our setup,
-% and we only use torque to define CoM anyway.
-Look at the diagram above and convince yourself that this convention makes sense:
-\begin{itemize}
-	\item $m_1$ is positive \note{(masses are usually positive)}
-	\item $d$ is negative \note{($m_1$ is \textit{behind} the pivot)}
-	\item therefore, $m_1 \times d$ is negative.
-\end{itemize}
-
-
-\definition{Center of mass}
-The \textit{center of mass} of a physical system is the point at which one can place a pivot \par
-so that the total torque the system experiences is 0. \par
-\note{In other words, it is the point at which the system may be balanced on a pin.}
-
-
-\problem{}
-Consider the following physical system:
-we have a massless rod of length $1$, with a mass of size 3 at position $0$
-and a mass of size $1$ at position $1$.
-Find the position of this system's center of mass. \par
-
-\begin{center}
-	\begin{tikzpicture}[scale=2]
-		\draw[line width = 0.25mm, pattern=north west lines] (0, 0) -- (-0.15, -0.3) -- (0.15, -0.3) -- cycle;
-
-		\draw[-, line width = 0.5mm] (-0.5, 0) -- (1.5, 0);
-
-
-		\fill[color = black] (-0.5, 0) circle[radius=0.1];
-		\node[above] at (-0.5, 0.2) {$3$};
-
-		\fill[color = black] (1.5, 0) circle[radius=0.08];
-		\node[above] at (1.5, 0.2) {$1$};
-	\end{tikzpicture}
-\end{center}
-
-\vfill
-
-\problem{}
-Do the same for the following system, where $m_1$ and $m_2$ are arbitrary masses.
-
-\begin{center}
-	\begin{tikzpicture}[scale=2]
-		\draw[line width = 0.25mm, pattern=north west lines] (0.7, 0) -- (0.55, -0.3) -- (0.85, -0.3) -- cycle;
-
-		\draw[-, line width = 0.5mm] (-0.5, 0) -- (1.5, 0);
-
-
-		\fill[color = black] (-0.5, 0) circle[radius=0.1];
-		\node[above] at (-0.5, 0.2) {$m_1$};
-
-		\fill[color = black] (1.5, 0) circle[radius=0.08];
-		\node[above] at (1.5, 0.2) {$m_2$};
-	\end{tikzpicture}
-\end{center}
-
-\vfill
-\pagebreak
-
-\problem{}<massline>
-Consider a massless horizontal rod of infinite length. \par
-Attach $n$ masses $m_1, m_2, ..., m_n$ to it, placing each $m_i$ at position $x_i$. \par
-Find the resulting system's center of mass.
-
-\vfill
-
-\problem{}
-Extend \ref{massline} into two dimensions: \par
-Place $n$ masses $m_1, m_2, ..., m_n$ at positions $(x_1, y_1),~ (x_2, y_2),~ (x_3, y_3)$
-on a massless plane. \par
-Find the coordinates of the resulting system's center of mass. \par
-\hint{If a plane balances on a pin, it does not tilt in the $x$ or $y$ direction.}
-
-\vfill
-\pagebreak

Advanced/COM/parts/1 continuous.tex

@@ -1,64 +0,0 @@
-\section{Continuous mass}
-
-Now let's extend this idea to a \textit{continuous distribution} of masses rather than discrete point masses. This isn't so different; a continuous distribution of mass is really just a lot of point-masses, only that there are so many of them so close together that you can't even count them\footnote{For example, your pencil might seem like a continuous distribution of mass, but it's really just a whole lot of atoms.}. In general, finding the CoM requires integral calculus, but not always...\footnote{Many of the following problems can be solved with integration even though you're meant to solve them without it. But remember, in math, whenever you accomplish the same task two different ways, that really means that they're somehow the same thing.}
-
-
-\problem{}
-You are given a cardboard cutout of Figure \ref{seahorse} and some office supplies. How might you determine the CoM? Does your strategy also work in 3D?
-
-\vfill
-
-\problem{}
-Where is the CoM of a right isosceles triangle? What about any isosceles triangle?
-
-\vfill
-
-\problem{}
-How can you easily find the CoM of any triangle? Why does this work?
-
-\vfill
-
-\pagebreak
-
-
-\problem{}
-Consider Figure \ref{soda} depicting a simplified soda can.
-If you leave just the right amount,
-you can get it to balance on the beveled edge, as seen in Figure
-\ref{soda filled}.
-
-\begin{figure}[htp]
-\centering
-\begin{minipage}{0.5\textwidth}
-    \centering
-    \includegraphics[width=0.6\linewidth]{img/soda.png}
-    \caption{}
-    \label{soda}
-\end{minipage}\hfill
-\begin{minipage}{0.5\textwidth}
-    \centering
-    \includegraphics[width=0.6 \linewidth]{img/soda_filled.png}
-    \caption{}
-    \label{soda filled}
-\end{minipage}
-\end{figure}
-
-\problem{}
-See Figure \ref{soda filled}. Let's take the can to be massless and initially empty. Let's also assume that we live in two dimensions. We start slowly filling it up with soda to a vertical height $h$. What is $h$ just before the can tips over?
-
-\vfill
-
-\problem{}<3D soda>
-Think about how you might approach this problem in 3D. Does $h$ become larger or smaller?
-
-\vfill
-
-\pagebreak
-
-So far we've made the assumption our shapes have mass that is \textit{uniformly distributed}. But that doesn't have to be the case.
-
-\problem{}
-A mathemagical wizard will give you his staff if you can balance it horizontally on your finger. The strange magical staff has unit length and it's mass is distributed in a very special way. It's density decreases linearly from $\lambda_0$ at one end to $0$ at the other. Where is the staff's balancing point?
-
-\vfill
-\pagebreak

Advanced/COM/parts/2 pappus.tex

@@ -1,57 +0,0 @@
-\section{Pappus's Centroid Theorem}
-
-\begin{figure}[htp]
-	\centering
-	\includegraphics[width=0.6\linewidth]{img/pappus_1.png}
-	\caption{}
-	\label{pappus1}
-\end{figure}
-
-\textit{Centroids} are closely related to, and often synonymous with, centres of mass. A centroid is the geometric centre of an object, regardless of the mass distribution. Thus, the centroid and centre of mass coincide when the mass is uniformly distributed.
-
-Figure \ref{pappus1} depicts three different surfaces constructed by revolving a line segment (in red) about a central axis. These are often called \textit{surfaces of revolution}.
-
-\textit{Pappus's First Centroid Theorem} allows you to determine the area of a surface of revolution using information about the line segment and the axis of rotation.
-
-	Can you intuitively come up with Pappus's First Centroid Theorem for yourself? Figure \ref{pappus1} is very helpful. It may also help to draw from surface area formulae you already know. What limitations are there on the theorem?
-
-\vfill
-
-\pagebreak
-
-\textit{Pappus's Second Centroid Theorem} simply extends this concept to \textit{solids of revolution}, which are exactly what you think they are.
-
-	Now that you've done the first theorem, what do you think Pappus's Second Centroid Theorem states?
-
-\vfill
-
-	The centroid of a semi-circular line segment is already given in Figure \ref{pappus1}, but what about the centroid of a filled semi-circle? (Hint: For a sphere of radius $r$, $V=\frac{4}{3}\pi r^3$)
-
-\begin{figure}[htp]
-	\centering
-	\includegraphics[width=0.4\linewidth]{img/arc.png}
-	\caption{}
-	\label{arc}
-\end{figure}
-
-\label{arc centroid} Given arc $AB$ with radius $r$ and subtended by $2\alpha$, determine $OG$, the distance from the centre of the circle to the centroid of the arc.
-
-\vfill
-
-\pagebreak
-
-	Using your answers from Problem \ref{isosceles centroid} and Problem \ref{arc centroid}. Where is the centroid of the \textit{sector} of the circle in Figure \ref{arc}. (Hint: Cut it up.)
-
-\vfill
-
-	Seeing your success with his linear staff, the wizard challenges you with another magical staff to balance. It looks identical to the first one, but you're told that the density decreases from $\lambda_0$ to $0$ according to the function $\lambda(x) = \lambda_0\sqrt{1-x^2}$.
-
-\vfill
-
-	Infinitely many masses $m_i$ are placed at $x_i$ along the positive $x$-axis, starting with $m_0 = 1$ placed at $x_0 = 1$. Each successive mass is placed twice as far from the origin compared to the previous one. But also, each successive mass has a quarter the weight of the previous one. Find the CoM if it exists.
-
-\vfill
-
-	(Bonus) Try to actually find $h$ from Problem \ref{3D soda}. Good luck.
-
-%\item (Bonus, not related to the packet) Spongebob, Patrick, and Squidward are all hiding from the Sea Bear. Initially, Spongebob and Patrick are keeping watch 100 yards apart with Squidward halfway in between them. When Spongebob gets scared, he runs to hide halfway between Squidward and Patrick. Then Patrick, not wanting to be the farthest from the centre, runs to be halfway between Squidward and Patrick. Squidward, seeing this, quickly finds the new halfway point between Spongebob and Patrick. This pattern keeps repeating until all three of them are pointlessly clambering on top of each other. Where do they end up relative to their initial positions?

Advanced/Origami/main.tex

@@ -4,6 +4,7 @@
 	solutions,
 	%shortwarning
 ]{../../resources/ormc_handout}
+\usepackage{../../resources/macros}
 
 \graphicspath{ {./images/} }
 

Intermediate/Slide Rules/main.tex

@@ -1,7 +1,7 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	nosolutions
+	solutions
 ]{../../resources/ormc_handout}
 \usepackage{../../resources/macros}