Advanced handouts

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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>

src/Advanced/Geometry of Masses/main.tex

@@ -0,0 +1,34 @@
+% use [nosolutions] flag to hide solutions.
+% use [solutions] flag to show solutions.
+\documentclass[
+	solutions,
+	singlenumbering,
+	shortwarning
+]{../../../lib/tex/ormc_handout}
+\usepackage{../../../lib/tex/macros}
+
+\usepackage{tikz}
+\usetikzlibrary{patterns}
+\usetikzlibrary{shapes.geometric}
+
+\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 1d.tex}
+	\input{parts/1 balance 2d.tex}
+	\input{parts/1 continuous}
+	\input{parts/2 pappus}
+
+\end{document}

src/Advanced/Geometry of Masses/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Geometry of Masses"
+
+[publish]
+handout = true
+solutions = false

src/Advanced/Geometry of Masses/parts/0 balance 1d.tex

@@ -0,0 +1,200 @@
+\section{Balancing a line}
+
+\example{}
+Consider a mass $m_1$ on top of a pin. \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}
+
+\begin{solution}
+	The CoM will be such that the rod is split into $d_1+d_2=1$
+	according to the relation $m_1 d_1 = m_2 d_2$.
+	This is sufficient, but if you want to solve for one of the $d$,
+	you get $d_1 = \frac{m_2}{m_1+m_2}$.
+
+	\vspace{2mm}
+
+	This should be intuitive; the distance of each mass from the CoM
+	is proportional to the other mass's share of the total mass.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+
+
+\definition{}
+Consider a massless, horizontal rod of infinite length. \par
+Affix a finite number of point masses to this rod. \par
+We will call the resulting object a \textit{one-dimensional system of masses}:
+\begin{center}
+	\begin{tikzpicture}[scale=1]
+		\draw[<->, line width = 0.5mm] (-4, 0) -- (4, 0);
+		\node[left] at (-4, 0) {$...$};
+		\node[right] at (4, 0) {$...$};
+
+
+		\fill[color = black] (-2.5, 0) circle[radius=0.12];
+		\node[above] at (-2.5, 0.15) {$m_1$};
+
+		\fill[color = black] (-0.5, 0) circle[radius=0.1];
+		\node[above] at (-0.5, 0.15) {$m_2$};
+
+		\fill[color = black] (1.5, 0) circle[radius=0.15];
+		\node[above] at (1.5, 0.15) {$m_3$};
+	\end{tikzpicture}
+\end{center}
+\vspace{5mm}
+
+
+
+\problem{}<massline>
+Consider a one-dimensional system of masses consisting of $n$ masses $m_1, m_2, ..., m_n$, \par
+with each $m_i$ positioned at $x_i$. Show that the resulting system always has a unique center of mass. \par
+\hint{Prove this by construction: find the point!}
+
+\begin{solution}
+	\begin{equation*}
+		x_0 = \frac{1}{M}\sum_{i=1}^n m_i x_i
+	\end{equation*}
+	where $M = \sum_{i=1}^n m_i$
+\end{solution}
+
+\vfill
+\pagebreak

src/Advanced/Geometry of Masses/parts/1 balance 2d.tex

@@ -0,0 +1,155 @@
+\section{Balancing a plane}
+
+\definition{}
+Consider a massless two-dimensional plane. \par
+Affix a finite number of point masses to this plane. \par
+We will call the resulting object a \textit{two-dimensional system of masses:}
+
+
+\begin{center}
+	\begin{tikzpicture}[scale = 0.5]
+
+		%\draw[
+		%	line width = 0mm,
+		%	pattern = north west lines,
+		%	pattern color = blue,
+		%]
+		%	(1, 0)
+		%	-- (0.5, 0.866)
+		%	-- (-0.5, 0.866)
+		%	-- (-1, 0)
+		%	-- (-0.5, -0.866)
+		%	-- (0.5, -0.866)
+		%	-- cycle;
+		%\draw[
+		%	line width = 0.5mm,
+		%	blue
+		%]
+		%	(1, 0)
+		%	-- (0.5, 0.866)
+		%	-- (-0.5, 0.866)
+		%	-- (-1, 0)
+		%	-- (-0.5, -0.866)
+		%	-- (0.5, -0.866)
+		%	-- cycle;
+		%\fill[color = blue] (0, 0) circle[radius=0.3];
+
+
+		\fill[color = black]
+			(-3, 3) circle[radius = 0.5]
+			node[above] at (-3, 3.5) {$m_1$ at $(x_1, y_1)$};
+
+		\fill[color = black]
+			(-5, -1.5) circle[radius = 0.4]
+			node[above] at (-5, -1.0) {$m_2$ at $(x_2, y_2)$};
+
+		\fill[color = black]
+			(3, -3) circle[radius = 0.35]
+			node[above] at (3, -2.5) {$m_3$ at $(x_3, y_3)$};
+
+		\draw[line width = 0.5mm]
+			(-7.5, -4.2)
+			-- (6, -4.2)
+			-- (6, 5)
+			-- (-7.5, 5)
+			-- cycle;
+
+	\end{tikzpicture}
+\end{center}
+\vspace{5mm}
+
+
+
+\problem{}
+Show that any two-dimensional system of masses has a unique center of mass. \par
+\hint{
+	If a plane balances on a pin, it does not tilt in the $x$ or $y$ direction. \par
+	See the diagram below.
+}
+\begin{center}
+	\begin{tikzpicture}[scale = 0.5]
+
+		% Horizontal
+		\draw[line width = 0.5mm, dotted, gray] (-3, 3) -- (-3, -5);
+		\draw[line width = 0.5mm, dotted, gray] (-5, -1.5) -- (-5, -5);
+		\draw[line width = 0.5mm, dotted, gray] (3, -3) -- (3, -5);
+		\draw[line width = 0.5mm, dotted, gray] (0, 0) -- (0, -5);
+		\draw[line width = 0.5mm] (-7, -5) -- (6.5, -5);
+
+		\fill[color = gray] (-3, -5) circle[radius = 0.3];
+		\fill[color = gray] (-5, -5) circle[radius = 0.3];
+		\fill[color = gray] (3, -5) circle[radius = 0.3];
+
+		\draw[line width = 0.25mm, pattern=north west lines]
+			(0, -5) -- (-0.6, -6) -- (0.6, -6) -- cycle;
+
+
+		% Vertical
+
+		\draw[line width = 0.5mm, dotted, gray] (-3, 3) -- (8, 3);
+		\draw[line width = 0.5mm, dotted, gray] (-5, -1.5) -- (8, -1.5);
+		\draw[line width = 0.5mm, dotted, gray] (3, -3) -- (8, -3);
+		\draw[line width = 0.5mm, dotted, gray] (0, 0) -- (8, 0);
+		\draw[line width = 0.5mm] (8, 4) -- (8, -4);
+
+		\fill[color = gray] (8, 3) circle[radius = 0.3];
+		\fill[color = gray] (8, -1.5) circle[radius = 0.3];
+		\fill[color = gray] (8, -3) circle[radius = 0.3];
+
+		\draw[line width = 0.25mm, pattern=north west lines]
+		(8, 0) -- (9, -0.6) -- (9, 0.6) -- cycle;
+
+
+		\draw[
+			line width = 0mm,
+			pattern = north west lines,
+			pattern color = blue,
+		]
+			(1, 0)
+			-- (0.5, 0.866)
+			-- (-0.5, 0.866)
+			-- (-1, 0)
+			-- (-0.5, -0.866)
+			-- (0.5, -0.866)
+			-- cycle;
+		\draw[
+			line width = 0.5mm,
+			blue
+		]
+			(1, 0)
+			-- (0.5, 0.866)
+			-- (-0.5, 0.866)
+			-- (-1, 0)
+			-- (-0.5, -0.866)
+			-- (0.5, -0.866)
+			-- cycle;
+		\fill[color = blue] (0, 0) circle[radius=0.3]
+			node[above] at (0, 1) {Pivot};
+
+		\fill[color = black]
+			(-3, 3) circle[radius = 0.5]
+			node[above] at (-3, 3.5) {$m_1$ at $(x_1, y_1)$};
+
+		\fill[color = black]
+			(-5, -1.5) circle[radius = 0.4]
+			node[above] at (-5.5, -1.0) {$m_2$ at $(x_2, y_2)$};
+
+		\fill[color = black]
+			(3, -3) circle[radius = 0.35]
+			node[above] at (3, -2.8) {$m_3$ at $(x_3, y_3)$};
+	\end{tikzpicture}
+\end{center}
+
+\begin{solution}
+	\begin{equation*}
+		x_0 = \frac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}
+		\qquad
+		y_0 = \frac{\sum_{i=1}^n m_i y_i}{\sum_{i=1}^n m_i}
+	\end{equation*}
+\end{solution}
+
+\vfill
+\pagebreak
+
+\vfill
+\pagebreak

src/Advanced/Geometry of Masses/parts/1 continuous.tex

@@ -0,0 +1,136 @@
+\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.}
+
+\begin{figure}[htp]
+	\centering
+	\includegraphics[width=0.3\linewidth]{img/seahorse.jpg}
+	\label{seahorse}
+\end{figure}
+
+\problem{}
+You are given a cardboard cutout of a seahorse and some office supplies. \par
+How might you determine its center of mass? Does your strategy also work in 3D?
+
+\begin{solution}
+	Many correct answers. One example:
+	\begin{enumerate}[label={(\arabic*)}]
+		\item Stick a thumb tack into the horse and let it come to equilibrium
+		\item Use a ruler or string to draw a straight line through that point along the direction of gravity
+		\item Repeat (1) and (2) at a different point
+		\item The intersection of the two lines marks the CoM
+	\end{enumerate}
+\end{solution}
+
+\vfill
+
+\definition{Centroid}
+Centroids are closely related to, and often synonymous with, centers of mass.
+A centroid is the geometric center of an object, regardless of the mass distribution.
+Thus, the centroid and center of mass are the same when the mass is uniformly distributed.
+
+\problem{}<rightiso>
+Where is the center of a right isosceles triangle? What about any isosceles triangle?
+
+\begin{solution}
+	There are probably some other clever ways of doing this without calculus but here's one way:
+	\begin{center}
+		\includegraphics[width=0.3\linewidth]{img/right_isos.png}
+	\end{center}
+	Clearly, the centroid $O$ must be somewhere along $SC$.
+	Now we just need to find $s$ in terms of $x$, that is, the balancing point along either of the shorter sides.
+	To do this, we split the triangle into three regions: $\triangle{AVT}$ on the left,
+	the rectangle $TUCV$ on the right, and $\triangle{TUB}$ also on the right.
+
+	Each region exerts a torque proportional to its area times the horizontal distance from $VT$ of that region's
+	centroid. Note that, even though we don't know what $x$ is yet, we can use it to find itself.
+	By similar triangles, the centroids of $\triangle{AVT}$ and $\triangle{TUB}$ are both located $x(1-x)$ away from $VT$.
+	The centroid of $TUVC$ is trivially $\frac{1-x}{2}$. So we get the following equation:
+	\begin{align*}
+		\frac{1}{2}x^2 \cdot x(1-x) &= x(1-x) \cdot \frac{1-x}{2} + \frac{1}{2}(1-x)^2 \cdot x(1-x) \quad .
+	\end{align*}
+	We easily find that $x=\frac{1}{3}$. Remarkably, the ratio $\frac{SO}{SC}$ is also $\frac{1}{3}$.
+
+	Any right triangle is just an isosceles right triangle that's been scaled along some axis, so the centroid scales with it and this one-third rule still applies.
+
+	Any isosceles triangle is just two right triangles, so \textit{its} centroid will be in between its two "sub-centroids" from each right triangle, that is, one-third the altitude.
+\end{solution}
+
+\vfill
+
+\problem{}
+How can you easily find the center of mass of any triangle? Why does this work?
+
+
+\begin{solution}
+    It turns out that all three medians of a triangle always intersect at a single point. That point is the centroid. You could feasibly guess this by taking what you learned from Problem 13 and applying Cavalieri's Theorem. Otherwise, I'm interested to see what students come up with.
+\end{solution}
+
+\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?
+
+\begin{solution}
+	Similar to our solution to \ref{rightiso}, we draw a vertical line from the
+	desired balancing point and split the regions into triangles and rectangles.
+	Using symmetry and simple trigonometry, we find that:
+	$h = \sqrt{\sqrt{2}+\frac{8}{9}} + 3\sqrt{2}$
+\end{solution}
+
+\vfill
+
+\problem{}<3D soda>
+Think about how you might approach this problem in 3D. Does $h$ become larger or smaller?
+
+\begin{solution}
+	This is a pretty open-ended question and is meant simply to make students think about
+	how the problem would change in 3D. I believe that $h$ gets smaller.
+\end{solution}
+
+
+\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.
+
+\begin{solution}
+	This problem is really just \ref{rightiso} again in disguise.
+	So, the balancing point is at $1/3$ the length of the staff measured from the dense-end.
+\end{solution}
+
+\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

src/Advanced/Geometry of Masses/parts/2 pappus.tex

@@ -0,0 +1,95 @@
+\section{Pappus's Centroid Theorem}
+
+\begin{figure}[htp]
+	\centering
+	\includegraphics[width=0.6\linewidth]{img/pappus_1.png}
+	\caption{}
+	\label{pappus1}
+\end{figure}
+
+\remark{}
+\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.
+
+\remark{}
+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}.
+
+\problem{}
+\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?
+
+\begin{solution}
+	\url{https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem}
+	\url{https://mathworld.wolfram.com/PappussCentroidTheorem.html}
+\end{solution}
+
+\vfill
+\pagebreak
+
+\problem{}
+\textit{Pappus's Second Centroid Theorem} simply extends this concept to \textit{solids of revolution}, which are exactly what you think they are.
+
+
+\problem{}
+Now that you've done the first theorem, what do you think Pappus's Second Centroid Theorem states?
+
+\vfill
+
+
+\problem{}
+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}
+
+\problem{}
+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.
+
+\begin{solution}
+	\url{https://mathspanda.com/A2FM/Lessons/Centres_of_mass_of_standard_shapes_LESSON.pdf}
+\end{solution}
+
+\vfill
+
+\pagebreak
+
+\problem{}<sector>
+Where is the centroid of the \textit{sector} of the circle in Figure \ref{arc}?
+\hint{cut it up.}
+
+
+\vfill
+
+\problem{}
+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 $\lambda(x) = \lambda_0\sqrt{1-x^2}$.
+
+\begin{solution}
+	This is equivalent to finding the x-coordinate of the centroid of a quarter-circle. See \ref{sector}.
+\end{solution}
+
+\vfill
+\problem{}
+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.
+
+\begin{solution}
+	We have $m_i = 1/4^i$ and $x_i = 2^i$ so
+	\begin{align*}
+		\sum_{i=0}^\infty m_i x_i &= \sum \frac{2^i}{4^i} \\
+		&= \sum \frac{1}{2^i} \\
+		&= \frac{1}{1-1/2} = 2 \qquad \text{as this is just a geometric series} \\
+	\end{align*}
+	\[ M = \sum m_i = \frac{1}{1-1/4} = 4/3 \]
+	Then
+	\[ x_{CM} = \frac{\sum m_i x_i}{M} = 3/2 \]
+\end{solution}
+
+\vfill
+
+\problem{Bonus}
+Try to actually find $h$ from Problem \ref{3D soda}. Good luck.
+