\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
