Advanced handouts

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

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.
+