\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