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Added masses intro

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2024-10-10 14:32:58 -07:00
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Advanced/Geometry of Masses/main.tex Executable file
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% 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}
\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}

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Advanced/Geometry of Masses/parts/0 balance 1d.tex Normal file
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\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}
\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!}
\vfill
\pagebreak

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Advanced/Geometry of Masses/parts/1 balance 2d.tex Normal file
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\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}
\vfill
\pagebreak
\vfill
\pagebreak