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Strings and Governors.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Strings and Governors} \begin{align*} \text{\bf General Configuration:}\quad&\text{A light and smooth string of length $L$ is attached to a vertical pole. One end of the string is}\\ &\text{attached to point $A$, which is fixed at the top of the pole, and another end to point $B$, $h$ units}\\ &\text{of length under $A$. A ring of mass $m$ has been threaded into the string at position $P$, so $AP$}\\ &\text{is $l_1$ units of length, and $PB$ is $l_2$. $AP$ is at angle $\theta_1$ to the vertical, and $PB$ at $\theta_2$ .}\\ \\ &\text{Point $P$ describes a horizontal circle of radius $r$ in Uniform Circular Motion at angular speed}\\ &\text{of $\omega$. While $A$ is always fixed, $P$ and $B$ can be loose in some configurations. When $B$ is loose,}\\ &\text{a mass $M$ will be attached to it. The tension on $AP$ is $T_1$ and that on $PB$ is $T_2$.}\\ \\ \text{\bf Strings of Equal Length:}\quad&\text{Let $l=l_1=l_2$ ,\quad and\quad$\theta=\theta_1=\theta_2$;\quad $P$ and $B$ are fixed.\quad$r=l\sin\theta$ ,\quad$h=2l\cos\theta$ .}\\ \text{Horizontal:}\;&T_1\sin\theta+T_2\sin\theta=mr\omega^2=ml\sin\theta\cdot\omega^2\:,\quad T_1+T_2=ml\omega^2\quad\ldots\:(1)\\ \text{Vertical:}\;&T_1\cos\theta-T_2\cos\theta=mg\:,\quad T_1-T_2=\frac{mg}{\cos\theta}=\frac{2mgl}{h}\quad\ldots\:(2)\\ \tfrac{(1)+(2)}{2}:\;&\boxed{T_1=\frac{ml}{2}\left(\omega^2+\frac{2g}{h}\right)}\:,\quad \tfrac{(1)-(2)}{2}:\;\boxed{T_2=\frac{ml}{2}\left(\omega^2-\frac{2g}{h}\right)}\:,\quad \boxed{\frac{T_1}{T_2}=\frac{\omega^2 h+2g}{\omega^2 h-2g}=\frac{l\omega^2\cos\theta+g}{l\omega^2\cos\theta-g}}\\ &\text{When}\quad\omega^2\geq\frac{2g}{h}\:,\quad T_2\geq 0\quad\text{and the second string will be taut.}\\ \text{Tensions in terms of $\theta$:}\;&\because\:h=2l\cos\theta\:,\quad\therefore\:\boxed{T_1=\frac{m}{2}\left(l\omega^2+\frac{g}{\cos\theta}\right)}\:,\quad\boxed{T_2=\frac{m}{2}\left(l\omega^2-\frac{g}{\cos\theta}\right)}\\ % \text{Analysis:}\;&\text{Since $P$ and $B$ are fixed, $l, h$ and $\theta$ are constants. So the faster the angular speed, the higher}\\ &\text{the value of $\omega^2$ and therefore higher tension on the strings ($T_1$ and $T_2$). While $T_1$ is always}\\ &\text{positive, $T_2$ can be negative numerically, which means the lower string is not taut. In such case,}\\ &\text{the configuration becomes a Conical Pendulum. If the strings are replaced by sticks, when $T_2$ is}\\ &\text{negative, the lower stick has an inward compress force in it.}\\ \\ \text{\bf Equal Length \& Collar:}\quad&\text{Let $l=l_1=l_2$,\quad and\quad$\theta=\theta_1=\theta_2$;\quad $P$ is fixed, but $B$ is a loose collar of mass $M$.}\\ &\text{For $P$, the previous formulae of a fixed $B$ still hold.}\\ &\text{For a loose $B$ with mass $M$, its vertical balance is }\:T_2\cos\theta=Mg\:,\quad T_2=\frac{Mg}{\cos\theta}=\frac{2Mgl}{h}\quad\ldots\:(3)\\ &\text{Compare with the $T_2$ formula before:}\quad \frac{2Mgl}{h}=\frac{ml}{2}\cdot\left(\omega^2-\frac{2g}{h}\right)\:,\quad \frac{2Mgl}{h}=\frac{ml\omega^2}{2}-\frac{mgl}{h}\:,\\ &\frac{g}{h}\left(2M+m\right)=\frac{m\omega^2}{2}\:,\quad\frac{\omega^2 h}{2g}=\frac{2M+m}{m}=\frac{2M}{m}+1\quad\ldots\:(4)\qquad\boxed{\frac{M}{m}=\frac{\:1\:}{2}\left(\frac{\omega^2 h}{2g}-1\right)}\\ \text{From (2):}\;&mg=T_1\cos\theta-T_2\cos\theta=T_1\cos\theta-Mg\:,\quad T_1\cos\theta=(M+m)g\:,\quad\frac{T_1\cos\theta}{T_2\cos\theta}=\frac{(M+m)g}{Mg}\\ &\therefore\boxed{\frac{T_1}{T_2}=\frac{M+m}{M}}=1+\frac{m}{M}=1+2\cdot\frac{2g}{\omega^2 h-2g}=\frac{\omega^2 h-2g+4g}{\omega^2 h-2g}=\frac{\omega^2 h+2g}{\omega^2 h-2g}\quad\text{(same as before)}\\ \text{From (3):}\;&T_2=\frac{2Mgl}{h}=Ml\omega^2\cdot\frac{2g}{\omega^2 h}\:,\quad\boxed{T_2=\frac{Ml\omega^2}{\frac{2M}{m}+1}}\:,\quad T_1=T_2\cdot\frac{M+m}{M}\:,\quad\boxed{T_1=\frac{(M+m)l\omega^2}{\frac{2M}{m}+1}}\\ \end{align*} \begin{align*} \text{\bf Loose Ring of Mass:}\quad&\text{$B$ is fixed and $P$ is a ring threaded loosely on the string, so $T=T_1=T_2$ .}\\ &r=l_1\sin\theta_1=l_2\sin\theta_2\:,\quad h=l_1\cos\theta_1+l_2\cos\theta_2\:.\\ \text{Horizontal:}\;&T\sin\theta_1+T\sin\theta_2=T(\sin\theta_1-\sin\theta_2)=mr\omega^2\quad\ldots\:(5)\\ \text{Vertical:}\;&T\cos\theta_1-T\cos\theta_2=T(\cos\theta_1-\cos\theta_2)=mg\quad\ldots\:(6)\\ (6)\div(5):\;&\frac{g}{r\omega^2}=\frac{\cos\theta_1-\cos\theta_2}{\sin\theta_1+\sin\theta_2} =\frac{-2\sin\left(\frac{\theta_1+\theta_2}{2}\right)\sin\left(\frac{\theta_1-\theta_2}{2}\right)}{2\sin\left(\frac{\theta_1+\theta_2}{2}\right)\cos\left(\frac{\theta_1-\theta_2}{2}\right)}=-\tan\left(\frac{\theta_1-\theta_2}{2}\right)\:,\quad \boxed{\tan\left(\frac{\theta_2-\theta_1}{2}\right)=\frac{g}{r\omega^2}}\\ \text{From (5):}\;&mr\omega^2=T\sin\theta_1+T\sin\theta_2=T\left(\frac{r}{l_1}+\frac{r}{l_2}\right)=Tr\left(\frac{l_2+l_1}{l_1 l_2}\right)\:,\quad\boxed{T=m\omega^2\frac{l_1 l_2}{L}} \end{align*} \end{document}