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Vertical Circular Motion.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Vertical Circular Motion} \begin{align*} \text{Simple Pendulum:}\\ &\text{A string of $l$ metres fixed at one end $O$, with a mess of $m$ kg attached to the other end $P$, and}\\ &\text{making oscillations in a vertical plane. The tension in the string is $T$ newtons and the inclination}\\ &\text{of the string to the vertical is $\theta$. ($P$ is swinging upwards when $\theta$ increases from 0 to $\pi$ .)}\\ \\ &\text{Tangental Force }F_T=-mg\sin\theta\:,\quad\text{Norminal (Radial) Force }F_N=T-mg\cos\theta\:.\\ &\text{Tangental Acceleration }a_T=-g\sin\theta=l\ddot{\theta}\:,\quad\text{Norminal (Radial) Acceleration }a_N=\tfrac{\:T\:}{m}-g\cos\theta=l\dot{\theta}^2\\ \\ &\text{From the formula of $a_T$ ,}\quad\ddot{\theta}=-\frac{\:g\:}{l}\sin\theta\\ &\text{When }\theta\to 0\:\text{(e.g. $-10^\circ\leq\theta\leq 10^\circ$)},\quad\sin\theta\to\theta\:,\quad\text{and }\quad\boxed{\ddot{\theta}\approx-\frac{\:g\:}{l}}\\ &\text{This is close to a simple harmonic motion ($\ddot{x}=-n^2x$) with $n=\sqrt{\frac{\:g\:}{l}}$ . The period is }\frac{2\pi}{n}=2\pi\sqrt{\frac{l}{g}}\\ \\ \text{Vertical Circular }&\text{Motion:}\\ &\text{Bike in a cage: The radius of the cage is $R$ . The normal inward force is $N$ .}\\ &\text{Tangental Force }F_T=-mg\sin\theta\:,\quad\text{Norminal (Radial) Force }F_N=N-mg\cos\theta\:.\\ &\text{Tangental Acceleration }a_T=-g\sin\theta=R\ddot{\theta}\:,\quad\text{Norminal (Radial) Acceleration }a_N=\tfrac{\:N\:}{m}-g\cos\theta=R\dot{\theta}^2\\ \\ &\text{As a side note, recall }\frac{d}{d\theta}\left[\frac{\:1\:}{2}\dot{\theta}^2\right]=\dot{\theta}\cdot\frac{d}{d\theta}\left(\frac{d\theta}{dt}\right)=\frac{d}{d\theta}\left(\frac{d\theta}{dt}\right)\cdot\frac{d\theta}{dt}=\frac{d}{dt}\left(\frac{d\theta}{dt}\right)=\ddot{\theta}\:,\quad\text{so }\boxed{\ddot{\theta}=\frac{d}{d\theta}\left[\frac{\:1\:}{2}\dot{\theta}^2\right]}\\ \\ &\text{From the formula of $a_T$ , }-g\sin\theta=R\ddot{\theta}=R\cdot\frac{d}{d\theta}\left[\frac{\:1\:}{2}\dot{\theta}^2\right]\:,\quad \int_0^\theta-g\sin\theta\:d\theta=\int_0^\theta R\cdot\frac{d}{d\theta}\left[\frac{\:1\:}{2}\dot{\theta}^2\right]\:d\theta\:,\\ &g\big[\cos\theta\big]_0^\theta=\frac{1}{2R}\left[(R\dot{\theta})^2\right]_0^\theta=\frac{v^2-u^2}{2R}\:,\quad v^2-u^2=2Rg(\cos\theta-1)\:,\quad \boxed{v^2=u^2-2Rg(1-\cos\theta)}\quad\ldots\quad(1)\\ \\ &\text{From the formula of $a_N\times R$ ,}\quad R\tfrac{N}{m}-Rg\cos\theta=R^2\dot{\theta}^2=v^2\quad\ldots\quad(2)\\ &(1)+(2):\quad v^2+R\tfrac{N}{m}-Rg\cos\theta=u^2-2Rg+2Rg\cos\theta+v^2\:,\quad R\tfrac{N}{m}=u^2-2Rg+3Rg\cos\theta\:,\quad\\ &\boxed{N=m\frac{u^2}{R}-mg(2-3\cos\theta)}\\ \\ &\text{$N$ must remain positive for all $\theta$ to hold the bike on the track and avoid an accident.}\\ &N=m\frac{u^2}{R}-mg(2-3\cos\theta)>0\:,\quad u^2>Rg(2-3\cos\theta)\:.\\ \\ &\text{RHS reaches its maximum of $5Rg$ when $\theta=\pi$ . If $u^2<5Rg$ , the bike will fall before reaching the top.}\\ &\text{On the other hand, if it falls when $\theta<\tfrac{\pi}{2}$, it simply rolls back safely. So }u^2