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Banked Track.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Banked Track} \begin{align*} \text{\bf Configuration:}\quad& \text{A car of mass $m$ on a curved banked track of radius $r$ is in uniform circular motion. The inclination}\\ &\text{of the track to the horizontal is $\theta$ . The track supports the car by a normal force of $N$ perpendicular}\\ &\text{to the surface, and a sideways frictional force of $F$ towards the inner rim (lower side) of the track.}\\ % \text{Horizontal:}\quad&N\sin\theta+F\cos\theta=mr\omega^2=m\frac{v^2}{r}\quad\ldots\:(1)\\ \text{Vertical:}\quad&N\cos\theta-F\sin\theta=mg\quad\ldots\:(2)\\ &(1)\cos\theta-(2)\sin\theta:\quad F\:(\cos^2\theta+\sin^2\theta)=m\frac{v^2}{r}\cos\theta-mg\sin\theta\:,\quad F=mg\cos\theta\left(\frac{v^2}{rg}-\tan\theta\right)\quad\ldots\:(3)\\ &(1)\sin\theta+(2)\cos\theta:\quad N\:(\sin^2\theta+\cos^2\theta)=m\frac{v^2}{r}\sin\theta+mg\cos\theta\:,\quad N=mg\cos\theta\left(\frac{v^2}{rg}\tan\theta+1\right)\quad\ldots\:(4)\\ % \text{Parallel to track:}\quad&\text{From (3):}\quad F+mg\sin\theta=m\frac{v^2}{r}\cos\theta\\ \text{Perpendicular:}\quad&\text{From (4):}\quad N-mg\cos\theta=m\frac{v^2}{r}\sin\theta\\ % \text{Friction $F$:}\quad& \text{In (3), $m, g, \theta$ and $r$ are constants, so there is a critical speed $v=V$ to make $F=0$ (no sideways friction).}\\ &F=mg\cos\theta\left(\frac{V^2}{rg}-\tan\theta\right)=0\:,\quad \boxed{\tan\theta=\frac{V^2}{rg}}\:,\quad\boxed{V=\sqrt{rg\tan\theta}}\:.\\ &\text{The road is designed such that friction is minimal at the ``usual'' speed, which is $V$. If the road is designed}\\ &\text{for higher speed, the inclination should be higher. The track is to be banked inwards by }\theta=\tan^{-1}\left(\tfrac{V^2}{rg}\right)\:.\\ % &\text{From (3): }F=mg\cos\theta\left(\frac{v^2}{rg}-\tan\theta\right)=mg\cos\theta\left(\frac{v^2}{rg}-\frac{V^2}{rg}\right)\:,\quad\boxed{F=m\frac{(v^2-V^2)}{r}\cos\theta}\quad\ldots\:(5)\\ &\text{For a small $\theta$, }\boxed{F\simeq m\frac{(v^2-V^2)}{r}}\quad\ldots\:(6)\\ &\frac{V^2}{rg}=\tan\theta\simeq\sin\theta=\frac{\:h\:}{d}\:,\quad\text{where $h$ is the outer rim height and $d$ is the width of the track. So}\;\;\boxed{h\simeq\frac{V^2 d}{rg}}\:.\\ % \text{Inclination $\theta$:}\quad& \text{On an unlaned track, the car may run on any part of the track. From (5), when the car is running below the}\\ &\text{critical speed, $v
V$ and $F$ becomes positive, pushing the car inwards while it is drifting}\\ &\text{outwards. So $V$ needs to be higher to get $F$ back to neutral (close to zero). A higher $V$ means a larger $\theta$, so}\\ &\text{the track is more inclined at the outer rim (and (6) no longer applies).}\\ \\ &\text{A well designed track should allow a free skidding car (when out of control) to stay within the track without}\\ &\text{much help from friction (as the tyres are most likely not gripping). The steep inclination at the outer rim}\\ &\text{should push the car back in when it is too fast for the curvature. When the car slows down, the flattened}\\ &\text{inner rim should steadily reduce its sideway motion until it eventually grips again and stops safely.}\\ \\ &\text{Also from (5), if a track is designed for a constant $V$, at the section of sharp turn, $r$ becomes small, and}\\ &\text{therefore requires a small $\cos\theta$ to balance the friction. A smaller cosine value means a larger $\theta$ . That is why}\\ &\text{the track is more inclined at sharp turn (e.g. a skeleton sled sliding down the curved track at high speeds).} \end{align*} % % % \begin{align*} \text{Co-efficient $\mu$:}\quad& \text{The friction $F$ is proportional to the normal force $N$ up to a limit before the object slips. The upper limit}\\ &\text{of $\left|\tfrac{F}{N}\right|$ is $\mu$, which is called the co-efficient of friction. When $\left|\tfrac{F}{N}\right|$ is larger than $\mu$ , the car starts to slip.}\\ % &\because N>0\:,\quad\therefore |F|\leq\mu N\:,\quad -\mu N\leq F\leq \mu N\:.\quad\boxed{F_{max}=\mu N ,\quad F_{min}=-\mu N}\quad\ldots\:(7)\:.\\ % (3)\div(4):\quad&\frac{F}{N}=\frac{\frac{v^2}{rg}-\tan\theta}{\frac{v^2}{rg}\tan\theta+1}\:,\quad\boxed{\frac{F}{N}=\frac{v^2-rg\tan\theta}{v^2\tan\theta+rg}}\quad\ldots\:(8)\\ \\ &\text{When }F=0\:,\quad V^2-rg\tan\theta=0\:,\quad\tan\theta=\frac{V^2}{rg}\:,\quad\text{which reconfirm the previous result.}\\ \\ \\ &\text{When $v$ is at its maximum, $F$ is at its maximum too, and the car's weight is on the outer wheels.}\\ &\text{From (7), (8):}\quad\mu=\frac{F_{max}}{N}=\frac{v_{max}^2-rg\tan\theta}{v_{max}^2\tan\theta+rg}\:,\quad \mu v_{max}^2\tan\theta+\mu rg=v_{max}^2-rg\tan\theta\:,\\ &v_{max}^2(1-\mu\tan\theta)=rg(\tan\theta+\mu)\:,\quad \boxed{v_{max}^2=rg\cdot\frac{\tan\theta+\mu}{1-\mu\tan\theta}}\:.\quad \text{Both sides need to be positive, so}\quad\mu<\frac{1}{\tan\theta}\:.\\ &\text{The closer $\mu$ is to $\tfrac{1}{\tan\theta}$, the larger $v_{max}$ can be. When $\mu>=\frac{1}{\tan\theta}$, the car will never slip even at high speed.}\\ \\ \\ &\text{When $v$ is at its minimum, $F$ is at its minimum too, and the car's weight is on the inner wheels.}\\ &\text{From (7), (8):}\quad\mu=-\frac{F_{min}}{N}=\frac{rg\tan\theta-v_{min}^2}{v_{min}^2\tan\theta+rg}\:,\quad \mu v_{min}^2\tan\theta+\mu rg=rg\tan\theta-v_{min}^2\:,\\ &v_{min}^2(1+\mu\tan\theta)=rg(\tan\theta-\mu)\:,\quad \boxed{v_{min}^2=rg\cdot\frac{\tan\theta-\mu}{1+\mu\tan\theta}}\:.\quad \text{Both sides need to be positive, so}\quad\mu<\tan\theta\:.\\ &\text{The closer $\mu$ is to $\tan\theta$, the smaller $v_{min}$ can be. When $\mu>=\tan\theta$, the car will never slip even when it stops.}\\ \\ \\ &\text{In civil engineering, when a road is designed, $r$ is determined by the geography and the average speed $V$ is}\\ &\text{determined by the traffic (e.g. freeway, local street). The inclination can therefore be determined from}\\ &\text{$\theta=\tan^{-1}\left(\tfrac{V^2}{rg}\right)$ . While $\mu$ is largely determined by the tyres of vehicles and the road surface material and}\\ &\text{condition, it needs to satisfy $\mu>=\tan\theta$ so the car can stop when required without rolling down the banked}\\ &\text{track. Also, $v_{max}$ needs to be high enough to prevent reckless drivers from killing themselves and others,}\\ &\text{literally along the way. Therefore, $\theta$ needs to be engineered to make sure all these criteria are met. If a}\\ &\text{solution for $\theta$ cannot be found, that means $r$ needs to be enlarged (e.g. by bridging around a hairpin) or}\\ &\text{reduce the recommended speed (e.g. using signs and speed cameras).}\\ \end{align*} \end{document}