Question

A racing track of curvatures 9.9 m is banked at tan-¹ 0.5 coefficients of statis friction between the track and tyres of vehicle is 0.2. Determine the speed limit with 10 % margin.​

Answer 1

Given :

Radius of circular path = 9.9m

Angle of banking = tan‾¹ 0.5

Coefficient of friction = 0.2

To Find :

Speed limit with 10% margin.

Solution :

❒ Banking of tracks (roads) : The maximum velocity with which a vehicle (in absence of friction) can negotiate a circular road of radius r and banked at an angle θ is given by

$\dag\:\underline{\boxed{\bf{\red{v=\sqrt{rg\tan\theta}}}}}$

When the frictional forces are also taken into account,

A] The maximum safe velocity :-

$\bigstar\:\underline{\boxed{\bf{v_{max}=\sqrt{rg\bigg(\dfrac{\mu+\tan\theta}{1-\mu\tan\theta}\bigg)}}}}$

B] The minimum safe velocity :-

$\bigstar\:\underline{\boxed{\bf{v_{min}=\sqrt{rg\bigg(\dfrac{\tan\theta-\mu}{1+\mu\tan\theta}\bigg)}}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){10}}\put(0,0){\vector(-1,0){0}}\put(10,0){\vector(1,0){0}}\put(1,0){\circle*{0.2}}\put(9,0){\circle*{0.2}}\put(2.5,0){\circle*{0.2}}\put(7.5,0){\circle*{0.2}}\put(1,0.5){ \sf +10\% }\put(7.6,0.5){ \sf -10\% }\put(0.7,-0.5){ \bf v_{min} }\put(2.2,-0.5){ \bf v'_{min} }\put(7.2,-0.5){ \bf v'_{max} }\put(8.6,-0.5){ \bf v_{max} }\end{picture}$

❖ Maximum safe velocity :-

$\sf:\implies\:v_{max}=\sqrt{rg\bigg(\dfrac{\mu+\tan\theta}{1-\mu\tan\theta}\bigg)}$

$\sf:\implies\:v_{max}=\sqrt{(9.9)(10)\bigg(\dfrac{0.2+0.5}{1-(0.2)(0.5)}\bigg)}$

$\sf:\implies\:v_{max}=\sqrt{99\bigg(\dfrac{0.7}{1-0.1}\bigg)}$

$\sf:\implies\:v_{max}=\sqrt{99\bigg(\dfrac{0.7}{0.9}\bigg)}$

$\sf:\implies\:v_{max}=\sqrt{99\times 0.77}$

$\sf:\implies\:v_{max}=\sqrt{77}$

$\sf:\implies\:v_{max}=8.775\:ms^{-1}$

We are asked to find $\sf{v'_{max}}$.

$\sf:\implies\:v'_{max}=v_{max}-10\%\:of\:v_{max}$

$\sf:\implies\:v'_{max}=8.775-\dfrac{10}{100}(8.775)$

$\sf:\implies\:v'_{max}=8.775-(0.1)(8.775)$

$\sf:\implies\:v'_{max}=8.775-0.8775$

$\bf:\implies\:v'_{max}=7.897\:ms^{-1}$

❖ Minimum safe velocity :-

$\sf:\implies\:v_{min}=\sqrt{rg\bigg(\dfrac{\tan\theta-\mu}{1+\mu\tan\theta}\bigg)}$

$\sf:\implies\:v_{min}=\sqrt{(9.9)(10)\bigg(\dfrac{0.5-0.2}{1+(0.2)(0.5)}\bigg)}$

$\sf:\implies\:v_{min}=\sqrt{99\bigg(\dfrac{0.3}{1+0.1}\bigg)}$

$\sf:\implies\:v_{min}=\sqrt{99\bigg(\dfrac{0.3}{1.1}\bigg)}$

$\sf:\implies\:v_{min}=\sqrt{99\times 0.27}$

$\sf:\implies\:v_{min}=\sqrt{27}$

$\sf:\implies\:v_{min}=5.196\:ms^{-1}$

We are asked to find $\sf{v'_{min}}$.

$\sf:\implies\:v'_{min}=v_{min}+10\%\:of\:v_{min}$

$\sf:\implies\:v'_{min}=5.196+\dfrac{10}{100}(5.196)$

$\sf:\implies\:v'_{min}=5.196+(0.1)(5.196)$

$\sf:\implies\:v'_{min}=5.196+0.5196$

$\bf:\implies\:v'_{min}=5.716\:ms^{-1}$

❖ Speed limit with 10% margin :-

$\dag\:\underline{\boxed{\bf{\gray{v'_{min}-v'_{max}=5.716-7.897}}}}$

Answer 2

Answer:

❖❖I hope it's helpful for you dear

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