Question

A car is negotiating a curved road of radius $\text{R}$ . The road is banked at an angle $\theta$ . The coefficient of friction between the tyres of the car and the road is ${\mu }_s$ . The maximum safe velocity on this road is:

• A. $\sqrt{g{R}^2\frac{{\mu }_s+tan\theta }{1-{\mu }_{s}tan\theta }}$
• B. $\sqrt{gR\frac{{\mu }_s+tan\theta }{1-{\mu }_{s}tan\theta }}$
• C. $\sqrt{\frac{g}{R}\frac{{\mu }_s+tan\theta }{1-{\mu }_{s}tan\theta }}$
• D. $\sqrt{\frac{g}{R^2}\frac{{\mu }_s+tan\theta }{1-{\mu }_{s}tan\theta }}$

Answer 1

Answer: (B)

$N=mgcos\theta +\frac{m{v}^2}{R}sin\theta$
$f_{max}=\mu N\Rightarrow$ $f_{max}={\mu }_{s}mgcos\theta +\frac{{\mu }_{s}m{v}^2}{R}sin\theta$
$mgsin\theta +f_{max}=\frac{m{v}^2}{R}cos\theta$
$mgsin\theta +{\mu }_{s}mgcos\theta +\frac{{\mu }_{s}m{v}^2}{R}sin\theta =\frac{m{v}^2}{R}cos\theta$
$gsin\theta +{\left(\mu \right)}_{s}gcos\theta =\frac{v^2}{R}\left(cos\theta -{\left(\mu \right)}_{s}sin\theta \right)$
$gR\left[\frac{tan\theta +{\mu }_s}{1-{\mu }_{s}tan\theta }\right]=v^2$