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. $\sqrt{g R \frac{\mu _{s} + tan \theta }{1 - \mu _{s} tan ⁡ \theta }}$
• C. $\sqrt{\frac{g}{R} \frac{\mu _{s} + tan \theta }{1 - \mu _{s} tan ⁡ \theta }}$
• D. $\sqrt{\frac{g}{R^{2}} \frac{\mu _{s} + t a n \theta }{1 - \mu _{s} tan \theta }}$

Answer 1

Answer: (B)

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