Question

A car is negotiating a curved road of radius $R$. 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. $\sqrt{\frac{g}{R}\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$
• B. $\sqrt{\frac{g}{R^2}\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$
• C. $\sqrt{g{R}^2\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$
• D. $\sqrt{gR\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$

Answer 1

The correct option is D $\sqrt{gR\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$


For vertical equilibrium on the road,
$N\cos \theta =mg+f\sin \theta$
$mg=N\cos \theta -f\sin \theta$ ....$\left(i\right)$
Centripetal force for safe turning,
$\frac{m{v}^2}{R}=N\sin \theta +f\cos \theta$ …… $\left(ii\right)$
Dividing equation $\left(ii\right)$ by $\left(i\right)$, we get,
$\frac{v^2}{Rg}=\frac{N\sin \theta +f\cos \theta }{N\cos \theta -f\sin \theta }$
$\Rightarrow \frac{v^2}{Rg}=\frac{N\sin \theta +{\mu }_{s}N\cos \theta }{N\cos \theta -{\mu }_{s}N\sin \theta }$

$v_{\text{max}}=\sqrt{Rg\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)}$

Hence, option $\left(D\right)$ is correct.