First slide

Banking of road

Question

A car is negotiating a curved road of radius $R$ . The road is banked at an angle $θ$ . The coefficient of friction between the tyres of the car and the road is $μ_{s}$ . The maximum safe velocity on this road is

Moderate

A

$\sqrt{\frac{g}{R} \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$
B

$\sqrt{\frac{g}{R^{2}} \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$
C

$\sqrt{g R^{2} \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$
D

$\sqrt{g R \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$

Solution

For vertical equilibrium on the road,
$N \cos θ = m g + f \sin θ$
$m g = N \cos θ - f \sin θ$ ……….(1)
Centripetal force for safe turning,
$N \sin θ + f \cos θ = \frac{m v^{2}}{R}$ …….(2)
From eqns. (i) and (ii), we get
$\begin{array}{r} \frac{v^{2}}{R g} = \frac{N \sin θ + f \cos θ}{N \cos θ - f \sin θ} \\ \Rightarrow \frac{v_{\max}^{2}}{R g} = \frac{N \sin θ + μ_{s} N \cos θ}{N \cos θ - μ_{s} N \sin θ} \\ v_{\max} = \sqrt{R g (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})} \end{array}$

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