A car is negotiating a curved road of radius R. The road is banked at an angle . The coefficient of friction between the tyres of the car and the road is . The maximum safe velocity on this road is [2016]

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Solution

Q.
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 [2016]

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

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

3 $\sqrt{g R^{2} \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$

4 $\sqrt{g R \frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ}}$

Ans.
(4)

For vertical equilibrium on the road,

$N \cos θ = m g + f \sin θ \Rightarrow m g = N \cos θ - f \sin θ$ ...(i)

Centripetal force for safe turning, $N \sin θ + f \cos θ = \frac{m v^{2}}{R}$ ...(ii)

From eqns. (i) and (ii), we get

$\frac{v^{2}}{R g} = \frac{N \sin θ + f \cos θ}{N \cos θ - f \sin θ} \Rightarrow \frac{v_{max}^{2}}{R g} = \frac{N \sin θ + μ_{s} N \cos θ}{N \cos θ - μ_{s} N \sin θ}$

$v_{max} = \sqrt{R g (\frac{μ_{s} + \tan θ}{1 - μ_{s} \tan θ})}$

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