Question

What is meant by banking of road? What is need of banking of roads? Obtain an expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle $\theta$. The coefficient of friction between the wheels is and the road is $\mu$.

Answer 1

Banking of road means raising the outer edge of the road with respect to inner edge so that the road makes an angle with the horizontal.

When a vehicle moves along a curve, the force of friction provides the necessary centripetal force. But this force has a limit = $\mu mg$.When the speed of the vehicle increases beyond this maximum value, the banking of roads is necessary so that the component of normal reaction of the road towards the centre provides the necessary centripetal force.

From the figure, for vertical equilibrium :

Total upwards force = total downwards force

$Ncos\theta =mg+fsin\theta mg=Ncos\theta -fsin\theta ...\left(1\right)$

As the forces $Nsin\theta ,fcos\theta$ provide the necessary centripetal force.

$\frac{m{v}^2}{r}=Nsin\theta +fcos\theta ...\left(2\right)$

Thus dividing (2) by (1),

$\frac{\frac{m{v}^2}{r}}{mg}=\frac{Nsin\theta +fcos\theta }{Ncos\theta -fsin\theta }$

As $f={\mu }_{s}N$

Thus,

$\frac{v^2}{rg}=\frac{sin\theta +{\mu }_{s}cos\theta }{cos\theta -{\mu }_{s}sin\theta }=\frac{tan\theta +{\mu }_s}{1-{\mu }_{s}tan\theta }=\frac{tan\theta +tan\lambda }{1-tan\lambda tan\theta }=tan(\theta +\lambda )$

$v=\sqrt{rgtan(\theta +\lambda )}$ where $tan\lambda ={\mu }_s$