Question

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.

Answer 1

Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.
Consider a car taking a left turn along a road of radius $r$ banked at an angle $\theta$ for a designed optimum speed $V$. Let $m$ be the mass of the car. In general, the forces acting on the car are:
(a) Its weight $\overrightarrow{mg}$, acting vertically down
(b) The normal reaction of the road $\overrightarrow{N}$, perpendicular to the road surface
(c) The frictional force $\overrightarrow{f_s}$ along the inclined surface of the road.
Resolve $\overrightarrow{N}$ and $\overrightarrow{f_s}$ into two perpendicular components $N\cos \theta$ vertically up and $\overrightarrow{f_s}\sin \theta$ vertically down, $N\sin \theta$ and $\overrightarrow{f_s}\cos \theta$ horizontally towards the centre of the circular path.
If $v_{max}$ is the maximum safe speed without skidding.

$\frac{m{v}_{max}^2}{r}=N\sin \theta +f_s\cos \theta$

$=N\sin \theta +{\mu }_{s}N\cos \theta$

$\frac{m{v}_{max}^2}{r}=N(\sin \theta +{\mu }_s\cos \theta )....\left(1\right)$

and

$N\cos \theta =mg+f_s\sin \theta$
$=mg+{\mu }_{s}N\sin \theta$
$\therefore mg=N(\cos \theta -{\mu }_s\sin \theta )...\left(2\right)$

Dividing eq. $\left(1\right)$ by eq. $\left(2\right)$,
$\frac{m{v}_{max}^2}{r.mg}=\frac{N(\sin \theta +{\mu }_s\cos \theta )}{N(\cos \theta -{\mu }_s\sin \theta )}$

$\therefore \frac{v_{max}^2}{rg}=\frac{\sin \theta +{\mu }_s\cos \theta }{\cos \theta -{\mu }_s\sin \theta }=\frac{\tan \theta +{\mu }_s}{1-{\mu }_s\tan \theta }$

$\therefore v_{max}=\sqrt{\frac{rg(\tan \theta +{\mu }_s)}{1-{\mu }_s\tan \theta }}...,.\left(3\right)$

This is the expression for the maximum safe speed on a banked road.
At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting ${\mu }_s=0$ in eq. $\left(3\right)$, the optimum speed on a banked circular road is
$v=\sqrt{rg\tan \theta }...\left(4\right)$
$\therefore \tan \theta =\frac{v^2}{rg}$ or $\theta ={\tan }^{-1}\left(\frac{v^2}{rg}\right)$
From this eq. we see that $\theta$ depends upon $v,r$ and $g$. The angle of banking is independent of the mass of a vehicle negotiating the curve.