Question

What is the angle of banking? Derive the expression for the angle of banking on a curved road with a certain coefficient of friction.

Answer 1

Step 1: Angle of banking:

1. The angle of banking is defined as the angle at which a vehicle becomes inclined with respect to the horizontal around its longitudinal axis.
2. The outer edge of the vehicle is raised above the inner edge to make it safer on a curved road.

Step 2: Expression for an angle of banking on a curved road with a certain coefficient of friction:

Consider a vehicle of mass $m$ with moving speed $v$ on the banked road with a radius $r$. Let $\theta$ be the angle of friction force acting between the road and the tires of the vehicle.

Total upwards force = Total downward force

$N\cos \theta =mg+f\sin \theta$

Where $N\cos \theta$ is one of the components of normal reaction along the verticle axis, $mg$ is the weight of the vehicle acting vertically downward and $f\sin \theta$ is one of the components of frictional force along the verticle axis.

Therefore,

$mg=N\cos \theta -f\sin \theta$ (1)

$\frac{m{v}^2}{r}=N\sin \theta +f\cos \theta$ (2)

Step 3:Divide eq (1) in eq (2)

$\frac{{\displaystyle \frac{m{v}^2}{r}}}{mg}=\frac{N\sin \theta +f\cos \theta }{N\cos \theta -f\sin \theta }$ (3)

Step 4: Substitute ${\mu }_{s}N$ for $f$ in equation(3).

$$\begin{gathered} \frac{{\displaystyle \frac{m{v}^2}{r}}}{mg}=\frac{N\sin \theta +{\mu }_{s}N\cos \theta }{N\cos \theta -{\mu }_{s}N\sin \theta } \\ \frac{v^2}{rg}=\frac{N\tan \theta +{\mu }_{s}N}{N-{\mu }_{s}N\tan \theta } \\ \theta ={\tan }^{-1}\left(\frac{v^2-{\mu }_{s}rg}{1+v^2{\mu }_s}\right) \end{gathered}$$

Hence, the Angle of banking is ${\tan }^{-1}\left(\frac{v^2-{\mu }_{s}rg}{1+v^2{\mu }_s}\right)$.