Worked example 7.1: A banked curve

Question: Civil engineers generally bank curves on roads in such a manner that a car going around the curve at the recommended speed does not have to rely on friction between its tires and the road surface in order to round the curve. Suppose that the radius of curvature of a given curve is $r=60 {\rm m}$, and that the recommended speed is $v=40 {\rm km/h}$. At what angle $\theta$ should the curve be banked?

Answer: Consider a car of mass $m$ going around the curve. The car's weight, $m g$, acts vertically downwards. The road surface exerts an upward normal reaction $R$ on the car. The vertical component of the reaction must balance the downward weight of the car, so

$$R \cos\theta = m g.$$

The horizontal component of the reaction, $R \sin\theta$, acts towards the centre of curvature of the road. This component provides the force $m v^2/r$ towards the centre of the curvature which the car experiences as it rounds the curve. In other words,

$$R \sin\theta = m \frac{v^2}{r},$$

which yields

$$\tan\theta = \frac{v^2}{r g},$$

or

$$\theta =\tan^{-1}\left(\frac{v^2}{r g}\right).$$

Hence,

$$\theta = \tan^{-1}\left(\frac{(40\times 1000/3600)^2}{60\times9.81}\right)= 11.8^\circ.$$

Note that if the car attempts to round the curve at the wrong speed then $m v^2/r\neq m g \tan\theta$, and the difference has to be made up by a sideways friction force exerted between the car's tires and the road surface. Unfortunately, this does not always work--especially if the road surface is wet!