A highway curve of radius $500m$ is designed for traffic moving at a speed of $90km/h$. What is the correct banking angle of the road?

A highway curve of radius $500m$ is designed for traffic moving at a speed of $90km/h$.

To analyze the situation, we must draw a free-body diagram for the car as shown below. In the vertical axis, the weight of the car is against the $y$component of the normal force, so the net force in the $y$- direction is

$F_{y,\text{net}}=N\cos \theta -mg$

$\cos \theta =\frac{mg}{N}$

Note that, the net force is zero to keep the car moves in a circular way. In the horizontal axis. the centripetal force which is given by equation (8.6) acts on the car in opposite direction to the $x$component of the normal force, so the net force in the $x$-direction is

$F_{x,\text{net}}=N\sin \theta -F_c$

$F_{x,\text{net}}=N\sin \theta -\frac{m{v}^2}{r}$

$0=N\sin \theta -\frac{m{v}^2}{r}$

localid="1647835741868" $\sin \theta =\frac{m{v}^2}{Nr}$

Our target is to find the angle $\theta$, so we divide equation (2) by equation (1), but first, let us convert the unit of the velocity from$(\mathrm{km}/\mathrm{h})$where $1\mathrm{km}=1000\mathrm{m}$and $1\mathrm{h}=3600\mathrm{s}$.

$v=\left(90\frac{\mathrm{km}}{\mathrm{h}}\right)\left(\frac{1000\mathrm{m}}{1\mathrm{km}}\right)\left(\frac{1\mathrm{h}}{3600\mathrm{s}}\right)=25\mathrm{m}/\mathrm{s}$

Now, divide both equations (2) and (1)

$\sin \theta =\frac{m{v}^2}{Nr}$

$\cos \theta =\frac{mg}{N}$

$\frac{\sin \theta }{\cos \theta }=\frac{\left(m{v}^2\right)/\left(Nr\right)}{(mg/N)}$

$\tan \theta =\frac{v^2}{rg}$

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

Now, plug the values for $v,r$and $g$into equation (3) to get $\theta$

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

$={\tan }^{-1}\left(\frac{(25\mathrm{s}{)}^2}{(500\mathrm{m})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\right)$

$=7.3°$

The correct banking angle of the road $\theta =7.3°$.