# Suppose the designated speed for an exit ramp on a highway is to be $13.4\; m/s$ and the radius of the curve is $50.0\; m$. At what angle should the curve be banked? (i.e, the roadway is tilted towards inside of the curve).

$\begin{array}{1 1} 12.2^{\circ} \\ 20.1^{\circ} \\ 4.8^{\circ}\\ 2.9^{\circ}\end{array}$

Answer :$20.1^{\circ}$
On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between car and road.
However, if the road is banked at an angle $\theta$, Only the component $n= n \sin \theta$ causes the centripetal acceleration.
Hence , Newton's second law for the radical direction gives
$\sum F_r = n \sin \theta= \large\frac{mv^2}{r}$-----(1)
The car is in equilibrium in the vertical direction.
Thus, from $\sum F_y =0$ we have
$n \cos \theta =mg$ ----------(2)
Dividing (1) and (2) gives
(3) $\tan \theta =\large\frac{v^2}{rg}$
$\theta = \tan ^{-1} \bigg( \large\frac{(13.4\;m/s)^2}{(50.0\;m)(9.80\;m/s^2)} \bigg)$$=20.1^{\circ}$