Question 5.15: Banking angle. (a) For a car traveling with speed v around a......
Banking angle. (a) For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required. (b) What is this angle for an expressway off-ramp curve of radius 50 m at a design speed of 50 km/h?
APPROACH Even though the road is banked, the car is still moving along a horizontal circle, so the centripetal acceleration needs to be horizontal. We choose our x and y axes as horizontal and vertical so that a_R, which is horizontal, is along the x axis. The forces on the car are the Earth’s gravity mg downward, and the normal force F_N exerted by the road perpendicular to its surface. See Fig. 5-24, where the components of F_N are also shown. We don’t need to consider the friction of the road because we are designing a road to be banked so as to eliminate dependence on friction.
(a) Since there is no vertical motion, \Sigma F_{y}=m a_{y} gives us
F_{\mathrm{N}} \cos \theta-m g=0 \text {. }
Thus,
F_{\mathrm{N}}=\frac{m g}{\cos \theta} .
[Note in this case that F_{\mathrm{N}} \geq m g since \cos \theta \leq 1.]
We substitute this relation for F_{\mathrm{N}} into the equation for the horizontal motion,
F_{\mathrm{N}} \sin \theta=m \frac{v^{2}}{r}
and obtain
\frac{m g}{\cos \theta} \sin \theta=m \frac{v^{2}}{r}
or
\tan \theta=\frac{v^{2}}{r g} \text {. }
This is the formula for the banking angle \theta : no friction needed at speed v.
(b) For r=50 \mathrm{~m} and v=50 \mathrm{~km} / \mathrm{h} (or 14 \mathrm{~m} / \mathrm{s} ),
\tan \theta=\frac{(14 \mathrm{~m} / \mathrm{s})^{2}}{(50 \mathrm{~m})\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.40
so \theta=22^{\circ}.