# Banked turns

When a car is driving along a level turn, the only thing keeping it in its lane is friction. This must be large enough to provide the centripetal force—the turn is only safe if ${\overrightarrow{F}}_{\text{f}}\ge {\overrightarrow{F}}_{\text{c}}\text{,}$ which limits speed by $v\le \sqrt{r\mu g}\text{.}$

Instead of relying on friction, we can incline the road towards the centre of the curve; this is called a banked turn. A banked turn with friction is pretty complicated, so we will consider *frictionless* banked turns. In this case, the normal force has to provide the centripetal force by itself.

The normal force can be split into vertical and horizontal components:

${F}_{\text{N,v}}=\left|{\overrightarrow{F}}_{\text{N}}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}\theta $ and ${F}_{\text{N,h}}=\left|{\overrightarrow{F}}_{\text{N}}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta \text{.}$

The horizontal component must supply the centripetal force, so we can set them equal:

$\left|{\overrightarrow{F}}_{\text{N}}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta =\frac{m{v}^{2}}{r}\text{.}$

Since the vehicle doesn’t move vertically, the vertical component has to balance gravity with $\left|{\overrightarrow{F}}_{\text{N}}\right|\phantom{\rule{0.1667em}{0ex}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}\theta =mg\text{.}$ We can solve this for the normal force and substitute it into our first equation to get

$\frac{mg}{\phantom{\rule{0.1667em}{0ex}}\mathrm{cos}\phantom{\rule{0.08335em}{0ex}}\theta}\mathrm{sin}\phantom{\rule{0.08335em}{0ex}}\theta =\frac{m{v}^{2}}{r}\text{,}$

and solving for speed gives us the maximum safe speed on a frictionless banked curve,

$v=\sqrt{rg\phantom{\rule{0.1667em}{0ex}}\mathrm{tan}\phantom{\rule{0.08335em}{0ex}}\theta}\text{.}$

We can also find the minimum safe radius at a given speed:

$r=\frac{{v}^{2}}{g\phantom{\rule{0.1667em}{0ex}}\mathrm{tan}\phantom{\rule{0.08335em}{0ex}}\theta}\text{.}$