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:

$F_{\text{N,v}}=\left|{\overrightarrow{F}}_{\text{N}}\right|\cos \theta$ and $F_{\text{N,h}}=\left|{\overrightarrow{F}}_{\text{N}}\right|\sin \theta \text{.}$

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

$\left|{\overrightarrow{F}}_{\text{N}}\right|\sin \theta =\frac{m{v}^2}{r}\text{.}$

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

$\frac{mg}{\cos \theta }\sin \theta =\frac{m{v}^2}{r}\text{,}$

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

$v=\sqrt{rg\tan \theta }\text{.}$

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

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