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 `vec F_"f" >= vec F_"c"`, which limits speed by `v <= sqrt(rmug)`.

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_"N,v" = |vec F_"N"|cos theta qquad and qquad F_"N,h" = |vec F_"N"|sin theta`.

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

`|vec F_"N"|sin theta = (mv^2)/r`.

Since the vehicle doesn’t move vertically, the vertical component has to balance gravity with `|vec F_"N"|cos theta = mg`. We can solve this for the normal force and substitute it into our first equation to get

`(mg)/(cos theta)sin theta = (mv^2)/r`,

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

`v = sqrt(rg tan theta)`.

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

`r = v^2/(g tan theta)`.