What banking angle (in degrees) will prevent cars from sliding off the road, assuming everyone travels at the speed limit and there is no friction present?

Question

A given highway turn has a 115 km/h speed limit and a radius of curvature of 1.1 km.

Answer 1

v = 115 km/hr = 31.9 m/s, r = 1.1 km = 1100 m

The only forces acting on the car are gravitational force (mg)

and the normal force from the banked curve (F_N).

F_N has two components:

The vertical component of F_N (F_N cosθ)

F_N cosθ counterbalances the gravitational force.

So F_N cosθ = mg

which means that F_N = mg/cosθ

The horizontal component of F_N (F_N sinθ)

F_N sinθ is the source of centripetal force F_c = mv²/r that keeps the car on the road.

F_N sinθ = (mg/cosθ)sinθ = mg(sinθ/cosθ) = mg tanθ = mv²/r

which means g tanθ = v²/r,

so tanθ = v²/rg

The required banking angle is given by

θ = tan⁻¹(v²/rg) = tan⁻¹[(31.9)²/(1100)(9.8)] = tan⁻¹(0.09466) = 5.4°