An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,
• the normal force with magnitude n
• the car's weight with magnitude w
and the net force points toward the center of the circle made by the turn, with centripetal acceleration
a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²
Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,
∑ F (⟂) = N + W (⟂) = m a (⟂)
and
∑ F (//) = W (//) = m a (//)
Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to
N - m g cos(θ) = m a sin(θ)
and
m g sin(θ) = m a cos(θ)
The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get
tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615
→ θ ≈ 3.52°
