Banking angle of the curve

Problem

To avoid safety depending solely on friction, motorways feature banked curves. Calculate the banking angle that would allow a curve of radius $r = 40\;\text{m}$ (the same as the previous problem) to be taken at $v = 90\;\text{km/h}$ with no friction.

Solution

Data. $r = 40\;\text{m}$, $v = 90\;\text{km/h}$, $g = 9{,}8\;\text{m/s}^2$, zero friction.

Speed conversion: $v = 90\;\text{km/h} = \frac{90}{3{,}6} = 25\;\text{m/s}$

Set-up. With no friction, only the weight $\vec{P}$ (vertical, downwards) and the normal reaction $\vec{R}$ (perpendicular to the road surface, inclined at $\theta$) act on the vehicle. Resolving $\vec{R}$: $\begin{cases} R\sin\theta = \dfrac{m\,v^2}{r} & \text{(horizontal: centripetal force)} \\ R\cos\theta = mg & \text{(vertical: balances the weight)} \end{cases}$

Dividing the first equation by the second eliminates $R$ and $m$: $\tan\theta = \frac{v^2}{g\,r}$

Numerical substitution: $\tan\theta = \frac{25^2}{9{,}8 \cdot 40} = \frac{625}{392} = 1{,}59 \quad\Rightarrow\quad \theta = \arctan(1{,}59) = 57{,}9^\circ$

Result. $\ev{\tan\theta = \frac{v^2}{g\,r} \;\Rightarrow\; \theta \approx 58^\circ}$

Such a steep angle signals that $90\;\text{km/h}$ on a radius of only $40\;\text{m}$ is an extreme curve: on real motorways the radii are much larger and the banking angles are only a few degrees.

Links

Topics: Dynamics Concepts: Uniform circular motion · Centripetal force · Normal force · Weight force Skills: Vector resolution · Use of trigonometry Objects: Banked curve