Flat bend, dry and wet

Problem

A car travels round a flat bend of radius $r = 50\;\mathrm{m}$. The coefficient of static friction between tyres and asphalt is $\mu_s = 0{,}40$. What is the maximum speed at which it can take the bend without slipping? What happens if the bend is wet and $\mu_s$ drops to $0{,}15$?

Solution

What provides the centripetal force. On a flat bend (not banked) the only horizontal force available for cornering is the static friction between tyres and asphalt. It must supply the entire centripetal force: $f_s = \frac{m v^2}{r}$ Static friction has a maximum $f_{s,\max} = \mu_s N = \mu_s\,mg$ (on a flat road $N = mg$). The maximum speed occurs when friction is at its limit: $\mu_s\,mg = \frac{m v_\text{max}^2}{r} \;\Rightarrow\; v_\text{max} = \sqrt{\mu_s\,g\,r}$ The mass cancels: the limiting speed does not depend on how heavy the car is.

Dry ($\mu_s = 0{,}40$). $v_\text{max} = \sqrt{0{,}40\cdot 9{,}8\cdot 50} = \sqrt{196}\;\mathrm{m/s}$

$\ev{v_\text{max} = 14\;\mathrm{m/s}\ (\approx 50\;\mathrm{km/h})}$

Wet ($\mu_s = 0{,}15$). $v_\text{max} = \sqrt{0{,}15\cdot 9{,}8\cdot 50} = \sqrt{73{,}5} \approx 8{,}6\;\mathrm{m/s}$

$\ev{v_\text{max}^\text{wet} \approx 8{,}6\;\mathrm{m/s}\ (\approx 31\;\mathrm{km/h})}$

Comment. Since $v_\text{max}\propto\sqrt{\mu_s}$, reducing the friction from $0{,}40$ to $0{,}15$ (a factor of $\approx 2{,}7$) lowers the limiting speed by $\sqrt{2{,}7}\approx 1{,}6$ times, from $14$ to about $8{,}6\;\mathrm{m/s}$. On a wet road, taking the same bend at $14\;\mathrm{m/s}$ would make the car slide outward: that is why on a bend, in the wet, you must slow down a lot.

Links

Topics: Dinamica · Attrito Concepts: Attrito statico · Forza centripeta · Moto circolare uniforme · Forza normale Skills: Applicazione delle leggi di Newton · Impostazione simbolica Objects: Curva sopraelevata