Flat curve with strong friction

Problem

A car takes a flat curve of radius $r = 50\;\mathrm{m}$ on asphalt with static friction coefficient $\mu_s = 0{,}80$. Calculate the maximum speed beyond which the car skids outward.

Static friction is the only horizontal force and provides the centripetal force directed towards the centre of the curve.

Solution

On the horizontal plane the weight is balanced by the normal force ($N = mg$), so the only available horizontal force is static friction, which must supply the whole centripetal force. The limiting condition occurs when friction reaches its maximum value $\mu_s N = \mu_s m g$: $\mu_s\, m g = \frac{m\, v_\text{max}^2}{r}$ The mass cancels; isolating the speed: $v_\text{max} = \sqrt{\mu_s\, g\, r}$ Substituting the numerical values (with $g = 9{,}81\;\mathrm{m/s^2}$): $v_\text{max} = \sqrt{0{,}80 \cdot 9{,}81 \cdot 50} = \sqrt{392{,}4} \approx 19{,}8\;\mathrm{m/s}$ In kilometres per hour: $v_\text{max} \approx 19{,}8 \cdot 3{,}6 \approx 71\;\mathrm{km/h}$ Beyond this speed static friction is no longer enough to curve the trajectory and the car slides outward. $\ev{v_\text{max} = \sqrt{\mu_s\, g\, r} \approx 19{,}8\;\mathrm{m/s} \approx 71\;\mathrm{km/h}}$

Linked items

Topics: Dinamica · Attrito Concepts: Moto circolare uniforme · Attrito statico · Forza centripeta · Accelerazione centripeta Skills: Applicazione delle leggi di Newton