Answer to Question #197180 in Physics for gibson

Question

A cyclist is riding at a speed v on the level road. As he reaches a sharp circular corner of

radius r, he maintains the same speed. If the coefficient of friction between the tires and

the road is µ

(a) Show that the maximum permissible speed of the cyclist is

𝒗 = √𝝁𝒓𝒈

(b) If the cyclist’s speed was 20km/h, radius of the circular corner 15m and coefficient of

friction 0.35, show that the cyclist will skid away.

Answer 1

(a) As the cyclist makes the turn, the friction force provides necessary centripetal force:

$$F_c=F_{fr},$$ $$\dfrac{mv^2}{r}=\mu mg,$$ $$v=\sqrt{\mu rg}.$$

(b) Let's find the maximum permissible speed of the cyclist:

$$v=\sqrt{\mu rg}=\sqrt{0.35\cdot15\ m\cdot9.8\ \dfrac{m}{s^2}}=7.17\ \dfrac{m}{s}.$$

The cyclist’s speed as he reaches a sharp circular corner was $20\ \dfrac{km}{h}$ or $5.5\ \dfrac{m}{s}.$ As we can see, $v_{cyclist}<v_{max},$ therefore, the cyclist will not skid away.