Answer to Question #155370 in Physics for Andu

Question

The motorcyclist crosses a horizontal turn with radius R = 25 m at a speed of V = 50 km / h, being on the slip threshold. Calculate the coefficient of sliding friction and the angle of stability.

Answer 1

a) In order to motorcycle not to slip the force of friction should provide the necessary centripetal acceleration:

$$F_{fr}=F_c,$$ $$\mu mg=\dfrac{mv^2}{R},$$ $$\mu=\dfrac{v^2}{gR},$$ $$\mu=\dfrac{(50\ \dfrac{km}{h}\cdot\dfrac{1000\ m}{1\ km}\cdot\dfrac{1\ h}{3600\ s})^2}{9.8\ \dfrac{m}{s^2}\cdot 25\ m}=0.78$$

b) We can find the angle of stability (the angle at which the motorcyclist doesn't slip) from the formula:

$$tan\theta=\mu,$$ $$\theta=tan^{-1}\mu=tan^{-1}(0.78)=38^{\circ}.$$

Answer:

a) $\mu=0.78$

b) $\theta=38^{\circ}.$