Answer to Question #155370 in Physics for Andu

Question #155370

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.


1
Expert's answer
2021-01-19T07:10:29-0500

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


Ffr=Fc,F_{fr}=F_c,μmg=mv2R,\mu mg=\dfrac{mv^2}{R},μ=v2gR,\mu=\dfrac{v^2}{gR},μ=(50 kmh1000 m1 km1 h3600 s)29.8 ms225 m=0.78\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:


tanθ=μ,tan\theta=\mu,θ=tan1μ=tan1(0.78)=38.\theta=tan^{-1}\mu=tan^{-1}(0.78)=38^{\circ}.

Answer:

a) μ=0.78\mu=0.78

b) θ=38.\theta=38^{\circ}.


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