The exit ramp from a highway is roughly circular and is not banked (tilted). The

Question

The exit ramp from a highway is roughly circular and is not banked (tilted). The curve has a radius of 225.0m . A motorcycle exits the highway on to the ramp at a speed of 25.0m/s .

Part A

What is the minimum coefficient of friction between the tires and the ramp that will prevent the motorcycle from skidding?

Part B

Suppose the ramp is icy and the coefficient of friction between the motorcycle tires and the ramp is only one-third what you found in part A. What is the maximum safe speed at which the motorcycle can round the curve safely?

Explanation / Answer

(a)
The force that the ramp must apply to the tyres is F = mv2/R
The maximum force of friction between the tyres and the road, Ff = mg * coefficient of friction
Equating above two equations, we get
mv2/R = mg * coefficient of friction
v2/R = g * coefficient of friction
252 / 225 = 9.8 * coefficient of friction
coefficient of friction = 0.2834

(b)
now as v^2 is proportional to the coefficient then v is proportional to the SQUARE ROOT of the coefficient.
ie v2/v1 = sqrt(c2/c1)
v2 = v1 * sqrt(c2/c1)
v2 = 25 * sqrt(1/3)
v2 = 43.3 m/s

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