You’re late for class and entering the freeway via an unbanked curve to the righ

Question

You’re late for class and entering the freeway via an unbanked curve to the right with radius 40 m. Assume the mass of your car is 2000 kg. The coefficient of friction between the tire rubber and the roadway is s = 0.90

a. What is the highest speed you can drive through this curve without slipping?

b. Now that you are on the freeway, you have to drive over a banked, curved bridge with radius 70 m. What is the maximum speed you can drive on this curve without slipping, if the curve is banked at 15°?

c. The coefficient of friction between the tire rubber and the roadway is a much lower s-wet = 0.60 when the roadway is wet. What are the maximum speeds for parts a and b if the roadway is wet?

Explanation / Answer

here,

mass of car, m = 2000 kg

coefficient of friction, u = 0.90

radius of path, r = 40 m

From newton Second Law, Fnet = 0-------------(1)
For x axis,

mv^2/r = mg*SinA + us*mgCosA

For y Axis,
0 = mg*CosA - us*mgSinA - mg -------------(2)

upon rearranging eqn 1 and 2 we get,

Vmax = sqrt( r*g(SinA + us*CosA)/(CosA - us*SinA) ) ----------------(3)

Part A:
When path is flat, maximum velocity is given as:
as A = 0

V = sqrt(us*r*g) -----------------------(4)
V = sqrt(0.9 * 40 * 9.8)
v = 18.783 m/s

Part B:
The maximum speed attained by car without slipping, Vmax is given as,

Angle , A = 15 degrees
ardius of path, r = 70

Vmax = sqrt( r*g(SinA + us*CosA)/(CosA - us*SinA) ) -----------------(5)
Vmax = sqrt( 70*9.8(Sin15 + 0.9*Cos15)/(Cos15 - 0.9*Sin15) )
Vmax = 34.494 m/s

Part C:

Case 1 :
us = 0.60
R = 40

Using Eqn 4,
V = sqrt(0.6 * 40 * 9.8)
V = 15.336 m/s

Case 2:
us = 0.6
r = 70

Using eqn 5
Vmax = sqrt( 70*9.8(Sin15 + 0.6*Cos15)/(Cos15 - 0.6*Sin15) )
Vmax = 26.636 m/s

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