Question

A curve of radius 78 m is banked for a design speed of 100 km/h .

Angle θ of the embankment: 45.3

If the coefficient of static friction is 0.39 (wet pavement), at what range of speeds can a car safely make the curve?

Answer 1

Radius of the curve, r = 78 m

Design speed = 100 km/h = (100 x 1000) / (60x60) = 27.8 m/s

Angle of embankment, $\Theta$ = 45.3 deg

Suppose, v is the speed of the car on the circular path.

To determine the range of the speed, we have to determine the maximum and minimum value of speed around the circular path.

Now,

Normal acceleration a_n = gcosΘ + v²sinΘ/r

and friction acceleration a_f = µ*a_n

At the maximum speed, friction and gravity point down-slope, so

gsinΘ + µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r

Plugging in knowns (and dropping units for ease),

9.8sin45.3 + 0.39*(9.8cos45.3 + v²sin45.3/78) = v²cos45.3/78

=> 6.96 + 0.39*(6.89 + 0.0091v²) = 0.00902v²

=> 6.96 + 2.69 + 0.0035v² = 0.00902v²

=> 0.00552v² = 9.65

=> v² = 9.65 / 0.00552 = 1748.2

=> v = 41.8 m/s = (41.8 x 3600 / 1000) km/h = 150.5 km/h

This is the maximum speed. But the design speed of the road is 100 km/h.

Therefore, the maximum speed = 100 km/h

Now, determine the minimum speed.

At the minimum speed, friction points up-slope, so

gsinΘ - µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r

Plugging in knowns (and dropping units for ease),

9.8sin45.3 - 0.39*(9.8cos45.3 + v²sin45.3/78) = v²cos45.3/78

=> 6.96 - 0.39*(6.89 + 0.0091v²) = 0.00902v²

=> 6.96 - 2.69 - 0.0035v² = 0.00902v²

=> 0.01252v² = 4.27

=> v² = 4.27 / 0.01252 = 341.05

=> v = 18.47 m/s = (18.47 x 3600) / 1000 km/h = 66.5 km/h

This is the minimum speed.

Hence, the required speed range for the car is 66.5 km/h to 100 km/h (Answer)