Question

A curve of radius 90 m is banked for a design speed of 100 km/h. If the coefficient of static friction is 0.30 (wet pavement), at what range of speeds can a car safely make the curve?

Answer 1

This means that the downslope component of gravity is equal to the upslope component of centripetal acceleration.

100 km/hr = 27.77m/s { to convert from km/hr to m/s just multiply the spped with 5/18 and vice versa}
gsinΘ = v²cosΘ / r
sinΘ / cosΘ = tanΘ = v² / gr = (27.77m/s)² / (9.8m/s² * 90m) =0.874
Θ = 41.16º

normal acceleration a_n = gcosΘ + v²sinΘ/r
and friction acceleration a_f = µ*a_n

At the maximum speed, friction and gravity point downslope, so
gsinΘ + µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r
Plugging in knowns (and dropping units for ease),
9.8sin41.6 + 0.3(9.8cos41.6 + v²sin41.6/90) = v²cos41.6/90
8.70+ 0.00221v² = 0.0083v²
0.00609v² = 8.70
v² = 1428.57
v = 37.79 m/s = 136.044 km/h

At the minimum speed, friction points upslope, so
gsinΘ - µ(gcosΘ + v²sinΘ/r) = v²cosΘ/r
Plugging in knowns (and dropping units for ease),
9.8sin41.6 - 0.3(9.8cos41.6 + v²sin41.6/90) = v²cos41.6/90
4.3 - 0.00221v² = 0.0083v²
0.01051v² = 4.3
v² = 409.13
v = 20.22 m/s = 72.792 km/h