Physics: If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the...?
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).
(a) Calculate the ideal speed to take a 120-m radius curve banked at 13.0°.
(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
- oubaasLv 76 years agoFavorite Answer
banked curve data :
r = 120 m
Θ = 13°
aequilibrium in ideal conditions requires that centripetal force Fc = m*ac = m*V^2/r must equate inner sliding force Fi = m*g*tang Θ....so :
m*V^2/r = m*g*tang Θ
V = √9.8*0.231*120 = 16.482 m/sec (59.34 km/h ; 36.87 mph)
if real speed Vr differs from the ideal speed V, then aequilibrium is maintaines by tires transversal friction μk such that :
m*Vr^2/r±m*g*μk = m*g*tang Θ
in case Vr<V , then m*g*μk is a positive quantity and must sum up to centripetal force and vice-versa for Vr>V
μk = (g*tanΘ-Vr^2/r)/g = (9.8*0.231-20^2/(3.6^2*120))/9.8 = 0.205
- NCSLv 76 years ago
(a) "Ideal speed" means there is no friction, and the downslope component of gravity is equal to the upslope component of centripetal acceleration.
gsinΘ = (v²/r)cosΘ
v = √(grtanΘ) = √(9.8m/s² * 120m * tan13º) = 16.5 m/s = 59.3 km/h
(b) At speeds less than the ideal speed, the friction points upslope.
normal acceleration a_n = gcosΘ + (v²/r)sinΘ
upslope friction acceleration a_f = µ*a+n = µgcosΘ + µ(v²/r)sinΘ
upslope centrip accel a_c = (v²/r)cosΘ
downslope gravity a_g = gsinΘ
Then gsinΘ = µgcosΘ + µ(v²/r)sinΘ + (v²/r)cosΘ
gsinΘ - (v²/r)cosΘ = µ(gcosΘ + (v²/r)sinΘ
µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ)
v² = (20km/h * 1m/s / 3.6km/h)² = 30.9 m²/s², so
µ = (9.8sin13 - (30.9/120)cos13) / (9.8cos13 + (30.9/120)sin13)
µ = 0.203
Hope this helps!