A curve of radius 23 m is banked so that a 900 kg car traveling at 38.5 km/h can
ID: 1418797 • Letter: A
Question
A curve of radius 23 m is banked so that a 900 kg car traveling at 38.5 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them? v_min = km/h v_max = km/hExplanation / Answer
Forces acting on the car (Fx,Fz):
(0, -mg) gravity
(Rx, Rz) reaction from the road
(includes normal reaction and friction)
Newton's second law:
m(v²/r, 0) = (0, -mg) + (Rx, Rz)
Rz = mg
Rx = mv²/r
In absence of friction the force of reaction must be
perpendicular to the road.
Angle of banking of the road :
Ao = arctan (Rx/Rz) = arctan(v²/gr) = 21.47 degrees
If fricion is non-zero, than angle between the road and
the force of rection must be less than critiacal angle
|A(v) - Ao| <= Ac = arctan() = 11.31 degrees.
Minumum velocity:
tan(A(v_vmin)) = Rx/Rz = v_min²/gr = tan(Ao - Ac)
v_min² = gr tan(Ao - Ac)
v_min² = gr [v² - gr] / [gr + v²]
v_min² = 9.8 x 23[(38.5)² - 0.30 x 9.8 x 23 ] / [9.8 x 23 + 0.30 x 38.5²]
Maximum velocity:
v_max² = gr tan(Ao + Ac)
v_max² = gr [v² + gr] / [gr - v²]
v_max² = 9.8 x 23[(38.5)² - 0.30 x 9.8 x 23 ] / [9.8 x 23 - 0.30 x 38.5²]
Note that if was sufficiently lagre,
maximum velocity could become infinite.