Question

A curve of radius 68m is banked for a design speed of 80km/h .

f the coefficient of static friction is 0.36 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]

Express your answers using two significant figures separated by a comma. And express to Vmin, Vmax with km/h.

Answer 1

A curve of radius 68 m is banked for a design speed of 80 km/h. This means, the force should be such that the car should be able to maintain a speed of 80 km/h while moving over the curve.

Radius of the curve is, R = 68 m

Designed speed of the car is, v_d = 80 km/h = 22.22 m/s

Coefficient of static friction is, = 0.36

Centripetal force needed to turn the car around the curve depends on the speed of the car as the mass of the car (m) and the radius of the curve are fixed and is given by,

F = mv² / R

Centripetal force available to turn the car, which is the horizontal component of the normal force, is also fixed as the mass of the car and the angle of inclination of the bank () are fixed.

F = mg tan

For the designed speed v, both the needed centripetal force and the available centripetal force will be equal. Thus, we have,

mv_d² / R = mg tan

tan = v_d² /gR = 22.22 * 22.22 / (9.8 * 68)

tan = 0.740888956

= 36.5343383683 = 36.53⁰

When the car goes too slow, that is when the speed of the car v < v_d, the horizontal component of the normal force is greater than the required centripetal force and hence the car tends to move down the incline towards the center of the curve and frictional force acts away from the center to oppose this movement of the car and keeps the car on track. Thus, in this case,

mv² / R - mg tan >= - mg cos

v² / 68 - 9.8 * 0.7409 >= - 0.36 * 9.8 * 0.8035

v² / 68 >= 4.426072

v²>= 300.972896

v >= 17.348570431 m/s = 17.35 m/s is the minimum speed of the car with which it can safely cover the curve, without slipping.

When the car goes too fast, that is when the speed of the car v > v_d, the horizontal component of the normal force is smaller than the required centripetal force and hence the car tends to move up the incline away from the center of the curve and frictional force acts towards the center of the curve to oppose this movement of the car and keeps the car on track. Thus, in this case,

mv² / R - mg tan <= mg cos

v² / 68 - 9.8 * 0.7409 <= 0.36 * 9.8 * 0.8035

v² / 68 <= 10.095568

v²<= 686.498624

v <= 26.201118755 m/s = 26.20 m/s is the maximum speed of the car with which it can safely cover the curve, without slipping.

Hence, the range of speeds at which car can safely make the curve is,

17.35 m/s < v < 26.20 m/s