Page 118 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 9] CIRCULAR MOTION AND GRAVITATION 103

Hence Centripetal force = frictional force
mv 2
= µ s mg
r
v 2
r =
µ s g
The car’s velocity is

ft/s
v = (20 mi/h) 1.47 = 29.4 ft/s
mi/h
and so the minimum turning radius is
v 2 (29.4 ft/s) 2
r = = 2 = 34 ft
µ s g (0.8)(32 ft/s )

SOLVED PROBLEM 9.6
Highway curves are usually banked (tilted inward) at an angle θ such that the horizontal component of
the reaction force of the road on a car traveling at the design velocity equals the required centripetal force.
Find the proper banking angle for a car making a turn of radius r at velocity v.
In the absence of friction, the reaction force F of the road on the car is perpendicular to the road surface. The
vertical component F y of this force supports the weight w of the car, and its horizontal component F x provides the
2
centripetal force mv /r. From Fig. 9-1 we have
mv 2
F x = F sin θ =
r
F y = F cos θ = w = mg

Dividing F x by F y gives
F sin θ mv 2 v 2
= tan θ =
F cos θ mgr gr

Fig. 9-1

The proper banking angle θ depends only on v and r, not on the mass of the car. If a car goes around a banked turn
more slowly than at the design velocity, friction tends to keep it from sliding down the inclined roadway; if the car
goes faster, friction tends to keep it from skidding outward. If the car’s speed is too great, however, friction may not
be sufficient to keep the car on the road.