Chapter 2: Space Curves
Section 2.9: Applications to Dynamics
Together, all four tires of a car in uniform motion around a circular track can exert towards the center of the track, a total frictional force of no more than 450 pounds. The radius of the track is 200 feet and the car weighs 3200 pounds. What is the fastest constant speed the car can sustain without skidding?
Taking the gravitational constant to be 32 ft/sec2, the mass of the car is
m = wg = 320032 = 100 slugs
The curvature of the track is the reciprocal of its radius, so
κ=1r = 1200
The speed v is constant, so v.=0. Hence, the acceleration becomes
a = 0 T + 1200 v2 N
The acceleration is directed along N, which itself points towards the center of the circular track. The force of friction also points towards the center of the track, and so must be along N. Hence, F = 450 N. But F = m a, and a = v2200 N, so
450 N = 100 v2200 N
is to be solved for v. Thus, write 450=v2/2, so v=±900=±30. The positive answer, 30 feet per second, is the appropriate answer. Note that the units in the example were fixed as soon as the gravitational constant g was taken as 32 ft/sec2.
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