A car enters a level,unbanked semi-circular hairpin turn of 300 m radius at a speed of 40 m/s.The coefficient of friction between the tires and the road is $\mu$ = 0.25.If the car maintains a constant speed of 40 m/s,it will
A)attempt to dig into the road surface.
B)tend to veer toward the center of the semicircle.
C)arrive safely at the end of the semicircle.
D)tend to veer toward the outside of the circle.
E)veer toward the center for the first quarter-circle,then veer toward the outside for the second quarter-circle.

Question

Quizplus · Accepted Answer

The car will experience a centripetal force towards the center of the semicircular turn, given by F = mv^2/r. The frictional force acting on the car, in the direction opposite to its motion, is given by f = μmg, where μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity. The frictional force must be large enough to provide the necessary centripetal force and keep the car moving in a circular path.

In this case, the maximum frictional force that can act on the car is f = μmg = 0.25 x m x g. If the car enters the hairpin turn at a speed of 40 m/s, it will require a centripetal force of F = mv^2/r = m x (40)^2/300 = 1600m/300 = 5.33mN (newton).

Since the frictional force available is only 0.25 x m x g, which is less than the required centripetal force of 5.33mN, the car will tend to slide towards the outside of the circle. This means the correct choice is option D, which states that the car will tend to veer towards the outside of the circle.

A Car Enters a Level,unbanked Semi-Circular Hairpin Turn of 300 μ\muμ

A Car Enters a Level,unbanked Semi-Circular Hairpin Turn of 300 $\mu$