Question

What is the maximum speed that a 2200 kg car can go around a level circular track with a radius of 30.0 m without slipping if the coefficient of static friction between the tires and the road is 0.900?

Answer 1

Answer:

The maximum speed of turn on the given circular track is 16.27 m/s.

Explanation:

Given;

mass of the car, m = 2200 kg

radius of the track, r= 30 m

coefficient of static friction between the tires and the road, μ = 0.9

The net vertical force on the car = N = mg

The net horizontal force on the car = Centripetal force

The coefficient of static friction is given as;

$\mu = \frac{F_c}{N}$

$F_c = \mu N\\\\\frac{mv^2}{r} = \mu mg\\\\ \frac{v^2}{r}= \mu g\\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr}$

where;

v is the maximum speed of turn

$v = \sqrt{\mu gr} \\\\v = \sqrt{0.9 \times 9.8 \times 30 } \\\\v = 16.27 \ m/s$

Therefore, the maximum speed of turn on the given circular track is 16.27 m/s.