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

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

Step-by-step 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 = (F_c)/(N) \\\\`

`F_c = \mu N\\\\(mv^2)/(r) = \mu mg\\\\ (v^2)/(r)= \mu g\\\\v^2 = \mu gr\\\\v = √(\mu gr)`

where;

v is the maximum speed of turn

`v = √(\mu gr) \\\\v = √(0.9 * 9.8 * 30 ) \\\\v = 16.27 \ m/s`

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