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;
[tex]\mu = \frac{F_c}{N} \\\\[/tex]
[tex]F_c = \mu N\\\\\frac{mv^2}{r} = \mu mg\\\\ \frac{v^2}{r}= \mu g\\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr}[/tex]
where;
v is the maximum speed of turn
[tex]v = \sqrt{\mu gr} \\\\v = \sqrt{0.9 \times 9.8 \times 30 } \\\\v = 16.27 \ m/s[/tex]
Therefore, the maximum speed of turn on the given circular track is 16.27 m/s.
