Question

a 1,730 Kg car traveling at a rate of 166Kph is racing around a flat circular track, maintaining a perfectly circular 2 Km path.

any help would work; explanations, formulas, steps..etc

a. what is the centripetal force acting on the car?

b. what is the minimum coefficient of statistic friction between the tires and the track that will maintain this situation?

c. how many rotations do the tires make on a single loop around the track.

Answer 1

Centripetal force
F=mv^2/r_=?

Answer 2

Given
m=1730kg
v=166kph=?m/s

Answer 3

First, note that 166 kph = 46 m/s. Also, the 2 km = 2000 m mentioned in the question is the circumference of the circular path being traveled. By C = 2πr, you know the radius of this circle has to be 318 m. Now for your questions:

For question A, you probably know the equation for centripetal force is:

\[F_{c} = \frac{mv^2}{r}\]
You can plug in the values for m, v, and r to calculate F(c).

Question B is asking you what force of friction (acting radially OUTWARD from the centre of the circular path) will exactly balance the centripetal force (which acts radially INWARD towards the centre of the circular path), to prevent the car from sliding radially inward. You just calculated the centripetal force in question A, so you set that equal to friction:

\[F_{c} = f\]\[F_{c} = \mu_{s}N\]\[F_{c} = \mu_{s}mg\]
You can plug in the values for F(c), m, and g to calculate μ(s).

For question C...we would have to know the size of the tires to be able to answer this question. Is this information given somewhere else?

Answer 4

Here
2*pi*r=20000
r=radius of circular path