A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the tr

Question

A NASCAR racecar rounds one end of the Martinsville Speedway. This end of the track is a turn with radius approximately 187 ft. If we approximate the track to be completely flat and the racecar is traveling at a constant 64.00 mph around the turn, what is the racecar's centripetal (radial) acceleration?

Convert this result to SI units.

What is the force responsible for the centripetal acceleration in this case?

To keep from skidding into the wall on the outside of the turn what is the minimum coefficient of static friction between the racecar's tires and track?

Explanation / Answer

We'll assume that there is such thing as a racetrack that has a flat curve, and that it's safe for a car to drive along that curve at such a high speed. We'll also assume that there is no oil on the road, as that would affect friction.

Let's go ahead and get everything provided into SI units.

Radius = 187ft (12 inches/ft)(2.54cm/in)(.01m/cm) = 57.0 meters.
Speed = 64mph(1.60934km/mi)(1000m/km)(1hr/3600s) = 28.61m/s.

ac = v2/r = 28.612/57.0 = 14.36m/s2

We'll assume the mass of the car is m, in which case the force due to the centrepetal acceleration is 14.36 times m, and is caused by the inward acceleration of the car as it goes around the curve.

In order to keep the car on the track, the coefficient of static friction needs to be at least enough so the force due to friction equals the centripetal force.

We need to know the normal force applied by the car on the road. Because it is a horizontal road, the normal friction is just the weight of the car, which is mass times gravity, or mg.

Fc = FN = mg

So...

14.36 x mass = g x mass

Canceling out the mass, and solving for , we get:

14.36 = g

= 14.36/g 14.36/9.81 = 1.464

So the coefficient of static friction needs to be at least 1.464 to keep the car from skidding off the track.

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