Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn, what is the race car's centripetal (radial) acceleration?

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

friction, normal, gravity, or weight?

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?

a) The race car's centripetal (radial) acceleration = 16.32 m/s²

b) The force responsible for the centripetal acceleration in this case is the frictional force between the tires of the NASCAR and the surface of the tracks.

c) μ = 1.67

Explanation:

The centripetal acceleration for circular motion is given as

α = (v²/r)

v = velocity of the NASCAR = 30.5 m/s

r = radius of the circular motion = 57.0 m

α = (v²/r) = (30.5²/57) = 16.32 m/s²

b) The forces acting on the NASCAR include the weight of the car acting downwards, the normal reaction of the surface on the NASCAR, the frictional force between the NASCAR & the surface of the race track acting horizontally together with the centripetal force that keeps the NASCAR in circular motion.

The vertical and horizontal forces balance out.

For the vertical forces, the weight of the NASCAR balances the normal rwaction of the race track surface on the NASCAR.

For the horizontal forces, the centripetal force is balanced by the frictional force between surface and the NASCAR tires.

The force responsible for the centripetal acceleration in this case is the frictional force between the tires of the NASCAR and the surface of the tracks.

c) 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?

Centripetal force = frictional force

Centripetal force = (mv²/r)

Frictional force = μN

N = normal reaction = weight = mg

Frictional force = μmg

(mv²/r) = μmg

μ = (v²/gr) = [30.5²/(9.8×57)] = 1.67

A coefficient of friction that is more than one just means that the frictional force is stronger than the normal force.

Hope this Helps!!!

answer: i don't know the answer good luck

explanation: