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 57.0 m. If we approximate the track to be completely flat and the racecar is traveling at a constant 27.5 m / s (about 62 mph) around the turn, what is the racecar's centripetal (radial) acceleration? What is the force responsible for the centripetal acceleration in this case? (Normal, Friction, Weight, or Gravity)? 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

Centipetal acceleration = V^2 / R

= 27.5^2 / 57

= 13.26 m/sec^2

Force responsile for providing centripetal acceleration if frictional force.

Centripetal force required = m*V^2 / R

Which is provided by the term = mu*N = mu*m*g

So,

  m*V^2 / R = mu*m*g

mu = V^2 / (R*g)

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