1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Frictionless banked turn? What is wrong with my thinking?

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    The Daytona 500 is the major event of the NASCAR season. It is held at the Daytona International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius (at the top) of r = 316, and are banked steeply, with θ = 31°. Suppose these maximum-radius turns were frictionless. At what speed would the cars have to travel around them?


    2. Relevant equations
    Fc = mv^2/r



    3. The attempt at a solution
    εfx = Fac = Nx
    (mv^2)/r = sinθN
    (mv^2)/r = sinθ(-wcosθ)
    (mv^2)/r = sinθ(-mgcosθ)
    v = √[sinθ(-gcosθ)r]
    v = √[sin31°(-(-9.8))(cos31°)(316)]
    v = 36.9m/s

    answer is actually 43m/s
     
  2. jcsd
  3. Oct 19, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Omit the minus signs, as you use magnitudes.

    Check the figure. Is N=mgcosθ?



    ehild
     
  4. Oct 19, 2013 #3
    yes, on an incline the normal force is equal to the magnitude of the y component of the weight. the y component of the weight is equal to mgcosθ
     
  5. Oct 19, 2013 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is true if an object slides on a slope. Now, the car does not slide but stays at a constant hight. The vertical component of the normal force opposes the weight. The horizontal component of the normal force provides the centripetal force.
    ehild
     
  6. Oct 19, 2013 #5
    why is it only true if the object slides on a slope?
     
  7. Oct 19, 2013 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The normal force is a force of constraint. It ensures that the body performs the constrained motion. Its magnitude depends on the motion.

    In case of sliding along a slope, the body can not rise from the surface, neither can it penetrate into the surface: there is a force balancing the normal component of gravity: the normal force is equal to -mgcosθ.

    Here the car has to travel along a horizontal circle with speed V. If the normal force was equal to -mgcosθ, the resultant force would be parallel to the slope, and downward, with magnitude mgsinθ. But the circular motion to be maintained, the resultant force has to be horizontal, and equal to mV2/R.

    ehild
     
    Last edited: Oct 19, 2013
  8. Oct 19, 2013 #7
    ah okay thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Frictionless banked turn? What is wrong with my thinking?
  1. Frictionless Bank (Replies: 3)

Loading...