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Radius of curvature car turn

  1. Nov 13, 2009 #1
    An automobile moves at a constant speed over the crest of a hill traveling at a speed of 85.7km.h. At the top of the hill a package on a seat in the rear of the car barely remains in contact with the seat. What is the radius of curvature (m) of the hill?

    85.7km/h=2.38m.s v=2.38m/s ac=v^2/r r=v^2/a

    I know that the only force acting on the package is gravity. Does this mean that the mass of the package is 9.80? mg=m*ac

    I am not sure where to go from here
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

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    Error in "85.7km/h=2.38m.s"
    In 85.7 km/h you must replace the km with 1000 m and the h with 3600 s.

    You are so close when you say "only force acting on the package is gravity"!
    Put this into F = ma and realize that the acceleration is centripetal so put that formula in for a. Solve for R.
     
  4. Nov 13, 2009 #3
    So....
    F=m*v^2/r

    9.80=m*23.8^2/r

    How do I solve for r if I don't know the mass of the package?
     
  5. Nov 13, 2009 #4

    Delphi51

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    F = ma
    mg = m*v^2/r
    and the m's cancel.
     
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