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Grade 12 Physics Thru ILC, Please help!

  1. Dec 3, 2008 #1
    Physics SPH4U-A Lesson 4 Question #15

    A race-car driver is driving her car at a record-breaking speed of 225km/h. The first turn on the course is banked at 15 degrees, and the car's mass is 1450kg.

    a) Calculate the radius of the curvature for this turn.

    b) Calculate the centripetal acceleration of the car.

    c) If the car maintains a circular track around the curve (does not move up or down the bank), what is the magnitude of the force of static friction?

    d) What is the coefficient of static friction necessary to ensure the safety of this turn?

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    a) Fnsin(theta)=(mv^2)/r
    r=(mv^2)/Fnsin(theta)
    r=(1450kgx62.5m/s)/(14,711Nxsin15)
    r=1488m

    b.) a=v^2/r
    a=(62.5m/s)^2/1488m
    a=2.63m/s^2

    c.) Fs=Fnsin(theta)
    =(14,711Nxsin15)
    =3,807N

    d.) Us=Fs/Fn
    =3,807N/14,711N
    =0.26


    Can anyone verify that I used the correct formulas and that this information is correct?
    Thanks for your time,
    Jayker
     
  2. jcsd
  3. Dec 3, 2008 #2

    turin

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    I think this problem is missing something ...
     
  4. Dec 4, 2008 #3
    okay, any ideas?
     
  5. Dec 4, 2008 #4

    turin

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    I don't think this is possible from the given information:

    - speed of 225km/h
    - turn banked at 15 degrees
    - mass is 1450kg

    If it had said something like, "in order to eliminate lateral contact force" ...

    It is vaguely possible that I just haven't learned the basic physics principle needed (though I doubt it). The other parts of the problem rely on the result to (a), I believe. It is probably just poorly worded; ask your instructor if you need to assume anything else.
     
  6. Dec 4, 2008 #5
    But we also know Fn

    Fn=(mg)/cos15

    which only leaves r out of the equation to be solved for.
     
  7. Dec 4, 2008 #6

    turin

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    What equation containing r are you thinking of?
     
  8. Dec 4, 2008 #7
    Fnsin(theta)=mv^2

    where Fn=mg/cos15
    =(1450kgx9.8N/kg)/cos15
    m=1450kg
    v=225km/h=62.5m/s
    theta=15 degrees
     
  9. Dec 5, 2008 #8

    turin

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    The problem doesn't give you enough information to know this - especially since it is asking you, in part (c), to calculate the static friction by assuming that the car is in circular motion, implying that these are extra assumption that do not apply to parts (a) and (b). I would tell the instructor that I need more information to solve this problem.
     
  10. Dec 5, 2008 #9
    Yeah, not enough info. Obviously, you can travel around a curve with the same radius of curverature at different speeds. Does the geometry of the road uniquely determine your velocity?

    A higher velocity would simply require a larger centripital force vector. If both the normal force and frictional force increase as to coninually cancel the vertical component, you can have any velocity untill you break your coeficient of static friction.
     
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