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Homework Help: Centrifugal and centripetal force question: race-car and banked curve question

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A race-car driver is driving her car at a record-breaking speed of 225kh/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?

    2. Relevant equations

    vector v^2 = (vector gravity)(radius)(tan incline of bank)
    centripetal acceleration = vector v ^2 / radius

    3. The attempt at a solution

    First of all, I don't really understand centrifugal force. Centripetal force is for when an object has uniform circular motion, right? How is it affected by centrifugal force?

    So my basic understanding of this question is, when a car makes a turn it doesn't slide off laterally because of the friction between tires and the road, and if the curve has a bank it would increase force of friction when a car goes on it.

    V = 225km/h = 225000m/h = 62.5m/s
    Incline of bank = 15 degrees
    M = 1450kg

    vector v^2 = (vector gravity)(radius)(tan incline of bank)
    (radius) = (vector v)^2 / (vector gravity)(tank incline of bank)
    (radius) = (62.5m/s)^2 / (9.8m/s^2)(tan15)
    (radius) = (3906.25m^2/s^2) / (2.625902086m/s^2)
    (radius) = 1487.58m

    This is strange because if I divide (62.5m/s)^2 / (9.8m/s^2) first and then multiply it with (tan15) I get 106.80m

    So I'm not sure which answer is correct.
    I think all the other questions need a correct radius value so I can't go further :/
  2. jcsd
  3. Mar 1, 2009 #2

    Doc Al

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    Staff: Mentor

    All you need is centripetal force. (Centrifugal force is a "fictitious" force used when analyzing things from an accelerating frame--no need for that here.)

    As far as this problem goes, you are not given enough information to solve it. Can you please tell me the textbook this is from and the problem number. (In case I have the textbook.)
  4. Mar 1, 2009 #3
    This is from an independent learning course from the Independent Learning Center in Ontario. It's in Lesson 4 of Unit 1.

    What other information do I need to solve it?
  5. Mar 1, 2009 #4

    Doc Al

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    Staff: Mentor

    I don't have that one. This exact problem has cropped up before. The problem is flawed.

    Generally a road is banked for a given speed so that a car can make the turn without needing any friction. You are not given any information about the speed for which the curve is designed.
  6. Mar 2, 2009 #5
    Can i sue them for giving flawed questions?
  7. Mar 15, 2009 #6
    you better read the contents before you go to this question.
    firstly, you have to draw a FBD.
    b) a=Fc/m
  8. Mar 15, 2009 #7

    Doc Al

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    Staff: Mentor

    On what basis did you calculate r? You don't know the speed for which the track is banked. (You cannot assume that it's banked for the given speed of 225 km/h.)

    (And your equation for r is upside down.)
  9. May 13, 2010 #8
    ok so i just did that 2 weeks ago, got full marks for it XD.

    so for a) u use v= sqrt ( g r tan(angle))

    v= velocity
    g= earth's gravitational force

    v= 225km/h = 62.5 m/s
    angle= 15
    m= 1450kg

    so i'm not going to draw a FBD, you do that yourself, now continueing with a)

    v^2= g r tan(angle)
    62.5^2=9.8 r tan 15
    r=3906.25 / 2.63

    b) ac = centripetal acceleration

    ac= v^2 / r , meh i'm not going to write the rest of the problem to this because i believe you can finish this.
    solution = 2.63 m/s^2

    c) Fs max = m g sin(angle)

    Fs max= magnitude of the force of static friction

    solution = 3677.8N

    d) (can't make the mu symbol for coefficient of friction >< )

    mu s = tan (angle)
    mu s = tan 15
    mu s = 0.268

    you're very welcome, please send me a fruit basket for x-mas =D
  10. May 13, 2010 #9

    Doc Al

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    That equation assumes that the track is banked for the given speed and angle and zero friction.

    Sorry, no fruit basket.
  11. May 13, 2010 #10
    darn it i really wanted that fruit basket XD .

    well i say we should complain to the teachers, after all im only the poor little student that is using the equations that are given in the book >.> .

    but i did get 3/3 marks for that question, therefore i assume it was meant to be a flawed example for the simple purpose of teaching the student?
  12. May 13, 2010 #11

    Doc Al

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    Ask your instructor to explain how he did part a. (Do you understand how the equation you used is derived?)
  13. May 13, 2010 #12
    yes i understand how the equation was obtained, but i'm in the process of learning so i might think i know but i might not know at all.

    anyways what books on classical physics for around grade 12 and perhaps beyond would you recommend that explain it well and are reasonable to understand??
  14. May 14, 2010 #13

    Doc Al

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    How would you derive it? Start with a free body diagram.

    This should be in just about every textbook, so perhaps poking around your library will turn up one to your liking.

    Here are some web resources: http://www.batesville.k12.in.us/physics/phynet/mechanics/Circular%20Motion/banked_no_friction.htm" [Broken]
    Last edited by a moderator: May 4, 2017
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