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Banked Curve - Minimum Turn Radius

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A car travelling 10 m/s is moving along a track banked at 5 degrees. The tire-road friction coefficient is .3 What is the minimum radius it can travel without slipping?

    v0= 10 m/s
    Bank Angle Θ = 5°
    μ=0.3

    Note I am working through prep material for the exam. The solution in the book skips steps and doesn't make sense.

    2. Relevant equations
    Normal and Tangential Kinetics for Planar Problems - FE Exam Reference Guide Page 70 - Dynamics
    ∑Ft=mat
    ∑Fn=mvt/ρ , where ρ is the radius of curvature

    3. The attempt at a solution
    ∑Fx=Fc-Fμ
    ∑Fx=mv2/ρ-μmgcosΘ=0, where Θ=5°
    ρ=v2/μgcosΘ=100/.3⋅*9.81⋅cos(5°)=34.1 m


    Note:

    Why is the review manual saying to use the following solution? I do not understand the FΘ term or how it was derived.

    Fc=Fμ+Fθ
    mv2/ρ=μmg+mg⋅tanΘ
    ρ=26m


    Snapshot.jpg


    Note: Thanks for the help. I don't understand the solution and my attempt is the wrong answer.
     
  2. jcsd
  3. Aug 11, 2015 #2

    olivermsun

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    ##F_c = F_\mu + F_\theta## appears to decompose the centripetal force into a part caused by friction with coefficient ##\mu## and a part caused by the inward radial component of the normal force exerted by banked track.

    So you know that the banked track is pushing normal to its force with some force ##F## (acting at angle ##\theta##relative to vertical). The vertical component of ##F## balances downward gravity, so
    $$F \cos\theta = mg.$$
    Meanwhile, the inward radial component of ##F## is what's left, namely
    $$F_\theta = F\sin\theta.$$
    Now combine them to get an expression for ##F_\theta## as a function of ##mg## and ##\theta##.

    Also, I wonder if you are supposed to use the small angle approxation to get ##F_\mu = \mu mg##.
     
  4. Aug 11, 2015 #3
    I am confused. Is my FBD messed up? I see a gravitational force in the direction of the centripetal force.

    Snapshot.jpg
     
  5. Aug 11, 2015 #4

    ehild

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    The centripetal force must be horizontal, as the car moves along a horizontal circle.
     
  6. Aug 11, 2015 #5
    Car is on a 5 degree bank. Did I still mess up the FBD?

    Snapshot.jpg
     
  7. Aug 11, 2015 #6

    ehild

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    There is a normal force FN which is perpendicular to the road and static friction Fs that is parallel with it. The force of friction acts inward in case of smallest possible radius, to prevent the car sliding out.
    bankedfr.JPG
     
  8. Aug 11, 2015 #7
    I am confused because the solutions states:

    Fc = Ff + Fθ


    The overhead view looks like this. Right?
    Snapshot.jpg
     
  9. Aug 12, 2015 #8

    ehild

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    I do not understand your notations.
    The friction is not horizontal.
    Decompose all forces into horizontal and vertical components.
     
  10. Aug 12, 2015 #9
    First of all. Thanks for your help. I clearly am having a physics fundamental issue here. I have a fundamental question before I rehash the scenario here.


    Centripetal force is sucking the car in towards the instantaneous center of rotation right?

    Snapshot.jpg
     
  11. Aug 12, 2015 #10

    ehild

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    The centripetal force points towards the centre of curvature. But it does not "suck the car". It is not an operating force: it is the resultant of all forces acting on the car.
    What do you mean on IoCp???
     
  12. Aug 12, 2015 #11
    I got sloppy with the paint tool. Sorry. It was supposed to be I.C. , the acronym for instantaneous center.

    Okay next fundamental question, there are 3 additional forces acting on the particle. The friction force, force due to bank angle and the normal force due to gravity.

    1) Is the force of friction preventing the car from slipping towards or away from the center of curvature?

    2) What does the book mean when it states "The bank angle is the centrifugal force contributing to the normal force."
     
  13. Aug 12, 2015 #12

    ehild

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    There is no force due to bank angle, and the normal force is not due to gravity.
    The forces are gravity, normal force (due to the road) and friction.

    You have to figure it out. The force of static friction is less or equal to μN.

    That statement has no sense. The bank angle is not a force.
     
  14. Aug 12, 2015 #13
    How can I thank you for your help?
     
  15. Aug 12, 2015 #14

    ehild

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    Show your (correct) solution :)
     
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