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Circular Motion,Uniform motion

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data[

    An Aston Martin V8 Vantage sports car has maximum "lateral acceleration" of 0.96"g's" ,after which the car begins to skid out of its turn.

    Calculate the coefficient of static friction between the car's tyres and the road.


    2. Relevant equations



    3. The attempt at a solution

    where Us is coefficient of friction

    Fnet(vertical)=n-mg=o
    Fnet(radial)=fs= Us*N=m(v^2/r)
    Us*mg=m(v^2/r)
    Us=V^2/r*g
    =a(r)/g
    = (0.96*9.81)m/s^2/9.81m/s^2
    =0.96
     
  2. jcsd
  3. Apr 18, 2012 #2
    It's correct.
     
  4. Apr 18, 2012 #3
     
  5. Apr 18, 2012 #4
    "No friction is needed in a banked road because the horizontal component of the normal force helps the car to turn?"

    Not a true statement. If centrifugal force component exceeds the sum of frictional force and component of force down the slope, the car slides.
     
  6. Apr 18, 2012 #5
    For a range of speeds, what is the minimum speed the vehicle can travel without sliding down the slope? That is the lower limit.
     
  7. Apr 18, 2012 #6
    V=√rgtanθ is only valid when the car is not sliding I think,so this "v" will be the minimum speed.But for the upper limit I am not sure.
     
  8. Apr 18, 2012 #7
    "V=√rgtanθ is only valid when the car is not sliding I think,so this "v" will be the minimum speed.But for the upper limit I am not sure. "

    The above formula is the speed when centrifugal force balances the force of the vehicle wanting to slide down the hill. It's the speed when there is no friction.

    Determine the forces on the tires where they meet the pavement. You have 3 forces that you consider in order to determine the maximum speed given the banking and friction.
     
  9. Apr 18, 2012 #8

    In the questions that I did so far on banked curves the centrifugal force is the horizontal component of the normal.And won't the frictional force be a one of centrifugal forces thus if its present?
    Okay.So is it true to make an assumption that in banked curves the is never friction?,for some reason I cannot figure out since my statement is false.
     
  10. Apr 18, 2012 #9
    How do we know that a present centrifugal force in a particular problem balances the force of the vehicle wanting to slide down the hill.So that you do not include friction?


    forces on the tires
    Fnet(Vertical)=nsin(theta)-mg=0
    Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
    v=sqrt(grtan(theta)+rUs/sin(theta))
     
    Last edited: Apr 18, 2012
  11. Apr 18, 2012 #10
    The formula V=sqrt(rg)tan(theta) is derived as follows.

    Force up slope due to centrifugal force = mv^2/r * cos(theta)
    Force down slope due to weight = mgsin(theta)

    Equate them and you get the velocity for the zero friction case. You have friction in your problem so the force balance has another term.
     
  12. Apr 18, 2012 #11
    I have added the extra term which is friction in my case.Is my equation correct now?
     
  13. Apr 18, 2012 #12
    Where is the equation?
     
  14. Apr 18, 2012 #13
    forces on the tires
    Fnet(Vertical)=nsin(theta)-mg=0
    Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
    v=sqrt(grtan(theta)+rUs/sin(theta))
     
  15. Apr 18, 2012 #14
    v=sqrt(grtan(theta)+rUs/sin(theta))

    Are the units of gr the same as rUs? If not, they cannot be added and you have an error.
     
  16. Apr 18, 2012 #15
    No.
    Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
    So I took the friction to be towards the center thus I added it to the horizontal forces.
     
  17. Apr 18, 2012 #16
    Do you mind spotting my error because I cannot?
     
  18. Apr 18, 2012 #17
    Just equate the forces up and down the slope. The component down the slope due to weight plus the friction force must equal to component of centrifugal force up the slope. That's the equation for maximum velocity.

    Whenever deriving a formula, always check units. They will reveal errors if any exist. Correct units are a necessary but not sufficient condition for correctness.
     
  19. Apr 18, 2012 #18
    OOOOOOhhhhh!....I see my mistake
    Forces on the tires
    Fnet(Vertical)=nsin(theta)-mg=0
    Fnet(horizontal)=ncos(theta) + Us*n=mv^2/r
    v=sqrt(grtan(theta)+grUs/sin(theta))
     
  20. Apr 18, 2012 #19
    "v=sqrt(grtan(theta)+grUs/sin(theta))"

    Look at your formula. Suppose the slope were zero. Your formula says the speed would be infinite. Your second term is the problem.
     
  21. Apr 18, 2012 #20
    So along with unit checking, look at formulas you derive at the extremes to see if they provide realistic results and results that make sense.
     
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