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Centripetal acceleration of car around curve

  1. Feb 16, 2005 #1
    A highway curves to the left with radius of curvature R = 25m. The highway's surface is banked at 27 degrees so that cars can this curve at higher speeds. Consider a car of mass 1800kg whose tires have a static friction coefficient of 0.62 against the pavement. Take g=9.8. How fast can the car take this curve without skidding to the outside of curve? answer in m/s.


    I tried this problem and got it to be 15.7 m/s, but when I submitted it, it was wrong. I treat the direction going down the bank to be the positive direction. I said that the forces acting in the positive direction where the force due to static friction (9.8*1800*cos(27)*.62) and the force due to gravity (9.8*1800*sin(27)). I said for the car not to go off the outside of the curve, the forces going in the positive direction have to be equal to mv^2/R, which is centripetal acceleration. So I solved for v and got 15.7 but its wrong. What did I do wrong?
     
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  3. Feb 16, 2005 #2

    xanthym

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    The Centripetal Force vector is horizontal and directed towards the center of the circular highway. It is not pointed down the bank.




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    Last edited: Feb 16, 2005
  4. Feb 16, 2005 #3

    dextercioby

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    Really weird...With the data given,it seems the correct answer.I think the radius of 25m refers to the inclined circle.

    Daniel.


    EDIT:Xanthym may be right though.Perhaps the 25 m radius are not referred to the inclined circle after all..
     
    Last edited: Feb 16, 2005
  5. Feb 16, 2005 #4

    Doc Al

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    The centripetal acceleration is towards the center of the circle: horizontally to the left, not down the bank. So, the sum of the horizontal components of the forces (normal force, friction, weight) must equal mv^2/R. And the vertical components must add to zero. Also: Do not assume the normal force equals mg cos(27).
     
  6. Feb 16, 2005 #5

    Andrew Mason

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    The horizontal component of the normal force also contributes to the centripetal acceleration. So the forces acting in the 'positive' direction along the road surface don't have to provide all the centripetal acceleration. Take the horizontal components of friction force and the normal force and set them equal to v^2/r.

    AM
     
  7. Feb 16, 2005 #6
    Still need help

    I tried using some of the suggested methods, and got a smaller answer of 11.17, but it is still not correct. I do not know what else to try. If anyone can perhaps emphasize a post that they know is correct in case I messed up, or provide a totally new approach, it would be greatly appreciated. Thanks!
     
  8. Feb 16, 2005 #7

    Andrew Mason

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    11.17 m/sec is the speed that the car could make the curve without any friction (ie. just the horizontal component of the normal force providing the centripetal acceleration). What is the maximum horizontal component of the friction force?

    AM
     
  9. Feb 16, 2005 #8

    Andrew Mason

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    The horizontal forces are:

    (1)[tex]F_{xfriction} = \mu_sF_Ncos\theta[/tex]

    (2)[tex]F_{Nx} = F_Nsin\theta[/tex]

    Therefore:

    (3)[tex]F_Nsin\theta + \mu_sF_Ncos\theta = mv^2/r[/tex]

    Correction: I see that it is a bit more complicated. It is not quite right that the normal force is: [itex]F_N = mgcos\theta[/itex] as I said before, because both the centripetal acceleration and gravity push the car toward the road.

    One has to look at the vertical components of the forces to find the normal force. These have to sum to zero (since there is no vertical acceleration). The friction force has a downward vertical component and this, together with gravity, equals the vertical component of the normal force:

    [tex]mg + \mu_sF_Nsin\theta = F_Ncos\theta[/tex]

    So:
    (4)[tex]F_N = mg/(cos\theta - \mu_ssin\theta)[/tex]

    So substituting into (3):

    [tex]\frac{mg}{(cos\theta - \mu_ssin\theta)}(sin\theta + \mu_scos\theta) = mv^2/r[/tex]

    (5)[tex]v = \sqrt{\frac{rg(sin\theta + \mu_scos\theta)}{(cos\theta - \mu_ssin\theta)}[/tex]

    Substituting values cos(27) = .89 and sin(27) = .45

    [tex]v = \sqrt{25*9.8*(.45 + .62*.89)/(*(.89 - .62*.45)} = 20 m/sec[/tex]

    AM
     
    Last edited: Feb 17, 2005
  10. Feb 16, 2005 #9
    Amazing

    Unbelievable!! 14.77 was not correct!! The method seemed to be completely correct. Thanks for all the help, I really appreciate it. Perhaps there is an error in the online hw server .
     
  11. Feb 17, 2005 #10

    Andrew Mason

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    How about 20 m/sec? See corrected post above.

    AM
     
  12. Feb 17, 2005 #11

    If we think about the a car not moving on the curved road, there are components of Normal Force and gravity which will make the car slide towards the inner bank. So for a stationary car static frictional force is towards the outer bank.

    But when the car is at motion, there is the additional centrifugal component of the force towards the outer bank plus above mentioned forces toward inner bank. What direction do we choose for frictional force? You have chosen it to be towards inner bank. What made you to choose this direction? This seems to be right because choosing the opposite direction make the signs switch leading to an imaginary v.


    Regards.
     
  13. Feb 17, 2005 #12
    Yes! 20 m/s seconds was correct. Thank you for clarifying the fact that the normal force is not just mgcos(thetat). I noticed that someone had mentioned that previously, but I had no idea how to explain it. Makes sense now. Thanks to all who helped!
     
  14. Feb 17, 2005 #13

    Andrew Mason

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    The actual static friction force will vary in magnitude and direction depending on the motion of the car. What we are trying to find is the maximum speed. At that speed, we know that the static friction force is exactly [itex]\mu_sF_N[/itex] and directed toward the inside bank because it is providing part of the centripetal acceleration. We know this because exceeding that speed will exceed the static friction force and will result in the car leaving the road by going off the outer bank (ie. the static friction force and gravity are not enough to supply the needed centripetal acceleration).

    AM
     
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