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Circular Motion of a car around a curve

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A car rounds a banked curve as discussed in Example 6.4 and shown in Figure 6.5.The radius of curvature of the road is R,the banking angle is θ,and the coefficient of static friction is μs.
    Determine the range of speeds the car can have without slipping up or down the road.

    2. Relevant equations



    3. The attempt at a solution
    I've just drawn two free-body diagram representing the situation that the centripetal acceleration is large that the car tends to sliding up and the opposite situation.
    But I don't know how to find the range.
     

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    Last edited: Dec 29, 2013
  2. jcsd
  3. Dec 29, 2013 #2

    Doc Al

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    Good.

    Apply Newton's 2nd law to each extreme. That will allow you to solve for the speeds.
     
  4. Dec 29, 2013 #3
    What forces are there?

    Observe that the net force on the car must be strictly horizontal.
     
  5. Dec 29, 2013 #4
    Is the friction accounts for the centripetal acceleration?
     
  6. Dec 29, 2013 #5

    Doc Al

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    What accounts for the centripetal acceleration is the net force on the car. That includes the friction.
     
  7. Dec 29, 2013 #6
    There is my attempted solution:
    tanθ=F/mg
    F=mgtanθ
    mgtanθ+fcosθ=mvmax2/R
    gtanθ+μsg=vmax2/R
    vmax=[Rg(tanθ+μs)]1/2
    But it's not the correct answer.
    Could you tell me the mistakes?
     
  8. Dec 29, 2013 #7

    Doc Al

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    What is F?

    Do this: List all forces acting on the car. There are three. Then write Newton's 2nd law for vertical and horizontal components.
     
  9. Dec 30, 2013 #8
    Like this?
     

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  10. Dec 30, 2013 #9

    ehild

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    Where is the force of friction?

    ehild
     
  11. Dec 30, 2013 #10

    Doc Al

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    In your diagram, I'd like to see three vectors only--representing the three forces acting on the car.

    Then I'd like you to list the horizontal and vertical components of each force. You'll use those components to write Newton's 2nd law.
     
  12. Dec 30, 2013 #11
    How to find the horizontal component of the weight?
    I think Wb in the picture is not parallel to the centripetal acceleration,so it may not be a component of the centripetal force.
     

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    Last edited: Dec 30, 2013
  13. Dec 30, 2013 #12

    ehild

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    W is vertical, it does not have horizontal component.


    ehild
     
  14. Dec 30, 2013 #13
    So is the weight giving the centripetal acceleration?
     
    Last edited: Dec 30, 2013
  15. Dec 30, 2013 #14

    ehild

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    Why do you think so?

    ehild
     
  16. Dec 30, 2013 #15

    Doc Al

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    As ehild explained, since the weight acts vertically, the horizontal component will be zero.

    For some reason, you found components of the weight parallel (Wb) and perpendiculur (Wa) to the incline surface. This was not necessary--stick to horizontal and vertical components.

    (FYI: If you took the horizontal components of both Wa and Wb, you'll find that they cancel out.)
     
  17. Dec 30, 2013 #16
    Finally I've found that the vmax=[Rg(tanθ+μs)]
    But the right answer is [Rg(tanθ+μs)/(1-μstanθ)].
    How (1-μstanθ) comes out?
     
  18. Dec 31, 2013 #17

    haruspex

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    You still have not clearly explained your working.
    If N is the normal force and F the frictional force, what is the net force in the vertical direction, and what is the net force in the horizontal direction? (You can take F as being either up slope or down slope as you wish, so long as you are consistent.)
     
  19. Dec 31, 2013 #18
    Is the net force in vertical direction 0N?
    Is the net force in horizontal direction Nsinθ+fcosθ?
     
  20. Dec 31, 2013 #19

    Doc Al

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    Again, don't just give answers, show your work step by step. Once again I request that you give the vertical and horizontal components of each force.
     
  21. Dec 31, 2013 #20

    Doc Al

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    Yes, the net force in the vertical direction will equal 0, but express this as an equation: ∑Fy = 0. What is ∑Fy?

    Yes, for one of the extremes.
     
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