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Circular motion of a road curve

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data
    A concrete highway curve of radius 70m is banked at a 15 degree angle.

    What is the max speed with which a 1500kg rubber-tired car can take this curve without sliding?


    2. The attempt at a solution
    Concrete and rubber Uk= 0.8

    I tried mv^2/r = umgcos15, but thats not correct. I think I am missing mgsin15, but how should i apply to it? The answer is 34.5m/s

    Thanks
     
    Last edited: Nov 1, 2007
  2. jcsd
  3. Nov 1, 2007 #2
    Assuming by sliding the question means skidding outside the circle (if its the opposite simply reverse what Im about to show you) this is how you'd go about it


    We know Fnet=MAc

    What we also know is that MAc is comprised of every vector in the direction of the center. Meaning you must decompose your Fn. Find the Horizontal value for Fn and equate Fnet to Fn + Ff. That is again equated to MAc

    We know Fnet = Mv2/R and Fn + Ff = MV2/R

    We have all variables except for V. Isolate V and solve.

    Note: FF = UkFN

    EDIT- Sorry, my mistake. I forgot to add in you must decompose your Ff as well. Since FF is going down the slope you decompose it into vertical and horizontal values.

    Do the exact same for your Fn (Get Fn Via decomposing your FG). Take the Horizontal component of FF

    So Fn(H) - FF(h) = MV^2/R if the sliding means down. FN + FF if it means up. Sorry about that. Use Sin/Cost/Tan to get the horizontal value for FF.

    Also note FF= UKFN. So find that first, than decompose, and thats the FF you use in the above equation. Once gain sorry.
     
    Last edited: Nov 1, 2007
  4. Nov 1, 2007 #3
    Were you able to get 34.5m/s?? Because I still can't. Maybe i'm not fully understand your solution.
     
  5. Nov 1, 2007 #4
    I never actually did the question. But that's the solution set if the Ff is keeping the car from skidding outside. If by sliding the question means sliding down the ramp than the Fnet= Fn(H) H= horizontal - Ff = MV^2/R. Try that.
     
  6. Nov 1, 2007 #5
    Ok, i tried

    mgsin(x) - umgcos(x) = mv^2/r

    It does not work because you'd have to squareroot a negative. But after ignoring it, the value is only 18.77.

    I tried takiing nsin(x) = mg, then n=mg/sin(x)

    So,

    mg/sin(x) - umgcos(x) = mv^2/r

    But the answer is larger than 34.5m/s.
     
  7. Nov 1, 2007 #6
    Sorry, my mistake. I forgot to add in you must decompose your Ff as well. Since FF is going down the slope you decompose it into vertical and horizontal values.

    Do the exact same for your Fn (Get Fn Via decomposing your FG). Take the Horizontal component of FF

    So Fn(H) - FF(h) = MV^2/R if the sliding means down. FN + FF if it means up. Sorry about that. Use Sin/Cost/Tan to get the horizontal value for FF.

    Also note FF= UKFN. So find that first, than decompose, and thats the FF you use in the above equation. Once gain sorry.
     
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