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Uniform Circular Motion Hills and Cars

  1. Sep 26, 2009 #1
    1a. The problem statement, all variables and given/known data
    You are riding in a car over a hill that has a radius of curvature R. What speed should you be going if you want to feel one-third of your normal weight. Express your answer in terms of the radius R and the acceleration due to gravity g.
    1b. You continue your journey at the same speed down the other side of the hill until you reach a valley that has a radius of curvature of two-thirds R. In terms of your true weight (mg) what is your apparent weight at the lowest point of the valley?


    2. Relevant equations
    F=(GMm)/(r^2)
    [tex]\sum[/tex]F=ma
    a=(v^2)/(r)
    v=(2[tex]\pi[/tex]r)/(T)
    v=(2pi r)/(T(period))

    3. The attempt at a solution
    1a.) I know that the sum of the forces in the y direction are F=(mg)(1/3)-N=mv^2/r
    so v=[tex]\sqrt{rg/3-rN/m}[/tex]

    1b.) Your weight should be more than your actual weight because mg-N=(mv^2)/r
    I just don't know if i should put 2/3r in the equation. Then it would be mg=N-(mv^2)/(2/3r)

    Sorry if i used the symbols incorrectly.
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 26, 2009 #2

    rock.freak667

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    Homework Helper

    for the first question, the weight you feel as you go over the curve is the normal reaction N.

    So the centripetal force is given by mv2/R=mg-N. You want N to be (1/3)mg, so what is v in terms of R?

    For the second part now: If at the lowest point, the weight points downwards and the normal reaction points upwards (your 'apparent' weight). In what direction should the centripetal force point and what is it equal to in terms of your weight and normal reaction?
     
  4. Sep 26, 2009 #3
    So v2=[tex]\sqrt{r(g-1/3g}[/tex] which means v=[tex]\sqrt{r2/3g}[/tex] for 1a.

    for 1b.) the centripetal force points in the direction the normal force does, upwards. the equation would be N-mg=mv[tex]^{2}[/tex]/(2r/3)
    if i replace v with the equation from 1a.) then the equation would be:

    mg=N-(m(2/3)gr)/(2/3 r) which simplifies to

    mg=N-m
    Is this correct?
    Thank you
     
  5. Sep 26, 2009 #4

    rock.freak667

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    Homework Helper

    1a) the 'g' should be in the numerator

    b) Make N the subject of the formula and you'll get your answer. (see the part in red where you left out the 'g', I'll assume that is a typo)
     
  6. Sep 26, 2009 #5
    Thanks!
    for b i ended up getting mg=N/2
     
  7. Sep 26, 2009 #6
    Doesn't centripetal acceleration/force always point to the center of the circle? In 1a) I know your apparent weight would be less but wouldn't you have N upwards, mg downwards, and mv2/r downwards also?
     
  8. Sep 26, 2009 #7

    rock.freak667

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    Homework Helper

    well right N=2mg. Which means that you 'feel' as if you weight twice as much.
     
  9. Sep 26, 2009 #8
    To Chris
    You are right, centripetal force always point to the center of the circle.
    I just chose a coordinate system where the centripetal acceleration is on the positive y-axis.
     
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