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How much further does a ring roll up a hill

  1. Nov 19, 2013 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/5nLnI.png [Broken]

    2. Relevant equations

    I = mr2

    3. The attempt at a solution

    Moment of inertia:

    Mass density = mass / area
    = 2.59 / (π*9.52 - π*7.52)
    = 0.02424772368 kg / cm2


    I = (areaA)(density)(radiusA) - (areaB)(density)(radiusB)
    = π*9.52*0.02424772368*0.0952 - π*7.52*0.02424772368*0.0752
    = 0.0379435

    Ek = 0.5*m*v2
    = 0.5*2.59*2.72
    = 9.44055

    Er = 0.5*I*ω2
    = 0.5*0.0379435*(v/r)2
    = 0.5*0.0379435*(2.7/0.095)2
    = 15.32455

    Eg = Ek + Er
    mgh = 9.44055 + 15.32455
    (2.59*9.8)h = 9.44055 + 15.32455
    h = (9.44055 + 15.32455) / (2.59*9.8)
    = 0.975695

    sin(37.8) = 0.975695 / d
    d = 0.975695 / sin(37.8)
    = 1.59m
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 20, 2013 #2

    haruspex

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    You don't get the MoI of a disc by multiplying its mass by its radius. (That would be dimensionally wrong.) Besides, computing the density is rather a long way round.
    Treat the annulus as the difference of two discs. What is the MoI of a disc about its centre?
    Also, please don't plug in numbers until the final step. It's much easier to follow the logic and spot mistakes if you keep everything symbolic as long as possible. It can also avoid some calculation steps, so reduce numerical error.
     
  4. Nov 20, 2013 #3
    Oh wow I left out the 1/2 on the disks that was dumb. I fixed that and got the right answer. My initial post was treating the annulus as a difference of two disks. How could I do this without finding the mass density? I need the mass of the two disks.
     
  5. Nov 21, 2013 #4

    haruspex

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    Yes, I see. I suppose what I would have done effectively does that, but not so explicitly:
    M(Router2 - Rinner2*(Rinner2/Router2))/2.
    More importantly, I urge you to get into the habit of working entirely symbolically, as mentioned.
     
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