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Solid disk rolling

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A solid disk of mass m and radius R rolls without slipping with a velocity v. Assuming it doesn't slip, how far vertically will it roll up an incline?

    2. Relevant equations
    I=0.5mr2
    E=0.5Iω2
    KE=0.5mv2
    PE=mgh

    3. The attempt at a solution
    I'm thinking that we need to find the height h when the kinetic energy is converted to potential energy. So:
    0.5Iω2+mv2=mgh
    0.5*0.5mr2ω2+mv2=mgh
    0.25r2ω2+v2=gh
    h=(0.25r2ω2+v2)/g
    Is that right? I feel weird because I still have ω in there.
     
  2. jcsd
  3. Apr 30, 2013 #2

    SammyS

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    What's the relationship between v and ω ?
     
  4. Apr 30, 2013 #3
    Ah, either you're a genius or I'm dumb. Probably both. Thank you.

    I got a final answer of (1.2*v^2)/g = h.
     
  5. Apr 30, 2013 #4
    Oops, 1.25... ((5/4)(v^2))/g = h
     
  6. May 1, 2013 #5

    SammyS

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    Probably neither.

    I considered a lengthier reply to your Original Post, but then though I'd see what you could do with a fairly subtle hint/question. From that, you completed the exercise. I commend you for that.

    You might be surprised at how many people need to be led by the nose, step-by-step to an answer.

     
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