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Linear and Rotational Kinetic Energy

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

    A solid cylinder is rolling without slipping. What fraction of its kinetic energy is linear?

    2. Relevant equations

    Ke=[itex]\frac{mv^2}{2}[/itex]+[itex]\frac{I(v/r)^2}{2}[/itex]

    3. The attempt at a solution


    Ke=[itex]\frac{mv^2}{2}[/itex]+[itex]\frac{(v/r)^2}{2}[/itex]*[itex]\frac{(mr)^2}{2}[/itex]

    Ke=[itex]\frac{3(mv)^2}{4}[/itex]

    Linear Ke =[itex]\frac{(mv)^2}{4}[/itex]

    Fraction of Linear Ke = [itex]\frac{linear Ke}{Ke}[/itex] = [itex]\frac{(1/4)}{(3/4)}[/itex] = [itex]\frac{1}{3}[/itex]
     
  2. jcsd
  3. Oct 29, 2013 #2

    tiny-tim

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    hi woaini! :smile:
    uhh? :confused:
     
  4. Oct 29, 2013 #3

    haruspex

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    The m should not be getting squared. You need to be more careful with the parentheses.
     
  5. Oct 29, 2013 #4
    Nvm it should be 1/2. Therefore I should get a final answer of 2/3 linear kinetic energy?
     
  6. Oct 30, 2013 #5

    tiny-tim

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    yup! :biggrin:

    btw, a way of checking this is to say the mass is m, the "rolling mass" (= I/r2) = m/2, so the "effective mass" is 3m/2, and eg the acceleration down a slope must be multiplied by 2/3 (but don't use that in the exam :wink:)

    (for a sphere, it's m + 2m/5 = 7m/5, and the acceleration must be multiplied by 5/7)
     
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