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Conservation of energy (rotation+translation)

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data

    A small solid sphere, with radius 0.25 cm and mass 0.68 g rolls without slipping on the inside of a large fixed hemisphere with radius 23 cm and a vertical axis of symmetry. The sphere starts at the top from rest. The moment of inertia of a sphere is I = 2/5 MR2.
    (a) What is its kinetic energy at the bottom?
    (b) What fraction of its kinetic energy at the bottom is associated with rotation about an axis through its center of mass?

    2. Relevant equations

    E=U+K=constant

    U=mgh

    [tex]K=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex]

    3. The attempt at a solution

    For part (a) only:
    I started to write up the energy conservation law in great and glorious detail, but then I realized that since K_i=0 and U_f=0, and the question only wants to know K at bottom, I could say
    [tex]mgh=K[/tex]
    [tex](.00068kg)(9.81m/s^2)(0.23m)=K[/tex]
    K=0.00153J

    But the correct answer is 0.152J, a rather striking difference. Where did I go wrong?
     
  2. jcsd
  3. Dec 20, 2006 #2

    Doc Al

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    Your method and answer seem correct to me. (Could be a typo in the answer key or the problem statement.)
     
  4. Dec 20, 2006 #3
    Thanks! (the night before the final. Who to trust? My professor's answer key or Doc Al? You got it!)
     
  5. Dec 20, 2006 #4

    vanesch

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    If you really want to nitpick you can say that the height over which the small sphere is travelling, is not 2R but 2(R-r). It won't make much difference given that R >> r.
     
  6. Dec 21, 2006 #5
    That actually occured to me briefly; but to make a difference of 1e2!!
     
  7. Dec 21, 2006 #6

    Doc Al

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    Since its a hemisphere, the height traveled would be R, not 2R. And using a height of R-r, instead of R, would make the KE even smaller--so that won't help!
     
  8. Dec 21, 2006 #7

    vanesch

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    There's clearly a problem with the orders of magnitude, but I thought the problem was that the book said something of 152 instead of 153...

    And concerning the hemisphere, you don't know how it is oriented :tongue2: :blushing:
     
  9. Dec 21, 2006 #8

    Doc Al

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    I wouldn't worry about the third significant figure in the answer, considering that the data given only has two significant figures. :wink:

    Lots of luck rolling that ball down the inside of a bowl with a horizontal axis of symmetry! :biggrin: And the problem did state that the axis of symmetry is vertical. :tongue:
     
    Last edited: Dec 21, 2006
  10. Dec 21, 2006 #9

    vanesch

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    Damn ! I'm not going to talk me out of this one :blushing:
     
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