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Why does work have the wrong units?

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string that passes through a tiny hole of negligible radius in the table, and I am holding the other end of the string underneath the table. Initially the particle is moving in a circle of radius r0 with angular velocity ω0, but I now pull the string down until the radius reaches r. Assuming that I pull the string so slowly that we can approximate the particle’s path by a circle of slowly shrinking radius, calculate the work I did by pulling the string, and compare it to your answer in (5.1).

    2. Relevant equations
    W = ∫F⋅dr
    Answer for KE is [(mωo2ro2)/2]((ro/r)2-1)

    3. The attempt at a solution
    W = ∫F⋅dr = ∫mv2/r dr= ∫mω2rdr = (mω2/2)(r2-ro2)

    From conservation of angular momentum I get that ω = (ωoro2)/r. replacing this with the result gives (mωo2ro4)/2[((ro/r)2-1)] and the units don't check out. Shouldn't this be the same as KE?
     
    Last edited: Feb 26, 2017
  2. jcsd
  3. Feb 26, 2017 #2

    Doc Al

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    This equation is not dimensionally consistent and thus cannot be true.
     
  4. Feb 26, 2017 #3
    Oh that is v not omega. Crap, stupid mistakes. I got (mωo2ro4)/2r2[(1-(ro/r)2)]
     
  5. Feb 26, 2017 #4
    To check this answer I set it equal to KE to see if it turned to 0 but instead it gave me that r=-ro which isn't true. Is this answer still correct or is this not a good way to check your answer?
     
  6. Feb 26, 2017 #5
    The problem is in here is ##F## is not constant as you can see ##F=\frac {mv^2} {r}## but here ##v## depends on ##r## (angular momentum conservation).So you have to find first ##F(r)##
     
  7. Feb 26, 2017 #6

    haruspex

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    Please post your steps there. How do you handle the fact that r and ω are both varying?
     
  8. Feb 26, 2017 #7
    Typo. It should have been W = ∫F⋅dr = ∫mv2/r dr= ∫mω2rdr = (mω2/2)(r2-ro2)
     
  9. Feb 26, 2017 #8

    haruspex

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    But that is wrong. In the integration you treated ω as constant.
     
  10. Feb 26, 2017 #9
    Got it, thanks
     
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