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CD spinning - Centripetal Acceleration

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A computer is reading data from a rotating CD-ROM. At a point that is 0.0244 m from the center of the disk, the centripetal acceleration is 283 m/s2. What is the centripetal acceleration at a point that is 0.0856 m from the center of the disc?

    2. Relevant equations

    Ac = v^2/r

    3. The attempt at a solution

    So for the first point:
    283 = v^2/(.0244)
    283 *.0244 = v^2
    v^2 = 6.9052

    Since v is constant, then plug it into the next equation:

    6.9052/(.0856) = Ac
    Ac = 80.668 m/s^2

    However, this does not agree with the answer I should be getting. I don't understand where my math went wrong. This seems like a such an easy problem. Any suggestions? Thanks!
  2. jcsd
  3. Sep 15, 2007 #2

    Doc Al

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    Staff: Mentor

    V is not constant, but [itex]\omega[/itex] (angular speed) is. Find or derive an alternate expression for centripetal acceleration in terms of [itex]\omega[/itex].
  4. Sep 15, 2007 #3
    We haven't dealt with w in my class. What is it?
  5. Sep 15, 2007 #4


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    Homework Helper

    angular velocity. v = Rw
  6. Sep 15, 2007 #5
    Ok so solving for w gives me 107.696. Then:
    v = Rw
    v = .0856*107.696
    v = 9.219
    v^2/r = 992.8197 m/s^2

    This can't be right, because as radius increases, the centripetal acceleration decreases. The second acceleration should be less than 283 m/s2.
  7. Sep 15, 2007 #6


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    Homework Helper

    Your work looks good to me. As radius increases, cent. acc. decreases only for constant velocity... but velocity changes with R here...

    cent acc = v^2/r = (rw)^2/r = rw^2, so as r increases cent. acc. increases.
  8. Sep 15, 2007 #7
    Ah, gotcha. Thanks so much!
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