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Homework Help: Rotational Energy

  1. Nov 9, 2005 #1
    I am having problems with a rotational energy problem.

    It reads...
    a think stick of length 1.6m is dense at one edn than at teh other: it's mass desity is p=.4kg/m-.070kg/m^2x, where x is the distance from the heavier end of teh stick. The stick rotates about an axis perpindicular to the heavier end with a period of 1.1s. Determine teh Rotational kinetic energy of teh stick.

    I used K=1/2Iw^2. I found w by using the period. However, when I try to use the integral I=(integral)pR^2 dV, I cannot find R becuase it is not stated. How do I find R or is there another way around the problem?
     
  2. jcsd
  3. Nov 9, 2005 #2

    Fermat

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    R is the length of the stick.

    And your integral is wrong. It should be I=(integral)pr² dr, not I=(integral)pR^2 dV,
     
    Last edited: Nov 9, 2005
  4. Nov 9, 2005 #3

    Päällikkö

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    I disagree: it should be dV (I can't think why it shouldn't, enlighten me :smile:).

    (The distance from the rotational axis) R2 = x2 + y2. As the stick's thin: y,z [itex]\to[/itex] 0.
     
    Last edited: Nov 9, 2005
  5. Nov 9, 2005 #4
    R is the radius of the stick, I am sure of this. And the integral has to be with respect to volume, it's a 3d object
     
  6. Nov 9, 2005 #5

    Fermat

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    We have a stick rotating about one end. A small element δm at a distance r giving a small inertia δI = r²δm. Puting δm = ρ*δr and substituting gave me my final result. Hmm, how is it derived with dV, I'm not familiar with that ?
     

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    Last edited: Nov 10, 2005
  7. Nov 9, 2005 #6

    Päällikkö

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    I feel this should be:
    δm = ρ*δV

    Then again as the stick's thin, δm = ρ*δV basically reduces to δm = y*z*ρ*δr, where y and z are the thicknesses into different directions (I don't think you can jus cancel them out).

    The attachement will take a while before it is approved (thus I cannot view it), maybe host the file on http://www.imageshack.us" [Broken]? (EDIT: You already did :smile:)

    If you use your method to derive the I for a uniform thin rod, you will come up with (1/3)ρR3, which is wrong. Either of the methods I've mentioned should give the correct (1/3)MR2

    EDIT: Lots and lots of edits :smile:, small typos of all kinds.
     
    Last edited by a moderator: May 2, 2017
  8. Nov 9, 2005 #7

    Fermat

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    ρ is the linear density, in kg/m
     
  9. Nov 9, 2005 #8
    The integral is still from 0 to 1.6 though right?
     
  10. Nov 9, 2005 #9

    Fermat

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    Yeah, I think we would both agree on that :smile:
     
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