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Moment of inertia of an irregularly shaped object

  1. Feb 24, 2005 #1
    This problem describes one experimental method of determining the moment of inertia of an irregularly shaped objectt such as the payload for a satelite. Picture shows a mass m suspended by a cord wound around a spool of radius r, forming part of a turntable supporting the object. When the mass is released from rest, it descends through a distance h, acquiring a speed v. Show that the moment of inertia I of the equptment (including the turntable) is [tex]mr^2(2gh/v^2-1)[/tex].

    here's what I tried, by using this fromula I found in the book:
    [tex]v=(\frac{2mgh}{m+\frac{I}{r^2}})^{1/2}[/tex]

    [tex]v^2(m+\frac{I}{r^2})=2mgh[/tex]

    [tex]v^2m+\frac{v^2I}{r^2}=2mgh[/tex]

    [tex]\frac{v^2I}{r^2}=2mgh-v^2m[/tex]

    [tex]v^2I=2ghmr^2-v^2mr^2[/tex]

    [tex]I=\frac{mr^2(2gh-v^2)}{v^2}[/tex]

    I'm not sure how to get the right answer.
     
    Last edited: Feb 24, 2005
  2. jcsd
  3. Feb 24, 2005 #2
    Divide each of the terms in the bracket by v^2.
     
  4. Feb 24, 2005 #3
    [tex]I=\frac{mr^2(2gh-v^2)}{v^2}[/tex]

    [tex]I=\frac{mr^2}{v^2}*(\frac{2gh}{v^2}-1)[/tex]

    [tex]I=\frac{2ghmr^2}{v^4}-\frac{mr^2}{v^2}[/tex]

    [tex]I=\frac{mr^22gh-mr^2v^2}{v^4}[/tex]

    [tex]I=mr^2\frac{2gh-v^2}{v^4}[/tex]

    [tex]I=mr^2\frac{2gh}{v^4}-\frac{1}{2}[/tex]

    This does not equal
    [tex]mr^2\frac{2gh}{v^2}-1[/tex]

    what did I do wrong?
     
  5. Feb 24, 2005 #4
    [tex]I=\frac{mr^2(2gh-v^2)}{v^2}[/tex]

    [tex]I=mr^2 * \frac{2gh-v^2}{v^2}[/tex]

    [tex]\frac{2gh-v^2}{v^2} = \frac{2gh}{v^2} - 1[/tex]

    [tex]I=mr^2(\frac{2gh}{v^2} - 1)[/tex]
     
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