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Moment of Inertia of a Pendulum

  1. Jul 3, 2009 #1
    1. The problem statement, all variables and given/known data

    The 20 cm-long wrench in the figure swings on its hook with a period of 0.92s. When the wrench hangs from a spring of spring constant 350 N/m, it stretches the spring 3.5 cm.

    What is the wrench's moment of inertia about the hook?

    http://session.masteringphysics.com/problemAsset/1070606/9/14.EX25.jpg

    2. Relevant equations

    I = m*g*L*T^2/2pi

    3. The attempt at a solution

    Fsp = Fg
    kx = mg
    m = kx/g
    m = (350)(0.035)/9.8 = 1.25

    I = (1.25)(9.8)(0.14)(0.92)^2/2pi = 0.23 kg*m^2

    This isn't the correct answer however. Can someone explain what I did wrong?
     
  2. jcsd
  3. Jul 3, 2009 #2

    alphysicist

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    Hi MyNewPony,

    This formula does not look quite right to me. Do you see what it needs to be?
     
  4. Jul 3, 2009 #3
    Ah. I forgot to square the 2pi.

    So the equation becomes:

    I = mgLT^2/4pi^2

    Is that correct?
     
  5. Jul 3, 2009 #4

    Doc Al

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    Staff: Mentor

    This equation isn't quite right. It's off by a factor of 2pi.
     
  6. Jul 3, 2009 #5

    alphysicist

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    That looks right to me.
     
    Last edited: Jul 3, 2009
  7. Jul 3, 2009 #6

    Doc Al

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    Yes.
     
  8. Jul 3, 2009 #7
    Thanks a bunch!
     
  9. Jul 3, 2009 #8

    alphysicist

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    Glad to help!
     
  10. Dec 13, 2010 #9
    Sorry for bumping up an old thread, but thought it'd be better than making a new one about the same problem. Can someone tell me where equation for I is derived from? From my knowledge I know that I = cMR^2 (as an estimated value), but what exactly do you plug in to get to that point? (I = mgLT^2/4pi^2 )
     
  11. Dec 14, 2010 #10

    alphysicist

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    Hi ElTaco,

    It's a standard form for the period of a physical pendulum. The derivation should be in your book, and you might also look at:

    http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html
     
  12. Dec 14, 2010 #11
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