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Physical Pendulum inertia

  1. Oct 24, 2004 #1

    cb

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    I need hlep with this question...

    A 1.8-kg monkey wrench is pivoted 0.250-m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle osciallations is 0.940 s. a.) What is the moment of inertia of the wrench about an axis through the pivot? b.) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes throught the equilibrium position?

    I think it did this right, but hopefully someone can check my work.

    [tex]d\sin(\theta) = 0.250 m[/tex]

    [tex]\theta\rightarrow 0 \therefore d= 0.250m[/tex]

    [tex]T = 2\pi\sqrt{I / mgd}[/tex]

    [tex]I = -9.87 X 10^{-1}[/tex].

    Now, I'm not sure how to start part b. I would use energy methods, but I'm not sure how the use it in this situation. Could someone help me with that part.
     
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2

    Doc Al

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    Staff: Mentor

    I don't know what you are doing here. d in the formula below for period is not a displacement, but the distance from pivot to center of mass. Nonetheless, d = 0.25m, as given.

    Does a negative rotational inertia make sense? And don't forget units.

    Mechanical energy is conserved. As the pendulum swings down from its initial point, gravitational PE is transformed to rotational KE.
     
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