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Oscillation and Moment of inertia

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A pendulum is constructed of a solid metal sphere of
    mass M = 4.00 kg, attached to a thin metal rod of mass
    m = 1.00 kg and length L = 40.0 cm. The pivot point for
    the pendulum is at the upper end of the thin rod. The
    pendulum oscillates through a small angle with a period
    of T = 1.40 s. Find the radius of the sphere.

    2. Relevant equations

    [tex]\omega^2=\frac{I}{mgd}\hspace{15pt}\omega=\frac{2\∏}{T}\\I=\frac{1}{3}M_{rod}L^2+\frac{2}{5}M_{sphere}R^2(L+R)^2[/tex]Center of Mass[tex]=\frac{m_1d_1+m_2d_2....m_nd_n}{m_{total}}[/tex]
    For d I used the distance from the center of mass to the pivot point[tex]\frac{L}{2}+R[/tex]
    3. The attempt at a solution
    I have filled in all known values leaving only R, then solved for R using the quadratic formula.
    My professor gave use 10.8 cm as an answer. I am getting ~30cm. This should be straight forward right? Chug and Plug?
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 29, 2012 #2

    ehild

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    Check your equations. Neither the one for the moment of inertia nor that for d (distance of CM from the pivot) are correct.

    ehild
     
  4. Feb 29, 2012 #3
    Okay so this is my final equation, but I'm still not getting it right. What am I doing wrong.
    [itex]\omega^2=\frac{\frac{1}{3}L^2+\frac{2}{5}M_{sph}R^2+M_{sph}(L+R)^2}{M_{total}g(\frac{\frac{1}{2}LM_{rod}+RM_{sph}}{M_{total}})}[/itex]
     
  5. Mar 1, 2012 #4

    ehild

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    The CM is still wrong. How far is the centre of the sphere from the pivot?
    The mass of the rod is missing in the numerator.


    ehild
     
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