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I need help with finding moment of inertia for pendulum!

  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data
    A grandfather clock uses a physical pendulum to keep time. The pendulum consists of a uniform thin rod of mass M and length L that is pivoted freely about one end, with a solid sphere of the same mass, M, and a radius of L/2 centered about the free end of the rod.
    Obtain an expression for the moment of inertia of the pendulum about its pivot point as a function of M and L.

    2. Relevant equations
    Irod = ML2/3
    Isphere = 2MR2/5


    3. The attempt at a solution
    I substituted L/2 in for R, But, the sphere's inertia isn't around the central axis. I can't seem to figure that out. I put L/2 in for R, which becomes L2/4, and I get rid of the fractions. What else am I supposed to do?

    I got to:
    I = ML2/3 + ML2/10 + 3L2/4. I simplified and got it wrong. Should I get rid of the fractions first? Am I missing something?
     
  2. jcsd
  3. Nov 7, 2011 #2

    ehild

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    Apply the Parallel axis theorem to get the moment of inertia of the sphere, with its CM at the end of the rod.
    What is the last term in your formula? It is even dimensionally incorrect.

    ehild
     
  4. Nov 8, 2011 #3
    Would the distance from the axis be L, or L+L/2, which would be 3L/2? I'm not too good with PAT (Parallel-Axis Theorem) :(

    My formula is Itotal = Irod + Isphere , which would be:

    Itotal = Irod + (ICM + MD2) . R is L/2, thus giving me:

    (1/3) ML2 + [ (2/5) M(L/2)2 + ML2]. This is where I have trouble. Could it also be: (1/3) ML2 + [(2/5) M(L/2)2 + M [L + (L/2)]2] ?

    Plus, it's the distributing and simplifying that's giving me trouble.

    P.S. Is there a better way to write fractions on here?
     
    Last edited: Nov 8, 2011
  5. Nov 8, 2011 #4

    ehild

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    According to the Parallel Axis Theorem, you add the term MD2 to the moment of inertia with respect to the CM, where D is the distance of the CM from the pivot.

    As the CM of the sphere is at the end of the rod, D=L. Your first version is true, the red one is wrong.

    The formulae will look nicer with tex. Going to "advanced" and click on the Ʃ, you can find way under "Math" to write fractions: [itex]\frac{a}{b}[/itex]

    Writing in tex, your derivation will look like

    [tex]I=\frac{1}{3}ML^2+\frac{2}{5}M(\frac{L}{2})^2+ML^2[/tex]
    Factor out ML^2

    [tex]I=ML^2(\frac{1}{3}+\frac{2}{5} \cdot \frac{1}{4}+1)=ML^2(\frac{1}{3}+\frac{1}{10}+1)[/tex]

    Find common denominator
    [tex]\frac{1}{3}+\frac{1}{10}+1=\frac{10+3+30}{30}[/tex]

    ehild
     
  6. Nov 8, 2011 #5
    Thank you so much, ehild. Now, I can't seem to get the formula for the period for small oscillations. I know it's 2∏√I/mgd, but I put the I in, cancel the masses, and it says it's wrong. Do the masses cancel?

    EDIT: Never mind, the d in the formula for period isn't the same d in the moment of inertia. For the d in the expression for period, it WOULD be L + L/2.
     
    Last edited: Nov 8, 2011
  7. Nov 8, 2011 #6

    ehild

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    That d in the expression for the period is the distance of the CM of the whole pendulum from the pivot. It is d=[itex]\frac{M\frac{L}{2}+ML}{2M}=\frac{3}{4}L[/itex].
    But I have doubts about the position of the sphere, if "centred about the end of the rod " means the rod half inside the sphere. If your result is wrong again try to calculate I and d with D=3L/2.

    ehild
     
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