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Math in Maxwell needle method- the formula the moment of inertia

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data[
    How do I get I1( the moment of inertia of the picture b) ?is just a math problem but really no idea.

    2. Relevant equations
    I=mr^2


    3. The attempt at a solution
    maybe centre of mass .

    Guide will do . please help , thank
     

    Attached Files:

  2. jcsd
  3. Mar 6, 2013 #2

    haruspex

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    I cannot work out what your picture is showing, and there's no definition of the variables in the second attachment.
    Edit: Just noticed there's a tiny 3rd attachment explaining the variables...
     
  4. Mar 6, 2013 #3

    haruspex

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    What part of the equation do you not understand? Is it the 3L/8?
     
  5. Mar 6, 2013 #4
    yes , exactly that part.
     
  6. Mar 6, 2013 #5

    haruspex

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    It's just the parallel axis theorem. The centre of mass of the cylinder is 3L/8 from the axis of rotation.
     
  7. Mar 7, 2013 #6
    Thank you. One more to ask,
    I know that moment of inertia I=mr^2
    For a system, I= m(r1)^2 + m(r2)^2 + m(r3)^2 +....
    or we can write I=Ʃmr^2
    where the m= mass of a particle , r1 and r2 are not the same value.
    This is the only thing that I understand but I=∫r^2 dm=mr^2 from this link
    http://easycalculation.com/theorems/parallel-axis-moment-of-inertia.php

    How can I=∫r^2 dm=mr^2 ?
    here mean the r is constant ? m is the total mass?

    Thank you.
     
  8. Mar 7, 2013 #7

    haruspex

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    That link is saying that if the MoI about the CoM is Ic = ∫x2.dm, where x varies, and you want the MoI Ir about a an axis displaced by an amount r then Ir = ∫(x+r)2.dm = ∫(x2+2xr+r2).dm = ∫x2.dm+2∫xr.dm+∫r2.dm = Ic + 2∫xr.dm + ∫r2.dm
    Since r is a constant, Ir = Ic + 2r∫x.dm + r2∫.dm
    Since x is displacement from centre of mass, by definition ∫x.dm = 0, and ∫.dm is just the total mass, M:
    Ir = Ic + r2M
     
  9. Mar 7, 2013 #8
    Ic = ∫x2.dm is same as
    For a system, I= m(r1)^2 + m(r2)^2 + m(r3)^2 +....
    or we can write I=Ʃmr^2
    where the m= mass of a particle ?
    Why I=integral x^2 dm , x varies. Then for I =∫x dm ,it become zero.
     
  10. Mar 7, 2013 #9

    haruspex

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    Yes. Strictly, I= m1(r1)^2 + m2(r2)^2 + ...
    In the analysis, x is defined to be the displacement (in the x coordinate) of the element dm from the centre of mass. By definition of centre of mass, ∫x dm =0.
     
  11. Mar 7, 2013 #10
    Really thank you^^
     
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