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Moment of Inertia question?

  1. Aug 29, 2004 #1
    How would you find the moment of inertia of a circle with a square inscribed in it, or vice versa?

    I thought about subtracting the inner object's I from the outer's, but that didn't seem to make sense.
     
  2. jcsd
  3. Aug 29, 2004 #2

    Doc Al

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    Done properly, that would work. For example: The rotational inertia of a circle with a square cut out equals the rotational inertia of the circle minus the rotational inertia of the square.
     
  4. Aug 29, 2004 #3
    My textbook's example seems to use the equation M=(4-Pi)DR^2, where D is density in the example with the circle cut out of a sqaure, but i don't follow. Wouldn't it just be:

    [tex]Itotal=Isquare-Icircle[/tex]

    But that leaves you with some problems... how do you find the indivual mass of each?
     
  5. Aug 30, 2004 #4
    If the material is homogeneous, the density of the square cut out will equal that of the circle (its been quite a while since I did this, so forgive me if you find an inaccuracy in my explanation...do correct it though). You can compute the square and circle's moments of inertia separately and subtract using the idea you mentioned yourself and Doc Al's post. The individual mass can be calculated using the idea mentioned in the first line of this post.
     
  6. Aug 30, 2004 #5
    But, the way i see it, the square has an area of 4r^2, and the circle has an area of pir^2
     
  7. Aug 30, 2004 #6

    Doc Al

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    square minus circle

    That would be the case for a square (of side 2r) with a circle (of radius r) cut out. In which case, [itex]I = I_{square} - I_{circle}[/itex].
     
  8. Aug 30, 2004 #7
    With random numbers, (R=2 and M=2), i got: I=7.57
     
  9. Aug 30, 2004 #8
    I don't beleive thats right, because i kind of ad libbed a step where i used the relation i mentioned before. i'd show my work but its very long.



    Here's how it worked:
    M_s=Massofsqaure
    [tex]I_s=1/6*M_S(2R)^2[/tex]
    [tex]M_s=M+M_c[/tex]
    [tex]M_c=(4-pi)R^2[/tex]


    That last part I am unsure of.
     
  10. Aug 30, 2004 #9

    Doc Al

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    OK.
    Right, where M is the mass of the piece in question (square - circle).
    No. If I understand you properly, that should be [itex]M = (4 - \pi) D R^2[/itex], where D is density and R is the radius of the circle.

    What mass are you actually given? And what about [itex]I_c[/itex]?
     
  11. Aug 30, 2004 #10
    I just forgot to add what i put for [tex]I_c[/tex]
    It was:

    I_c=1/2*M_c*R^2[/tex]

    The example in the textbook derives it in a general form, with mass M, and Radius R, and a constant density. I Tried it with M=2 and R=2 And got the above number.
     
  12. Aug 30, 2004 #11

    Doc Al

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    Right.
    Show your work so we can check it over.
     
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