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Calculating the moment of Interia

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the moment of inertia of the same region about the x-axis, Ixx, is given by bh^3/21

    where y = h(x/b)^2


    2. Relevant equations

    [tex]\int(y^2*dA)[/tex]


    3. The attempt at a solution

    so I first figure that dA has to equal
    [tex] h(x/b)^2 * dx [/tex]
    and that y has to equal
    [tex] h^2 * (x/b)^4[/tex]

    so when I put it into the intergral, I get something along the lines of
    [tex]\int(h(\frac{x}{b})^2 * h(\frac{x}{b}) ^ 2 * h(\frac{x}{b})*dx)[/tex]
    from 0 to b

    except when I calculate it out, I get a solution of
    [tex]\frac{h^3*b}{7}[/tex]

    instead of

    [tex]\frac{h^3*b}{21}[/tex]

    somehow I am missing a 1/3.

    if it at all helps, I've already calculated the centroids to be <0.75b, .3h>
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 10, 2009 #2
    Set dA = dydx and evaluate a double integral over the given region (you have the limits of x as well as y). Remember to first evaluate with respect to the variable limits (your "inner integral") and then with respect to constants (your "outer integral" should have constant limits). That will give you the right answer. :smile:
     
  4. Nov 10, 2009 #3
    Ah uh, would my limit for dx be from 0 to b, and for dy to go from 0 to y=h(x/b)^2?
     
  5. Nov 10, 2009 #4
    Think you made a typo (should be h(x/b)^2) but that is exactly right :smile:

    EDIT: As I was replying, you fixed it :biggrin:
     
  6. Nov 10, 2009 #5
    Holy ****, I get this now.
    Thank you much.
     
  7. Nov 10, 2009 #6
    HAHAHA!!! Glad to be of help! You see, you can do it :wink:
     
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