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Find the moment of inertia

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data

    heres a very messy drawing for an idea of what were looking at: http://img89.imageshack.us/img89/1764/untitledyi0.png

    area moment of inertia Ix0 of a rectangle about the axis x0 that passes through the centroid is: Ix = 1/12bh^3

    the moment of inertia about an axis x that is parallel to the x0 is given by: Ix = Ix0 + Ad^2x

    where A is the area of the rectangle, and dx is the distance between the two axes.

    the inputs are width w, height h, and thickness t.

    w = 200mm h = 300 mm, and t = 22mm

    find the moment of inertia:

    2. Relevant equations

    Ix0 = 1/12bh^3

    Ix = Ix0 + Ad^2x

    3. The attempt at a solution

    my questions are..

    is t the what plugs into b in this problem?

    also, is area of the rectangle just Base * height, so 200 * 300?

    not sure how to find the distance between the two axis.

    please help.
  2. jcsd
  3. Mar 8, 2007 #2
    help please?
  4. Mar 9, 2007 #3

    Gib Z

    User Avatar
    Homework Helper

    Re draw your diagram. The reason no one is helping is because instantly, when they see something thats too hard to read, they wont bother. I know thats what I thought when I saw the picture..
  5. Mar 13, 2007 #4
    I didnt get your drawing, but from the wording it seems you need to use the parallel axis theorem. If [tex]I_{cm}=I_0[/tex] about a certain axis, then [tex]I_x=I_0+mx^2[/tex], where x is the distance of the shifted axis from the axis of the center of mass (With moment of inertia [tex]I_0[/tex]).
  6. Mar 14, 2007 #5
    how were you able to draw that. Is that found on image shack or somewhere on your computer. sorry for the off-topic comment
  7. Apr 18, 2007 #6
    Thanks anyways, I was able to figure this out eventually.

    Sorry for the awkward drawing. I used paintbrush to draw it, then hosted it through imageshack. However, i recommend hosting a bmp file instead of a low quality jpeg.
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