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Homework Help: Second Moment of Area for rotating rectangle

  1. Apr 10, 2010 #1
    Hi all, first post here but I've long browsed these forums for answers in the past.

    I couldn't find an answer to my question through searching but I appologise in advance if it has been asked before.

    1. The problem statement, all variables and given/known data

    I want to find an expression which relates second moment of area to the angle that the shape in question has rotated through.

    If you picture a rectangle with (for example) sides a = 1 and b = 10, with the neutral axis running through the centre of the rectangle, parallel to the long sides.

    At 0 and 90 degrees of rotation about the centre of the rectangle you can use bd^3/12 to work out the second moment of area. However between these two I know you need to integrate, but I can't quite picture how this would work.

    2. Relevant equations



    3. The attempt at a solution

    I know that I can use:

    (int) y^2 da
    (int)(int) y^2 dydx

    The problem is how to define the limits and then to solve the integration as the limits are both functions of theta (angle that the rectangle makes with the x - axis) rather than just x & y as in a static problem.

    If the width of the rectangle (a) was zero then the two limits would be on top of each other (obviously the area would then be zero).

    However, in that case trig would suggest that x = cos(theta)*b and y = sin(theta)*b,

    As the rectangle has a width there needs to be two pairs of functions with terms which determine their y-intersect in terms of the rectangles rotation.

    I've attached an image showing the problem to make it a little clearer. This problem isn't urgent but I will keep thinking about it until I've solved it so any help would be appreciated!
    Thanks in advance, Tom
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    When you draw the new set x' and y' axes, you need to measure how the new value changes with the angle.

    You should find that x' changes such that

    x'=xcosθ+ysinθ

    and y' changes such that

    y'=ycosθ-xsinθ


    then just put that into your expression for Ix'-x'= ∫A y12 dA
     
  4. Apr 13, 2010 #3
    Thanks for your reply, I will have a go at this.
     
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