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Moment of Inertia of helicopter blades.

  1. Dec 29, 2009 #1
    I have built a model of rotating blades of a helicopter and want to mathematically calculate it's moment of inertia. I am unsure on how to do that.
    Its a simple model:
    a square wooden base with uniform thickness and 4 equally long blades made of styrofoam from the midpoints of the 4 sides of the square.
    The rotational axis is through the centre of the square

    Lets just say i know the density, mass and thickness of the base and the blades, how do i get moment of inertia for a specific length of blades ?

    Thank you
     
  2. jcsd
  3. Dec 29, 2009 #2

    rcgldr

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  4. Dec 29, 2009 #3
    What if i have something like this:
    http://img706.imageshack.us/img706/418/moin.png [Broken]
    I have those extra triangles at the center to deal with. can i just get the MOI of the triangles and add it to the MOI of the rectangular blades ?
    yeah, i know I'm approximating a LOT, but it'll have to do.
    Can anyone help ? Can anyone show me the derivation of it as well ?
     
    Last edited by a moderator: May 4, 2017
  5. Dec 29, 2009 #4
    The moment of inertia of a rectangular plate might get you even closer depending how you want to set it up. Its usually given in text books. I personally dont like to do the integration on some of these figures.

    The moment of inertia of a rectangular plate is 1/12m(a^2 + b^2) where a is the width and b the length of the plate. Note that this is the moment of inertia taken from the center of mass of the plate.

    I see your picture. So a and b are b and l...

    So really one plate would have length 2l +b and width b... and there are two of them.
     
    Last edited: Dec 29, 2009
  6. Dec 30, 2009 #5
    From what you've said, if i take two rectangular plates, then after adding the two MOIs i' have an extra square in between. And i don't think i can just subtract that...

    Also, do you know the derivation of 1/12 M(a2+b2) or a link to it ? I can't find it in my textbook.

    I'm not sure if this is correct, but i tried my best to do the integration based on the baterials densities. I havn't added limits yet, but can you verify this for me ?:


    For the triangle in the middle for each blade:
    {σ : density of triangle}
    m=1/2 r2hσ
    r2=2m/hσ
    therefore I=∫2m/hσ
    =m2/hσ


    dm=dx*b*h*rho where dx is a small distance on the wing.
    {ρ: density of blade}
    therefore ∫r2 dm
    substituting for dm:
    =∫x2*b*h*ρ dx
    =1/3 b*h*ρ*x3

    The dimensions are right [ML2] but i'm really not too sure.......
    Thanks
     
  7. Dec 30, 2009 #6
    You dont have to worry about the middle square its taken care of if you use the sides I gave you.

    And I am going to admit I dont like doing the math on these. So here is an old discussion on this board.
    https://www.physicsforums.com/showthread.php?t=57119
     
  8. Jan 2, 2010 #7
    Really ? So i just multiply it by 2 ? i.e I= 1/6 M(a2+b2)
     
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