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Physics of Boomerangs

  1. Nov 28, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Im new at this Physics forum and i dont quite know anyone here.
    I came here because I'm doing a paper about "The Physics of Boomerangs" and so i found a topic about it where you explained a guy how to calculate the moment of inertia on a simple model of a boomerang.

    https://www.physicsforums.com/threa...rtia-for-wierdly-shaped-objects.503476/page-2

    However, I didnt understand the last part, about the end of the calculus. Gneill said I new = I + M.r^2 . So I use that formula 1/12 x M(a^2+b^2) and then add M x r^2 ? And after that you multiply by 3 ? In that case scenario r = 5.4 ?

    Greetings from Portugal
     
  2. jcsd
  3. Nov 28, 2016 #2

    gneill

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    Hi Ricardo.

    In future please use the formatting template that is provided in the edit window when a new thread is started in a homework forum. Don't delete it!

    The model boomerang in question consisted of three identical rectangular vanes that were arranged at 120° angle separation, and overlapped at the center (presumably glued together). Here's the picture provided by the original poster of the thread in question:

    dscf7654-1-jpg.36095.jpg

    Yes that's the idea. The r had units of cm, and was, I believe, an approximation based on the described dimensions of the model boomerang. The parallel axis theorem was applied to find the moment of inertia of one vane about the center of of rotation of the boomerang as a whole. The result was multiplied by three to account for the three vanes.

    The method relies on the fact that the vanes are in fact whole rectangles. If the vanes weren't overlapped and the boomerang was cut from a single sheet of material then another method would have to be found.
     
  4. Nov 28, 2016 #3
    I'll have that in mind next time I create a post

    Thank you so much !!
     
  5. Nov 28, 2016 #4
    One more thing, adding to the past one...

    I calculated the moment of inertia right ? Now I want to know the angular momentum, i only have to multiply this for angular velocity and its done ?

    Also, If I want to prove that the boomerang returns using this calculus what would you counsel me to do ? Adding torque calculus would help ?
     
  6. Nov 28, 2016 #5

    gneill

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    Yes.
    I fear that proving that a given boomerang returns, or simply determining any part of its trajectory would be a much more difficult problem involving aerodynamics. Angular momentum would be just a small factor in a much larger problem.

    Edit:
    Have a look at the Hyperphysics entry on the boomerang for an overview.
     
  7. Nov 28, 2016 #6
    What if I was allowed to consider only angular momentum ? Would I be able to prove it ? And how ?

    This is a small paper, I think I can assume that other stuff dont matter, or simply not consider them in my calculus
     
  8. Nov 28, 2016 #7

    gneill

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    See my Edit update in my previous post.
     
  9. Nov 28, 2016 #8

    CWatters

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