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Roational Inertia of a triangle

  1. Apr 25, 2009 #1
    I want to find the I of an equialteral triangle about the middle of it's base.
     

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  2. jcsd
  3. Apr 25, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi joserse46! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Apr 25, 2009 #3
    Well what I am doing now is splitting the triangle up into two right triangles and finding the RI of on about it's tip (the tip that makes the right anlge), from there I could use the rod about the tip the intergrate from 0 to root 3 over 2A (the length og the triangle) But I sure there is an easier way to do this>
     
  5. Apr 25, 2009 #4
    First, calculate the moment of inertia about its center of mass, which is at the intersection of the three medians (drawn from each vertex to the center of the opposite side). Now add to this moment of inertia the product of the triangle's mass times 2/3 the length of the median from the vertex to the base. This latter term is called the Principal Axis Theorem.
     
  6. Apr 26, 2009 #5
    well i came to 7/12 MA^2 but that doesn't seem right since I know for a fact that it's 5/12MA^2 about it's tip and 1/12MA^2 about it's center. so shouldn't it be between that or I'm assuming too much
    Hope someone can confirm this
     
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