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Moment of Inertia of a Regular Hexagon

  1. Mar 30, 2015 #1
    A equilateral triangular lamina, has a moment of Inertia of I if the axis of rotation passes through the centroid of the triangle, perpendicular to it's plane. What is the moment of inertia of a regular hexagon(Again, through it's geometrical centre, perpendicular to the plane), provided that, the length of the side of the triangle is equal to the sides of the regular hexagon. (I have assumed the side of the eq. triangle to be a. And m to be the mass of the eq. triangle. M of the hexagon M=6m)

    The equations, I have used to solve this,
    i)I' = I + Md^2
    ii)The cosine rule.
    iii)Basic Trigonometry.
    iv)The radius of gyration is directly proportional to the length of the side of the triangle.

    I did not receive any answer which had only I in it.

    The solution provided in the book, is something I did not understand. It's given that it could be taken as two sides of eq. triangle of sides 2a. Hence using the the fourth point. (Since the hexagon is like 6 eq. triangle) The gave the moment of inertia as 32 I. Then the subtracted 2I, as they remove the two extra triangle due to the us adding two extra triangles, while constructing. But is it not fundamentally flawed, doing such a thing. By the parallel axis theorem their moment of inertia should be 2(I + Md^2) where d = a + 3^0.5/2 a

    The answer provided is 30I.

    Thank you.
     
  2. jcsd
  3. Mar 30, 2015 #2

    SteamKing

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    Why don't you show us your calculations, not just the results? The problem may be there.
     
  4. Mar 30, 2015 #3

    andrevdh

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    This might help...
     

    Attached Files:

  5. Mar 30, 2015 #4

    BvU

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    Hello Prannoy,

    Your observation is correct, except that the distance d of the centroids to the rotation axis is not what you mention.

    If you make a drawing (I see andrev did that in the time I need to type this, I'm slow...) , you'll see that you need to use the parallel axis theorem twice: the triangles with sides 2a are not rotating around the axis through their centroids, so their combined moment of inertia is bigger than just 2 x 16 ##I##.

    The 30 ##I## answer is correct. It's kind of nice to see the md2 terms cancel (coincidence ?) !
     
  6. Mar 30, 2015 #5
    I have accounted for the distances. After cross checking my calculations, I found that I forgot to make take the sides as 2a instead of a while performing trignometric operations, with the triangle of 2a.
    I tried attaching my calculations, but I could not upload it, for some unknown reason. If you still require it, I'll attach the solution I prepared.
     
  7. Mar 30, 2015 #6

    BvU

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    It's not a matter of requiring and you don't have to attach any more pictures. I take it you checked and corrected your work the right way. What did you find for d ? And for the distance from the cross in andrev's drawing to the center dot ?
     
  8. Mar 30, 2015 #7
    d = 2a/(3)^0.5 and the cross from the cross in Andrev's drawing to the center dot is 1/3^0.5.

    Thanks a lot for all your help and support.
     
  9. Mar 30, 2015 #8

    BvU

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    Excellent work. Very nice exercise: I'm still impressed that these md2 terms cancel so elegantly.
     
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