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Moment of Inertia of equilateral triangle about vertex

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.

    2. Relevant equations
    Slender rod, axis through one end: I=[itex] \frac 1 3[/itex]ML2
    Parallel axis theorem: Ip=Icm+Md2
    Slender rod, axis through center: I=[itex] \frac {1} {12}[/itex]ML2

    3. The attempt at a solution
    First I drew the figure as an equilateral triangle with the axis at the top point and set each side equal to B.


    34qlqms.jpg

    Then I considered the 2 sides to be slender rods with the axis through one end.
    So, ΣI=[itex]\frac 1 3[/itex]Mb2+[itex]\frac 1 3[/itex]Mb2

    The third side I figured would be a slender rod with the axis in the middle but moved up a distance d which would equal [itex] \sqrt{ b^2- {\frac 1 4} b^2 } [/itex].

    Ip=[itex] \frac {1} {12} [/itex]Mb2+M[itex] \sqrt{ b^2- {\frac 1 4} b^2 } [/itex]2

    Then I added them all up to get the moment of inertia for the whole triangle.

    ΣI=[itex]\frac 1 3[/itex]Mb2+[itex]\frac 1 3[/itex]Mb2+[itex] \frac {1} {12} [/itex]Mb2+Mb2-[itex]\frac 1 4[/itex]Mb2

    But I get [itex] \frac 3 2 [/itex]Mb2 when the answer is supposed to be [itex] \frac 1 2 [/itex]Mb2.

    Am I maybe missing a negative or calculating one of the moments wrong?
     
  2. jcsd
  3. Mar 20, 2017 #2

    BvU

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    Hello there, haithere, :welcome:
    Is M the mass of one side, or of the whole thing ?
     
  4. Mar 20, 2017 #3
    M is the mass of the whole wire so should I be using M/3 for the value of M in each moment I'm adding?
    edit: I tried it and it works thanks for pointing that out
     
  5. Apr 13, 2017 #4

    muz

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    can you show how you found the distance d?
    '
     
  6. Apr 13, 2017 #5

    BvU

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    Hello Muz, :welcome:

    Thing to do in PF is start your own thread instead of continuing someone else's solved thread.

    Never mind: if you cut an equilateral triangle (sides length b) in half you can use Pythagoras to find ##d^2 = b^2 - ({1\over 2}b)^2 \ \ ##
     
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