Moment of Inertia of equilateral triangle about vertex

[oohaithere]

Homework Statement

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.

Homework Equations

Slender rod, axis through one end: I=\frac 1 3ML²
Parallel axis theorem: I_p=I_cm+Md²
Slender rod, axis through center: I=\frac {1} {12}ML²

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.[/B]

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

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 \sqrt{ b^2- {\frac 1 4} b^2 }.

I_p=\frac {1} {12}Mb²+M\sqrt{ b^2- {\frac 1 4} b^2 }²

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

ΣI=\frac 1 3Mb²+\frac 1 3Mb²+\frac {1} {12}Mb²+Mb²-\frac 1 4Mb²

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

Am I maybe missing a negative or calculating one of the moments wrong?

 

[BvU]

Hello there, haithere,
Is M the mass of one side, or of the whole thing ?

 

[oohaithere]

Hello there, haithere,
Is M the mass of one side, or of the whole thing ?

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

 

[muz]

can you show how you found the distance d?
'

 

[BvU]

Hello Muz,

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 \ \$