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oohaithere
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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=[itex] \frac 1 3[/itex]ML2
Parallel axis theorem: Ip=Icm+Md2
Slender rod, axis through center: I=[itex] \frac {1} {12}[/itex]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=[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?