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## Homework Statement

The attachment shows an equilateral triangle of side length "

**2d**"

It is a uniform triangle in the 2-D space.

Mass of triangle = M

I have to find the moment of inertia about one of its side . I am taking the side [itex]\overline{AB}[/itex] as the axis of rotation (hence i would be finding the Moment of inertia about this line)

The pink lines inside the triangle show an attempt to this problem.

## Homework Equations

>Integration

>area of triangle

>[itex]I_{total}[/itex] = [itex]\sum I_{1} + I_{2} + I_{3}... I_{n}[/itex]

>β = 60 degrees

## The Attempt at a Solution

Let the mass per unit area of the triangle be = m

The triangle can be considered (approximated) to be composed of many small rods of width Δy and length "x" (where x varies as the equation of the line)

Now, the area of this rod is = 2xΔy

Hence mass of the rod = 2mxΔy

moment of inertia of one of the rod about [itex]\overline{AB}[/itex] = 2mx[itex]y^{2}[/itex]Δy

The equation of the line [itex]\overline{AC}[/itex] is : y = tan(60)x =

**x**[itex]\sqrt{3}[/itex]

**x**=

**y**/[itex]\sqrt{3}[/itex]

Approximate moment of inertia = [itex]\frac{2m}{\sqrt{3}}[/itex][itex]\sum y^{3} Δy[/itex]

we use integration to find the exact moment of inertia :

[itex]I_{total}[/itex] = [itex]\frac{2m}{\sqrt{3}}[/itex][itex]\int (y)^{3} dy[/itex]

the limits of integration are from 0[itex]\rightarrow\sqrt{3}d[/itex]

Area of triangle = [itex]\sqrt{3}[/itex][itex]d^{2}[/itex]

Moment of inertia of the triangle = [itex]\frac{3Md^2}{2}[/itex]

The problem is that this answer is wrong, and the correct answer is : [itex]Md^2/2[/itex] [which i have deduced by flipping the triangle (i will post an attachment of the way the triangle was oriented which lead me to the correct answer] However the method for both was same, and the orientation of the triangle should never affect the Moment of inertia as long as it is taken about the same axis .

#### Attachments

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