Moment of inertia of triangle about centroidal axis

[mbigras]

Homework Statement

There is the moment of inertia about an x and a y-axis named, $I_{x}, I_{y}$. Then there is the moment of inertia about the centroidal x and y-axis named, $\overline{I}_{x}, \overline{I}_{y}$. Often we can look up these values in a table (like the figure included) and apply another idea called the parallel-axis therom to find moments about different axes. My question is: What is $I_{y}$ and $\overline{I}_{y}$ for the triangle? It seems to be missing from the picture. This also could illustrate my lack of understanding about what a moment of inertia really is.

Homework Equations

$$I = \overline{I} + A*d^{2}$$
where, $A$ is the area and $d$ is the distance between parallel axes.

 

[mbigras]

It seems like the key is that for the y-axis all you do is switch the location of $b$ and $h$. So what could be next to the triangle in the included figure is:
$$\overline{I}_{x'} = \frac{1}{36}bh^{3}\\ \overline{I}_{y'} = \frac{1}{36}hb^{3}\\ I_{x} = \frac{1}{12}bh^{3}\\ I_{y} = \frac{1}{12}hb^{3}\\$$

I am onto something here?

 

[haruspex]

It seems like the key is that for the y-axis all you do is switch the location of $b$ and $h$. So what could be next to the triangle in the included figure is:
$$\overline{I}_{x'} = \frac{1}{36}bh^{3}\\ \overline{I}_{y'} = \frac{1}{36}hb^{3}\\ I_{x} = \frac{1}{12}bh^{3}\\ I_{y} = \frac{1}{12}hb^{3}\\$$

I am onto something here?

No, you can't just switch b and h. b is the length of a side, while h is the distance from a side to a vertex.
There is no 'missing' equation. Rather, the rectangular case has a superfluous equation. In that case you can switch them around by symmetry to obtain the Iy from the Ix.
If you want an Iy equation for the triangle, where are you putting the y axis?

 

[mbigras]

I was going to put it right down the center so $I_{y} = \overline{I}_{y}$ but I'm not sure what do to now. What I really want is $\overline{I}_{y}$ so i can use the parallel axis theorem

 

[haruspex]

I was going to put it right down the center

Would that be through C and perpendicular to one side, or through C and a vertex?

 

[SteamKing]

You could always use the definition and calculate what the inertia would be about the y-axis.

 

[mbigras]

I think this picture describes where I'm trying to go with this question. Notice that I'm assuming that this triangle's "regular" y-axis is the same as the centroidal y axis. I can't explain why I think this except that it seems like that's how it should be because of symmetry. I'm not sure what equation in terms of $a$ and $b$ I should use, for the moments about the y axes.

also another key assumption: Let's assume that it's an equilateral triangle.

 

[SteamKing]

I repeat post #6.

 

[mbigras]

What I'm saying is I don't know how to do that. The situation is this: I know the moment of inertia with respect to the x-axis and with respect to the centroidal x-axis because its in the table. Now based on symmetry you can apply the definition of the moment of inertia to calculate the moment of inertia about the y-axis which equals the cendroidal y axis. But I don't know how to do that.

So the question is: How do I calculate the moment of inertia about the y-axis for an equilateral triangle, using the definition of the moment of inertia?

 

[haruspex]

I think this picture describes where I'm trying to go with this question. Notice that I'm assuming that this triangle's "regular" y-axis is the same as the centroidal y axis. I can't explain why I think this except that it seems like that's how it should be because of symmetry. I'm not sure what equation in terms of $a$ and $b$ I should use, for the moments about the y axes.

also another key assumption: Let's assume that it's an equilateral triangle.

The axis as shown in the picture will lead to a pretty complicated result. You will certainly need to know the exact shape of the triangle. But if you take it to be equilateral, or even just isosceles, then the axis you've drawn will also happen to pass through a vertex. This makes it much simpler.
With the axis through a vertex, it is equivalent to two triangles, the axis being along one side of each. Now all you need to know is the distance from that axis to the far vertex of each triangle, because that will allow you to apply the formula in the book. You do not need another formula.

 

[SteamKing]

And you need to study how the formulas in your book were derived.

 

[mbigras]

Thank you haruspex, that was the missing piece of information! I can break up the triangle into two triangles and apply information in the included figure to each. Thanks also steamking for the reminder, winter break is around the corner and there will be time spent trying to wrap my mind around this dog gone moment of inertia thing.