Calculate the rotational inertia of a solid hexagonal

[Yalanhar]

Homework Statement

Calculate the rotational inertia of a solid hexagonal (side = R) about an axis perpendicular through its center.

Relevant Equations

$I=\int r^2 dm$

What I did:

$\frac{dm}{dA} = \frac{M}{\frac{3\sqrt3 R^2}{2}}$

$dm = \frac{2M}{3\sqrt3 R^2} dA$ (1)

$dA=3\sqrt3 rdr$ (2)

(2) in (1)

$dm = \frac{2M}{3\sqrt3 R^2} 3\sqrt3 rdr$

Now in the integral

$I = \int \frac{r^2 2Mrdr}{R^2}$

How can I solve the integral interval? I think I need to change respect to $\theta$ and integrate over 2$\pi$. Or can I change r in the integral and integrate considering dA? Then the interval would be 0 to A ($A = \frac{3\sqrt3 R^2}{2}$.

I don't quite understand how i choose the interval. In a rod to it's center is -l/2 to +l/2, but in a ring it goes from 0 to R. Why not -R to +R?
Thanks

 

[haruspex]

You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?

 

[Yalanhar]

You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?

Yes, i see now that if I use dr or d$\theta$ it won't help me. I think I'm confusing r (distance of a infinitesimal part to the axis) to r (infinitesimal side of hexagon)

 

[haruspex]

Yes, i see now that if I use dr or d$\theta$ it won't help me. I think I'm confusing r (distance of a infinitesimal part to the axis) to r (infinitesimal side of hexagon)

Well, you can use dθ and integrate along the strip, but there is an easier way using a couple of standard results, as I hinted.

 

[Yalanhar]

You appear to have taken the integration element to be a rectangular strip, so the points along it are not equidistant from the hexagon's centre.

If you only had a thin rectangular strip to worry about, what standard results would you employ?

I would use $dA = dxdy$ The rectangles are for summing them from one side and then complete the hexagon?

 

[BvU]

Are you familiar with the parallel axis theorem ?

 

[haruspex]

I would use $dA = dxdy$ The rectangles are for summing them from one side and then complete the hexagon?

They are very thin rectangles, effectively rods.
Plus what BvU asked.

 

[Yalanhar]

They are very thin rectangles, effectively rods.
Plus what BvU asked.

I know the theorem. I saw someone doing this theorem for 6 triangles. He calculed the moment on the center for each triangle with the theorem and summed them up. But I didnt understand the rectangle, the rectangle itself can't fill the hexagon, for the x changes between the differents dy

 

[tnich]

I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon. Is that the case, or is the axis perpendicular to the hexagon? The problem as you have stated it in the original post leaves that question unanswered.

 

[Yalanhar]

[QUOperpendicularmr post: 6260489, member: 639870"]
I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon. Is that the case, or is the axis perpendicular to the hexagon? The problem as you have stated it in the original post leaves that question unanswered.
[/QUOTE]
Its perpendicur. Sorry

 

[tnich]

Suppose you divide the hexagon into $12$ congruent triangles, each with one vertex at the center of the hexagon. Then you could integrate $\rho r^2$ over one triangle and multiply by $12$ to get the result.

 

[haruspex]

But I didnt understand the rectangle

You did not post a diagram, but I presumed from your algebra that your integration element consisted of a narrow trapezoidal strip parallel to one side of the hexagon (or maybe six such, forming a hexagon).
Concentrating one one such strip, being so narrow, it is effectively a rod. What is the moment of a rod about its centre?

I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon.

It's useful regardless.