# Three principal axis moments of inertia for hexagonal prism?

Homework Statement
I'm trying to figure out the mass moments of inertia for a hexagonal prism, with the z-axis being longitudinal. I'm trying to recall my calc 3 from 2 years ago and am failing miserably.

I know the height of the prism is h. Each hexagonal side is length a. The prism has constant density ρ. I can easily find the area and volume via relationships with equilateral triangles.

With the above dimensions and constants

ρ = m/V
m = ρ*V = 3*√3 /2 *ρ*a2*h

I initially tried to find the 3 moments of inertia just using a combination of 6 equilateral triangles. My second attempt divided up the hexagon into rectangles and then the remaining triangles. I didn't get the results I had seen via internet searches, though some of the internet searches had different results for the moments of inertia so they may have been in error.

I'm not exactly sure what I am doing right or wrong. I think my errors are in defining the dx, dy, and dz differentials when I am trying to relate dm from the general inertia formula.

∫r2 dm

I'd appreciate any help you can provide.

Thanks

John

## Answers and Replies

andrewkirk
Science Advisor
Homework Helper
Gold Member
Wiki has a list of moments of inertia. The plane polygon one can give you the prism's moment around the longitudinal axis. The rod one can give you a reasonable approximation to the other two. The rod in wiki is cylindrical. If you want to get an exact answer for the prism for the two non-longitudinal axes, I suggest you first post your attempts along those lines, and people can point out where/if it has gone wrong.

Simon Bridge
Simon Bridge
Science Advisor
Homework Helper
Dividing the shape into simpler bits is what you are doing right.
Trouble is that you don't seem to know what to do with the simple bits... and, since you are so uncertain, you have not been brave enough to detail what you tried here. It is difficult to help you if we don't know how you were using the equilateral triangles etc.
Remember that you are dividing up a volume... and triangles do not have a volume. So what did you do?

The trick is to divide into bits that you already know the moment of inertia for, and use the parallel and perpendicular axis theorems. Is that what you tried? Or did you just get stuck after sketching out the shapes?

The calc approach is to realise that the moment of inertia about (say) the z axis, of the mass element dm at position (x,y,z) is
$$dL = \rho(x^2+y^2)dxdydz$$
... for uniform density object.

Thanks folks for replies. I'll take a photo of some of the work I've done later tonight, hopefully, to show the beginnings of putting the triple integral together for one of the axes. I think my most promising approach is dividing the hexagon into a rectangle and triangle and then going from there.

Simon Bridge