Three principal axis moments of inertia for hexagonal prism?

In summary: I'll post an update when I have something more substantive.In summary, John was trying to figure out the mass moments of inertia for a hexagonal prism. He was having trouble recalling from calc 3 from 2 years ago and was failing miserably. He attempted to find the moments of inertia using a combination of equilateral triangles and then rectangles. He was not sure what he was doing right or wrong and was looking for help. He attempted to find the moments of inertia by dividing the hexagon into pieces and using the parallel and perpendicular axis theorems. He was having trouble getting the results he wanted, so he planned to take a photo of his work later to show the beginnings of putting the triple integral together.
  • #1
John Darvish
2
1
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
 
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  • #2
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.
 
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  • #3
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 realize 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.
 
  • #4
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.
 
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1. What is the definition of "principal axis moments of inertia"?

The principal axis moments of inertia refer to the moments of inertia of a solid object around its three principal axes. These axes are the three mutually perpendicular axes that pass through the center of mass of the object and are used to calculate the rotational inertia of the object.

2. How are the principal axis moments of inertia calculated for a hexagonal prism?

The principal axis moments of inertia for a hexagonal prism can be calculated using the formula I = (m/12)(a^2 + b^2), where m is the mass of the prism and a and b are the lengths of the two sides of the base of the prism. This formula can be used for all three principal axes of the prism.

3. Why are the principal axis moments of inertia important?

The principal axis moments of inertia are important because they help determine the rotational behavior of an object. They are used in calculations for torque, angular acceleration, and angular momentum, which are all crucial in understanding the motion of a rotating object.

4. How do the principal axis moments of inertia differ for a hollow hexagonal prism compared to a solid one?

The principal axis moments of inertia for a hollow hexagonal prism will be smaller than those of a solid one, as the hollow center reduces the mass and therefore the rotational inertia of the object. The exact values will depend on the dimensions and the thickness of the walls of the hollow prism.

5. Can the principal axis moments of inertia change for a given hexagonal prism?

Yes, the principal axis moments of inertia can change for a hexagonal prism if the dimensions or mass of the prism are altered. Additionally, the moments of inertia can also vary depending on the orientation of the prism with respect to the principal axes. However, they will remain constant for a given prism as long as these factors remain unchanged.

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