August 24, 2014

Moments of inertia of triangular prisms

Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. These bodies, with mass density ρ, can be seen as stacks of infinitesimally thin triangles of thickness dh and surface density dμ=ρdh (we preserve the notations from the previous posts and introduce the height of the stack, H. The inertia moments of the prisms are denoted by P, instead of I.)
The centers of mass of these sheets are all situated on z, so the total moment of inertia about this axis is simply the sum of the individual ones. We must simply replace μ by ρH in (1) and (3):
Pz(L,H)=ρHL4348Ptrz(L,a,H)=ρH348[L43a412a2(La)2]

The derivation is slightly more complicated for axis y, since we need to account for the variable distance between it and the centers of mass of the sheets (using, of course, the overworked parallel axis theorem!) Fortunately, we only need the integral H/2H/2dhh2=H312 to get:
Py(L,H)=ρHL4396[1+2(HL)2]Ptry(L,a,H)=ρH396[L43a412a2(La)2+2H2(L23a2)]=ρHL4396[1+2(HL)212x2((1x)2+x24+x22(HL)2)]
where x=a/L.

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