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 \(\rho\), can be seen as stacks of infinitesimally thin triangles of thickness \(\text{d}h\) and surface density \(\text{d} \mu = \rho \text{d}h\) (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 \(\mu\) by \(\rho H\) in (1) and (3):

\[\begin{array}{ll}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 \(\mu\) by \(\rho H\) in (1) and (3):

P_z(L,H) &= \rho H L^4 \frac{\sqrt{3}}{48}\\

P^{\text{tr}}_z(L,a,H) &= \rho H \frac{\sqrt{3}}{48} [L^4 - 3 a^4 - 12 a^2 (L-a)^2]

\end{array}\tag{5}\]

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 \( \displaystyle \int_{-H/2}^{H/2} \text{d}h \, h^2 = \frac{H^3}{12}\) to get:

\[\begin{array}{ll}

P_y(L,H) &= \rho H L^4 \frac{\sqrt{3}}{96} \left [ 1 + 2 \left ( \frac{H}{L} \right )^2 \right ] \\

P^{\text{tr}}_y(L,a,H) &= \rho H \frac{\sqrt{3}}{96} \left [ L^4 - 3 a^4 - 12 a^2 (L-a)^2 + 2 H^2 (L^2 - 3 a^2) \right ]\\

&= \rho H L^4 \frac{\sqrt{3}}{96} \left [ 1 + 2 \left ( \frac{H}{L} \right )^2 - 12 x^2 \left ( (1-x)^2 + \frac{x^2}{4} + \frac{x^2}{2} \left ( \frac{H}{L} \right )^2 \right ) \right ]

\end{array}\tag{6}\]where \(x = a /L\).

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