# Moment of Inertia of a Triangular Prism

1. Mar 15, 2010

### Oijl

1. The problem statement, all variables and given/known data
A triangular prism of mass M, whose two ends are equilateral triangles parallel to the xy plane with side 2a, is centered on the origin with its axis on the z axis. Find its moment of inertia for rotation about the z axis. Without doing any integrals write down and explain its two products of inertia for rotation about the z axis.

2. Relevant equations

3. The attempt at a solution
I've arrived at a soultion, but I'd like to see what other people think, if it is right or not.

I know that the moment of inertia (here, just I), is this:

I = integral of [ (x^2 + y^2) dm]
and
p = M/V, so
dm = pdV, so
I = p * integral of [ (x^2 + y^2) dV], so
I = p * triple integral of [ (x^2 + y^2) dx dy dz]

This will need to be done as the sum of two integrals, since there is a discontinuity in the change in y.

For the first integral:
The limits of the x integral are -a to 0.
The limits of the y integral are -a/√3 to 2a/√3.
The limits of the z integral are, let's say, 0 to h.

I evaluated this to be p*h*(-4/3)(a^4)/(√3)
Since p = M/V and V = (h*a^2)/(√3), the first integral equals
M*(-4/3)(a^2)

For the second integral:

The limits of the x integral are 0 to a.
The limits of the y integral are 2a/√3 to -a/√3.
The limits of the z integral are 0 to h.

I evaluated this to be, of course, the same thing,
M*(-4/3)(a^2)

So,
I = (-8/3)(a^2)M

That should be the moment of inertia for rotation about the z axis.

The products of inertia should both be zero (because of symmetry).

Does this seem right? More importantly, is it right? Thank you for looking at it.