# Moment of Inertia about axis through body diagonal of a Cuboid

1. The problem statement, all variables and given/known data

Consider a cuboid of lengths a, b and c along the x, y and z axes respectively, centred at the origin.

The task is to show that the moment of inertia of the cuboid of mass M and mass density ρ about an axis along the body diagonal, from (-a/2, -b/2, -c/2) to (a/2, b/2, c/2), assumes the form:

##I = \frac{M}{6}\frac{(ab)^2+(ac)^2+(bc)^2}{a^2+b^2+c^2}##

2. Relevant equations

This is supposed to be done by making use of previously determined results. These are, that the moments of inertia assume the form:

##Θ_{xx}=\frac{M}{12}(b^2+c^2)##
##Θ_{yy}=\frac{M}{12}(a^2+c^2)##
##Θ_{zz}=\frac{M}{12}(a^2+b^2)##

and also, that all the products of inertia vanish.

N.B. I proved these results by integrating between -a/2 and a/2, -b/2 and b/2, -c/2 and c/2, which is different to what I’m doing below – I don’t know if this is correct!

3. The attempt at a solution

I attempted to calculate the principal moments as follows.

##I_{11}=ρ\int_v y^2+z^2 dv##
##I_{11}=ρ\int_v y^2+z^2 dxdydz##

I’m not sure that these limits are correct. I’m integrating from 0 to a, 0 to b, etc because I’m starting at one corner and moving ‘a’ in the x direction, ‘b’ in the y direction, etc. Nevertheless this comes out as:

##I_{11}=\frac{M}{3}(b^2+c^2)##

I then attempted to calculate the products of inertia as follows.

##I_{12}=-ρ\int_v xydv##
##I_{12}=-ρ\int_v xydxdydz##
##I_{12}=-ρ\int_0^c dz \int_0^b\int_0^a xydxdy##

Again I’m not sure if these limits are correct, but this comes out as:

##I_{12}=-\frac{M}{4}ab##

Now, the other principal moments and products of inertia will be different than these. I could calculate all 9 terms in the moment of inertia tensor in a similar way, but I cannot see how this will bring me closer to obtaining a simple solution of the form:

##I = \frac{M}{6}\frac{(ab)^2+(ac)^2+(bc)^2}{a^2+b^2+c^2}##

Any help as to how to approach this problem would be appreciated.

http://ift.tt/1exnW2I