Question

Find the moments of inertia for a rectangular brick with dimensions a ,b, and c , mass M, and constant density if the centre of the brick is situated at the origin and the edges are parallel to the coordinate axes.

Short answer

The moments of inertia for a rectangular brick and edges are parallel to the coordinate axes are:

\(\begin{aligned}{{\rm{I}}_{\rm{x}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{b}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}{\rm{,}}\;\;\\\;{{\rm{I}}_{\rm{y}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}{\rm{,}}\;\;\\\;{{\rm{I}}_{\rm{z}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}\end{aligned}\).

Step-by-step solution

Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

Find the moments of inertia for a rectangular brick

Consider the given values and simplify,

Where \({\rm{M = k a b c}}\)

Find the value of \({{\rm{I}}_{\rm{y}}}\)and\({{\rm{I}}_{\rm{z}}}\)

By the Law of Symmetry:

\({{\rm{I}}_{\rm{y}}}{\rm{ = }}\frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}\)

And

\({{\rm{I}}_{\rm{z}}}{\rm{ = }}\frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}\)

Therefore, the moments of inertia for a rectangular brick and edges are parallel to the coordinate axes are:

\(\begin{aligned}{{\rm{I}}_{\rm{x}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{b}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}{\rm{,}}\;\;\\\;{{\rm{I}}_{\rm{y}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{c}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}{\rm{,}}\;\;\\\;{{\rm{I}}_{\rm{z}}}\rm &= \frac{{{\rm{M}}\left( {{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{\rm{2}}}} \right)}}{{{\rm{12}}}}\end{aligned}\).