Moment of inertia of a cube

In case you're interested, this is what we did in class today in finding a moment of inertia of a cube with respect to any axis passing through its center of mass, provided that the mass density, $\varrho$ , is constant.

Let $\vec{r}$ be the position vector of any infinitesimal mass form the centre of the cube, $\hat{u}$ be a unit vector pointing along an arbitrary axis, and $\rho$ be the distance from the axis to the infinitesimal mass.

$\begin{array}{rcl} \vec{r} &=& x \hat{i} + y \hat{j} + z \hat{k} \\[10pt] \hat{u} &=& \dfrac{a \hat{i} + b \hat{j} + c \hat{k}}{\sqrt{a^2 + b^2 + c^2}} \\[10pt] I &=& \int \varrho \; \mathrm{d} V \; \rho^2 \end{array}$

$\begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left[ r^2 - (\vec{r} \cdot \hat{u})^2 \right] \\[10pt] &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left[ x^2 + y^2 + z^2 - \left( \frac{xa}{\sqrt{a^2 + b^2 + c^2}} + \frac{yb}{\sqrt{a^2 + b^2 + c^2}} + \frac{zc}{\sqrt{a^2 + b^2 + c^2}} \right)^2 \right] \\[10pt] &=& \varrho \int \int \int \mathrm{d} x \mathrm{d} y \mathrm{d} z \; \left \{ x^2 + y^2 + z^2 - \frac{1}{a^2 + b^2 + c^2} \cdot \right. \\[10pt] && \qquad \qquad \left. \left[ x^2 a^2 + y^2 b^2 + z^2 c^2 + 2xyab + 2yzbc + 2xzac \right] \right \} \end{array}$

It is clear that the integrals of the cross terms vanish.

$\begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \varrho \left[ \int_{- \frac{L}{2}}^{\frac{L}{2}} x^2 \left( 1 - \frac{a^2}{a^2 + b^2 + c^2} \right) \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} y \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} z \right. \\[10pt] && \left. + \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} y^2 \left( 1 - \frac{b^2}{b^2 + b^2 + c^2} \right) \mathrm{d} y \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} z \right. \\[10pt] && \left. + \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} x \int_{- \frac{L}{2}}^{\frac{L}{2}} \mathrm{d} y \int z^2 \left( 1 - \frac{c^2}{a^2 + b^2 + c^2} \right) \mathrm{d} z \right] \\[10pt] &=& \varrho \left[ \frac{2}{3} \frac{L^3}{8} L^2 \right] \left[ \left( 1 - \frac{a^2}{a^2 + b^2 + c^2} \right) + \left( 1 - \frac{b^2}{b^2 + b^2 + c^2} \right) + \left( 1 - \frac{c^2}{a^2 + b^2 + c^2} \right) \right] \end{array}$

$\begin{array}{rcl} \int \varrho \; \mathrm{d} V \; \rho^2 &=& \frac{1}{6} \varrho L^5 \\[10pt] &=& \frac{1}{6} \frac{M}{L^3} L^5 \\[10pt] &=& \frac{1}{6} M L^2 \end{array}$

Notice that this is independent of an axis chosen.

Wink

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