Rectangular Slab

25 Nov 2024

Parameterizing the Volume

We can do this with just standard Cartesian coordinates

\[\b{r}(x, y, z) = x \; \u{x} + y \; \u{y} + z \; \u{z} \\[10pt] V = \{ \b{r}(x, y, z) \ : \ -a/2 \leq x \leq a/2 \quad -b/2 \leq y \leq b/2 \quad -c/2 \leq z \leq c/2 \}\] \[\frac{\partial \b{r}}{\partial x} dx = dx \; \u{x} \qquad \frac{\partial \b{r}}{\partial y} dy = dy \; \u{y} \qquad \frac{\partial \b{r}}{\partial z} dz = dz \; \u{z}\]

Therefore

\[dV = \left \lvert \begin{array}{ccc} dx & 0 & 0 \\ 0 & dy & 0 \\ 0 & 0 & dz \end{array} \right \rvert = dx \; dy \; dz\]

Mass

We use Cartesian coordinates to evaluate this integral. Since it is a solid rectangular slab, all parameters $a$, $b$, and $c$ vary. Therefore, $dV = d x \; dy \; d z$.

\[\begin{align} M &= \int \; dm \\[10pt] &= \rho \int \; dV \\[10pt] &= \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} dx \; dy \; dz \\[10pt] &= \rho \left ( \int_{-a/2}^{a/2} dx \right ) \left ( \int_{-b/2}^{b/2} dy \right ) \left ( \int_{-c/2}^{c/2} dz \right ) \\[10pt] &= \rho (a) (b) (c) \\[10pt] &= \rho \cdot abc \end{align}\]

Moment of Inertia About Perpendicular Axis

\[\begin{align} I &= \int r_{axis}^2 \; dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} \int_{-c/2}^{c/2} (x^2 + y^2) \; dx \; dy \; dz \\[10pt] &= \rho \int_{-a/2}^{a/2} \int_{-b/2}^{b/2} (x^2 + y^2)c \; dx \; dy \\[10pt] &= \rho \int_{-a/2}^{a/2} (b x^2 + \tfrac{1}{12} b^3)c \; dx \\[10pt] &= \rho \left ( \tfrac{1}{12}a^3 b + \tfrac{1}{12} a b^3 \right ) c \\[10pt] &= \rho \cdot \tfrac{1}{12} abc (a^2 + b^2) \\[10pt] &= \tfrac{1}{12} M (a^2 + b^2) \end{align}\]

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Inertia Tensor

\[I = \begin{bmatrix} \tfrac{1}{12} M(b^2 + c^2) & 0 & 0 \\ 0 & \tfrac{1}{12} M(c^2 + a^2) & 0 \\ 0 & 0 & \tfrac{1}{12} M(a^2 + b^2) \end{bmatrix} = \tfrac{1}{12} M \begin{bmatrix} b^2 + c^2 & 0 & 0 \\ 0 & c^2 + a^2 & 0 \\ 0 & 0 & a^2 + b^2 \end{bmatrix}\]