Elliptical Disc

20 Nov 2024

Parametarizing the Surface

    

This is very similar to the parameterization of a circular disc in the previous post. Let the parameter $t$ denote the percentage of the “radius” of the ellipse at a given “angle”. These are in quotes because their definitions are more complicated than that of a circle. For example

\[\b{r}(t, \phi) = at \cos \phi \; \u{x} + bt \sin \phi \; \u{y} \\[10pt] A = \{ \b{r}(t, \phi) \ : \ 0 \leq t \leq 1 \quad 0 \leq \phi < 2\pi \}\]

Now, just as before we find the value of an infinitesimal section of area. In this case, it is much harder to derive it using geometry, so we rely on the rigor of calculus. We are cashing in all of the hard work from the previous posts.

\[\frac{\partial \b{r}}{\partial t} = a \cos \phi \; \u{x} + b \sin \phi \; \u{y} \qquad \frac{\partial \b{r}}{\partial \phi} = - at \sin \phi \; \u{x} + bt \cos \phi \; \u{y}\] \[d \b{A} = \left ( \frac{\partial \b{r}}{\partial t} dt \right ) \times \left ( \frac{\partial \b{r}}{\partial \phi} d\phi \right ) = (ab \; t \cos^2 \phi + ab \; t \sin^2 \phi) \; dt \; d\phi \; \u{z} = ab \; t \; dt \; d\phi \; \u{z}\] \[dA = \abs{ d \b{A} } = ab \; t \; dt \; d\phi\]

Before moving on to the integration, I want to pause and talk about $\phi$. We see that I’ve labeled the diagram $\phi’$ but I’m using $\phi$ as my argument in $\b{r}(t, \phi)$. This is because $\phi$ actually has no geometric meaning. Like the parameter $t$, it’s just a parameter we can use for integration. Using the right figure, we see that

\[\abs{\b{r}} \tan \phi' = \tfrac{b}{a} \tan \phi\]

Mass

\[\begin{align} M &= \int dm \\[10pt] &= \sigma \int dA \\[10pt] &= \sigma \int_{0}^{1} \int_{0}^{2 \pi} ab \; t \; d t \; d \phi \\[10pt] &= \sigma \cdot ab \left ( \int_{0}^{1} t \; d t \right ) \left ( \int_{0}^{2 \pi} d \phi \right ) \\[10pt] &= \sigma \cdot ab \left ( \frac{1}{2} \right ) \left ( 2 \pi \right ) \\[10pt] &= \sigma \cdot \pi ab \end{align}\]

Moment of Inertia About Central Axis

\[\begin{align} I &= \int r_{axis}^2 dm \\[10pt] &= \sigma \int r_{axis}^2 dA \\[10pt] &= \sigma \int_{0}^{1} \int_{0}^{2 \pi} (a^2 t^2 \cos^2 \phi + b^2 t^2 \sin^2 \phi) \cdot ab \; t \; d t \; d \phi \\[10pt] &= \sigma \cdot ab \left ( \int_{0}^{1} t^3 \; d t \right ) \left ( \int_{0}^{2 \pi} (a^2 \cos^2 \phi + b^2 \sin^2 \phi) \; d \phi \right ) \\[10pt] &= \sigma \cdot ab \left ( \frac{1}{4} \right ) \left ( \pi a^2 + \pi b^2 \right ) \\[10pt] &= \sigma \cdot \tfrac{1}{4} \pi ab(a^2 + b^2) \\[10pt] &= \tfrac{1}{4} M (a^2 + b^2) \end{align}\]

Moment of Inertia About Central Diameter

\[\begin{align} I &= \int r_{axis}^2 dm \\[10pt] &= \sigma \int r_{axis}^2 dA \\[10pt] &= \sigma \int_{0}^{1} \int_{0}^{2 \pi} t^2 b^2 \sin^2 \phi \cdot ab \; t \; d t \; d \phi \\[10pt] &= \sigma \cdot ab \left ( \int_{0}^{1} t^3 \; d t \right ) \left ( \int_{0}^{2 \pi} b^2 \sin^2 \phi \; d \phi \right ) \\[10pt] &= \sigma \cdot ab \left ( \frac{1}{4} \right ) \left ( \pi b^2 \right ) \\[10pt] &= \sigma \cdot \tfrac{1}{4} \pi ab^3 \\[10pt] &= \tfrac{1}{4} M b^2 \end{align}\]

Inertia Tensor

\[I = \begin{bmatrix} \tfrac{1}{4} M b^2 & 0 & 0 \\ 0 & \tfrac{1}{4} M a^2 & 0 \\ 0 & 0 & \tfrac{1}{4} M (a^2 + b^2) \end{bmatrix} = \tfrac{1}{4} M \begin{bmatrix} b^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 + b^2 \end{bmatrix}\]