Circular Disc

19 Nov 2024

We are moving onto $2\text{D}$ surfaces, which unintuitively are easier in some respects. A topologist would call a “flat disc” a $2\text{D}$ ball.

Parameterizing the Surface

    

We can parameterize the flat disc using cylindrical coordinates.

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

Now, we find the differential area element. We can prove this both geometrically (see the right diagram above) and rigorously using calculus.

\[\frac{\partial \b{r}}{\partial s} = \cos \phi \; \u{x} + \sin \phi \; \u{y} = \u{s} \qquad \frac{\partial \b{r}}{\partial \phi} = - s \sin \phi \; \u{x} + s \cos \phi \; \u{y} = s \; \u{\phi}\] \[d \b{A} = \left ( \frac{\partial \b{r}}{\partial s} ds \right ) \times \left ( \frac{\partial \b{r}}{\partial \phi} d\phi \right ) = (ds \; \u{s}) \times (s \; d\phi \; \u{\phi}) = s \; ds \; d\phi \; \u{z}\] \[dA = \abs{ d \b{A} } = s \; ds \; d\phi\]

Mass

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

Moment of Inertia About Central Axis

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

Moment of Inertia About Central Diameter

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

Inertia Tensor

\[I = \begin{bmatrix} \frac{1}{4} M R^2 & 0 & 0 \\ 0 & \frac{1}{4} M R^2 & 0 \\ 0 & 0 & \frac{1}{2}M R^2 \end{bmatrix} = \tfrac{1}{4} M R^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}\]