Spherical Ball

26 Sep 2023

Parameterizing the Volume

We can do this with just standard spherical coordinates

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

Therefore

\[\frac{\partial \b{r}}{\partial r} dr = (\sin \theta \cos \phi \; \u{x} + \sin \theta \sin \phi \; \u{y} + \cos \theta \; \u{z}) dr = dr \; \u{r} \\[10pt] \frac{\partial \b{r}}{\partial \theta} d\theta = (r \cos \theta \cos \phi \; \u{x} + r \cos \theta \sin \phi \; \u{y} - r \sin \theta \; \u{z}) d\theta = r \; d\theta \; \u{\theta} \\[10pt] \frac{\partial \b{r}}{\partial \phi} d\phi = (- r \sin \theta \sin \phi \; \u{x} + r \sin \theta \cos \phi \; \u{y}) d\phi = r \; \sin \theta \; d\phi \; \u{\phi}\] \[dV = \left \lvert \begin{array}{ccc} dr & 0 & 0 \\ 0 & r \; d\theta & 0 \\ 0 & 0 & r \; \sin \theta \; d\phi \end{array} \right \rvert = r^2 \sin \theta \; dr \; d\theta \; d\phi\]

Mass

\[\begin{align} M &= \int \; dm \\[10pt] &= \rho \int \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{\pi} \int_{0}^{2 \pi} r^2 \sin \theta \; dr \; d\theta \; d\phi \\[10pt] &= \rho \left ( \int_{0}^{R} r^2 \; dr \right ) \left ( \int_{0}^{\pi} \sin \theta \; d\theta \right ) \left ( \int_{0}^{2 \pi} d\phi \right ) \\[10pt] &= \rho \left ( \tfrac{1}{3} R^3 \right ) \left ( 2 \right ) \left ( 2 \pi \right ) \\[10pt] &= \rho \cdot \tfrac{4}{3} \pi R^3 \end{align}\]

Moment of Inertia About Any Diameter

We use spherical coordinates to evaluate this integral. Since it is a spherical ball, all parameters $r$,$\theta$, and $\phi$ vary. Therefore, $dV = r^2 \sin \theta \; d r \; d \theta \; d \phi$. Also, from the diagram we see that $r_{axis} = r \sin \theta$.

\[\begin{align} I &= \int r_{axis}^2 \; dm \\[10pt] &= \rho \int r_{axis}^2 \; dV \\[10pt] &= \rho \int_{0}^{R} \int_{0}^{\pi} \int_{0}^{2 \pi} (r \sin \theta)^2 r^2 \sin \theta \; dr \; d\theta \; d\phi \\[10pt] &= \rho \left ( \int_{0}^{R} r^4 \; dr \right ) \left ( \int_{0}^{\pi} \sin^3 \theta \; d\theta \right ) \left ( \int_{0}^{2\pi} d \phi \right ) \\[10pt] &= \rho \left ( \tfrac{1}{5} R^5 \right ) \left ( \frac{4}{3} \right ) \left ( 2 \pi \right ) \\[10pt] &= \rho \cdot \tfrac{8}{15} \pi R^5 \\[10pt] &= \tfrac{2}{5} M R^2 \end{align}\]

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Inertia Tensor of a Spherical Ball

\[I = \begin{bmatrix} \frac{2}{5} M R^2 & 0 & 0 \\ 0 & \frac{2}{5} M R^2 & 0 \\ 0 & 0 & \frac{2}{5} M R^2 \end{bmatrix} = \tfrac{2}{5} M R^2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]