moment of inertia - 10

[logistic_guy]

Find the moment of inertia of a uniform solid sphere of radius $\displaystyle R$ and mass $\displaystyle M$ when the rotation is through its center.

 

[khansaheb]

Find the moment of inertia of a uniform solid sphere of radius $\displaystyle R$ and mass $\displaystyle M$ when the rotation is through its center.

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[logistic_guy]

Let us start this problem with a sketch. We will rotate the solid sphere around the $\displaystyle z$-axis. Let us draw a line $\displaystyle \textcolor{purple}{r_0}$ on the $\displaystyle x$-$\displaystyle y$ plane that is perpendicular to the axis of rotation ($\displaystyle z$-axis). If we move this line $\displaystyle \textcolor{purple}{r_0}$ in space, it is still perpendicular and has the same distance to the axis of rotation as shown in the sketch. If the point $\displaystyle (x,y,z)$ is on the edge of a sphere (centered at the origin) of radius $\displaystyle r$, then we can express the distance $\displaystyle \textcolor{purple}{r_0}$ in terms of $\displaystyle r$ and $\displaystyle z$, as shown below:



$\displaystyle \textcolor{purple}{r_0^2} = r^2 - z^2$

Or

$\displaystyle \textcolor{purple}{r_0} = \sqrt{r^2 - z^2}$

This is the perpendicular distance between any mass inside the solid sphere and the axis of rotation.

Then, we have:

$\displaystyle I_z = \int R^2 \ dm = \int r_0^2 \ dm = \int (r^2 - z^2) \ dm$

Finding $\displaystyle \textcolor{purple}{r_0}$ was the most difficult part. Now, the rest is trivial.

$\displaystyle I_z = \int (r^2 - z^2) \ dm = \int (r^2 - z^2) \rho \ dV = \int\int\int (r^2 - z^2) \rho \ r^2 \sin \phi \ dr \ d\phi \ d\theta$

$\displaystyle z = r\cos \phi$ in the spherical coordinate.

$\displaystyle I_z = \int\int\int (r^2 - r^2\cos^2\phi) \rho \ r^2 \sin \phi \ dr \ d\phi \ d\theta = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R} r^4 \rho \ (1 - \cos^2 \phi)\sin \phi \ dr \ d\phi \ d\theta$


$\displaystyle = \int_{0}^{2\pi}\int_{0}^{\pi} \frac{R^5}{5} \rho \ (1 - \cos^2 \phi)\sin \phi \ d\phi \ d\theta = \int_{0}^{\pi} 2\pi\frac{R^5}{5} \rho \ (1 - \cos^2 \phi)\sin \phi \ d\phi$


$\displaystyle = 2\pi\frac{R^5}{5} \rho \left(\frac{\cos^3\phi}{3} - \cos \phi\right)\bigg|_{0}^{\pi} = 2\pi\frac{R^5}{5} \rho \left(\frac{\cos^3\pi}{3} - \cos \pi - \frac{\cos^{3}0}{3} + \cos 0\right)$


$\displaystyle = 2\pi\frac{R^5}{5} \rho \left(-\frac{1}{3} - (-1) - \frac{1}{3} + 1\right) = 2\pi\frac{R^5}{5} \rho \left(-\frac{2}{3} + 2\right) = 8\pi\frac{R^5}{15} \rho$

We know that the volume density $\displaystyle \rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}$, then

$\displaystyle I_z = 8\pi\frac{R^5}{15} \rho = 8\pi\frac{R^5}{15}\frac{M}{\frac{4}{3}\pi R^3} = \textcolor{blue}{\frac{2}{5}MR^2}$