Question

Using the formula for the moment of inertia of a uniform sphere, the moment of inertia of a thin spherical layer of mass $m$ and radius $R$ relative to the axis passing through its centre is $I=\frac{2}{x}m{r}^2$. Find the value of $x$.

Answer 1

Moment of inertia of the shaded portion, about the axis passing through it's centre,
$I=\frac{2}{5}\left(\frac{4}{3}\pi R^3\rho \right)R^2-\frac{2}{5}\left(\frac{4}{3}\pi R^3\rho \right)r^3$
$=\frac{2}{5}\cdot \frac{4}{3}\pi \rho (R^5-r^5)$
Now, if $R=r+dr$, the shaded portion becomes a shell, which is the required shape to calculate the moment of inertia.
Now, $I=\frac{2}{5}-\frac{4}{3}\pi \rho \{{(r+dr)}^5-r^5\}$
$=\frac{2}{5}\cdot \frac{4}{3}\pi \rho (r^5+5r^{4}dr+\dots \cdots -r^5)$
Neglecting higher terms.
$=\frac{2}{3}\left(4\pi r^{2}dr\rho \right)r^2=\frac{2}{3}m{r}^2$