Calculate the moment of inertia of a uniform solid cone about an axis through its center (given figure). The cone has massM and altitudeh .

The radius of its circular base is R .

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force).

Here we have one uniform solid cone. It has mass M and altitude h

Also, we have the radius of base of the cone is R.

Here we have altitude of cone is $h$.

Taking a small disc at height $x$ from center of cone having width $dx$.

Radius of disk is $r$.

Now, we can see that $\Delta AO\text{'}C\text{'}~\Delta AOC$

So, we can write

$$\begin{gathered} \begin{array}{c}\frac{AO\text{'}}{AO}=\frac{O\text{'}C\text{'}}{OC}\\ \frac{h-x}{h}=\frac{r}{R}\end{array} \\ 1-\frac{x}{h}=\frac{r}{R}\left(1\right) \end{gathered}$$

Now, density of cone is given by $\frac{M}{V}$

Where c is mass of cone and $V$ is volume of cone.

So,

$\frac{M}{V}=\frac{M}{\frac{1}{3}\pi R^{2}h}$

Mass of disc$=dM=\text{density}\times \text{volumeofdisk}$

$=\frac{3M}{\pi R^{2}h}\times \pi r^{2}dx$ (2)

Now, inertia of disk $=dI=\frac{dM}{2}{r}^2$

From equation (2)

$\begin{array}{l}=\frac{3M{r}^{2}dx}{R^{2}h}\times \frac{r^2}{2}\\ =\frac{3Mdx}{R^{2}h}\times \frac{r^4}{2}\end{array}$

So, we get

$dI=\frac{3Mdx}{R^{2}h}\times \frac{r^4}{2}$ (3)

Now, differentiating equation (1), we get

$\begin{array}{c}-\frac{dx}{h}=\frac{dr}{R}\\ dx=-\frac{hdr}{R}\end{array}$

Now, integrating equation (3), we get

$\begin{array}{l}\underset{0}{\overset{I}{\int }}dI=\underset{R}{\overset{0}{\int }}\frac{3M}{2R^{2}h}\times r^4\left(-\frac{h}{R}dr\right)\\ I=\frac{3M}{10}{R}^2\end{array}$

So, moment of inertia of uniform cone is

$I=\frac{3M}{10}{R}^2$