(i) Moment of inertia of a solid cone about its axis of symmetry:

Let us consider a right circular solid cone ABC of mass M and height h, as shown in the above Fig.1. If r be the radius of base of the solid cone then the total volume of this solid cone is $\frac{1}{3}\pi{r^2}h$.

Mass density i.e., mass per unit volume of the solid cone is $\sigma=\frac{M}{\frac{1}{3}\pi{r^2}h}$ = $\frac{3M}{\pi{r^2}h}$.

In order to calculate the moment of inertia of the solid cone about the axis of symmetry AO, let us consider an elementary circular disc of radius $r_1$ and thickness $dz$ at a distance $z$ from the vertex A. So the volume of this disc is $(\pi{r_1}^2dz)$ and the mass of this disc is $\frac{3M}{\pi{r^2}h}\times{\pi{r_1}^2dz}$ = $\left(\frac{3M{r_1}^2}{r^2{h}}\right)dz$.

From Fig. 1 we can write,

$\displaystyle{\frac{r_1}{r}=\frac{z}{h}}$

$or,\ \displaystyle{r_1=\frac{z}{h}r}$

So the moment of inertia of the elementray disc about the axis of symmetry is

$dI=\frac{1}{2}\left(\frac{3M{r_1}^2}{r^2{h}}\right)dz\times{r_1}^2$
[ To read the full derivationn of moment of inertia of a solid circilar disc about a axis passing through the centre of mass of it and perpendicular to the plane of it, (CLICK HERE) ]

$or,\ dI=\displaystyle{\frac{3M}{2{r^2}h}{r_1}^4\ dz}$

$or,\ dI=\displaystyle{\frac{3M}{2{r^2}h}\frac{z^4r^4}{h^4}\ dz}$
[putting the value of $r_1$]

$or,\ dI=\displaystyle{\frac{3Mr^2}{2h^5}z^4\ dz}$

So the moment of inertia of the whole solid cone about the axis of symmetry AO is

$I=\displaystyle{\frac{3Mr^2}{2h^5}\int_0^h{z^4}\ dz}$

$or,\ I=\displaystyle{\frac{3Mr^2}{2h^5}\frac{1}{5}\left[{z^5}\right]_0^h}$

$or,\ I=\displaystyle{\frac{3Mr^2}{10h^5}h^5}$

$or,\ \displaystyle{I=\frac{3}{10}Mr^2}$

(ii) Moment of inertia of a solid cone about the axis passing through the vortex and perpendicular to the axis of symmetry:

Let us consider a right circular solid cone ABC of mass M and height h, as shown in the above Fig.2. If r be the radius of base of the solid cone then the total volume of this solid cone is $\frac{1}{3}\pi{r^2}h$.

Mass density i.e., mass per unit volume of the solid cone is $\sigma=\frac{M}{\frac{1}{3}\pi{r^2}h}$ = $\frac{3M}{\pi{r^2}h}$.

Let us consider an axis PQ passing through the vortex $A$ and perpendicular to the axis of symmetry $AO$.

In order to calculate the moment of inertia of the solid cone about the axis PQ, let us consider an elementary solid circular disc of radius $r_1$, thickness $dz$ at a distance $z$ from the vortex $A$.

From Fig. 2 we can write

$\displaystyle{\frac{r_1}{r}=\frac{z}{h}}$

$or,\ \displaystyle{r_1=\frac{z}{h}r}$

The volume of this disc is $(\pi{r_1}^2dz)$ and the mass of this disc is $\frac{3M}{\pi{r^2}h}\times{\pi{r_1}^2dz}$ = $\left(\frac{3M{r_1}^2}{r^2{h}}\right)dz$

So the moment of inertia of the disc about its diameter parallel to the axis PQ is,

$I_d=\frac{1}{4}\times(mass\ of\ the\ disc)\times{r_1}^2$

$or,\ I_d=\displaystyle{\frac{3M}{4r^2h}{r_1}^4\ dz}$

$or,\ I_d=\displaystyle{\frac{3Mr^2}{4h^5}{z}^4\ dz}$
[putting the value of $r_1$]

So by applying the theorem of parallel axes we can write that the moment of inertia of the elementary disc about the axis PQ is given by,

$dI_{PQ}=I_d+\left(\frac{3M{r_1}^2}{r^2{h}}\right)dz\times{z^2}$

$or,\ dI_{PQ}=I_d+\frac{3M}{h^3}z^4\ dz$
[putting the value of $r_1$]

$or,\ dI_{PQ}=\displaystyle{\frac{3Mr^2}{4h^5}{z}^4\ dz}+\displaystyle{\frac{3M}{h^3}z^4\ dz}$

So the moment of inertia of the solid cone about the axis $PQ$ is given by,

$I_{PQ}=\frac{3Mr^2}{4h^5}\displaystyle{\int_0^h}z^4\ dz+\frac{3M}{h^3}\displaystyle{\int_0^h}z^4\ dz$

$or,\ I_{PQ}=\frac{3Mr^2}{4h^5}\frac{h^5}{5}+\frac{3M}{h^3}\frac{h^5}{5}$

$or,\ I_{PQ}=\frac{3Mr^2}{20}+\frac{3Mh^2}{5}$

$or,\ \displaystyle{I_{PQ}=\frac{3M}{5}\left(\frac{r^2}{4}+h^2\right)}$