# Prove that the moment of inertia of a cone is I=3/10mr^2 with respect of its axis continuing through mass center? h=height; radius of base =r

Aug 9, 2017

See the proof below

#### Explanation:

The mass of the elemental disc is $\mathrm{dm} = \rho \cdot \pi {r}^{2} \mathrm{dz}$

The density of the cone is

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

Therefore,

$\mathrm{dm} = \frac{M}{\frac{1}{3} \pi {R}^{2} h} \pi {r}^{2} \mathrm{dz}$

$\mathrm{dm} = \frac{3 M}{{R}^{2} h} {r}^{2} \mathrm{dz}$

But

$\frac{R}{r} = \frac{h}{z}$

$r = R \frac{z}{h}$

dm=3M/(R^2h)*(R^2)/h^2*z^2dz=3M/h^3 z^2dz

The moment of inertia of the elemental disc about the $z -$axis is

$\mathrm{dI} = \frac{1}{2} \mathrm{dm} {r}^{2}$

$\mathrm{dI} = \frac{1}{2} \cdot 3 \frac{M}{h} ^ 3 {z}^{2} \cdot {z}^{2} {R}^{2} / {h}^{2} \mathrm{dz}$

$\mathrm{dI} = \frac{3}{2} \cdot M {R}^{2} / {h}^{5} {z}^{4} \mathrm{dz}$

Integrating both sides,

$I = \frac{3}{2} \cdot M {R}^{2} / {h}^{5} {\int}_{0}^{h} {z}^{4} \mathrm{dz}$

$I = \frac{3}{2} \cdot M {R}^{2} / {h}^{5} {\left[{z}^{5} / 5\right]}_{0}^{h}$

$I = \frac{3}{2} \cdot M {R}^{2} / {h}^{5} \cdot {h}^{5} / 5$

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