Question

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

Answer 1

See the proof below

Explanation:

The mass of the elemental disc is `dm=rho*pir^2dz`

The density of the cone is

`rho=M/V=M/(1/3piR^2h)`

Therefore,

`dm=M/(1/3piR^2h)pir^2dz`

`dm=(3M)/(R^2h)r^2dz`

But

`R/r=h/z`

`r=Rz/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

`dI=1/2dmr^2`

`dI=1/2*3M/h^3z^2*z^2R^2/h^2dz`

`dI=3/2*MR^2/h^5z^4dz`

Integrating both sides,

`I=3/2*MR^2/h^5int_0^hz^4dz`

`I=3/2*MR^2/h^5[z^5/5]_0^h`

`I=3/2*MR^2/h^5*h^5/5`

`=3/10MR^2`