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

Question

51967 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-09T10:34:58

See the proof below

enter image source here

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,

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

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

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

$=3/10MR^2$

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