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

51415 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#