what is the derivation for the moment of inertia of a hollow cone?

Question

consider the hollow cone to be made of a number of circular rings

Answer 1

Integrate dI from the top of the cone to the bottom.

For a ring of radius r,

dI = r^2 dm = r^2 (2 pi r) dz * density

For a cone, r is proportional to z (ratio of base radius to height), so

r = z R/h

Density (rho) is the surface area of the cone over the mass.
rho = ((pi) R sqrt (R^2 + h^2))/M

Plug in everything and integrate dI from z=0 to h.

Answer 2

Since the moment of inertia will not change, if
you move any of your 'rings' along the axis,

you can simply push all the 'rings' into one plane,
and replace the cone with a simple flat disc.

Answer 3

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