Question

find the moment of inertia of a hollow cone about it's axis of symmetry

Answer 1

Consider a hollow cone of slanting height l and base radius R. We wish to find moment of inertia( M.I.) about the axis op shown in the fig. As shown in figure, we consider ,first, a ring of radius x and slanting height dl. If d’ is surface density and M is mass of the cone then

d’=M/(surface area of the coin)=M/(pi Rl)…………………….(1)

The area of the considered ring =2pix(dl)

The mass of the ring=2pix(dl)(d’)…………………………….,(2)

M.I. of ring=2pixd’(dl)x^2. But, from the geometry, dl/dx=l/R or dl=(l/R)dx. Therefore ,

M.I. of the ring=2 pi d’x^3ldx/R. Integrating this from x=0 to x=R, we get,

(2 pi d’ l/R)(R^4/4)= M.I. of the hollow cone. Using value of d’ from (1), we have. M.I. of hollow cone about op=(2 pi l R^3/4)(M/pi R l)

=(1/2)MR^2