Calculate the moment of inertia
(a) of a copper uniform disc relative to the symmetry axis perpendicular to the plane of the disc, if its thickness is equal to `b=2.0mm` and its radius to `R=100mm`,
(b) of a uniform solid cone relative to its symmetry axis, if the mass of the cone is equal to m and the radius of its base to R.

Explanation

(a) Consider an elementary disc of thickness `dx`. Moment of inertia of this element about the z-axis, passing through its C.M.
`dI_z=((dm)R^2)/(2)=rhoSdx(R^2)/(2)`
where `rho`=density of the material of the plate and S=area of cross section of the plate.
Thus the sought moment of inertia
`I_z=(rhoSR)/(2)underset(0)overset(b)intdx=R^2/2rhoSb`
`=pi/2rhobR^4` (as `S=piR^2`)

putting all the values we get, `I_z=2*gm*m^2`
(b) Consider an element disc of radius r and thickness `dx` at a distance `x` from the point O. Then `r=xtanalpha` and volume of the disc.
`=pir^2tan^2alphadx`
Hence, its mass `dm=pix^2 tan alpha dx*rho` (where `rho`=density of the cone `=m//1/3piR^2h`)
Moment of inertia of this element, about the axis OA,
`dI=dm(r^2)/(2)`
`=(pix^2tan^2alphadx)(x^2tanalpha)/(2)`
`=(pirho)/(2)x^4tan^4alphadx`

Thus the sought moment of inertia `I=(pirho)/(2)tan^4alphaunderset(0)overset(h)intx^4dx`
`=(pirhoR^4*h^4)/(10h^4)` (as `tanalpha=R/h`)
Hence `I=(3mR^2)/(10)` (putting `rho=(3m)/(piR^2h)`)