Find the moment of inertia of a uniform ring of mass M and radius R about a diameter.

Explanation

The sphere may be divided into disks perpendicular to the given axis, as shown in fig. The disk at a distance x from the center of the sphere has a radius `r=(R^2-x^2)^(1/2)` and a thickness dr. If `rho=M/V` is the volume mass density (mass perunit volume), the mass of this elemental disk is `dm=rhodV=ppir^2 dx,` or `dm=rhopi(R^2-x^2)dx`From the last example, the moment of inertia of this elemental disk is `dI=1/2dmr^2=1/2rhopi(R^2-x^2)dx` The total moment of inertia is`I=1/2rhopiint_(-R)^(R)(R^4-2R^2x^2+x^4)dx` `=1/2rhopi[R^4x-2/3R^2x^3+1/5x^5]_(-R)^(R)=8/15rhopiR^5`The total mass of the sphere is `M = rho (4/3 piR^3)`, so the moment of inertia maybe written as : `I = 2/5 MR^2`.