From a complete ring of mass M and radius R, a sector angle is removed. What is the moment of inertia of the incomplete ring about axis passing through the center of the ring and perpendicular to the plane of the ring?

From a complete ring of mass M and radius R, a sector angle is removed

1.	From a complete ring of mass M and radius R, a sector angle is removed. What is the moment of inertia of the incomplete ring about axis passing through the center of the ring and perpendicular to the plane of the ring?
Answer» Solution :Let R be the radius of the ring and M be the total mass of the COMPLETE ring. Let m be the mass of the section removed from the ring thens, mass of the incomplete ring is M-m, Let us INTRODUCE a positive INTEGER (n), such that or `n=(360^(@))/( theta)` mass of incomplete ring =M=m `m=(M)/(3620^(@))xx theta` `:.` Mass of in completer ring `=M-(M)/(3602^(@))xx theta` Mass of incomplete ring `=M-(M)/(n)=M((n-1))/(n)` For example, (a) when `theta=60^(@), n=(360^(@))/(60^(@))=6` `:. n=1=5` Mass of incomplete ring `=(5)/(6)M` (b) When `0-30^(@), n=(360^(@))/(30^(@))=12` `n=1=11` Mass of incomplete ring `-(11)/(12)M` The moment of inertia of the incomplete ring is, `1-M((n-1)/(n))r^(2)`