A disc of mass `m` and radius `R` has a concentric hole of radius `r`. Its moment of inertia about an axis through its center and perpendicular to its plane is

Explanation

Let s = mass per unit area = `M/(pi(R^2 -r^2))` If we fill the hole with mass densities `+sigma-sigma,` then the system can be treated as a combination of two discs of radii r and R and densities `-sigma+sigma.` `I=I_1+I_2=1/2M_1R_(1)^(2)+1/2M_2R_(2)^(2)` `=1/2(-sigma)pir^2(r^2/2)+1/2(+sigma)piR^2(R^2/2)` `(sigmapi)/4(R^4-r^4)` `=(M(R^4-r^4))/(4(R^2-r^2))=M/4(R^2+r^2)`.