The surface density (mass/area) of a circular disc having a radius a varies with the distance from the centre according to (r) = A + Br . Determine its moment of inertia about a line passing through its centre and perpendicular to the plane of the disc.

Step-by-step solution

Consider an elemental ring of radius r and thickness dr in the circular disc. The area of this elemental ring is dA = 2 r dr . The mass of the elemental ring is dm = (r) dA = (A + Br) 2 r dr . The moment of inertia of this elemental ring about the axis passing through the centre and perpendicular to the plane is dI = r^2 dm . Substituting the expression for dm : dI = r^2 (A + Br) 2 r dr dI = 2 (A r^3 + B r^4) dr The total moment of inertia of the disc is obtained by integrating dI from r = 0 to r = a : I = ₀^ a 2 (