Question

derive expression for moment of inertia of uniform disc about an Axis passing through the centre and perpendicular to the plane​

Answer 1

Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is, dl = (dm) R2 As the mass is uniformly distributed, the mass per unit area (σ) is σ = massareamassarea = MπR2MπR2 The mass of the infinitesimally small ring is, dm = σ 2πr dr = MπR2MπR2 2πr dr