Question

Derivation of moment of inertia of an uniform circular disc

Answer 1

We have, moment of inertia of the disc about its diameter, Id =1/4MR²

Let us suppose that x and y-axis are the two perpendicular diameters of the disc. Then,

Ix = Iy = Id =1/4MR²

a.       If  is moment of inertia of the disc about an axis passing through its centre and normal to its plane, then according to the theorem of perpendicular axis,

            Iz = Ix + Iy = 1/4MR² +1/4MR² = 1/2MR²

b.       If ICD is the moment of inertia of the disc about an axis passing through a point on its edge and normal to its plane, then according to theorem of parallel axis,

            CD = z + Mh² and h = R

  I[CD] =   1/2MR² +MR² = 3/2MR²

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Answer 2

If  is moment of inertia of the disc about an axis passing through its centre and normal to its plane, then according to the theorem of perpendicular axis,

           Iz = Ix + Iy = 1/4MR² +1/4MR² = 1/2MR²