Question

Find the moment of inertia of a disc of mass M and radius R about an axis tangential to disc and lying on the plane of the disc.

Answer 1

Consider the disc in figure:

we know M.I. of disc about an axis perpendicular to disc and passing through its CM = \(\frac {MR^2}{2}\)

Using perpendicular axis theorem

l_AB + I_CD = ||⊥ = \(\frac {MR^2}{2}\)

⇒ 2 I_AB = \(\frac {MR^2}{2}\) [Since I_AB = I_CD] (symmetry)

∴ I_AB = \(\frac {MR^2}{2}\)

using parallel axis theorem