Q. The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is

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The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is

Solution:

We should use parallel axis theorem.

Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane

is $ {{I}_{AB}}=frac{1}{2}M{{R}^{2}} $

Using theorem of parallel axes, we have

$ {{I}_{CD}}={{I}_{AB}}+M{{R}^{2}} $

$ =frac{1}{2}M{{R}^{2}}+M{{R}^{2}} $

$ =frac{3}{2}M{{R}^{2}} $

NOTE: The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion.