Q. The moment of inertia of a disc of mass $M$ and radius $R$ about an axis, which is tangential to the circumference of the disc and parallel to its diameter, is
Solution:
Moment of inertia of disc about its diameter is $I_{d} = \frac{1}{4} MR^{2}$
MI of disc about a tangent passing through rim and in the plane of disc is
$I=I_{G}+MR^{2}=\frac{1}{4}MR^{2}+MR^{2}=\frac{5}{4}MR^{2}$
