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(B) The moment of inertia of a disc of mass $M$ and radius $R$ about an axis passing through its centre and lying in its own plane (diameter) is given

Vedclass · Accepted Answer

$M\left(\frac{R^2}{4}\right)$

Vedclass · Answer

(B) The moment of inertia of a disc of mass $M$ and radius $R$ about an axis passing through its centre and lying in its own plane (diameter) is given by $I_{diameter} = \frac{1}{4}MR^2$.In this problem,the disc lies in the $X-Y$ plane,and its centre is located at $(a, 0)$.The $X$-axis passes through the centre of the disc and lies in the plane of the disc.Therefore,the $X$-axis is a diameter of the disc.Since the moment of inertia of a disc about its diameter is $\frac{1}{4}MR^2$,the moment of inertia about the $X$-axis is simply $\frac{1}{4}MR^2$ or $M\left(\frac{R^2}{4}\right)$.

The adjoining figure shows a disc of mass $M$ and radius $R$ lying in the $X-Y$ plane with its centre on the $X$-axis at a distance $a$ from the origin. Then the moment of inertia of the disc about the $X$-axis is

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