Question

A disk of mass $M$, radius $R$ and thickness $\frac{R}{6}$ has a moment of intertia $I$ about an axis passing through the centre and perpendicular to the disk. The disk is recasted to form a sphere. Obtain the moment of inertia of the sphere about its diameter.

Answer 1

Initially, moment of inertia of disk,

$I=\frac{M{R}^2}{2}$ (for a disk) ....(1) [Ref. image 1]

and volume = $\pi R^2\frac{R}{6}=\frac{\pi }{6}{R}^3$

In the recasting process the volume of material remains constant volume of sphere : $\frac{4}{3}\pi R^{\prime 3}=\frac{\pi }{6}{R}^3$ [R' : Radius of sphere]

$\Rightarrow R^{\prime }=\frac{R}{2}$ ....(2) [Ref. image 2]

New moment of inertia of sphere will be :-

$I^{\prime }=\frac{2}{5}M{R}^{\prime 2}$

From (2) and (1) we get,

$I^{\prime }=\frac{2}{5}M{\left(\frac{R}{2}\right)}^2=\frac{1}{5}\left(\frac{M{R}^2}{2}\right)\Rightarrow I^{\prime }=\frac{I}{5}$