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  4. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be

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Q. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be

System of Particles and Rotational Motion

Solution:

Moment of inertia of solid sphere of mass M and radius R about an axis passing through the centre of mass is $I=\frac{2}{5}MR^{2}$.
Let the radius of the disc is $r.$
Moment of inertia of circular disc of radius $r$ and mass $M$ about an axis passing through the centre of mass and perpendicular to its plane $=\frac{1}{2}Mr^{2}$
Using theorem of parallel axes, moment of inertia of disc about its edge is
$I '=\frac{1}{2}Mr^{2}+Mr^{2}=\frac{3}{2}Mr^{2}$
Given, $I = I ′$
$\therefore \frac{2}{5}MR^{2}=\frac{3}{2}Mr^{2}$ or $r^{2}=\frac{4}{15}R^{2}$ or $r=\frac{2R}{\sqrt{15}}$