A solid sphere of mass M,radius R and having | vedclass

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Question

(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about an axis passing through its centre of mass is given by $I = \frac{2}{5}MR

Vedclass · Accepted Answer

$\frac{2R}{\sqrt{15}}$

Vedclass · Answer

(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about an axis passing through its centre of mass is given by $I = \frac{2}{5}MR^2$.Let the radius of the disc be $r$. The mass of the disc remains $M$ as it is recast from the sphere.The moment of inertia of a circular disc of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is $I_{cm} = \frac{1}{2}Mr^2$.Using the parallel axis theorem,the moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is $I' = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.Given that the moment of inertia remains $I$,we have $I = I'$.Substituting the expressions: $\frac{2}{5}MR^2 = \frac{3}{2}Mr^2$.Solving for $r^2$: $r^2 = \frac{2}{5} 	imes \frac{2}{3} R^2 = \frac{4}{15}R^2$.Therefore,$r = \frac{2R}{\sqrt{15}}$.

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