From a solid sphere of mass M and radius R , a cube of the maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:

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TestHub · Accepted Answer

Let a be length of edge for cube
with maximum possible volume diagonal length = 2R

$\Rightarrow \sqrt{3} a = 2 R \Rightarrow a = \frac{2 R}{\sqrt{3}}$

As densities of sphere and cube are equal. Let M' be mass of cube

$\frac{M}{\frac{4}{3} π R^{3}} = \frac{M^{'}}{a^{3}}$

$M^{'} = \frac{3 M a^{3}}{4 π R^{3}}$

Moment of inertia of cube about an axis passing through its centre is

$I= \frac{M^{'} (2 a^{2})}{12}$ $= \frac{3 M a^{3}}{4 π R^{3}} \times \frac{2 a^{2}}{12}$ $= \frac{M a^{5}}{8 π R^{3}}$

$Also, a = \frac{2}{\sqrt{3}} R$

$I = \frac{M \times 32 R^{5}}{8 π \times 9 \sqrt{3} R^{3}}$ $= \frac{4 M R^{2}}{9 \sqrt{3} π}$

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