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What is the moment of inertia of a hollow cylinder of radius $R$ inside which a cube of face diagonal $2R$ is placed about the axis passing through the axis of the cylinder? Both have same mass $m$.

I just added the individual moment of inertia of the two bodies as $$ I = mR^2 + \frac{m(\sqrt2R)^2}{6} = \frac{4mR^2}{3}$$Is this correct?

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Yes it is correct, since $I = \int{r^2 . dm}$(which is just a number), the moment of inertia of both objects can be added up.

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