The parallel axis theorem relates $I_{om},$ the moment of inertia of an object about an axis passing through its center of mass, to $I_p,$ the moment of inertia of the same object about a parallel axis passing through point p$.$ The mathematical statement of the theorem is $I_p=I_{cm}+M{d}^2,$ where $d$ is the perpendicular distance from the center of mass to the axis that passes through point $p,$ and $M$ is the mass of the object.

Part A

Part B

Now consider a cube of mass $m$ with edges of length $a.$ The moment of inertia $I_{am}$ of the cube about an axis through its center of mass and perpendicular to one of its faces is given by $I_{cm}=\frac{1}{6}m{a}^2.($Figure $1)$Find $I_{\text{e}\text{d}\text{g}\text{e}\text{,}\text{t}\text{h}\text{e}\text{m}\text{o}\text{m}\text{e}\text{n}\text{t}of\text{i}\text{n}\text{e}\text{r}\text{t}\text{i}\text{a}\text{a}\text{b}\text{o}\text{u}\text{t}an}$ axis $p$ through one of the edges of the cube

Express $I_{\text{e}\text{d}\text{g}\text{e}}$ in terms of $m$ and $a.$ Use fractions rather than decimal numbers in your answer.

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A$$

$?$

$I_{\text{e}\text{d}\text{g}\text{e}}=$

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