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(D) The length of the wire is $l$,so the radius $R$ of the semicircle is given by $\pi R = l$,which implies $R = \frac{l}{\pi}$.The moment of inertia

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$\frac{M l^2}{2 \pi^2}$

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(D) The length of the wire is $l$,so the radius $R$ of the semicircle is given by $\pi R = l$,which implies $R = \frac{l}{\pi}$.The moment of inertia of a thin wire (or ring) of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = MR^2$.For a semicircle,the moment of inertia about the diameter (the axis passing through the ends) is half of the moment of inertia of the full ring about its center,because the mass is distributed symmetrically.Thus,$I = \frac{1}{2} M R^2$.Substituting $R = \frac{l}{\pi}$ into the equation:$I = \frac{1}{2} M \left(\frac{l}{\pi}\right)^2 = \frac{M l^2}{2 \pi^2}$.

$A$ thin wire of length $l$ and mass $M$ is bent in the form of a semicircle. What is its moment of inertia about an axis passing through the ends of the wire?

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