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(D) The axis passing through the ends of the semicircular wire is the diameter of the circle.Let $r$ be the radius of the semicircle. The length of th

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

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(D) The axis passing through the ends of the semicircular wire is the diameter of the circle.Let $r$ be the radius of the semicircle. The length of the wire is $l = \pi r$,so $r = \frac{l}{\pi}$.The moment of inertia of a thin ring of mass $M$ and radius $r$ about its central axis (perpendicular to the plane) is $I_z = M r^2$.By the perpendicular axes theorem,$I_z = I_x + I_y$. Since the ring is symmetric,$I_x = I_y = I_d$,where $I_d$ is the moment of inertia about any diameter.Thus,$M r^2 = 2 I_d$,which gives $I_d = \frac{M r^2}{2}$.Substituting $r = \frac{l}{\pi}$ into the expression for $I_d$:$I_d = \frac{M}{2} \left( \frac{l}{\pi} \right)^2 = \frac{M l^2}{2 \pi^2}$.

$A$ 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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