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

Vedclass · Accepted Answer

$\frac{Ml^2}{2\pi^2}$

Vedclass · Answer

(D) The axis passing through the ends of the wire is the diameter of the semi-circle.Let the radius of the semi-circle be $r$. The length of the wire is $l = \pi r$,so $r = \frac{l}{\pi}$.Every point on the semi-circle is at a distance $r$ from the center of the circle.The moment of inertia of the wire about an axis passing through the center and perpendicular to the plane of the semi-circle is $I_z = \int r^2 dm = M r^2$.Let $I_x$ and $I_y$ be the moments of inertia about two perpendicular axes in the plane of the semi-circle,where one axis is the diameter passing through the ends of the wire $(I_x = I_d)$.By the perpendicular axes theorem,$I_z = I_x + I_y$. Since the wire is symmetric about the diameter,$I_x = I_y = I_d$.Therefore,$M r^2 = 2 I_d$,which gives $I_d = \frac{M r^2}{2}$.Substituting $r = \frac{l}{\pi}$,we get $I_d = \frac{M}{2} (\frac{l}{\pi})^2 = \frac{Ml^2}{2\pi^2}$.

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

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