The following geometry consists of a semicircle and a rectangle. (Hint: For an entire circle, the moment of inertia calculated from the center is $I_x = I_y = pi r^4/4$.)
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For the rectangle, the moment of inertia is calculated as I = bh^3/12, where b is the base and h is the height. In this case, the base is 2r and the height is r, so the moment of inertia for the rectangle is I = 2r*r^3/12 = r^4/3. Show more…
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Here, $\pi r=l \therefore r=1 / \pi$ Moment of inertia of a ring about its diameter $=\frac{1}{2} M r^{2}$ $\therefore$ Moment of inertia of semicircle $=\frac{1}{2}\left[m\left(\frac{l}{\pi}\right)^{2}\right]=\frac{m l^{2}}{2 \pi^{2}}$.
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