Four holes of radius R are cut from a thin sq | vedclass

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(C) Let $M$ be the mass of the square plate before cutting the holes. The area of the square plate is $(4R)^2 = 16R^2$. The mass per unit area is $\si

Vedclass · Accepted Answer

$\left( \frac{8}{3} - \frac{10\pi }{16} \right) M R^2$

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(C) Let $M$ be the mass of the square plate before cutting the holes. The area of the square plate is $(4R)^2 = 16R^2$. The mass per unit area is $\sigma = \frac{M}{16R^2}$.Each hole has an area of $\pi R^2$,so the mass of one hole is $m = \sigma 	imes \pi R^2 = \frac{M}{16R^2} 	imes \pi R^2 = \frac{\pi M}{16}$.The moment of inertia of the square plate about the $z-$ axis (passing through the center and perpendicular to the plane) is $I_{	ext{square}} = \frac{M}{12} (a^2 + a^2) = \frac{M}{12} (16R^2 + 16R^2) = \frac{32MR^2}{12} = \frac{8}{3} MR^2$.Each hole is a circular disc of radius $R$. The distance of the center of each hole from the $z-$ axis is $d = \sqrt{R^2 + R^2} = R\sqrt{2}$.Using the parallel axis theorem,the moment of inertia of one hole about the $z-$ axis is $I_{	ext{hole}} = I_{	ext{cm}} + md^2 = \frac{mR^2}{2} + m(R\sqrt{2})^2 = \frac{mR^2}{2} + 2mR^2 = \frac{5}{2} mR^2$.The moment of inertia of the remaining portion is $I = I_{	ext{square}} - 4 I_{	ext{hole}} = \frac{8}{3} MR^2 - 4 \left( \frac{5}{2} mR^2 ight) = \frac{8}{3} MR^2 - 10 mR^2$.Substituting $m = \frac{\pi M}{16}$,we get $I = \frac{8}{3} MR^2 - 10 \left( \frac{\pi M}{16} ight) R^2 = \left( \frac{8}{3} - \frac{10\pi}{16} ight) MR^2$.

Four holes of radius $R$ are cut from a thin square plate of side $4R$ and mass $M$. The moment of inertia of the remaining portion about the $z-$ axis is

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