A thin square plate of side 4R has a mass M. | vedclass

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(B) The mass per unit area of the square plate is $\sigma = \frac{M}{(4R)^2} = \frac{M}{16R^2}$.The mass of one circular disc of radius $R$ is $m = \s

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$\left[ \frac{8}{3} - \frac{10\pi}{16} \right] MR^2$

Vedclass · Answer

(B) The mass per unit area of the square plate is $\sigma = \frac{M}{(4R)^2} = \frac{M}{16R^2}$.The mass of one circular disc of radius $R$ is $m = \sigma (\pi R^2) = \frac{M}{16R^2} \cdot \pi R^2 = \frac{\pi M}{16}$.The moment of inertia of the original square plate about the $z$-axis (passing through the center) is $I_{square} = \frac{M}{12} (a^2 + a^2) = \frac{M}{12} ((4R)^2 + (4R)^2) = \frac{M}{12} (32R^2) = \frac{8}{3} MR^2$.Each circular hole has its center at a distance $d = \sqrt{2}R$ from the center of the square. Using the parallel axis theorem,the moment of inertia of one hole about the $z$-axis is $I_{hole} = I_{cm} + md^2 = \frac{mR^2}{2} + m(\sqrt{2}R)^2 = \frac{mR^2}{2} + 2mR^2 = \frac{5}{2} mR^2$.The moment of inertia of the remaining part is $I = I_{square} - 4 I_{hole} = \frac{8}{3} MR^2 - 4 \left( \frac{5}{2} mR^2 \right) = \frac{8}{3} MR^2 - 10 mR^2$.Substituting $m = \frac{\pi M}{16}$,we get $I = \frac{8}{3} MR^2 - 10 \left( \frac{\pi M}{16} \right) R^2 = \left[ \frac{8}{3} - \frac{10\pi}{16} \right] MR^2$.

$A$ thin square plate of side $4R$ has a mass $M$. Four circles of radius $R$ are cut from it as shown. Calculate the moment of inertia of the remaining part about the $z$-axis passing through the center of the square.

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