A square shaped hole of side l = frac{a}{2} i | vedclass

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(D) Let $\sigma$ be the surface mass density of the disk material.Mass of the complete disk,$M_1 = \sigma \pi a^2$,with its centre of mass at $O$ (coo

Vedclass · Accepted Answer

$23$

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(D) Let $\sigma$ be the surface mass density of the disk material.Mass of the complete disk,$M_1 = \sigma \pi a^2$,with its centre of mass at $O$ (coordinate $x_1 = 0$).Mass of the removed square portion,$M_2 = \sigma l^2 = \sigma (\frac{a}{2})^2 = \sigma \frac{a^2}{4}$,with its centre of mass at distance $d = \frac{a}{2}$ from $O$ (coordinate $x_2 = \frac{a}{2}$). The centre of mass of the remaining portion $X_{com}$ is given by:$X_{com} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}$$X_{com} = \frac{(\sigma \pi a^2)(0) - (\sigma \frac{a^2}{4})(\frac{a}{2})}{\sigma \pi a^2 - \sigma \frac{a^2}{4}}$$X_{com} = \frac{-\frac{\sigma a^3}{8}}{\sigma a^2 (\pi - \frac{1}{4})} = \frac{-\frac{a}{8}}{\frac{4\pi - 1}{4}} = -\frac{a}{2(4\pi - 1)}$Comparing this with $-\frac{a}{X}$,we get $X = 2(4\pi - 1) = 8\pi - 2$.Using $\pi \approx 3.14159$,$X \approx 8(3.14159) - 2 = 25.1327 - 2 = 23.1327$.The nearest integer value of $X$ is $23$.

$A$ square shaped hole of side $l = \frac{a}{2}$ is carved out at a distance $d = \frac{a}{2}$ from the centre $O$ of a uniform circular disk of radius $a$. If the distance of the centre of mass of the remaining portion from $O$ is $-\frac{a}{X}$,find the value of $X$ (to the nearest integer).

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