From a uniform circular disc of diameter d, a circular hole of diameter $\frac{\text{d}}{6}$ and having its centre at a distance of $\frac{\text{d}}{4}$ from the centre of the disc is scooped out. Find the centre of mass of the remaining portion.
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Let the mass per unit area of the disc be m. Then, Mass of the disc, $\text{M}=\pi\Big(\frac{\text{d}}{2}\Big)^2\text{m}=\frac{\pi\text{md}^2}{4}$ Mass of the scooped out portion of the disc, $\text{M}=\pi\Big(\frac{\text{d}}{12}\Big)^2\text{m}=\frac{\pi\text{md}^2}{144}$ Let us take the centre of the disc O as origin. The masses M and M' are supposed to be concentrated at their respective centres of the disc and scooped out portion. Since the portion is removed, its mass M' will be represented by negative sign. Then x-coordinate of the COM of the remaining portion of the disc is given by, $\text{x}=\frac{\text{Mx}_1-\text{M}'\text{x}_2}{\text{M}-\text{M}}$ $=\frac{\text{M}\times0-\text{M}'\times\frac{\text{d}}{4}}{\frac{\pi\text{md}^2}{4}-\frac{\pi\text{md}^2}{144}}$ $=\frac{-\text{M}'\text{d}}{4}\times\frac{144}{35\pi\text{md}^2}$ $=\frac{-\pi\text{md}^2}{144}\frac{\text{d}}{4}\frac{144}{35\pi\text{md}^2}$ $=\frac{-\text{d}}{140}$ i.e., the COM of the remaining portion of the disc is at a distance of $\frac{-\text{d}}{140}$towards left of the orign O.
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