A disc of radius $r c m$ of uniform thickness and uniform density $p$ has a square hole with sides of length $l = \frac{r}{\sqrt{2}} c m .$ One corner of the hole is located at the centre of the disc and centre of the hole lies on the y-axis as shown. Find the y-coordinate of the position of the centre of mass of disc with the hole (in cm).

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Solution

Mass of hole desk without cutting square $= P (π r^{2})$
Center is at $0.$
$\Rightarrow$ Mass of square $= σ {(\frac{r}{\sqrt{2}})}^{2}$
$= \frac{σ}{2} r^{2}$
Taking negative mass of square,
$\Rightarrow$ Com is at $= \frac{p (π r^{2}) \times 0 - p (\frac{r^{2}}{2})}{p (π r^{2}) + \frac{p}{2} r^{2}}$
$= \frac{- r^{2}}{r^{2} (1 + 2 π)}$ below from hole.
$= \frac{- 1}{(1 + 2 π)}$ below from hole.
Hence, solved.

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Mass of hole desk without cutting square =P(πr2)Center is at 0.⇒ Mass of square =σ(r√2)2 =σ2r2Taking negative mass of square, ⇒ Com is at =p(πr2)×0−p(r22)p(πr2)+p2r2 =−r2r2(1+2π) below from hole. =−1(1+2π) below from hole.Hence, solved.