Find the centre of mass of a uniform:
- Half-disc.
- Quarter-disc.
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Let mass of half disc is M. Area of half disc $=\frac{\pi\text{R}^2}{2}$ Mass per unit area $\text{m}=\frac{2\text{M}}{\pi\text{R}^2}$ 
Surface area of a semicircular strip $=\frac{\pi}{2}\big[(\text{r}+\text{dr})^2-\text{r}^2\big]$
$=\frac{\pi}{2}\big[\text{r}^2+\text{dr}^2+2\text{rdr}-\text{r}^2\big]$
$=\pi\text{r dr} $
$\therefore$ Mass of strip $\text{dm}=\frac{2\text{M}}{\pi\text{R}^2}\cdot\pi\text{r dr}$
$\text{dm}=\frac{2\text{M}}{\text{R}^2}\cdot\text{r dr}$
Let (x, y) are the co-ordinates of c.m. of this strip $(\text{x},\text{y})=\Big(0,\frac{2\text{r}}{\pi}\Big)$
$\text{x}=\text{x}_\text{cm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\text{x dm}=\int\limits^{\text{R}}_00\text{ dm}=0$
$\text{y}_\text{cm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\text{y dm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\frac{2\text{r}}{\pi}\times\frac{\text{2}\text{M}}{\text{R}^2}\text{r dr}$
$=\frac{1}{\text{m}}\cdot\frac{4\text{M}}{\pi\text{R}^2}\int\limits^{\text{R}}_0\text{r}^2\text{ dr}=\frac{4\text{M}}{\pi\text{R}^2}\Big[\frac{\text{r}^3}{3}\Big]_0^{\text{R}}=\frac{4}{3\pi\text{R}^2}\cdot\text{R}^3$
$\text{y}_\text{cm}=\frac{4\text{R}}{3\pi}$
So, centre of mass of circular half disc $=\Big(0,\frac{4\text{R}}{3\pi}\Big)$
Using symmetry,
For a half disc along y-axis c.m. will be at $\text{y}=\frac{4\text{R}}{3\pi}$
So, the c.m. of quarter disc $=\Big(\frac{4\text{R}}{3\pi},\frac{4\text{R}}{3\pi}\Big)$
- The half disc can be divided into a larger number of semi circular strips.
Surface area of a semicircular strip $=\frac{\pi}{2}\big[(\text{r}+\text{dr})^2-\text{r}^2\big]$
$=\frac{\pi}{2}\big[\text{r}^2+\text{dr}^2+2\text{rdr}-\text{r}^2\big]$
$=\pi\text{r dr} $
$\therefore$ Mass of strip $\text{dm}=\frac{2\text{M}}{\pi\text{R}^2}\cdot\pi\text{r dr}$
$\text{dm}=\frac{2\text{M}}{\text{R}^2}\cdot\text{r dr}$
Let (x, y) are the co-ordinates of c.m. of this strip $(\text{x},\text{y})=\Big(0,\frac{2\text{r}}{\pi}\Big)$
$\text{x}=\text{x}_\text{cm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\text{x dm}=\int\limits^{\text{R}}_00\text{ dm}=0$
$\text{y}_\text{cm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\text{y dm}=\frac{1}{\text{M}}\int\limits^{\text{R}}_0\frac{2\text{r}}{\pi}\times\frac{\text{2}\text{M}}{\text{R}^2}\text{r dr}$
$=\frac{1}{\text{m}}\cdot\frac{4\text{M}}{\pi\text{R}^2}\int\limits^{\text{R}}_0\text{r}^2\text{ dr}=\frac{4\text{M}}{\pi\text{R}^2}\Big[\frac{\text{r}^3}{3}\Big]_0^{\text{R}}=\frac{4}{3\pi\text{R}^2}\cdot\text{R}^3$
$\text{y}_\text{cm}=\frac{4\text{R}}{3\pi}$
So, centre of mass of circular half disc $=\Big(0,\frac{4\text{R}}{3\pi}\Big)$
- Mass per unit area of quarter disc centre of mass of a uniform quarter disc,
Using symmetry,
For a half disc along y-axis c.m. will be at $\text{y}=\frac{4\text{R}}{3\pi}$
So, the c.m. of quarter disc $=\Big(\frac{4\text{R}}{3\pi},\frac{4\text{R}}{3\pi}\Big)$
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