Find the centre of mass of a uniform :

$(a)$ half-disc,$(b)$ quarter-disc. 

$\mathrm{M}=$ mass of half disc $\mathrm{R}=$ radius of half disc

Mass per unit area of half disc

$\mu=\frac{\mathrm{M}}{\frac{1}{2} \pi \mathrm{R}^{2}}=\frac{2 \mathrm{M}}{\pi \mathrm{R}^{2}}$

 

$(a)$ Suppose, half disc consists of many semicircular rings of mass $d m$ and thickness $d r$ from $r$ $=0$ to $r=\mathrm{R}$.

Area of semicircular ring of radius,

$r=d a=\frac{1}{2} 2 \pi r \times d r=\pi r d r$

$\therefore$ Mass of this ring, $d m=d a \times \mu$

$d m=\pi r d r \times \frac{2 M}{\pi R^{2}}$

$d m=\frac{2 M}{R^{2}} r d r$

If $(x, y)$ are co-ordinates of centre of mass of this element, then $(x, y)=\left(0, \frac{2 r}{\pi}\right)$

Therefore, $x=0$ and $y=\frac{2 r}{\pi}$

Let $x_{\mathrm{CM}}$ and $y_{\mathrm{CM}}$ be the co-ordinates of the centre of mass of the semicircular disc.

Then, $x_{\mathrm{CM}}=\frac{1}{\mathrm{M}} \int_{0}^{\mathrm{R}} x d m=\frac{1}{\mathrm{M}} \int_{0}^{\mathrm{R}} 0 d m=0$

$\therefore y_{\mathrm{CM}}=\frac{1}{\mathrm{M}} \int_{0}^{\mathrm{R}} y d m=\frac{1}{\mathrm{M}} \int_{0}^{\mathrm{R}} \frac{2 r}{\pi} \times\left(\frac{2 \mathrm{M}}{\mathrm{R}^{2}} r d r\right)$