Find the centre of mass of a uniform :
(a) half-disc, (b) quarter-disc.

Explanation

M = mass of half disc
R = radius of half disc
Mass per unit area of half disc
`mu=(M)/((1)/(2)piR^(2))=(2M)/(piR^(2))`

(a) Suppose, half disc consists of many semicircular rings of mass dm and thickness dr from `r=0` to `r=R`.
Area of semicircular ring of radius,
`r=da=(1)/(2)2pirxxdr=pirdr`
`therefore` Mass of this ring, `dm=daxxmu`
`dm=pirdrxx(2M)/(piR^(2))`
`dm=(2M)/(R^(2))rdr`
If `(x,y)` are co-ordinates of centre of mass of this element, then `(x,y)=(0,(2r)/(pi))`
Therefore, `x=0 and y=(2r)/(pi)`
Let `x_(CM)andy_(CM)` be the co-ordinates of the centr of mass of the semicircular disc.
Then, `x_(CM)=(1)/(M)underset(0)overset(R)intxdm=(1)/(M)underset(0)overset(R)int0dm=0`
`therefore y_(CM)=(1)/(M)underset(0)overset(R)intydm=(1)/(M)underset(0)overset(R)int(2r)/(pi)xx((2M)/(R^(2))rdr)`
`(4)/(piR^(2))underset(0)overset(R)intr^(2)dr=(4)/(piR^(2))[(r^(3))/(3)]_(0)^(R)`
`=(4)/(piR^(2))xx((R^(3))/(3)-0)=(4R)/(3pi)`
`therefore` Centre of mass of the semicircular disc,
`(0,(4R)/(3pi))`
(b) Centre of mass of a uniform quarter disc.
Mass per unit area of the quarter disc,
`=(M)/((piR^(2))/(4))=(4M)/(piR^(2))`

Using symmetry.
For a half-disc along y-axis centre of mass will be at `x=(4R)/(3pi)`
For a half-disc along x-axis centre of mass will be at `x-(4R)/(3pi)`
Hence, for the quarter disc centre of mass,
`=((4R)/(3pi),(4R)/(3pi))`