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

Explanation

Fig. shown upper half of unifrom circular disc of radius `R` in `XY` plane. Let `m `be total mass of half disc. Mass per unit area of half
disc `= (M)/(piR^(2)//2) = (2M)/(piR^(2))`.
The half disc can be supposed to be made up of a large number of semicirular rings, each of mass `dm` and radii ranging from `r = 0` to `r = R`. consider one such semicircular ring of radius `r` and thickness `dr` as shown. Surface area of this semicirular ring `= (pi r) dr`.
If `(x,y)` are co-ordinats of c.m of this element, then `(x,y) = (0,(2r)/(pi)), i.e., x = 0, y = (2r)/(pi)`
Let `x_(cm), y_(cm)` be co-ordinates of c.m of semicircular disc.
`:. x_(cm) = (1)/(M)int_(0)^(R) x dm = (1)/(M)int_(0)^(R) 0 dm = 0`
and `y_(cm) = (1)/(M) int_(0)^(R) y dm = (1)/(M)int_(0)^(R) ((2r)/(pi))(2M)/(R^(2)) r dr = (4)/(piR^(2)) [(r^(3))/(3)]_(0)^(R) = (4)/(piR^(2)) [(R^(3))/(3) - 0] = (4R)/(3pi)`
`:. c.m` of uniform half disc `= (0, 4R//3 pi)`.
(b) Centre of mass of a unifrom quarter disc
Here, mass per unit area of quarter disc `= (M)/(piR^(2)//4) = (4M)/(piR^(2))`
Proceeding as in (a) above, we can shown that c.m of uniform quarter disc is `((4R)/(3pi),(4R)/(3pi))`