Find the location of centre of mass of a uniform semicirular plate of radius `R` and mass `M`.

Explanation

In Fig. we have chosen `XY` axes with the origin at `O`, the centre of semicirular disc. By origin symmetry, the centre of mass would lie on Y-axis. Thus `x_(cm) = 0`, and
`y_(cm) = (1)/(M) int y dm` …(i)
If `sigma` mass per unit area of the disc, then
`sigma = (M)/(piR^(2)//2) = (2M)/(piR^(2))`
The semicirular plate can be supposed to be made up of a large number of semicircular strips, each of mass `dm` and radii ranging from `r = 0` to `r = R`. Consider one such semicirular strip of radius `r` and thickness `dr` as shown in Fig.
Surface area of this element `= pi r(dr)`
`:.` Mass of this element, `dm = sigma xx pi r dr`
`= ((2M)/(piR^(2))) pi r dr`
`dm = (2M)/(R^(2)) r dr`
The co-ordinates of `CM` of this element
`(x,y) = (0,(2r)/(pi))`
Therefore, for semicirular plate
`y_(cm) = (1)/(M)int_(0)^(R) y dm`
`= (1)/(M) int_(0)^(R) (2r)/(pi).((2M)/(R^(2)) r dr)`
`y_(cm) = (4)/(piR^(2)) [r^(3)/(3)]_(0)^(R ) = (4R)/(3pi)`
Hence centre of mass of uniform semicirular plate is
`(0,(4R)/(3pi))`