From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
From a big uniform disc of radius R with centre O, a smaller circular hole of radius R/2 with its centre O1 (where OO1 = R/2) is cut out.
Let the centre of gravity of remaining flat body be C and OC= d.
If Ω is mass per unit area, then mass of the whole disk of Radius R is,
M1 = πR2Ω
and mass cut out part
M2 = π(R/2)2 Ω = πR2Ω/4 = M1/4
Let centre of mass of original disc be origin(O)
It can be considered as centre of mass of system consisting of smaller carved out disc and remaining portion.
Thus, we have,
M1 × (0) = (M1 -M2) × d + M2 × (OO1)
0 = (M1 – M2) × d + M2 × (- R/2)
(Since, M2 = M1/4)
d = (M2R/2)/(M1-M2)
= (M1/4) × (R/2) / (M1 - (M1/4))
= R/6
∴ Cis at a distance R/6 from the centre of disc on diametrically opposite side of the centre of hole.