Question

From a uniform disc of radius R , a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from centre of original disc . Locate the centre of mass of the resulting flat body

Answer 1

Answer by:- sanyam

Let the mass of the disc = M 
Therefore, the mass of the removed part of disc, m = (M/R^2)*(R/2)^2 = M/4 
Now the center of gravity of the resulting flat body, 
R= [M*0 – (M/4)*(R/2)]/(M-M/4) 
= -(MR/8)/(3M/4) 
= -R/6 
Negative sign shows that the center of gravity lies at opposite direction of the original COM at a distance R/6

Answer 2

Let us consider a disc of mass M and another disc which is cut out from these disc of mass -M/4.

So, m1=M=sigma *pie*R^2

m2=-M/4=sigma*pie* (R/2)^2

So, X=m1*x1+m2*x2/m1+m2

X=sigma*pie*R^2*0+(-M/4)*(R/2)/M+(-M/4)

X=-MR/8*4/3M = -R/6

I HOPE THESE IS HELP YOU!!!