Figure-4.11 shows a circular a disc of radius R from which a small disc is cut such that the periphery of the small disc touch the large disc and whose radius is `R//2`. Find the centre of mass of the remaining part of the disc.

Explanation

The disc will be like the one shown below

The axis of symmetry of the above body will pass through the centres of the complete disc and the hole.
The centre of mass of the remaining part of disc will tie on the aixs of symmetry and one can easily guess that it will be to the left of `O`. Let it be at `O"` at a distance `x` from the centre `O`.
Now , the initial complete disc can be regarded as a two component system, with one component the circular disc of radius `R//2` and the remaining disc. The complete disc can be shown like the one below.

Let `O` be the origin. So we can assume that there are two particles of masses `m_(1)` and `m_(2)` kept at `O"` and `O'` respectively. The centre of mass of a two particle, we can write
`x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))`
Now, `x_(cm)=0` (`:' O` is at origin)
The masses can be taken proportional to the area. the actual value of masses does not really matter.
So, `m_(1) prop pi[R^(2)-((R )/(2)^(2)]`
and `m_(2) prop pi[((R )/(2))^(2)]`
So, if `m_(2)=m`, `m_(1)=3m`. Substituting the values, we obtain
`0=(3mxx x_(1)+mxx(R )/(2))/(m_(1)+m_(2))`
`impliesx_(1)=-(R )/(6)`
The negative sign signifies that `O"` is located to left of `O`.