Chapter 6 : Centre of Gravity 91

6.11. CENTRE OF GRAVITY OF SECTIONS WITH CUT OUT HOLES

The centre of gravity of such a section is found out by considering the main section, first as a

complete one, and then deducting the area of the cut out hole i.e., by taking the area of the cut out hole

as negative. Now substituting a 2 (i.e., the area of the cut out hole) as negative, in the general equation

for the centre of gravity, we get

`11 2 2`

12

`ax a x`

x

aa

`= and^11 2 2`

12

`ay a y`

y

aa

`=`

Note. In case of circle the section will be symmeterical along the line joining the centres of the

bigger and the cut out circle.

Example 6.12. A square hole is punched out of circular lamina, the digonal of the square

being the radius of the circle as shown in Fig.6.22. Find the centre of gravity of the remainder, if r

is the radius of the circle.

Solution. As the section is symmetrical about X-X axis, therefore its centre of gravity will lie

on this axis. Let A be the point of reference.

`(i) Main circle`

a 1 = π r^2

and x 1 = r

(ii) Cut out square

2

2 2 0.5

`rr`

ar

`×`

==

`and 2 1.5`

2

`r`

xr=+ = r

We know that distance between centre of gravity of the section and A,

22

11 2 2

22

12

- ()–(0.51.5)
- –0.5

`ax a x rr r r`

x

aa rr

`π× ×`

==

π

`3`

2

`( – 0.75) ( – 0.75)`

(–0.5) –0.5

`rr`

r

`ππ`

==

π π^ Ans.

Example 6.13. A semicircle of 90 mm radius is cut out from a trapezium as shown in Fig 6.23

Fig. 6.23.

Find the position of the centre of gravity of the figure.

Solution. As the section is symmetrical about Y-Y axis, therefore its centre of gravity will lie

on this axis. Now consider two portions of the figure viz., trapezium ABCD and semicircle EFH.

Let base of the trapezium AB be the axis of reference.

(i) Trapezium ABCD

`1 2`

`200 300`

120 30 000 mm

2

`a`

+

=× =

`Fig. 6.22.`