Chapter 6 : Centre of Gravity „„„„„ 93

3

40

40 60 mm

2

x =+ =

and 3

4420

8.5 mm

33

r

y

×

== =

ππ

We know that distance between centre of gravity of the area and left face of trapezium,

11 2 2 3 3

123

ax a x a x

x

aa a

+

=

+

(2400 40) (1200 53.3) – (628.3 60)

2400 1200 – 628.3

×+ × ×

=

+

= 41.1 mm Ans.

Similarly, distance between centre of gravity of the area and base of the trapezium,

11 2 2 3 3

123

ay a y ay

y

aa a

+

=

+^

(2400 15) (1200 40) – (628.3 8.5)

2400 1200 – 628.3

×+ × ×

• = 26.5 mm Ans.
  Example 6.15. A circular sector of angle 60° is cut from the circle of radius r as shown in
  Fig. 6.26 :

Fig. 6.26.
Determine the centre of gravity of the remainder.
Solution: As the section is symmetrical about X-X axis, therefore its centre of gravity will lie
on this axis.
Let C be the reference point.
(i) Main circle
a 1 = π r^2
and x 1 = r
(ii) Cut out sector
22 2
2

60

360 360 6

rr r

a

πθ π × ° π

== =

°°

and 2

2 r

xr=+

π

We know that distance between the centre of gravity of the section and C

2

2

11 2 2

2

12 2

2

()–

-^6

• 
  
  
  
  • 6
  
  
  
  

rr

rr r

ax a x

x

aa r

r

⎡⎤π ⎛⎞

π× ⎢⎥× +⎜⎟

⎣⎦⎝⎠π

==

π

π

2

2

12 1 2

––

66

1 1

1– 1–

6 6

rr

rr r r r

r

⎡⎤⎡⎤⎛⎞ ⎛⎞

π+⎢⎥⎢⎥⎜⎟ ⎜⎟×+

⎣⎦⎣⎦⎝⎠ ⎝⎠ππ

==

π ⎛⎞

⎜⎟

⎝⎠