Engineering Mechanics

Chapter 6 : Centre of Gravity „„„„„ 89

We know that distance between centre of gravity of the body and bottom of hemisphere D,

32

11 2 2

12 32

25

383 4

2

33

⎛⎞⎛⎞⎛⎞ππ

⎜⎟⎜⎟⎜⎟×× + ×× +

+ ⎝⎠⎝⎠⎝⎠

==

+ ⎛⎞⎛ ⎞ππ

⎜⎟⎜ ⎟×+××

⎝⎠⎝ ⎠

rh

rrhr

vy v y

y

vv rrh

Now for stable equilibrium, we know that the centre of gravity of the body should preferably

be below the common face AB or maximum may coincide with it. Therefore substituting y equal to
r in the above equation,

32

32

25

383 4

2

3

⎛⎞⎛⎞⎛⎞ππ

⎜⎟⎜⎟⎜⎟×× + ×× +

⎝⎠⎝⎠⎝⎠

=

⎛⎞⎛ ⎞ππ

⎜⎟⎜ ⎟×+××

⎝⎠⎝ ⎠ 3

rh

rrhr

r

rrh

or

(^243)
33
rrh
⎛⎞⎛⎞ππ
⎜⎟⎜⎟×+×
⎝⎠⎝⎠
(^54322)
12 3 12
rrhrh
⎛⎞⎛ ⎞⎛ ⎞ππ π
=×+××+××⎜⎟⎜ ⎟⎜ ⎟
⎝⎠⎝ ⎠⎝ ⎠
Dividing both sides by π r^2 ,
25222 322
or
3312312 1212
rrhrrhh rh
+= ++ =
3 r^2 = h^2 or h = 1.732 r Ans.
Example 6.10. A right circular cylinder of 12 cm diameter is joined with a hemisphere of
the same diameter face to face. Find the greatest height of the cylinder, so that centre of gravity of the
composite section coincides with the plane of joining the two sections. The density of the material of
hemisphere is twice that the material of cylinder.
Solution. As the body is symmetrical about the vertical axis, therefore its centre of gravity
will lie on this axis. Now let the vertical axis cut the plane joining the two sections at O as shown in
Fig. 6.20. Therefore centre of gravity of the section is at a distance of 60 mm from P i.e., bottom
of the hemisphere.
Let h = Height of the cylinder in mm.
(i) Right circular cylinder
Weight (w 1 ) 1 2
4
dh
π
= ρ ×× ×
2
114 (120) hh3 600
π
=ρ ×× ×= πρ
and 1 60 60 0.5 mm
2
h
yh=+=+
(ii) Hemisphere
Weight (w 2 ) 2133
22
2 (60)
33
r
ππ
=ρ××=ρ×× ...(Q ρ 2 = 2 ρ 1 )
= 288 000 π ρ 1
and 2
5560300
37.5 mm
88 8
r
y
×
== = =
Fig. 6.20.