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.