(^90) A Textbook of Engineering Mechanics

We know that distance between centre of gravity of the combined body from Py(),

11 2 2

12

60

wy w y

ww

- =

# 11

11

3 600 (60 0.5 ) (288 000 37.5)

3 600 288 000

hh

h

πρ + + πρ ×

# πρ + πρ

216 000 1800 2 10 800 000

3 600 288 000

hh

h

++

# 216 000 h + 17 280 000 = 216 000 h + 1 800 h^2 + 10 800 000

1 800 h^2 = 17 280 000 – 10 800 000 = 6 480 000

6 480 000 3 600 60 mm

1 800

h=== Ans.

Example 6.11. Find the centre of gravity of a segment of height 30 mm of a sphere of radius

60 mm.

Solution. Let O be the centre of the given sphere

and ABC is the segment of this sphere as shown in

Fig. 6.21

As the section is symmetrical about X-X axis,

therefore its centre of gravity lies on this axis.

Let O be the reference point.

We know that centre of gravity of the segment of

sphere

3(2 – )^22 3(2 60 – 30)

4(3 – ) 4(3 60 – 30)

rh

x

rh

×

# ×

3(90)^2

4150

×

× = 40.5 mm. Ans.

EXERCISE 6.2

- A hemisphere of 60 mm diameter is placed on the top of the cylinder having 60 mm

diameter. Find the common centre of gravity of the body from the base of cylinder, if its

height is 100 mm. [Ans. 60.2 mm] - A solid consists of a cylinder and a hemisphere of equal radius fixed base to base. Find the

ratio of the radius to the height of the cylinder, so that the solid has its centre of gravity at

the common face. [Ans. 2:1]

Hint. For stable equilibrium, the centre of the body should be below the common face or

maximum lie on it. So take the centre of gravity of the body at a distance (a) from the

bottom of the hemisphere. - A body consisting of a cone and hemisphere of radius (r) on the same base rests on a table,

the hemisphere being in contact with the table. Find the greatest height of the cone, so that

the combined solid may be in stable equilibrium. [Ans. 1.732 r] - Find the centre of gravity of a segment of height 77 mm of a sphere of radius 150 mm.

[Ans. 100 mm]

`Fig. 6.21.`