Engineering Mechanics

Chapter 6 : Centre of Gravity „„„„„ 87

6. A figure consists of a rectangle having one of its sides twice the other, with an equilateral
   triangle described on the larger side. Show that centre of gravity of the section lies on the
   line joining the rectangle and triangle.


7. A plane lamina of radius 100 mm as shown in fig 6.16 given below:

Fig. 6.16.

Find the centre of gravity of lamina from the point O.

[Ans. 65 mm]

6.10. CENTRE OF GRAVITY OF SOLID BODIES

The centre of gravity of solid bodies (such as hemispheres, cylinders, right circular solid cones

etc.) is found out in the same way as that of plane figures. The only difference, between the plane
figures and solid bodies, is that in the case of solid bodies, we calculate volumes instead of areas. The
volumes of few solid bodies are given below :

1. Volume of cylinder = π × r^2 × h


2. Volume of hemisphere^3
   2
   3

r

π

=×

3. Volume of right circular solid cone^2
   3

rh

π

=× ×

where r = Radius of the body, and
h = Height of the body.
Note. Sometimes the densities of the two solids are different. In such a case, we calculate the
weights instead of volumes and the centre of gravity of the body is found out as usual.

Example 6.7. A solid body formed by joining the base of a right circular cone of height H to
the equal base of a right circular cylinder of height h. Calculate the distance of the centre of mass of
the solid from its plane face, when H = 120 mm and h = 30 mm.
Solution. As the body is symmetrical about the vertical axis, therefore its centre of gravity
will lie on this axis as shown in Fig. 6.17. Let r be the radius of the cylinder base in cm. Now let base
of the cylinder be the axis of reference.
(i) Cylinder
v 1 = π × r^2 × 30 = 30 π r^2 mm^3

and 1

30

15 mm

2

y ==

(ii) Right circular cone

223

vrhr 2 33 120 mm

ππ

=××=× ×

= 40 πr^2 mm^3

and 2

120

30 60 mm

4

y =+ = Fig. 6.17.