Engineering Mechanics

(^94) „„„„„ A Textbook of Engineering Mechanics
626
–––
566 563
rr rr
rr
⎡⎤⎛⎞⎡ ⎤
=+=⎢⎥⎜⎟⎢⎥
⎣⎦⎝⎠⎣ ⎦ππ
65 2
––
56 3 5
rr
rr
⎛⎞
==⎜⎟
⎝⎠ππ^ Ans.
Example 6.16. A solid consists of a right circular cylinder and a hemisphere with a cone cut
out from the cylinder as shown in Fig. 6.27.
Fig. 6.27
Find the centre of gravity of the body.
Solution. As the solid is symmetrical about horizontal axis, therefore its centre of gravity lie
on this axis.
Let the left edge of the hemispherical portion (E) be the axis of reference.
(i) Hemisphere ADE
33 3
1
22
(60) 144 000 mm
33
vr
ππ
=×=× = π
and 1
5560
37.5 mm
88
r
x
×
== =
(ii) Right circular cylinder ABCD
v 2 = π × r^2 × h = π × (60)^2 × 150 = 540 000 π mm^3
and 2
150
60 135 mm
2
x =+ =
(iii) Cone BCF
22 3
vrh 3 33 (60) 150 180 000 mm
ππ
=××=× × = π
and 3
3
60 150 172.5 mm
4
x =+ ×=
We know that distance between centre of gravity of the solid and left edge of the hemi-
sphere (E),
11 2 2 3 3
123

vx v x vx

x

vv v

+

=

+

(144 000 37.5) (540 000 135) – (180 000 172.5)

144 000 540 000 – 180 000

π× + π× π×

=

π+ π π

= 93.75 mm Ans.