(^88) „„„„„ A Textbook of Engineering Mechanics
We know that distance between centre of gravity of the section and base of the cylinder,
22
11 2 2
22
12
(30 15) (40 60) 2850
mm
30 40 70
vy v y rr
y
vv rr

• π× + π×
  == =


• π+ π
  = 40.7 mm Ans.
  Example 6.8. A body consists of a right circular solid cone of height 40 mm and radius
  30 mm placed on a solid hemisphere of radius 30 mm of the same material. Find the position of
  centre of gravity of the body.
  Solution. As the body is symmetrical about Y-Y axis, therefore its centre of gravity will lie on
  this axis as shown in Fig. 6.18. Let bottom of the hemisphere (D) be the point of reference.
  (i) Hemisphere
  333
  1
  22
  (30) mm
  33
  vr
  ππ
  =×=
  = 18 000 π mm^3
  and 1
  5530
  18.75 mm
  88
  r
  y
  ×
  == =
  (ii) Right circular cone
  223
  vrh 2 33 (30) 40 mm
  ππ
  =××=× ×
  = 12 000 π mm^3
  and 2
  40
  30 40 mm
  4
  y =+ =
  We know that distance between centre of gravity of the body and bottom of hemisphere D,
  11 2 2
  12
  (18 000 18.75) (12 000 40)
  mm
  18 000 12 000
  vy v y
  y
  vv


• 
  

  π× + π×

  
  

  +π+π
  = 27.3 mm Ans.
  Example 6.9. A body consisting of a cone and hemisphere of radius r fixed 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 body may stand upright.
  Solution. As the body is symmetrical about Y-Y axis, therefore its centre of gravity will lie on
  this axis as shown in Fig. 6.19. Now consider two parts of the body viz., hemisphere and cone. Let
  bottom of the hemisphere (D) be the axis of reference.
  (i) Hemisphere
  3
  1
  2
  3
  vr
  π
  =×
  and 1
  5
  8
  r
  y =
  (ii) Cone
  2
  vrh (^23)
  π
  =××
  and 2
  4
  h
  yr=+
  Fig. 6.18.
  Fig. 6.19.