This document discusses the center of gravity, center of mass, and centroid of rigid bodies. It defines these terms and presents methods to calculate them using integrals of differential elements. Examples are provided to demonstrate calculating the centroid for areas and lines using appropriate coordinate systems and differential elements. Centroids are found by taking moments of these elements and the document outlines the general procedure to perform these calculations.

1. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Center Mass
A rigid body is composed of an infinite
number of particles
Consider arbitrary particle
having a weight of dW
x= ∫ ; y = ∫ ;z = ∫z
~dW
x ~dW
y ~dW
∫dW ∫dW ∫dW

2. 9.2 Center of Gravity and Center
of Mass and Centroid for a Body
Center Mass
γ represents the specific weight and dW = γdV
∫x
~γdV
∫y
~γdV
∫z
~γdV
x =V ;y =V ;z =V
∫γdV ∫γdV ∫γdV
V V V

3. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Center of Mass
Density ρ, or mass per unit volume, is related
to γ by γ = ρg, where g = acceleration due to
gravity
Substitute this relationship into this equation
to determine the body’s center of mass
∫x
~γdV
∫y
~γdV
∫ ~γdV
z
x =V ;y =V ;z = V
∫ γdV ∫ γdV ∫ γdV
V V V

4. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Defines the geometric center of object
Its location can be determined from
equations used to determine the body’s
center of gravity or center of mass
If the material composing a body is uniform
or homogenous, the density or specific
weight will be constant throughout the body
The following formulas are independent of
the body’s weight and depend on the body’s
geometry

5. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Volume
Consider an object subdivided into volume
elements dV, for location of the centroid,
∫x
~dV
∫y
~dV
∫z
~dV
x =V ;y =V ;z =V
∫dV ∫dV ∫dV
V V V

6. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Area
For centroid for surface area of an object,
such as plate and shell, subdivide the area
into differential elements dA
∫x
~dA
∫y
~dA
∫z
~dA
x=A ;y = A ;z = A
∫dA ∫dA ∫dA
A A A

7. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Centroid
Line
If the geometry of the object such as a thin rod
or wire, takes the form of a line, the balance of
moments of differential elements dL about each
of the coordinate system yields
∫x
~dL
∫y
~dL
∫z
~dL
x=L ;y = L ;z = L
∫dL ∫dL ∫dL
L L L

8. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Line
Choose a coordinate system that simplifies as
much as possible the equation used to
describe the object’s boundary
Example
Polar coordinates are appropriate for area
with circular boundaries

9. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Symmetry
The centroids of some shapes may be partially or
completely specified by using conditions of
symmetry
In cases where the shape has an axis of
symmetry, the centroid of the shape must lie
along the line
Example
Centroid C must lie along the
y axis since for every element
length dL, it lies in the middle

10. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Symmetry
For total moment of all the elements about
the axis of symmetry will therefore be
cancelled ~dL = 0 x = 0
∫x
In cases where a shape has 2 or 3 axes of
symmetry, the centroid lies at the
intersection of these axes

11. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Differential Element
Select an appropriate coordinate system, specify
the coordinate axes, and choose an differential
element for integration
For lines, the element dL is represented as a
differential line segment
For areas, the element dA is generally a
rectangular having a finite length and differential
width

12. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Differential Element
For volumes, the element dV is either a
circular disk having a finite radius and
differential thickness, or a shell having a
finite length and radius and a differential
thickness
Locate the element at an arbitrary point (x,
y, z) on the curve that defines the shape

13. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Size and Moment Arms
Express the length dL, area dA or volume dV
of the element in terms of the curve used to
define the geometric shape
Determine the coordinates or moment arms
for the centroid of the center of gravity of
the element

14. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Procedure for Analysis
Integrations
Substitute the formations and dL, dA and dV into
the appropriate equations and perform
integrations
Express the function in the integrand and in
terms of the same variable as the differential
thickness of the element
The limits of integrals are defined from the two
extreme locations of the element’s differential
thickness so that entire area is covered during
integration

15. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.1
Locate the centroid of
the rod bent into the
shape of a parabolic arc.

16. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
Located on the curve at the arbitrary point (x, y)
Area and Moment Arms
For differential length of the element dL
2
 dx 
dL = (dx )2 + (dy )2 =   + 1 dy
 dy 
Since x = y2 and then dx/dy = 2y
dL = (2 y )2 + 1 dy
The centroid is located at
~ = x, ~ = y
x y

17. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫ ~dL ∫1 x 4 y 2 + 1 dy ∫1 y 2 4 y 2 + 1 dy
x
x=L = 01 = 01
∫ dL ∫0 4 y 2 + 1 dy ∫0 4 y 2 + 1 dy
L
0.6063
= = 0.410m
1.479
∫ ~dL ∫1 y 4 y 2 + 1 dy
y
y=L = 01
∫ dL ∫ 4 y 2 + 1 dy
0
L
0.8484
= = 0.574m
1.479

18. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.2
Locate the centroid of
the circular wire
segment.

19. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
A differential circular arc is selected
This element intersects the curve at (R, θ)
Length and Moment Arms
For differential length of the element
dL = Rdθ
For centroid,
~ = R cosθ
x ~ = R sin θ
y

20. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫ ~dL ∫π / 2 (R cosθ )R dθ
x
R2 ∫
π /2
cosθ dθ
x= L = 0
π /2
= 0
π /2
∫ dL ∫0 R dθ R∫
0
dθ
L
2R
=
π
∫ ~dL ∫π / 2 (R sin θ )R dθ R 2 ∫π / 2 sin θ dθ
y
y=L = 0 π /2 = 0
π /2
∫ dL
∫0 R dθ R∫
0
dθ
L
2R
=
π

21. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.3
Determine the distance
from the x axis to the
centroid of the area
of the triangle

22. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Differential element
Consider a rectangular element having
thickness dy which intersects the boundary
at (x, y)
Length and Moment Arms
For area of the element
b
dA = xdy = (h − y )dy
h
Centroid is located y distance from the x
axis ~=y
y

23. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
~dA
∫y b
∫0 h (h − y ) dy
h
y
y= A =
∫ dA ∫0h b (h − y ) dy
A h
1 2
bh
6 h
= =
1
bh 3
2

24. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.4
Locate the centroid for
the area of a quarter
circle.

25. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
Use polar coordinates for circular boundary
Triangular element intersects at point (R,θ)
Length and Moment Arms
For area of the element
1 R2
dA = ( R)( R cosθ ) = dθ
2 2
Centroid is located y distance from the x axis
~ = 2 R cosθ
x ~ = 2 R sin θ
y
3 3

26. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
π / 2 2 2
∫ ~dA
x  R dθ  2 R  π / 2 cosθ dθ
∫0  R cosθ  3  2  3  ∫0
 
x= A = = π /2
∫ dA π / 2 R2
∫0 dθ ∫0 dθ
A 2
4R
=
3π
π / 2 2 2 π /2
∫ ~dA
y
∫0  R sin θ 

R
dθ  R  ∫ sin θ dθ

2

y= A = 3  2 = 3  0
π /2
∫ dA π /2 R 2
A ∫0 2 dθ ∫0 dθ
4R
=
3π

27. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
Circular arc element
having thickness of dr
Element intersects the
axes at point (r,0) and
(r, π/2)

28. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
dA= (2πr / 4)dr
Centroid is located y distance from the x axis
~ = 2r /π ~ = 2r /π
x y

29. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫ ~dA ∫0R 2r  2πr  dr ∫ R r 2 dr
x
 
π 4  = 0
x=A = R 2πr π R
∫ dA
∫0 4 dr 2 ∫0 r dr
A
4R
=
3π
∫ ~dA ∫0R 2r  2πr  dr ∫ R r 2 dr
y
 
π 4  = 0
y= A = R 2πr π R
∫ dA ∫0 4 dr
2 ∫0 r dr
A
4R
=
3π

30. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.5
Locate the centroid of
the area.

31. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
Differential element of thickness dx
Element intersects curve at point (x, y), height y
Area and Moment Arms
For area of the element
dA = ydx
For centroid
~=x
x ~ = y/2
y

32. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫x
~dA 1 1 3
∫0 xy dx = ∫ x dx
x= A = 1 0
∫ dA
1 2
A
∫0 y dx ∫ 0
x dx
0.250
= = 0.75m
0.333
∫ ~dA ∫1( y / 2) y dx ∫1( x 2 / 2) x 2 dx
y
y=A = 0 1 = 0 1
∫ dA ∫ y dx
0 ∫ x 2 dx
0
A
0.100
= = 0.3m
0.333

33. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
Differential element
of thickness dy
Element intersects
curve at point (x, y)
Length = (1 – x)

34. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
dA= (1− x) dy
Centroid is located y distance from the x axis
~ = x + 1− x  = 1+ x
x  
 2  2
~= y
y

35. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫ ~dA ∫1[(1 − x ) / 2](1 − x ) dx
x 1 1
∫0 (1 − y ) dy
x= A = 0 = 21
∫ dA ∫ (1 − y )dy
1
A
∫0 (1 − x ) dx 0
0.250
= = 0.75m
0.333
∫ ~dA ∫1 y(1 − x ) dx ∫1 y − y3 / 2 dy
y
( )
y=A = 01 = 01
∫ dA ∫ (1 − x ) dx ∫ (1 − y )dy
0 0
A
0.100
= = 0.3m
0.333

36. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.6
Locate the centroid of
the shaded are bounded
by the two curves
y=x
and y = x2.

37. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
Differential element of thickness dx
Intersects curve at point (x1, y1) and (x2, y2), height y
Area and Moment Arms
For area of the element
dA = ( y2 − y1 ) dx
For centroid
~=x
x

38. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫ ~dA ∫1x( y2 − y1)dx ∫1x(x − x2)dx
x
x=A = 01 = 01
∫dA ∫0( y2 − y1) dx ∫0(x − x2)dx
A
1
= 12 = 0.5m
1
6

39. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
Differential element
of thickness dy
Element intersects
curve at point (x1, y1)
and (x2, y2)
Length = (x1 – x2)

40. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
dA= ( x1 − x2 ) dy
Centroid is located y distance from the x axis
~ = x +  x1 − x2  = x1 + x2
x 2  
 2  2

41. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
∫0 [(x1 + x2 ) / 2](x1 − x2 ) dx = ∫0 [( y + y )/ 2]( y − y )dy
~dA
∫x 1 1
x= A
=
∫ dA (x1 − x2 ) dx ∫0 ( y − y ) dx
1 1
A
∫0
1 1
2 ∫0
(
y − y 2 dy) 1
= 1 = 12 = 0.5m
(
∫0 y − y dx 6) 1

42. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.7
Locate the centroid for the
paraboloid of revolution,
which is generated by
revolving the shaded area
about the y axis.

43. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 1
Differential element
Element in the shape of a thin disk, thickness dy,
radius z
dA is always perpendicular to the axis of revolution
Intersects at point (0, y, z)
Area and Moment Arms
For volume of the element
( )
dV = πz 2 dy

44. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
For centroid
~=y
y
Integrations
~dV
∫y ∫ ( )
100
y πz dy
2
100
100π ∫ y 2 dy
y=V = 0
= 0
∫ dV ∫ (πz ) dy
100 100
2
100π ∫ y dy
0 0
V
= 66.7 mm

45. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Method 2
Differential element
Volume element in the form of thin
cylindrical shell, thickness of dz
dA is taken parallel to the axis of
revolution
Element intersects the
axes at point (0, y, z) and
radius r = z

46. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Area and Moment Arms
For area of the element
2πrdA = 2πz (100 − y )dz
Centroid is located y distance from the x axis
~ = y + (100 − y ) / 2 = (100 + y ) / 2
y

47. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Integrations
~dV
∫y ∫ [(100 + y ) / 2]2πz (100 − y ) dz
100
y= V
= 0
∫ dV ∫ 2πz (100 − y ) dz
100
0
V
π ∫ z (104 − 10 − 4 z 4 ) dz
100
= 0
2π ∫
0
100
(
z 100 − 10 − 2 z 2 dz)
= 66.7 m

48. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Example 9.8
Determine the location of
the center of mass of the
cylinder if its density
varies directly with its
distance from the base
ρ = 200z kg/m3.

49. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
For reasons of material symmetry
x = y=0
Differential element
Disk element of radius 0.5m and thickness dz since
density is constant for given value of z
Located along z axis at point (0, 0, z)
Area and Moment Arms
For volume of the element
dV = π (0.5)2 dz

50. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
For centroid
~=z
z
Integrations
∫z
~ρdV
∫0
1
z (200 z )π (0.5)2 dz
~=V
z = 1
∫ ρdV ∫0 (200 z )π (0.5)2 dz
V
1 2
=
∫
0
z dz
= 0.667m
1
∫0 zdz

51. 9.2 Center of Gravity and Center of
Mass and Centroid for a Body
Solution
Not possible to use a shell
element for integration
since the density of the
material composing the
shell would vary along the
shell’s height and hence the
location of the element
cannot be specified