6161103 9.2 center of gravity and center of mass and centroid for a body

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 � Consider arbitrary particle

having a weight of dW

∫∫

∫∫

∫∫ ===

dW

dWzz

dW

dWyy

dW

dWxx

~;

~;

~

9.2 Center of Gravity and Center of Mass and Centroid for a Body9.2 Center of Gravity and Center of Mass and Centroid for a Body

Center Mass

� γ represents the specific weight and dW = γdV

∫∫∫ dVzdVydVx γγγ ~~~

∫

∫

∫

∫

∫

∫===

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

xγ

γ

γ

γ

γ

γ ~

;

~

;

~



Center of Mass

� Density ρ, or mass per unit volume, is related to γ by γ = ρg, where g = acceleration due to gravitygravity

� Substitute this relationship into this equation to determine the body’s center of mass

∫

∫

∫

∫

∫

∫===

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

xγ

γ

γ

γ

γ

γ ~

;

~

;

~



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 masscenter 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



Centroid

Volume

� Consider an object subdivided into volume elements dV, for location of the centroid, elements dV, for location of the centroid,

∫

∫

∫

∫

∫

∫===

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

x

~

;

~

;

~



Centroid

Area

� For centroid for surface area of an object, such as plate and shell, subdivide the area such as plate and shell, subdivide the area into differential elements dA

∫

∫

∫

∫

∫

∫===

A

A

A

A

A

A

dA

dAz

zdA

dAy

ydA

dAx

x

~

;

~

;

~



Centroid

Line

� If the geometry of the object such as a thin rod or wire, takes the form of a line, the balance of or wire, takes the form of a line, the balance of moments of differential elements dL about each of the coordinate system yields

∫

∫

∫

∫

∫

∫===

L

L

L

L

L

L

dL

dLz

zdL

dLy

ydL

dLx

x

~

;

~

;

~



Line� Choose a coordinate system that simplifies as

much as possible the equation used to describe the object’s boundary

ExampleExample� Polar coordinates are appropriate for area

with circular boundaries



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 � 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

Symmetry

� For total moment of all the elements about the axis of symmetry will therefore be cancelled



00~ ==∫ xdLx

� In cases where a shape has 2 or 3 axes of symmetry, the centroid lies at the intersection of these axes

00 ==∫ xdLx



Procedure for Analysis

Differential Element

� Select an appropriate coordinate system, specify the coordinate axes, and choose an differential 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




Differential Element

� For volumes, the element dV is either a circular disk having a finite radius and 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




Size and Moment Arms

� Express the length dL, area dA or volume dV of the element in terms of the curve used to 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



Procedure for AnalysisIntegrations� Substitute the formations and dL, dA and dV into

the appropriate equations and perform integrationsintegrations

� 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



Example 9.1

Locate the centroid of

the rod bent into the

shape of a parabolic arc.shape of a parabolic arc.



Solution

Differential element

� Located on the curve at the arbitrary point (x, y)

Area and Moment ArmsArea and Moment Arms

� For differential length of the element dL

� Since x = y2 and then dx/dy = 2y

� The centroid is located at

( ) ( )

( )

yyxx

dyydL

dydy

dxdydxdL

==

+=

+

=+=

~,~

12

1

2

222



Solution

Integrations

dyyydyyxdLx

x L1414

~

1

1

022

1

1

02 +

=+

==∫∫

∫

∫

m

dyy

dyyy

dL

dLy

y

m

dyydyydLx

L

L

L

574.0479.18484.0

14

14~

410.0479.16063.0

1414

1

02

1

02

1

021

02

==

+

+==

==

+=

+==

∫∫

∫

∫

∫∫∫



Example 9.2


the circular wire

segment.segment.



Solution


� A differential circular arc is selected

� This element intersects the curve at (R, θ)� This element intersects the curve at (R, θ)

Length and Moment Arms

� For differential length of the element

� For centroid,

θθ

θ

sin~cos~ RyRx

RddL

==

=



Solution

Integrations

( ) θθθθππ

dRdRRdLx

x Lcoscos

~ 2/

022/

0 ===∫∫∫

( )

π

θ

θθ

θ

θθ

π

θθ

π

π

π

π

ππ

R

dR

dR

dR

dRR

dL

dLy

y

R

dRdRdLx

L

L

L

L

2

sinsin~

2

2/

0

2/

02

2/

0

2/

0

2/

0

02/

0

0

=

===

=

===

∫∫

∫∫

∫

∫

∫∫

∫∫

∫



Example 9.3

Determine the distance

from the x axis to the

centroid of the area centroid of the area

of the triangle



Solution


� Consider a rectangular element having thickness dy which intersects the boundary thickness dy which intersects the boundary at (x, y)


� For area of the element

� Centroid is located y distance from the x axis

( )

yy

dyyhhb

xdydA

=

−==

~



Solution

Integrations

( )~0

dyyhh

bydAy

y

h

A−

==∫∫

( )

32161 2

0

0

h

bh

bh

dyyhh

bh

dAy

h

A

A

==

−==

∫

∫∫

∫



Example 9.4

Locate the centroid for

the area of a quarter

circle.circle.



Solution

Method 1


� Use polar coordinates for circular boundary� Use polar coordinates for circular boundary

� Triangular element intersects at point (R,θ)




θθ

θθ

sin32~cos

32~

2)cos)((

21 2

RyRx

dR

RRdA

==

==



Solution

Integrations

θθθθ

π

ππcos

32

2cos

32~

2/

2/

02

2/

0

2dRd

RRdAx

x A

=

==∫∫

∫

∫

π

θ

θθ

θ

θθ

π

θθ

π

π

π

π

ππ

34

sin32

2

2sin

32~

34

2

2/

0

2/

0

2/

0

2

2/

0

2

2/

02/

0

2

R

d

dR

dR

dR

R

dA

dAy

y

R

ddRdA

x

A

A

A

=

=

==

=

===

∫

∫

∫

∫∫

∫

∫∫∫



Solution

Method 2


� Circular arc element � Circular arc element having thickness of dr

� Element intersects the

axes at point (r,0) and

(r, π/2)



Solution

Area and Moment Arms



( )

ππ

π

/2~/2~

4/2

ryrx

drrdA

==

=



Solution

Integrations

ππ

ππ

24

22~

02

0 drr

r

drrr

dA

dAx

xR

R

R

R

A =

==

∫∫

∫

∫∫

∫

π

ππ

ππ

π

ππ

34

242

422~

34

242

0

02

0

0

00

R

drr

drr

drr

drrr

dA

dAy

y

R

drrdrrdA

x

R

R

R

R

A

A

RR

A

=

=

==

=

===

∫∫

∫

∫∫

∫

∫∫∫



Example 9.5


the area.



Solution

Method 1


� Differential element of thickness dx� Differential element of thickness dx

� Element intersects curve at point (x, y), height y



� For centroid

2/~~ yyxx

ydxdA

==

=



Solution

Integrations

dxxdxxydAx

x A

~ 1

031

0 ===∫∫∫

m

dxx

dxxx

dxy

dxyy

dA

dAy

y

m

dxxdxydAx

A

A

A

A

3.0333.0100.0

)2/()2/(~

75.0333.0250.0

1

02

1

022

1

0

1

0

1

02

01

0

0

==

===

==

===

∫∫

∫∫

∫

∫

∫∫

∫∫

∫



Solution

Method 2


Differential element � Differential element of thickness dy

� Element intersects curve at point (x, y)

� Length = (1 – x)



Solution




( )

yy

xxxx

dyxdA

=

+=

−+=

−=

~2

12

1~

1



Solution

Integrations

( )[ ]( )

( )

( )

( )

dyydxxx

dA

dAx

x A1

21

12/1~

1

1

01

1

0−

=−−

==

∫

∫

∫∫

∫

∫

( ) ( )

( )

( )

( )( )

m

dyy

dyyy

dxx

dxxy

dA

dAy

y

m

dyydxxdAx

A

A

A

3.0333.0100.0

11

1~

75.0333.0250.0

11

1

0

1

02/3

1

0

1

0

1

0

1

0

==

−

−=

−

−==

==

−=

−==

∫∫

∫∫

∫

∫

∫∫∫



Example 9.6


the shaded are bounded the shaded are bounded

by the two curves

y = x

and y = x2.



Solution

Method 1


� Differential element of thickness dx� Differential element of thickness dx

� Intersects curve at point (x1, y1) and (x2, y2), height y



� For centroid

xx

dxyydA

=

−=

~

)( 12



Solution

Integrations

( ) ( )dxxxxdxyyxdAx

x A

~ 1

021

0 12 −=

−==

∫∫∫ ( )

( )

( )( )

m

dxxx

dxxxx

dxyy

dxyyx

dAx

A

A

5.0

61121

1

02

01

0 12

0 12

==

−

−=

−

−==

∫∫

∫∫

∫



Solution

Method 2


� Differential element � Differential element of thickness dy

� Element intersects curve at point (x1, y1) and (x2, y2)

� Length = (x1 – x2)



Solution




( )

22~ 2121

2

21

xxxxxx

dyxxdA

+=

−

+=

−=



Solution

Integrations

( )[ ]( ) ( )[ ]( )dyyyyydxxxxxdAx 2/2/~ 11

−+−+ ∫∫∫ ( )[ ]( )

( )

( )[ ]( )( )

( )( )

mdxyy

dyyy

dxyy

dyyyyy

dxxx

dxxxxx

dA

dAx

x

A

A

5.0

61

121

21

2/2/

1

0

1

0

2

1

0

01

0 21

0 2121

==−

−=

−

−+=

−

−+==

∫

∫

∫∫

∫∫

∫

∫



Example 9.7

Locate the centroid for the

paraboloid of revolution,

which is generated by which is generated by

revolving the shaded area

about the y axis.



Solution

Method 1


� Element in the shape of a thin disk, thickness dy, radius zradius z

� dA is always perpendicular to the axis of revolution

� Intersects at point (0, y, z)


� For volume of the element

( )dyzdV 2π=



Solution

� For centroid

Integrationsyy~ =

Integrations

( )( )

mm

dyy

dyy

dyz

dyzy

dV

dVy

y

V

V

7.66

100

100~

100

0

100

0

2

100

0

2

100

0

2

=

===

∫∫

∫∫

∫

∫

π

π

π

π



Solution

Method 2


� Volume element in the form of thin � 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



Solution




( )

( ) ( ) 2/1002/100~

10022

yyyy

dzyzrdA

+=−+=

−= ππ



Solution

Integrations

( )[ ] ( )

( )

dzyzydVy

y V10022/100

~ 100

0−+

==∫

∫

∫ π

( )

( )( )

m

dzzz

dzzz

dzyzdVy

V

7.66

101002

1010

1002

100

0

22

100

0

444

100

0

=

−

−=

−==

∫∫

∫∫

−

−

π

π

π



Example 9.8

Determine the location of

the center of mass of the the center of mass of the

cylinder if its density

varies directly with its

distance from the base

ρ = 200z kg/m3.



Solution

For reasons of material symmetry


yx 0==

� 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)


� For volume of the element

( ) dzdV 25.0π=



Solution

� For centroid

Integrations

zz~ =

Integrations

( )

( )

mzdz

dzz

dzz

dzzz

dV

dVz

z

V

V

667.0

5.0)200(

5.0)200(~

~

1

0

1

02

1

0

2

1

0

2

==

==

∫∫

∫∫

∫

∫

π

π

ρ

ρ



Solution

� Not possible to use a shell element for integration since the density of the 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

6161103 9.2 center of gravity and center of mass and centroid for a body

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