Mechanical Engineering Principles

FIRST AND SECOND MOMENT OF AREAS 87

=

1

2

∫ 3

0

16 x^2 dx

18

=

1

2

[

16 x^3

3

] 3

0

18

=

72

18

= 4

Hence the centroid lies at (2, 4).

C

A

B

D

y=4x

3 x

x

y

y

12

0

Figure 7.5

In Figure 7.5,ABDis a right-angled triangle. The
centroid lies 4 units fromABand 1 unit fromBD
showing that the centroid of a triangle lies at one-
third of the perpendicular height above any side
as base.

Now try the following exercise

Exercise 35 Further problems on cent-

roids of simple shapes

In Problems 1 to 5, find the position of the

centroids of the areas bounded by the given

curves, thex-axis and the given ordinates.

1. y= 2 x;x=0,x= 3 [(2, 2)]


2. y= 3 x+2;x=0,x=4 [(2.50, 4.75)]


3. y= 5 x^2 ;x=1,x=4 [(3.036, 24.36)]


4. y= 2 x^3 ;x=0,x= 2 [(1.60, 4.57)]


5. y=x( 3 x+ 1 );x=−1,x= 0
   [(−0.833, 0.633)]

7.6 Further worked problems on

centroids of simple shapes

Problem 4. Determine the co-ordinates of

the centroid of the area lying between the

curvey= 5 x−x^2 and thex-axis.

Figure 7.6

y= 5 x−x^2 =x( 5 −x).Wheny=0,x=0or

x=5. Hence the curve cuts thex-axis at 0 and 5

as shown in Figure 7.6. Let the co-ordinates of the

centroid be (x,y) then, by integration,

x=

∫ 5

0

xydx

∫ 5

0

ydx

=

∫ 5

0

x( 5 x−x^2 )dx

∫ 5

0

( 5 x−x^2 )dx

=

∫ 5

0

( 5 x^2 −x^3 )dx

∫ 5

0

( 5 x−x^2 )dx

=

[

5 x^3

3

−

x^4

4

] 5

0

[

5 x^2

2

−

x^3

3

] 5

0

=

625

3

−

625

4

125

2

−

125

3

=

625

12

125

6

=

(

625

12

)(

6

125

)

=

5

2

= 2. 5

y=

1

2

∫ 5

0

y^2 dx

∫ 5

0

ydx

=

1

2

∫ 5

0

( 5 x−x^2 )^2 dx

∫ 5

0

( 5 x−x^2 )dx

=

1

2

∫ 5

0

( 25 x^2 − 10 x^3 +x^4 )dx

125

6