CENTRE OF MASS. The centre of mass of a body is the point at which its weight acts. The centre of...

CENTRE OF MASS

The centre of mass of a body is the point at which its weight acts.

The centre of mass of a body lies on any line of symmetry.

The principle to find a centre of mass of a composite bodyis that, the sum of the moments of the composite parts of the body about a given axis is equal to the moment of thewhole body about the same axis.

Example 1: A uniform wire of length 24cm is bent to form a triangle ABC where AB = 6cm and BC = 8cm. Find the distanceof the centre of mass from AB and from BC.

A

B C

Body Weight Distance of G

from AB.

Distance of G

from BC.

AB

BC

AC

Whole

Since AB = 6, BC = 8 AC = 10

6m

Let the mass per unit length be m, and the centre of mass be at G.

The mass of each of the 3 sides acts at their centres.

Taking moments about AB: 6m(0) + 8m(4) + 10m(4) = 24m( x )

Taking moments about BC: 6m(3) + 8m(0) + 10m(3) = 24m( y )

x

x = 3

y = 2

y

8m

10m

24m

0

44

3

0

3

6m

8m

10mBy Pythagoras’ theorem, the triangle is right angled at B.

( 62 + 82 = 102 )

Problems involving laminas are treated in the same way, with the mass of each part acting at its centre of mass:


from AB.

Distance of G

from BC.

ABPF

PCDE

Whole

Example 2: For the lamina ABCDEF shown, find the distance of the centre of mass from AB and from BC.

A

C

D

B

E

F4

2

6 5

Let the mass per unit area be m, and the centre of mass be at G.

24m

Taking moments about AB:

10m

34m

2

6.5

3

1

x y

24m(2) + 10m(6.5) = 34m( x )x =

y =

Taking moments about BC: 24m(3) + 10m(1) = 34m( y )

3.32

2.41 (3 sig.figs.)

(3 sig.figs.)

P


from AB.

Distance of G

from BC.

ABPF

PCDE

Whole

Example 3: For the lamina ABCDEF shown, the section PCDE is made of a material which is twice as heavy as the material usedfor the section ABPF. Find the distance of the centre of mass from AB and from BC.

A

C

D

B

E

F4

2

6 5

P

Let the mass per unit area be m for ABPF, and2m for PCDE and the centre of mass be at G.

24m


20m44m

2

6.5

3

1

x y

24m(2) + 20m(6.5) = 44m( x )x =

y =


4.05

2.09 (3 sig.figs.)

(3 sig.figs.)


from AB.

Distance of G

from BC.

ABPF

PCDE

Whole

Example 4: For the lamina ABCDEF shown, the section ABPF is twice as heavy as the section PCDE. Find the distance of the centre of mass from AB and from BC.

A

C

D

B

E

F4

2

6 5

P

m


2m

3m

2

6.5

3

1

x y

2m(2) + m(6.5) = 3m( x )

x =

y =

Taking moments about BC: 2m(3) + m(1) = 3m( y )

3.5

2.33 (3 sig.figs.)

Certain shapes have centres of mass at points which can be quoted.

The centre of mass of a triangular lamina lies one third of the distance along the line from the centre of a side of the triangle to the opposite vertex.

d

2d

The centre of mass of a lamina in the form of a sector of a circle is another standard result.

OG

r

αα

A

B

If OAB is a sector of a circle, radius r, withangle 2α at the centre then the centre of massG lies on the line of symmetry such that:

O

G

r

Hence for a semicircle, α =π2

where α is measured in radians.

OG = 4 r3π

2r sin

3

2π

2π =

OG = 2r sin α

3 α

The centre of mass of a uniform wire in the form of an arc of a circle is one more standard result.

G

r

Oαα

A

B

If AB is an arc of a circle, radius r, withangle 2α at the centre, then the centre of mass G lies on the line of symmetry such that:

where α is measured in radians.

OG = r sin α α

Example 5: For the lamina ABCD shown, find the distance of the centre of mass from AB and from BC.


from AB.

Distance of G

from BC.

PBCD

APD

Whole


6

2

3

A

B C

DP

12m

9m

21m

3 1

2 3

x y

Taking moments about AB: 12m(3) + 9m(2) = 21m( x )

x =

y =

2.57

(3 sig.figs.)

(3 sig.figs.)


1.86

Example 6: A uniform circular lamina, radius 5a has a circular hole of radius 2a. The circumference of the hole lies on the centreof the lamina. Find the distance of the centre of mass of the lamina fromits centre.


from y-axis.

5a

x


4a2 π m

x

y

5a 2a

25a2 π m

21a2 π m

7a

Taking moments about the y-axis:

21a2 π m x + 4a2

π m ( 7a ) = 25a2 π m ( 5a )

The distance of G from its centre is:

a9721

821

ax =

x + 28a = 125a219721

a = 5a –

Note: G lies on the x-axisby symmetry.

Example 7: A lamina ABCDE is made by joining a rectangle to an isosceles triangle as shown: a) Find the distance of the

centre of mass from CD. The lamina is now suspended from point E. Find the angle that DE makes with the vertical.


from CD.Let the mass perunit area be m, and the centre of mass be at G.

CD

E B

A

3

5

852m

40m

12m 6

2.5

y

Taking moments about CD: 12m(6) + 40m(2.5) = 52m( y )

y = 3.308

The distance of the centre of mass from CD is 3.31 (3 sig.figs.)

G

y

5

8

4θ

Since the only forces acting areequal and opposite, the centre of mass, G must lie vertically below E.

By symmetry, G must also lieon the line joining A to themid-point of CD.

D

E

A

B

C

DE makes an angle 90 – θ with the vertical,

5 – 3.308 4

5 – y4

= θ = 22.9° tan θ =

= 67.1°.

Example 7a: The lamina ABCDE in example 7 is now placed on an inclined plane with DC on the plane as shown. Given that

the plane is sufficiently rough to prevent the lamina sliding down theplane, find the greatest angle of inclination of the plane to the horizontal such that the lamina will not topple.

C

D

BA

E

4

yThe lamina will be on the pointof toppling when the weight actsthrough the point D as shown:

θ

If the inclination of the plane is θ,then the angle between the line ofaction of the weight, and the line ofsymmetry of the lamina is

θ

also θ.

G

43.308

θ = 50.4°.tan θ = =4 y

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Sep.2011

The centre of mass of a body is the point at which its weight acts.

The centre of mass of a body lies on any line of symmetry.

The centre of mass of a triangular lamina lies one third of the distance along the line from the centre of a side of the triangle to the opposite vertex.

OG = 2r sin α

3 αO

G

r

αα

A

B

For a lamina in the form of a sector of a circle:

G

r

Oαα

A

B

For an arc of a circle:

OG = r sin α α

For both, α is measured in radians.

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