c entres of mass

Centres of Mass

Chapter objectives• Find the centre of mass of a system of particles

distributed in one dimension• Find the centre of mass of a system of particles

distributed in two dimensions• Use knowledge of standard results to find the

centre of mass of plane figures• Consider the equilibrium of a lamina which is

suspended from a fixed point or placed on an inclined plane

Keywords

centre of mass – the point at which the whole mass of the body can be considered to be concentrated

Lamina – an object which is very thin compared to it’s other two dimensions is modelled as a lamina e.g. a sheet of paper

Uniform – mass is evenly distributed Find the centre of mass of a system of particles distributed in one dimension

Find the centre of mass of a system of particles distributed in one dimension

This means your particles are arranged along a straight line

For example

Three particles of mass 6kg, 3kg and 2.5kg are attached to a light rod PQ of length 3m at the points P, Q and R, where PR = 0.9m. Find the position of the centre of mass of the system.

Draw a diagram or 2!

6g 3g 2.5g

P Q R

0.9m

Start by adding the centre of mass to the diagram and let the distance PG be x.

6g 3g 2.5g

P Q R

0.9m

G

11.5g

6g 3g 2.5g

P Q R

0.9m

G

11.5g

Taking moments about P gives: 11.5g × x = 3g × 0.9 + 3 × 2.5g

x = 0.89m

There is a general rule you can apply to this

∑𝑖=1

𝑛

𝑚𝑖 𝑦 𝑖=𝑚∑𝑖=1

𝑛

𝑦 𝑖

And our text book says you should learn it!

Find the centre of mass of a system of particles distributed in two dimensions

Now our particles are arranged in a plane

For example

Particles of mass 2kg, 4kg, 5kg and 6kg are attached to the corners of a light rectangular plate PQRS. Given that PQ = 5cm and QR = 12cm calculate the distance of the centre of mass of the system froma) PQb) PS

Draw a diagram

The question asks us to:

calculate the distance of the centre of mass of the system froma) PQb) PS


There are 3 different methods to solve this question

Separate Masses Total Mass

Mass 2 4 5 6 17x co-ord 0 0 12 12Y co-ord 0 5 5 0

Method 1: taking P as (0, 0) put the information in a table



Mass 2 4 5 6 17x co-ord 0 0 12 12Y co-ord 0 5 5 0

i i imx mX 60 72 17X

X 7.76

To find the distance of the centre of mass from PQ we use the formula:

i i imy mY 20 25 17Y

Y 2.65

To find the distance from PS we use the formula:

centre of mass is at (7.66, 2.65)

Method 2

Take moments about the Y axis PQ:

5 x 12 + 6 x 12 = 17 n.b. g has been cancelled = 7.66

Take moments about the X axis PS:

4 x 5 + 5 x 5 = 17 = 2.65


Method 3

Use position vectors taking P as (0, 0)

2 + 6 + 5 + 4 = 17

+ + + = 17

=


The diagram below shows a series of particles that make up a system. The centre of mass of the system is at the point (x,y). Find the coordinates of the centre of mass of the system

Pick a method and try this example

Y

X

3kg

4Kg

2Kg

2.5kg

(x,y)

2 + 3 + 4 + 2 = 1

+ + + = 1

= 1

=

centre of mass is at (2.22, -0.04)

There is a general rule that goes with all this.

It can be found on page 35.

There is an error in question 9 on page 39

The third side is CD not CA

Use knowledge of standard results to find the centre of mass of plane figures

There are some standard results to be taken for granted when it comes to the centre of mass. Uniform rectangular lamina:- at centre of the shape Uniform circular disc:- at centre of disc Uniform triangular lamina:-

o Equilateral:- at centreo Isosceles:- at the intersection of the medians

A median is a line that joins a vertex of a triangle to the centre of the side opposite to the vertex. The centre of mass of a scalene triangle is at a point one third of the way along the median (from the edge).

The centre of mass of the triangle ABC is at the point G, where EG = ⅓EC.

A

B

C

G E

F

D The centre of mass of the triangle ABC is at the point G, where EG = ⅓EC

If the coordinates of the three vertices of a uniform triangular lamina are (x1, y1), (x2, y2) and (x3, y3) then the

coordinates of the centre of mass are found by taking the mean of the coordinates of the 3 vertices.

Calculate the coordinates of the centre of mass of the uniform triangular lamina ABC if the point A is placed at the origin.

A

B

C

18m

12m

centre of mass G is at the point =

r

2α

Unif orm Sector:- centre of mass is on the axis of symmetry at a distance 2r sin

3

f rom the centre, where α is measured in radians.

(note the use of α in the f ormula and 2α in the diagram.)

This formulae is in the booklet

Find the centre of mass of the lamina with radius 5 cm

OA B

G

C

The centre of mass lies on the line of symmetry OCUsing the result for a sector with r = 5 and α =

Then OG = =

The centre of mass lies on the line OC a distance from O.

Application to composite figures

7.5cm F

A

D

E

C B

12cm

9cm

You can find the centre of mass for this shape by applying the ideas we have already come across:1. Work out the centre of mass for the rectangle2. Work out the centre of mass for the triangle3. Find the centre of mass of both ‘particles’

centre of mass for ABCE

Take A as the origin then(6, 4.5) is the centre of mass

centre of mass for CDE

= = (14.5, 4.5)

Because the shapes are uniform the area of each can be used to represent it’s mass

Mass of rectangle = 108 cm² mass of triangle = 33.75cm²

centre of mass of shape

= 108 141.75

centre of mass is 8.02 cm from AB and 4.5 cm from AE

The object below is formed by removing a uniform semi circular disc of radius 3m from a second uniform semi circular disc of radius 6m. Calculate the centre of mass of the object.

3m6m

By defi nition the centre of mass will lie on the mirror line. The f ormula for the centre of mass f or a unif orm sector is 2r sin

3

where the angle at the centre is 2α. Obviously in the above example 2α = ∏ .

Using the tabular approach

Separate Masses

Little disc Big disc

Total Mass

Mass

y co-ord

92 18 27

2

2 3 sin 42 sin3 3 22

2 6 sin 82 sin3 22

X

the disc we are interested in can be found by subtracting the small semi circular disc from the larger

one.

i i imx mX 8 9 4 2718 sin sin X2 2 2 2

27144 18 X2

X 2.97

So the centre of mass is 2.97cm from the centre along the axis of symmetry.

The object below is formed by removing two uniform circular discs from a uniform rectangular lamina. Calculate the centre of mass of the plate.

A

10cm

16cm

D

CB

1.5cm

6cm

12cm

3cm

The same principle of adding and subtracting parts can be applied to this question

Separate Masses Total Mass Circle 1 Circle 2 Rectangul

ar LaminaPlate

Mass 2.25∏ 2.25∏ 120 120-4.5∏x co-ord 3 12 6Y co-ord 6 6 5

X

Y

i i imx mX

X

X

Plate = Rectangular Lamina - Circle 1 - Circle 2

(120-4.5∏)

= 720 – (7.75∏ + 27∏)

= 5.77 cm

Looking at it vertically: i i imy mY (120-4.5∏ )Y = 600 – (13.5∏ + 13.5∏ ) Y = 4.87cm Theref ore the centre of mass is 5.77cm f rom AB and 4.87cm f rom AD.

2.5You can find the centre of mass of a framework by using the centre

of mass of each rod or wire that makes up the framework

The framework ABC is made up of three uniform rods and a semi circular arc. Find the centre of mass of the framework assuming that A is at the origin

20cm

A

B

C

16cm

12cm

Remembering that the f ormula f or the centre of mass of a uniform circular arc is r sin

and that 2

. Also note that the centre of mass

of the arc will have a negative x coordinate.


Rod AB Rod AC Rod BC Arc AB Framework

Mass 12m 16m 20m 6∏m (48 + 6∏)m

x co-ord 0 8 8

Y co-ord 6 0 6 6

6sin 122

2

X

Y

Looking at the problem horizontally: i i imx mX

12(16 8) (20 8) (6 ) (48 6 )X

X 3.23cm

And now vertically: i i imy mY

(12 8) (20 6) (6 6) (48 6 )Y

Y 4.56cm

Theref ore the centre of mass is at the point with coordinates (3.23,4.56).

2.6 Laminas in equilibrium

A suspended lamina will be in equilibrium when its centre of mass is directly below the point of suspension

It can be suspended from a fixed point or a pivot

G G

W W

O

O

Fixed point

Point of suspension

stringR

T pivot

Lets look at the example in the book on page 54

Equilibrium of a uniform lamina on an inclined plane. For an inclined lamina to remain in equilibrium on an inclined plane the line of action of the weight must fall within the side of the lamina that is in contact with the plane (as shown in the diagram below).

G

θ

In the second diagram the object will topple over

G

θ

A uniform rectangular lamina is placed on a plane inclined at an angle θ. Given that the lamina is in limiting equilibrium find the angle θ.

6cm

15cmG

θ

By simple trigonometry:3Tan 7.5

21.8

c entres of mass

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