Centres of mass of plane figures 2F

Centres of mass of plane figures 2F

1 a From question 1a in Exercise 2D,

1

13

x ?2 ;y ?

2

14

Vertical

In equilibrium, G will be vertically

below O i.e. OG is the vertical.

tan ? ?

y

?

x

13

14

5

2

13 2 13

? ?

14 5 35

? 13 ?

? ? tan ?1 ? ? ? 20.4? (3 s.f .)

? 35 ?

?

b From question 1a in Exercise 2E,

x?

11

5

;y ?

4

4

As above, tan ? ?

y

?

x

5

4

11

4

5

11

?5?

? ? tan ?1 ? ? ? 24.4? (3s.f .)

? 11 ?

i.e. tan ? ?

c From question 1b in Exercise 2D.

x ? 1.7; y ? 2.6

2.6 26

?

1.7 17

? 26 ?

? ? tan ?1 ? ? ? 56.8? (3 s.f .)

? 17 ?

tan ? ?

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2

From question 4 in Exercise 2D,

These are the coordinates

of the centre of mass, G,

referred to O as origin.

79

51

x? ; y?

26

26

A is the point of suspension.

? 79 51 ?

G is ? ,

?.

? 26 26 ?

When the lamina hangs in equilibrium from A,

AG will be the downward vertical.

Let N be the point on AJ such that GN is perpendicular

to AJ.

See diagram.

? ? ? is the required angle.

Then NAG

Since A is the

point (1, 3).

GN x ? 1

tan ? ?

?

AN 3 ? y

79

? 1 79 ? 26

? 26 51 ?

3 ? 26 78 ? 51

?

Multiply top and

bottom by 26.

53

? ? ? 63.0 (3 s.f .)

27

?7 ?

G, the centre of mass has coordinates ? , 2 ?

?3 ?

taking O as origin.

3

Since AG will be vertical in

equilibrium, the angle between

AC and the horizontal will be ?.

?? is the required angle

tan ? ?

? ?

7

3

2

?2

6

7?6

?6

?

Multiply top and bottom

by 3 to clear fractions.

?

? ? 80.5? (3 s.f .)

?

?

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G is the centre of mass if the framework

4

OG ? x ?

9( 3 ? 2)

(6 ? ¦Ð)

G is on the line

of symmetry.

(see question 2 in Exercise 2E)

AG will be vertical, when

the framework hangs in

equilibrium.

? (see diagram) is the required angle.

?

? ? 60 ? GAN

? ? GN ? 6 cos30? ? x

tan GAN

AN

6sin 30?

3 3?x

?

3

3( 3 ? 2)

? 3?

6?¦Ð

?

So, GAN ? 26.898?¡­

So, ? ? 33.1? (3 s.f .)

3

, where G is

2¦Ð

the centre of mass

so, AG will be vertical in

equilibrium.

OG ?

5

See question 4 in

Exercise 2E.

Since the angle with the

horizontal will be 90 ?

angle with the vertical.

? ??

Required angle is AGO

tan ? ?

?

AO

OG

3

3

2¦Ð

? 2¦Ð

? ? tan ?1 (2¦Ð)

? 81.0? (3 s.f .)

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Centre of mass of ?RNQ

has position vector

6

Taking PQ and PS

as axes.

1 ??? 4 ? ?10 ? ? 4 ? ? ? 6 ?

?? ? ? ? ? ? ? ? ? ? ? ?

3 ??? 0 ? ? 0 ? ? 6 ? ? ? 2 ?

? 2?

?6?

?x?

24 ? ? ? 18 ? ? ? 42 ? ?

? 3?

? 2?

? y?

? 48 ? ? 108 ?

?x?

? ???

? ? 42 ? ?

? 72 ? ? 36 ?

?y?

? 156 ?

?x?

?

? ? 42 ? ?

? 108 ?

? y?

? 26 ?

? 7 ? ?x?

? ? ?? ?

? 18 ? ? y ?

? ?

? 7 ?

Use

?m r ? r?m

i i

i

Simplify.

So the centre of mass of the lamina is

a

26

cm from PS and

7

b

18

cm from PQ.

7

c We have

tan ? ? ? 187 ? / ?10 ? 267 ? ?

9

22

So ? ? 22.2?

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7 We choose coordinates so that the origin is at C and that BC lies on the x-axis, by considering the

lamina as the union of a 2 ¡Á 6 rectangle and a 6 ¡Á 2 rectangle we see the centre of mass will satisfy

? x?

? ?1 ?

?3?

24 ? ? ? 12 ? ? ? 12 ? ?

? y?

? ?3 ?

? ?5 ?

So

? x? ? 1 ?

? ??? ?

? y ? ? ?4 ?

Now we attach a mass of 0.2M kg to F so the centre of mass of the whole system will satisfy

? x?

?1?

?6?

1.2M ? ? ? M ? ? ? 0.2 M ? ?

? y?

? ?4 ?

? ?6 ?

So

? x ? ? 2.2 ?

1.2 ? ? ? ?

?

? y ? ? ?5.2 ?

So

? x ? 1 ? 11 ?

? y ? ? ? ?26 ?

? ? 6?

?

Hence the angle ? satisfies

26

tan ? ?

11

So

? ? 67.1?

8 We choose coordinates so that the origin is at B and the x-axis is parallel to AC then the centre of

mass of the lamina is

0

?

?

? 32

¦Ð?

? ? sin ?

4?

? 3¦Ð

And the coordinates of C are

?2 2 ?

?

?

? ?2 2 ?

?

?

Hence the centre of mass of the system

Satisfies

0

?

?

? 2 2 ?

?x?

?

1.5M ? ? ? M

?

32

¦Ð ? ? 0.5M ??

?

? ? sin ?

?

2

2

? y?

?

?

4?

? 3¦Ð

Hence

?

?

2

1

?

?

? x? 2?

? 2 2?

?

? y ? ? 3 ? 16 2

? ? 3 ? ? (16 ? 3¦Ð) ?

? ?

? 2?

??

3¦Ð ?

?

? 3¦Ð

?

? 0.9428... ?

??

?

? ?2.5434... ?

Hence

? 0.9428 ?

? ? 45? ? tan ?1 ?

? ? 45? ? 20.3? ? 65.3?

? 2.5434 ?

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