Team of Slovakia International Young Physicists’ Tournament, Seoul 2007 4. Spring thread

Team of Slovakia

International Young Physicists’ Tournament, Seoul 2007

4. Spring thread

No. 4. Spring ThreadPresented by Tomas Bzdusek

Task

Pull a thread through the button holes as shown in the picture. The button can be put into rotating

motion by pulling the thread. One can feel some elasticity of the thread. Explain the elastic

properties of such a system.


Outline

• Theory– What is elasticity?– Deriving the motion equation

• Experiments

• Conlusion


What is elasticity?

• Elastic material acts according to Hooke‘s law

• In other notation

• Force is directly proportional to relative extension.

• Hooke‘s law can be used also for compressing an elastic material. Then is relative contraction.

E

fF


Our system

• Two extremal views of the situation:

1.) Fixed distance of ends of the thread.

- When rotating the button, the thread has to extend, so it acts due to Hooke‘s law

2.) The distance changes, we assume no extension in real lenght of the thread – It changes only due to entangling of the thread.


Where is the reality?

• Somewhere between the extremes.– The distance changes significantly.– There are some small changes in real lenght of

the thread.


Used threads

• We used three different types of thread– Thin thread (green, orange, yellow); r = 0.12 mm– Thick thread (white); r = 0.29 mm– Silon; r = 0.65 mm


Measuring the Young modulus


Young modulus of used threads

• Thin thread

{ } = 0.003 F

00.0050.01

0.0150.02

0.0250.03

0.0350.04

0.045

0 1 2 3 4 5 6 7 8F/[N]



• Thick thread

{ } = 0.001 F + 0.0015

00.0020.0040.0060.008

0.010.0120.0140.0160.018

0.02

0 2 4 6 8 10 12 14 16

F/[N]



• Silon

{ } = 0.0045 F - 0.0032

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0 2 4 6 8 10 12 14 16

F/[N]


Our model

• Extensibilities due to Hooke’s law are small – we will neglect them.

• The system is only being shortened due to convolution


Is this system elastic?

• Force F can be arbitrary, whatever the relative contraction of the system is. – Against Hooke’s law

• Therefore THE SYSTEM IS NOT ELASTIC.

• However, we can feel some “elasticity”.

,,fF


Parameters

• Half-lenght of the system in steady state (not shortened) – L0

• Actual half-lenght – L• Radius of the thread – r• Angle of rotation of the button (in comparison with

steady state) – • Angle of convolution of the thread -


Parameters shown in a picture

L

r


Basic equtions

• The thread is homogeneously rolled on a cyllindrical surface with radius r.

L

L0 r

rolled threadL

r tan


Length of the system

• According to Phytagorean theorem:

• For shortening of the system:

– N is number of windings.

• For small number of windings:

220 rLL

222200

2200 422 rNLLrLLS

2

0

224N

L

rS


Measuring the shortening

buttonfixed end free end


Shortening of used threads

• Thin thread

0

50

100

150

200

250

300

0 100 200 300 400 500 600

Number of windings N

Sh

ort

en

ing

S/[

mm

]

measuerd

predicted



• Thick thread

0

10

20

30

40

50

60

70

80

0 5 10 15 20 25 30


sh

ort

en

ing

S/[

mm

]

measuerd

predicted



• Silon

0

10

20

30

40

50

60

0 20 40 60 80 100


sh

ort

en

ing

S/[

mm

]

measured

predicted


Further equations

• In the thread there will be a strain T:

cos2TF

2

F2

F

tan2

F


Motion equation

• For torque we can obtain:

• For small number of windings– Direct proportion:

0LL

L

Frr

FM

22tan

24

0

22

L

FrM


Prooving the direct proportion

weight of mass F/gchangeable wieght


Prooving the direct proportion


Measuring the direct proportion

• In equilibrum:

• Theoretically:

– d = radius of a button– M = mass suspended at free end of the system– m = mass of changeable weight

mgdL

Fr

22

22 22 Mr

Ldm

F

g

r

Ldm


Results

• Thin threadM = 1kg ; d = 14 mm ; L0= 0.2 m ; m = 0.5 g

0

50

100

150

200

250

0 10 20 30 40 50

number of weights n

nu

mb

er

of

win

din

gs

N

theoretically

measured


Results

• Thick threadM = 1kg ; d = 14 mm ; L0= 0.6 m ; m = 0.5 g

0

5

10

15

20

25

30

35

40

0 10 20 30 40 50

number of weights n

nu

mb

er

of

win

din

gs

N

theoretically

measured


Results

• SilonM = 1kg ; d = 14 mm ; L0= 0.6 m ; m = 0.5 g

020406080

100120140160180200

0 10 20 30 40 50 60

number of weights n

nu

mb

er

of

win

din

gs

N

theoretically

measured


Motion equations

• We obtained

• Therefore

• Linear harmonic oscillator with period

0

22

L

FrM

LI

Fr

I

M 22

222

Fr

LIT


Motion equations

• If we assume some damping b:

(where )

• Well-known solution of this equation (for 0=0) is:

0 Kb

LI

FrK

22

Ct

C

bCte

tb

sin2

cos20

4

2bKC


Damped oscillations

-250

-200

-150

-100

-50

0

50

100

150

200

250

0 2 4 6 8 10time / [s]

nu

mb

er

of

win

din

gs


Spring thread as a toy

• When playing with the toy, we act – with larger force when the system is expanding– with smaller force when the system is shortening

• In our model, we suppose the forces to be F and F/2.


Simulation• We can see, that the system begins to rezonate

Spring Thread Resonation

-150

-100

-50

0

50

100

150

0 0,5 1 1,5 2 2,5 3 3,5 4

t/[s]

fi/[

rad

]


Conclusion

• Elasticity of the system is a dynamic property.– We can feel it only when the button is rotating.

• Spring thread is rezonating torsional oscillator - THIS IS THE ELASTICITY we can feel.

xm

kx

k

m

LI

Fr 22


Thank you for your attention

No. 4. Spring ThreadPresented by Tomas Bzdusek18. Appendix

The motion equation (1)

We suppose a solution . When substitued into the differential equation, we obtain:

We suppose that

tie 0

0

02

0002

Kib

eKeibe tititi

42

2

2;1

bK

bi

4/2bK



Substitution:

We have:

For 0:

Since 0 is real, and

Cb

K 4

2

iCtBiCtAe

tiCb

BtiCb

A

tb

expexp

2exp

2exp

2

BA0*AB RA20



For angular velocity:

In zero time:

iCtiCttb

BiCb

AiCb

e

tiCb

BiCb

tiCb

AiCb

t

exp2

exp2

2exp

22exp

2d

d

2

BAiCBAb

20



We have and

Therefore we can simplify:

Finally we have:

We will use:

IiAiCb

BAiCBAb

222 00

IR iAAA IR iAAA *

20

RAC

b

AI 22 00

xiee

xeeixix

ixix

sin2

cos2



By putting together and simplifying we obtain:

Under special condition :

Ct

C

b

Ctetb

sin2cos00

02

00

Ct

C

bCte

tb

sin2

cos20

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Team of Slovakia International Young Physicists’ Tournament, Seoul 2007 4. Spring thread