Rotating spring

IYPT 2010 Austria, I. R. IranIYPT 2010 Austria, I. R. Iran

Problem No.

1

ROTATING SPRING

Reporter: Reza M. Namin

16

IYPT 2010 Austria, I. R. Iran2

The problem• A helical spring is rotated about one of its ends around

a vertical axis.

• Investigate the expansion of the spring with and without an additional mass attached to it’s free end.


Main approach• Theory

– Background– Theory base– Developing the equations– Numerical solution

• Experiment– Setup– Parameters, results and comparison

• Conclusion


Theory - Background• Act of a spring due to tensile force:

– Hook's law: F = k ∆L• F: Force parallel to the spring• k: Spring constant• ∆L: Change of length

– A spring divided to n parts:• F = n k ∆L• μ = k L remains constant

• Circular motion– a = r ω2

• a: Acceleration• r: Distance from the rotating axis• ω: Angular velocity


Theory - Base

• Effective parameters:– ω: Angular velocity– λ : Spring liner density = m / l – M: Additional mass– μ : Spring module = k l– l, l1, l2: Spring geometrical properties

M

l1 l l2


Theory - Base

• Looking for the stable condition in the rotating coordinate system– Accelerated system → figurative force

• Acting forces:– Gravity– Spring tensile force– Centrifugal force


Theory – Developing the equations• Approximation in mass attached

conditions:– Considering the spring to be

weightless:

l M

ω

x

y

FsFc

Mg

lxFF

xmF

llkF

sc

c

s

2

0 )(

20

mkkllapx

k

mgll 0min),max( minlll apx


Theory – Developing the equations• Exact theoretical description:

– Problem: The tension is not even all over the spring…

– Solution: Considering the spring to be consisted of several small springs.

M


Theory – Numerical solution

• Numerical method– Finite-volume approximation:

• Converting the continuous medium into a discrete medium

– Transient (dynamic unsteady) method– Programming developed with QB.

M

dlllT

dlllTxxmcf

ygmw

iiii

iiii

ii

i

)(

)(ˆ

ˆ

,1,1

,1,1

2

w

Ti-1

Ti+1

fc

lmnll /


Theory – Numerical solutionMesh independency check

80 85 90 95 100 105 110 115 1200.15

0.2

0.25

0.3

0.35

0.4

0.45

n = 5n = 10n = 15

Angular velocity (s-1)

Spri

ng le

ngth

(m)

n: Number of mesh points

As n increases, the result will approach to the correct answer


Theory – Numerical solutionTension in different points of the spring with different additional mass amounts:

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4

m = 0m = 2.5gm = 5g

Position in the spring (cm)

Ten

sion

(N)


Experiment• Finding spring properties

– Direct measurement: Mass & lengths– Suspending weights with the spring to

measure k and μ• Changing the angular velocity,

measuring the expansion– Change of the angular velocity with different

voltages– Measuring the angular velocity with

Tachometer– Measuring the length of the rotating spring

using a high exposure time photo


Experiment setupThe motor, connection to the spring and the sensor sticker


Experiment setupThe rotating spring and tachometer


Experiment setupHold and base


Experiment setupAll we had on the table


ExperimentsSuspending weights with the springFinding k and using that to find μ

0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.120

0.5

1

1.5

2

2.5

f(x) = 33.782655313683 x − 1.84774791757864R² = 0.999226158045033

Spring length (m)

Att

ache

d w

eigh

t (N

)

→K = 33.78 N/m

→μ = K l = 1.824 N


ExperimentsExpansion increases with increasing angular velocity


ExperimentsMeasurement of length in different angular velocitiesComparison with the numerical theory

10 20 30 40 50 60 70 80 900

5

10

15

20

25

ExperimentsNumerical result

Angular velocity (Rad / s)

Spri

ng le

ngth

(cm

)

λ =0.148 kg/mμ =1.824 Nl1 =1.5 cml2 = 1.7 cm


ExperimentsComparing the shape of the rotating spring in theory and experiment

0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000

-0.1000-0.0900-0.0800-0.0700-0.0600-0.0500-0.0400-0.0300-0.0200-0.01000.0000

λ =0.103 kg/mμ =0.369 Nl = 16.3 cml1 =1 cmω = 120 RPM


ExperimentsInvestigation of the l-ω plot within different initial lengths

100 150 200 250 300 3500

5

10

15

20

25

30

35

40

45Experiment: l = 16.3Experiment: l = 14Experiment: l = 11.5Experiment: l = 7.2Numerical: l = 16.3Numerical: l = 14Numerical: l = 11.5Numerical: l = 7.2

Angular velocity (RPM)

Spri

ng le

ngth

(cm

)

λ =0.103 kg/mμ =0.369 Nl1 =1 cm


ExperimentsComparison between the physical experiments, numerical results and theoretical approximation within different additional masses

0 50 100 150 200 250 300 350 400 4500

5

10

15

20

25

30

35

40

45Exp M = 0

Exp M = 5g

Exp M = 10gExp M = 15g

Num M = 0

Num M = 5g

Num M = 10g

Num M = 15g

Appx M = 0

Appx M = 5g

Appx M = 10g

Appx M = 15g


Spri

ng le

ngth

(cm

)

λ =0.103 kg/mμ =0.369 Nl = 5.4cml1 =1 cm


Conclusion• According to the comparison

between the theories and experiments we can conclude:– In case of weightless spring

approximation:

20

mkkllapx

k

mgll 0min

),max( minlll apx


Conclusion

• In general, the numerical method may be used to achieve precise description and evaluation.

• Some of the results of the numerical method are as follows:


ConclusionNumerical solution resultsChange of the spring hardness

0 50 100 150 200 250 300-0.0999999999999994

5.82867087928207E-16

0.100000000000001

0.200000000000001

0.300000000000001

0.400000000000001

0.500000000000001

0.600000000000001miu = 0.1miu = 0.2miu = 0.3miu = 0.5


Spring

leng

th (m

)

λ =0.1 Nl = 10 cml1 =1 cm

IYPT 2010 Austria, I. R. Iran

0 50 100 150 200 250 300 3500

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5Landa = 0.05Landa = 0.1Landa = 0.15Landa = 0.2

ConclusionNumerical solution resultsChange of spring density

μ =0.3 Nl = 10 cml1 =1 cm


0 50 100 150 200 2500

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4l = 8cml = 10cml = 12cml = 14cm


Sprin

g le

ngth

(m)

ConclusionNumerical solution resultChange of initial length

λ =0.2 kg/mμ =0.3 Nl1 =1 cm


ConclusionNumerical solution resultsChange in additional mass

0 20 40 60 80 100 120 140 160 1800

0.050.1

0.150.2

0.250.3

0.350.4

0.450.5

m = 0m = 5gm = 10gm = 15g


Sprin

g le

ngth

(m

)

λ =0.2 kg/mμ =0.3 Nl = 10cml1 =1 cm

IYPT 2010 Austria, I. R. IranIYPT 2010 Austria, National team of I. R. Iran

Thank you

Documents

Rotating spring