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# Rotating spring

16. Rotating spring. Reporter: Reza M. Namin. 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

## Rotating spring

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### Presentation Transcript

1. 16 Rotating spring Reporter: Reza M. Namin

2. 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.

3. Main approach • Theory • Background • Theory base • Developing the equations • Numerical solution • Experiment • Setup • Parameters, results and comparison • Conclusion

4. 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 = kL remains constant • Circular motion • a = rω2 • a: Acceleration • r: Distance from the rotating axis • ω: Angular velocity

5. 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 l1 l2 l M

6. Theory - Base • Looking for the stable condition in the rotating coordinate system • Accelerated system → figurative force • Acting forces: • Gravity • Spring tensile force • Centrifugal force

7. Theory – Developing the equations • Approximation in mass attached conditions: • Considering the spring to be weightless: Fs ω Fc Mg y l M x

8. 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 severalsmall springs. M

9. 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 Ti-1 fc Ti+1 w

10. Theory – Numerical solution Mesh independency check n: Number of mesh points As n increases, the result will approach to the correct answer

11. Theory – Numerical solution Tension in different points of the spring with different additional mass amounts:

12. 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

13. Experiment setup The motor, connection to the spring and the sensor sticker

14. Experiment setup The rotating spring and tachometer

15. Experiment setup Hold and base

16. Experiment setup All we had on the table

17. Experiments Suspending weights with the spring Finding k and using that to find μ →K = 33.78 N/m →μ = K l = 1.824 N

18. Experiments Expansion increases with increasing angular velocity

19. Experiments Measurement of length in different angular velocities Comparison with the numerical theory

20. Experiments Comparing the shape of the rotating spring in theory and experiment λ=0.103 kg/m μ =0.369 N l = 16.3 cm l1 =1 cm ω = 120 RPM

21. Experiments Investigation of the l-ω plot within different initial lengths

22. Experiments Comparison between the physical experiments, numerical results and theoretical approximation within different additional masses

23. Conclusion • According to the comparison between the theories and experiments we can conclude: • In case of weightless spring approximation:

24. 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:

25. Conclusion Numerical solution results Change of the spring hardness

26. Conclusion Numerical solution results Change of spring density μ =0.3 N l = 10 cm l1 =1 cm

27. Conclusion Numerical solution result Change of initial length λ=0.2 kg/m μ =0.3 N l1 =1 cm

28. Conclusion Numerical solution results Change in additional mass

29. Thank you

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