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Finding the spring constant of a torsion spring.

  1. Feb 19, 2012 #1
    Hi,

    How would I go about finding the spring constant of a bundle of nylon rope acting as a torsion spring?

    I know the length. I dont know the width of the bundle. When twisted I wanted roughly 6cm diameter (is that a wrong method)?

    I ll be twisting it by 1080 degrees.


    When twisted as above ^ -

    how can I calculate how much force/energy it will project if an arm in the middle of the torsion spring is pulled against it by a certain length/angle.

    Im guessing that the longer the arm, the more it can pull back.

    Like a ballista.

    Thank you.
     
    Last edited: Feb 19, 2012
  2. jcsd
  3. Feb 22, 2012 #2
    I'm sorry, could you clarify your question? But here are some stuff I know about torsion.


    Torque = -[itex]\kappa[/itex][itex]\theta[/itex]
    Where Kappa is the torsional coefficient and theta the angle rotated.

    Period = 2[itex]\pi[/itex][itex]\sqrt{I/\kappa}[/itex]
     
  4. Feb 22, 2012 #3
    Yes, the question is a bit over the place.

    Im trying to find the spring constant (k) of a bundle of nylon-6 rope.

    The total diameter of the rope is 6cm roughly.

    I've twisted the rope bundle 3 full revolutions and have fixed it at equilibrium.

    The length of this rope bundle spring is 36cm

    What is the working out/steps/equation to find the spring constant?

    Furthermore, If I fix an arm in the middle of the spring, I wish to find the force needed to pull back 50 degrees (length 50cm of arm), and the forces the arm will project when it returns to equilibrium.

    I know how to find the spring constant of a metal spring coil, however.
    k = ( E*d ^3) / (8 * D^4 * n)
    n = number of coils
    d= wire diameter, D = diameter of spring, E = young's modulus
     
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