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Rotating wire tension

  1. Nov 12, 2015 #1
    Hello this is my first question here :) So here is my problem:

    We have a circle radius 50 (m), made out of iron wire
    mass 5.506206591207348477884587 Kg per meter of rope.
    Rope diameter 11/8 inces or 3.4925cm

    The wire rotates with 3 rpm in zero gravity

    1)How can i calculate the tension? Thx in advance!
  2. jcsd
  3. Nov 12, 2015 #2


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    Welcome to the forum.
    this is a homework type question so the homework guidelines should be followed;

    This thread is a similar problem as yours so may help:

    Good luck.
  4. Nov 12, 2015 #3
    Thnx you for posting the similar thread! I will look up more careful next time!
    Not a student btw! (kinda older than that) :D
  5. Nov 13, 2015 #4
    Hello again, i looked up the similar thread but i don't think i can understand.

    I found out a formula about stress in a ring (im not sure if this is for a torus)
    σz = ω2 ρ ( r12 + r1 r2 + r22) / 3

    Did the maths with the following values

    ω=0.314159265rad/s (converted from 3 rpm)
    r(in a slice of rope)= 1,74625cm or 0,0174625m
    p=7850 (steel)
    r1(outer)= 50,0174625 m
    r2(inner)= 49,9825375 m

    and produced this result

    σz=246,74011949543844353458599854297 N/m2

    Does this number have any relation with tension :? I want to test what loads this ring can receive later so i wanted to know what is the tension from self for a start. A simple tension force in newtons would be much better to my understanding. Cause i only know the breaking force in newtons.

    Can anyone guide me a little bit? Need to know the method basically, maybe it is just the similar thread not sure. (poor math skills detected)
    Last edited: Nov 13, 2015
  6. Nov 13, 2015 #5
    I also found this post, according to it:
    If we happen to know the total force applied on the circle, which is the entire centrifugal force for all the mass lets name that Fc(total)


    The total mass of the wire is 1729,82~Kg
    If i plug that on a centrifugal calculator with r=50m and 3rpm, i get a Centrifugal force = 8536,35 N = Fc(total)

    And then by using the formula in bold i get a tension of 5434,4~ N

    Is that approach accurate? That formula is rly simple and handy, can anyone confirm?
  7. Nov 13, 2015 #6


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    I don't think this post is modeling what you want.

    No it's not. According to the OP, the rope has a mass per meter of 5.506 kg/m (ignoring the ridiculous number of digits in the original figure)

    Since the rope is 50 m long, then its mass is m = 5.506 kg/m × 50 m = 275.3 kg
    I think that centrifugal calculator assumes that a concentrated mass M is being swung at a radius R at a given number of RPM, so any tension you obtain will be erroneous.

    By analyzing this problem from first principles, we can obtain the following:

    Rope L = 50 m
    Angular velocity ω = 3 RPM = 2π rad/rev. × 3 rev./min. × (1/60) min./s = 6π/60 = π/10 rad/s

    The density of iron is

    ρ = 7874 kg/m3

    and the diameter of the rope is

    d = 1.125 in = 1.125 in. × 25.4 mm/in. = 28.575 mm

    The area of the rope's cross section is

    A = πd2 /4 = π × 28.5752 / 4 = 641.3 mm2
    A = 641.3 mm2 × 10-6 m2/mm2 = 6.413 × 10-4 m2

    The volume of a piece of rope, V = A × L
    The mass of this piece, m = ρV = ρAL

    Since there is no gravity, then the tensile force created on each small piece of rope with mass dm will be proportional to its distance r from the end of the rope and the square of the angular velocity, ω, since the centripetal acceleration is defined

    a = ω2r m/s2

    A small section of rope with length dr will have a mass

    dm = 5.05 kg/m * dr

    since L = dr and ρ = 7874 kg/m3 and A = 6.413 × 10-4 m2

    then dm = 5.05 dr

    F = ma = m ⋅ ω2r

    So the centrifugal force acting on dm is

    dF = a × dm = ω2 r × 5.05 dr

    dF = a × dm = 5.05*π2/100 * r dr

    dF = .04984 * r dr

    The total tensile force on the rope at the end about which it is being swung is

    ##F = \int_0^{50 m} 0.04984\, r ⋅ dr ##

    F = 0.04984 * (50)2/2

    F = 0.04984 * 2500 / 2 = 623 N
  8. Nov 13, 2015 #7
    Hey Steam King that is really helpful i will need that too for my design, But i think that you just described a wire 50 m long being swung with L=50 around a center.

    Sorry but i meant a full circle made out of wire, r=50 is just an indication for the size of that circle.
    The L of the circle is 314.1592 that's why the mass of 1729,82~Kg :( Sorry if i put you in trouble with my first post, but i need that tension too anyway, and i appreciate alot your detailed response.Hope your not angry.

    In that case is my previous post correct if i may ask?
  9. Nov 14, 2015 #8
    Hello, i made an image, hope this helps, the 623 N are for the first case (straight wire swung around a center) right?

  10. Nov 14, 2015 #9


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    All right, I see what you want to calculate now.

    If you cut the loop across its diameter, you'll wind up with two ends of rope located 100 m apart. Due to the rotation of the loop about its center, the centrifugal force acting on the loop wants to pull it apart. The force acting on each end of the rope will be:

    F = m ⋅ ω2 ⋅ r2

    Here, m is the mass per unit length for the rope.

    So the force F = 5.05 ⋅ (π/10)2 ⋅ 502 = 1246 N at each end of the loop.
  11. Nov 14, 2015 #10
    Thank you :)

    I also get the same number with F = (L*dm*r*ω2)/2π , (L=circumference, dm= 5,05)

    And on that centrifugal calculator if i divide the centrifugal provided by the entire mass on a single point/ 2π

    Interesting stuff!
    Last edited: Nov 14, 2015
  12. Nov 15, 2015 #11
    Hey again i just made a calculation with this formula:
    Hoopstrees σH=ρ*r2*r2, with that density ρ=5747,6 (found that number by the volume and mass of 1m wire, and then i spread this to 1 m3)

    So in an area equal to a slice of wire i found a force that is exactly double than the F in the previews post.

    Should i use 1246 or 2492 as the real tensile force within the wire:? (Im i litle bit confused)
  13. Nov 15, 2015 #12


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    It's not clear why ρ = 5747.6 kg/m3 here, unless the rope is not solid iron. If you have a wire rope made by weaving many strands of smaller wire together, then the cross section of the rope will vary, depending on the method of construction.
    The tensile force is like this:
    Code (Text):

    1246 N   <--  o--------------------------------o  -->  1246 N
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