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Elasticity of a Wire spinning in a circle

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A 12.0-kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.70 m, is whirled in a vertical circle with a constant angular speed of 120 rev>min. The cross-sectional area of the wire is 0.014 cm2. Calculate the elongation of the wire when the mass is (a) at the lowest point of the path and (b) at the highest point of its path.

    2. Relevant equations
    F⊥ = mv^2/R
    F = m*a
    Y = (F⊥/A)/(Δl/lo)
    v = wR

    3. The attempt at a solution

    a) So if we consider the mass at the end of the string. We get that T - mg = m*v^2/R
    T - 12g = 12*(wR)^2/R
    T - 12g = 12*w^2/R

    R = Δl + lo = 0.7 + Δl

    w = 120 * 2π/ 60 = 4π

    T - 12g = 12*16π / (0.7 + Δl)

    I'm not sure where to go from here
     
  2. jcsd
  3. Sep 8, 2016 #2

    Nidum

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    What are the material properties of Aluminium ?
     
  4. Sep 8, 2016 #3
    Young's modulus for aluminum, according to google is 69 N/m^2
     
  5. Sep 8, 2016 #4

    haruspex

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    Look more closely at the web page. I think you'll find you've omitted quite a large power of 10.
    Indeed, I would be surprised if the elongation in this case made enough difference to the length that you need to take it into account in the expression for centripetal force.
     
  6. Sep 8, 2016 #5

    Nidum

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    Ok Young's Modulus is the property you need to help solve this problem . You got the magnitude wrong though - should be 69 GN/m^2 or 69 GPa .
     
  7. Sep 8, 2016 #6
    Right, yeah, so that would be 69*10^9 then.

    Okay, so if I plug in 69*10^9 * Δl/lo = T/0.014,

    then T = 69*10^9*0.014 * Δl / 0.7 = 1380,000,000*Δl

    SO

    1,380,000,000*Δl - 12g = 12*16π / (0.7 + Δl)

    (1,380,000,000*Δl - 12g) * (0.7 + Δl) = 603

    And I solve this quadratic?

    It seems like I'm doing something wrong
     
  8. Sep 8, 2016 #7
    Also, to harupex's point, should I ignore the elongation and just treat the radius of the circle as R = lo = 0.7m since it should be small in comparison to the length.
     
  9. Sep 8, 2016 #8

    Nidum

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    Yes .
     
  10. Sep 8, 2016 #9

    Nidum

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    It is a bit difficult to follow your calculations as written .

    For the wire under tension T what is :

    The stress ?
    The strain ?
    The change in length ?
     
  11. Sep 8, 2016 #10
    The change in length should be Δl

    Using the two equations I had:

    T - 12g = 12*16π / (0.7 + Δl)

    and

    Y * Strain = Stress ⇒ 6.9*10^9 * Δl/0.7 = T/0.014

    Solving for T from the first equation to get T = 12*16π / (0.7 + Δl) + 12g

    Plugging it into the second, to get

    12*16π / (0.7 + Δl) + 12g = 6.9*10^9 * Δl/0.7 * 0.014

    Now in order to solve this equation I would have to multiply out the (0.7 + Δl) and get a quadratic, which seems unnecessarily complicated, assuming I've done everything right to this point. So if I take your advice and ignore the Δl, I get:

    12*16π / 0.7 + 12g = 6.9*10^9 * Δl/0.7 * 0.014

    solving this huge mess gives me: 7.094 * 10^-6.

    Which doesn't seem right to me
     
  12. Sep 9, 2016 #11

    haruspex

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    Check that step.
    What happened to the power of 2 on w?
     
  13. Sep 9, 2016 #12
    I was rushing everything and made alot of mistakes. I corrected those mistakes since then and got the correct answer. Thanks for the help!
     
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