Spring of mass 'm' hung from ceiling.total elongation?

  1. spring of mass 'm' hung from ceiling.what is the total elongation?
    assume total spring constant 'k',length 'l' and of uniform density when in natural state.
    also where is the center of mass?
    can we express density at a point as a function of distance from the ceiling?
     
    Last edited: Nov 30, 2013
  2. jcsd
  3. PhanthomJay

    PhanthomJay 6,267
    Science Advisor
    Homework Helper
    Gold Member

    Well you can use calculus if that's your strength. Or you can reason it out.....If the spring is slowly lowered to its equilibrium position, and the mass was concentrated all at the lower end, its elongation would be mg/k, using Hooke's Law. If the mass was all concentrated at the upper end, the spring would not elongate at all! But the center of mass of the spring is not at the top or bottom , its in the middle. So, the elongation is not mg/k , nor 0, but rather, it is ?????
     
  4. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    Assuming the total elongation of the spring is the same as the movement of the center of mass, without any justification, is a big leap IMO, even if it is true.

    But sometimes, you get lucky and the wrong logic still gives the right answer :smile:
     
  5. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    Usually in this type of question, you assume the displacements are small, so you can ignore the change in density as the spring stretches.

    The top half and the bottom half of the spring will stretch by different amounts, so it is wrong to say the total extension (at the bottom) is the same as the extension of the center of mass.

    But sometimes, you get lucky and hit the right answer by an invalid argument :smile:
     
  6. PhanthomJay

    PhanthomJay 6,267
    Science Advisor
    Homework Helper
    Gold Member

    the weight of the spring varies linearly with length. Rather than resort to calculus, which is the ideal way, experience tells me that in calculating total deflection, you may assume the weight to be concentrated at the center of mass. This effectively doubles the spring stiffness, yielding half the the deformation of the top half while the bottom half goes along for the ride. Of course, this method does not yield correct deformation at points over the springs length, but yields correct results for total elongation. You told me I was lucky twice, but rather, sound reasoning on my part. You don't have to be genius to find averages. Smile.
     
  7. I did the calculus and got the elongation as 'mg/2k'.I think I am wrong because I am not so good at calculus.
    the case is much more like rope of young's modulus 'Y' hung from ceiling.slinky is the closest reference.elongation in slinky is 'mg/2k' according to wikipedia
     
  8. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    The double post was finger trouble. The second post was meant to be an edit of the first one.

    You can call it whatever sort of reasoning you want, but if you made that argument in a report that I had to sign off at work, I wouldn't sign it.
     
  9. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    There is no real difference in the math between specifying Young's modulus E and the spring stiffness k.

    For the rope, the equivalent "spring stiffness" k = EA/L where A is the cross section area. For a spring, it is harder to give a formula for k in terns of the diameter of the wire, diameter of the coils, number of coils per unit length, material properties of the wire, etc, so you just use the value of k.

    I think your answer is right.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted