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Homework Help: Elastic pendulum energy problem

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that the string in fig 8-12 (just a pendulum released from the side and falls to the bottom) is very elastic, made of rubber, say, and that the string is unextended at length l when the ball is released.

    a) explain why you would expect the ball to reach a low point greater than a distance l below the point of suspension.

    b) show, using dynamic and energy considerations, that if [tex]\Delta[/tex]l is small compared to l, the string will stretch by an amount [tex]\Delta[/tex]l = 3 mg/k, where k is the assumed force constant of the string.

    2. Relevant equations

    3. The attempt at a solution

    for a, I suppose that the weight would displace the mass further down before the force of the string -kx "pulls" it back. but I'm not entirely sure..

    for b) I tried to work out the change in K = Work from weight + work from string(spring like). I think if the change in l is small compared to l, then I can just say change in l + l = l, approximately. besides that, I'm not really sure.. any help please? thanks
  2. jcsd
  3. Jan 16, 2010 #2
    I stopped thinking about this but now came back to it.. please, can someone help? My argument is that the weight will pull it down until the force done by the rubber band (-kx) is >= than the weight.

    For part b, I got pretty close:

    if [tex]\Delta[/tex]L is relatively small to L, then L+[tex]\Delta[/tex]L is approximately L.

    So, just as it is released, the system has an energy level of mgL + [tex]\Delta[/tex]L, approximately mgL. Then the system at its lowest point will have K approx = mgL. At the point just as the rubber band is about to stretch, mgL = K + mg[tex]\Delta[/tex]L. Then, K = mgL - mg[tex]\Delta[/tex]L.

    So then we have

    [tex]\Delta[/tex]k = [tex]\Sigma[/tex] Wc

    or approximately,

    mgL - mgL + mg[tex]\Delta[/tex]L = -k[tex]\Delta[/tex]L^2 - mg[tex]\Delta[/tex]L

    2mg[tex]\Delta[/tex]L = -k[tex]\Delta[/tex]L^2

    then we have 2mg = -K[tex]\Delta[/tex]L

    and [tex]\Delta[/tex]L = 2mg/-K

    but the answer is 3 mg / K.. What went wrong?

    Last edited: Jan 16, 2010
  4. Jan 16, 2010 #3
    okay, I realized my error - there is no "At the point just as the rubber band is about to stretch, mgL = K + mgL. Then, K = mgL - mgL." the rubber band stretches delta L over the fall.
    back to the drawing board..
  5. Jan 17, 2010 #4
    can anybody help? please?
  6. Jan 18, 2010 #5
    still looking for help..
  7. Jan 19, 2010 #6


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    Write out Newton's second law for the radial (towards the center) direction. You'll see that the tension in the string has to balance out the radial component of gravity and provide the centripetal acceleration. What happens to both as the bob moves downwards?

    Try writing out the equation for K = Work from weight + work from string in terms of l, delta-l, k, m, etc.
  8. Jan 19, 2010 #7


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    I don't quite understand the calculations you did for b, so here's a hint: go back to the equation you wrote in part a) for the radial direction. You'll see that if you know the speed of the pendulum at the bottom, you can solve for tension. From there you can solve for delta-L. How do you use the conservation of energy to solve for speed?
  9. Jan 19, 2010 #8

    thanks a lot - I appreciate it. but don't we already know the kinetic energy at the bottom? and what's the idea behind "that if that if LaTeX Code: \\Delta l is small compared to l".. should we consider L + delta L to just be L? I tried to solve for delta-L by sum of work = change in K.. where we know the bottom K

    thanks again! :)
  10. Jan 19, 2010 #9


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    If you write out the equation I mentioned, you'd see that T=mg+mv^2/L at the bottom. delta-L=T/k, so you need to find v in order to find delta-L. Yes, you know the kinetic energy at the bottom, so finding v should be easy.

    Also, yes, you can consider L+delta-L to equal L when computing v.
  11. Jan 19, 2010 #10
    thanks for replying. when I calculate the work = change in K, I need to calculate the work done by the weight, the spring force and the tension right? is that why I need to know about the tension? (other than the first part of the question)
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