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Homework Help: Bungee Jumping and Hooke's law. Is something missing?

  1. Jan 14, 2007 #1
    Ok, here is a physics problem that I cannot figure out for the life of me. I feel as though I can't solve it without knowing either the mass of the bungee jumper or the spring constant for the cord. Is there a way to get it somehow? Perhaps an important number is missing? I would appreciate any and all help in trying to understand this problem.

    1. The problem statement, all variables and given/known data
    A daredevil plans to bungee jump from a balloon 53.0 m above a carnival midway (Fig. P8.19). He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's force law. In a preliminary test, hanging at rest from a 5.00 m length of the cord, he finds that his body weight stretches it by 1.30 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

    (a) What length of cord should he use?

    (b) What maximum acceleration will he experience?

    2. Relevant equations
    ka + ua = kb +ub
    f = -k*x
    w = f*d
    w = 1/2 *mv2^2 - 1/2 mv1^2

    3. The attempt at a solution
    The first part was rather easy, I simply found the point of equilibrium between 10m and 53m, and PE was 0 at that point.

    The second part makes no sense to me. I do not know where to start to find maximum acceleration.
  2. jcsd
  3. Jan 14, 2007 #2
    If the bungee cord obeys Hooke's law, that means it is essentially a spring. You should be able to figure out the spring constant from the data given (body weight stretches a 5.00 m length by 1.30 m). I haven't tried to solve it, but it seems to me you would need this for part (a) also.

  4. Jan 14, 2007 #3
    For part A, I found the midpoint between 10m and the initial height. I'm not sure how I can calculate k with just the length of the spring and its displacement.
  5. Jan 15, 2007 #4


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    Calculating the mid point will not be of any help to you. At the start of the jump, the jumper has gravitational potential energy only, and it is best to reference that PE to the very bottom of the jump. At the very bottom of the jump, the jumper then has no gravitational PE, no KE (since he comes momentarily to a stop), and therfore only has spring Potential Energy. Use conservation of energy principle. Note that k can be calculated using Hooke's law as a function of the jumper's mass, based on the given data.
  6. Jan 15, 2007 #5


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    Actually, finding the mid point is a way to do part (a). The motion of the jumper is simple harmonic motion. The mid point of that motion is half way between the starting point and 10m from the ground. When he finally comes to rest at the midpoint, the cord is stretched to (5+1.3)/5 times its initial length. That's all you need find the unstretched length.
  7. Jan 15, 2007 #6


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    The jumper is in free fall for most of the journey, until the cord reaches its unstretched length. The motion of the free fall portion of the journey is not harmonic. To calculate the unstretched length, I would set mg(L +x) = 1/2kx^2 where L + x = 43, L is the unstretcghed length of the rope, and x is the maximum stretched length beyond the original unstretched length. See if that compares with your method.
  8. Jan 15, 2007 #7
    There are more coils with a longer spring, that means it should stretch farther given the same force, and the spring constant will be proportionally less. So you can use this to get an expression for the k of the longer spring.

    For part b, that's just a straightforward application of the 2nd law.

  9. Jan 15, 2007 #8
    I don't have k or m, and after reading these suggestions, I'm still at a loss as to how to calculate k. I tried making graphs and finding the slope, but it's still dependent on m(which does not come with the problem). Without either k or m, I can't progress.

    I have finished part a already, the answer was the midpoint of 21.5m.
  10. Jan 15, 2007 #9


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    Find K in terms of the jumper's mass M. That's enough to solve the problem.

    K also depends on the length of the rope.

    You know that for a rope 5 meters long, a force of Mg produces an extension of 1.3 meters. You can write down what K is for that rope: 1.3 K = Mg.

    For a 1 meter long piece of rope, how much extension would be for the same force Mg? (Think about what you would see if there were 5 marks each 1 meter apart on 5 meter rope.) What is the stiffness for a 1 meter long piece of rope?

    Given those two results, what is the stiffness of a piece of rope of any length L?
  11. Jan 15, 2007 #10
    Consider a spring with 5 coils and F = mg hanging from the spring which causes a spring extenstion of 5 cm. That means that each coil will move 1 cm.

    Now double the number of coils while keeping the material exactly the same. Hang the same mass from the spring. Each coil will still move 1 cm, it is the same material, after all. But now the total spring extension will be 10 cm, since now there are 10 coils. The spring is effectively less stiff, the spring constant is less, in proportion to its length.

    Hopefully this will give you enough to go on.

    Last edited: Jan 15, 2007
  12. Jan 16, 2007 #11


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    :blushing: OOPS - you are right, of course. :blushing:

    Sorry if that confused anybody.
  13. Jan 24, 2007 #12
    bungee cord

    by chance I was looking at the section on bungee jumping in the Salters-Horner AS physics students book today.

    It shows a rough force /extension graph for bungee cord (without reference or other justification - im afraid).

    the plot is no way linear, so
    F IS NOT = -kx

    There are 3 broad regions on the graph - an initial fairly linear part, followed by a 'stiff' section where F increases rapidly with x, followed by a final section which goes back to F increasing gently with x.
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