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The Physics Behind Bungee Jumping

  1. Mar 18, 2004 #1
    Hey, this is my first time on the forum and I have to do a project about bungee jumping.

    I need to talk about Hooke's Law, calculate The Conservation of Energy and the Conservation of Momentum...

    I know that F=k*x, but I am still a little unsure on how to relate it to bungee jumping.

    Assuming that there is no loss of energy due to friction, air resistance etc etc, I am confused on how the conservation of momentum and conservation of energy relate to bungee jumping. I would like to know how to calculate these topics for bungee jumping.

    Of course, this is my homework, so I am not expecting you to do it for me, although it would be a great help!


  2. jcsd
  3. Mar 18, 2004 #2
    Before you jump, you're at a certain height above the ground. We can attribute to this height a certain amount of potential energy. When you jump off, this energy starts to be converted into kinetic energy, and you speed up while you fall. Then the bungee cord kicks in and starts transforming the kinetic energy and potential energy to elastic energy, which is stored in the cord and why it stretches. Once you stop, most (hopefully there'll still be SOME left so you don't smash into the ground) of the potential energy and all of the kinetic energy (hence, stopped) has been converted to elastic energy.

    It's all about the energy! If you idealize your world enough, the calculations are pretty simple, too.

    Does that help?

  4. Mar 18, 2004 #3
    Yes! Thankyou!

    Potential = m*g*h (mass * gravity * height) ?
    Kinetic = 0.5 * m * v^2 (mass * velocity^2) ?
    Elastic potential = ???
  5. Mar 18, 2004 #4
    [tex]\textrm{Elastic Energy} = \frac{1}{2}kx^2[/tex]
    where k is an empirically determined constant and x is the displacement off of the rest position of the cord.

  6. Mar 18, 2004 #5
    How do I find k?
  7. Mar 18, 2004 #6
    I imagine the manufacturers of bungee cords are required to publish them. If not, you just test it by putting a few weights of known mass on the cord and measure how far it stretches. The graph of displacement off of rest position vs. mass will yield a straight line, the slope of which can be used to calculate the constant.

  8. Mar 18, 2004 #7
    Wicked! Thanks! You're a great help!

    If I have any more questions, I'll come back, for now, I think I am good =)

    Thanks again
  9. Mar 18, 2004 #8
    Okay, I think I have things figured out so far. Could you check it for me please?

    Height of bridge = 100m
    Rope Length = 60m
    Elongation = 50% (30m)
    Mass = 75kg

    To find k in (F=kx)

    F=m*g F=(75kg)x(9.81m/s^2) F=735.75N

    F=k*x 735.75N=k(30m) k=24.525

    That good so far?

    As the rope is 60m, does this mean that the equilibrium is at 60m? And because the rope does not compress, the equilibrium is also at 60m to 100m high?

    The person is at his maximum speed at 60m? How do I find his maximum speed? I know speed=distance/time, but I do not know the time.

    I'd like to go bungee jumping and see, but I don't have time for that!

    ~Thanks for all your help!
    Last edited: Mar 18, 2004
  10. Mar 18, 2004 #9
    Yes, your calculations are correct.

    How are you getting your data? It's generally advisable to use many data points in order to minimize error, although it is quite possible to determine the constant from a single point, as you just demonstrated.

    Oh, and you might want to stick the units on the constant.

  11. Mar 18, 2004 #10
    I made my data up... It's a hypothetical experiment. If there is no air resistance/drag, then he will be at terminal velocity at 60m?

    The units for the constant are (N/m)?
    Last edited: Mar 18, 2004
  12. Mar 19, 2004 #11
    No, the maximum (or minimum) velocity is achieved when the acceleration is zero. This is because the acceleration is the derivative of the velocity, and as you know f(x) has a maximum (or minimum, or "twist") when f'(x) = 0.

    This happens when the force that the rope makes on the person is equivalent to the gravitational force, but in the opposite direction. As you probably realize, this is exactly the equilibrium point, 90 meters below the bridge. (Of course, in this case the person doesn't stay still at the equilibrium point, since he still has kinetic energy.)
    Last edited: Mar 19, 2004
  13. Mar 19, 2004 #12
    I think that now that we got this discussion of maximum velocity going on, we should more precisely define a few things.

    Imagine the guy isn't dropping. Imagine that he's simply hanging from the rope. The guy will be resting when the force is zero (thank you, Captain Obvious), and this will happen when the elastic force is exactly equal and opposite the gravitational force, i.e. kx = mg, x = mg/k, where x is the displacement from the cord if there were zero mass on it. So there is a difference between the rest position of the cord and the equilibrium position of the cord. They're different by a factor of mg/k. Note that this shift has no bearing on the remaining calculations, i.e. F still equals -kX, where X is defined as (x - mg/k).

    That being said, the maximum velocity will be as Chen said, at the equilibrium position. It's not difficult to see by writing the full differential equation (it's not as scary as it sounds) for the motion of the bungee jumper and then deriving the equations of motion.

    There's probably one other thing to keep in mind. I might be wrong about this, but I've never seen many bungee cords that compressed. So the elastic force should be F = -kX for X >= mg/k and F = 0 for X < mg/k, at which point the jumper is in free fall and the rope just kind of loops up. It complicates things, but it's not very difficult to deal with.

    And yes, the spring constant will have units N/m.

  14. Mar 19, 2004 #13

    mass = 75kg
    height, before = 200m, after = 140m

    Energy before = Energy After

    Potential + Kinetic = Potential + Kinetic
    mgh + 0.5mv^2 = mgh + 0.5mv^2

    (75)(9.81)(200) +(0.5)(75)(0^2) = (75)(9.81)(140) + (0.5)(75)(v^2)
    147150J + 0 = 103005J + 37.5v^2
    44145J = 37.5v^2
    v^2 = 1177.2
    v = 34.31m/s

    This is the velocity at 140m (a 60m dive)... I think. Can anyone see a flaw in my calculations?

    velocity = displacement / time
    time = displacement / velocity
    time = 60m / 34.31m/s
    time = 1.75 s

    The jumper is 40m above ground at 1.75 seconds.

    Now for when the rope has elastic potential energy at maximum displacement:

    Energy Before = Energy After
    147150J = Potential + Kinetic + Elastic
    147150J = mgh + 0.5mv^2 + 0.5kx^2
    147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(k)(x^2)

    We need the height, the constant in the rope, and the maximum displacement,x,

    Using F=kx and F=mg to determine k.


    The maximum elasticity of the rope is 50%, so a rope that is 60m will stretch a maximum of 30m, before it loses it's elasticity (Hooke's Law). So, x = 30m.


    147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(-k)(30^2)
    147150J = (735.75)(h) + 0 + -11036.25
    735.75h =158186.25
    h = 215m

    However, how can it be 215m? it would be 15 metres above where it started... This is where I am confused.
  15. Mar 19, 2004 #14
    That would be the first mistake. Generally speaking:

    [tex]\Delta V \Delta t = \Delta X[/tex]

    In our case, however, the person's velocity is not constant. So the equation above is not true for large time intervals (it can only be used to calculate the momentary displacement).

    You need to use the more general formula of:

    [tex]X = X_0 + V_0 t + \frac {1}{2} at^2[/tex]

    Luckily for you, the initial displacement is 0 and so is the initial velocity. So finding the time should be a problem. :)
  16. Mar 19, 2004 #15
    If the maximum elasticity of the rope is 50%, I'm afraid our jumper will not live to see another day. Because only when the rope is elongated by 50%, it exerts enough force to cancel the weight of the person. Until that point, the gravitational force was still stronger than the rope's force. This means that only at 50% the person's acceleration is zero, but as I explained above this is also the point in which the person has the maximum velocity. The rope must elongate by more than 50%, or it would tear apart. (In this specific case)

    The second mistake is that your equation implies that the person is resting at 50% elongation, which is not true as I said. What you need to do is express h as a function of x (or x as a function of h), so that your equation only has one variable. The answer for x is only the minimum elastic range of the rope, if you want your daredevil jumper to survive.
    Last edited: Mar 19, 2004
  17. Mar 19, 2004 #16
    What level of math can you do? Can you solve the differential equation

    [tex]m\ddot{y} = -mg - (y + \frac{mg}{k})[/tex]


  18. Mar 19, 2004 #17
    I am assuming that a is the acceleration, v is velocity, and t is time. Without the acceleration or time, I can't find the velocity.

    I am in High School Calculus. We have learnt how to differentiate equations, but not those with so many variables.

    Once the rope begins the elongate, the jumper loses velocity because his kinetic energy is transferred to elastic energy. His maximum velocity is at 140m, when there is no elastic potential energy. His maximum velocity is still 34.31m/s.

    Energy Before = Energy After.

    Susbsequent to a long night of thinking about it, my equation did not contain all the energies. Thermal (frictional) eenrgy was not included, which is huge because of the anture of bungee ropes. All those little stretchy things inside the rope rub against each other. In a rope of 60m, the thermal energy will be large? Of course, there are many other types of trace energies that are not counted for. But as they are so little, it would not alter my answer tremendously.
    Last edited: Mar 19, 2004
  19. Mar 19, 2004 #18
    But I do not know how much energy is Elastic energy because I do not know the ropes k value.
  20. Mar 20, 2004 #19
    That is not precise. Once the rope begins to elongate, it does acquire elastic potential energy. But at first that energy comes at the expense of gravitational potential energy, not at the expense of kinetic energy. In fact, even when the elastic energy is growing, so does the kinetic energy until some point. That point is the equilibrium, in our case 50% elongation. Only then does the force of the rope begin to outbalance the weight of the person, which means his acceleration is now negative (he is decelerating). Only then does the kinetic energy begin to decrease.

    His maximum velocity is not necessarily at the point where he has no elastic energy, it is at the point where he has the minimum of potential energy, which includes both elastic and gravitational energy. If you write down the equation of the person's potential energy (which is preserved, if we ignore friction and heat) you will get:

    [tex]\Sigma E_p = E_{gra} + E_{els} = mgh + \frac{1}{2}kx^2[/tex]

    If you express h as a function of x you will get:

    [tex]\Sigma E_p = mg(200 - x) + \frac{1}{2}kx^2[/tex]

    To find the minimal potential energy, get the derivative of the function and see where it is zero:

    [tex]{E'}_p = -mg + kx = 0[/tex]
    [tex]x = \frac{mg}{k}[/tex]

    This is the point where the potential energy of the diver is minimal, and therefore it is the point where his kinetic energy (and velocity) is maximal.

    Do you understand or should I explain it a different way? :smile:
  21. Mar 21, 2004 #20
    I understand... you're right

    However, because of the lack of time, I don't have time to do all those calculations.... :frown:

    My website is at http://ca.geocities.com/physicsroxmysox/, I am sure you can find lots of mistakes in my work, because you're all really intelligent at Physics!

    Thankyou for all your help! I'll be sure to come back if I need any help

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