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Bungee Jumper

  1. Aug 7, 2009 #1
    I'm trying to solve "Part A" on this page:
    http://www.star.le.ac.uk/IPhO-2000/problems/theory1.htm [Broken]

    I'm getting some wonky results, so I fear I may be doing something wrong, so I'd be glad if someone could point out my mistake, and even more glad if someone could confirm my findings. :)

    For the first question:
    Considerations of energy are all that are required.
    We'll set our plane of reference for the gravitational potential energy at the height of the bridge.

    [tex]E_i=U_g=0[/tex]

    We are asked about the moment when he comes instantaneously to rest for the first time. At this point, he is [tex]y[/tex] below the bridge, and has only gravitational potential energy, and elastic potential energy (By the definition of this state, his kinetic energy is 0)

    [tex]E_f=U_g+U_e=-mgy+\tfrac{1}{2}kx^2[/tex]

    [tex]x[/tex] is defined as the elongation or contraction of the rope from its unstretched length, so it is therefore equal to [tex]y-l[/tex]

    Conservation of energy holds for this system as all forces are conservative.
    [tex]E_i=E_f[/tex]

    [tex]mgy=\tfrac{1}{2}k(y^2-2ly+l^2)[/tex]

    Rearranging produces a quadratic equation in [tex]y[/tex] :

    [tex]y^2-2(l+\tfrac{mg}{k})+l^2=0[/tex]

    Whittling that quadratic equation down produces:
    [tex]y_{1,2}=l+\tfrac{mg}{k}\pm\sqrt{(l+\tfrac{mg}{k})^2-l^2}[/tex]

    Which further reduces to:

    [tex]y_{1,2}=l+\tfrac{mg}{k}\pm\sqrt{\tfrac{mg}{k}^2+2l\tfrac{mg}{k}}[/tex]

    And here I'm pretty much stuck, not quite sure on how to simplify it further.
    A quick check showed that the root expression cannot be larger than the non-root expression, this makes both the plus and the minus solutions viable. I suspect the minus solution is the height of the jumper above the point of equilibrium when he comes to an instantaneous halt for the second time, which of course doesn't happen, since the rope can only pull, but unlike a spring, cannot push.
    But I'm pretty sure I missed something big to be getting a result this nasty.

    Second question:
    Kinematics, forces and energy.
    The jumper is at his maximum velocity as he passes through the point of equilibrium for the rope. Once he passes it, the direction of his acceleration becomes opposite the direction of his velocity, this happens once the linearly increasing [tex]kx[/tex] overtakes the constant [tex]mg[/tex]

    At the point of equilibrium:
    [tex]kx_0=mg \rightarrow x_0=\tfrac{mg}{k}[/tex]

    Again using conservation of energy, at our plane of reference, the jumper has no energy:
    [tex]E_i=0[/tex]

    At the point of equilibrium, which is [tex]l+x_0[/tex] below the plane of reference:
    [tex]E_f=-mg(l+x_0)+\tfrac{1}{2}kx_0^2+\tfrac{1}{2}mv_{max}^2[/tex]
    Isolating for [tex]v_{max}^2[/tex] and substituting [tex]x_0=\tfrac{mg}{k}[/tex] provides the following result:

    [tex]v_{max}^2=2gl+\tfrac{mg^2}{k}[/tex]

    Again, these results are not very aesthetic, so I don't trust them too much.

    Question 3:
    Kinematics and harmonic motion.
    As this is a rope and not a spring, it can only exert pulling forces, but not pushing ones. From this we conclude that the motion of the jumper consisted of two kinds of motion.

    Free fall up to the point that the rope began to extend and SHM up to his instantaneous stop.

    SHM motion began at the moment the jumper was more than [tex]l[/tex] below the plane of bridge, when the rope started extending.
    [tex]v_f^2=2gl\rightarrow v_{SHM}=\sqrt{2gl}[/tex]
    [tex]t_{freefall}=\sqrt{\frac{2l}{g}}[/tex]

    For our vertical 'spring' we'll be using the harmonic function to measure the displacement relative to the point of equilibrium, with angular velocity [tex]\omega\equiv\sqrt\tfrac{k}{m}[/tex]

    [tex]x(t)=A\sin{(\omega t+\theta_0)}[/tex]

    [tex]v(t)=\omega A\cos{(\omega t+\theta_0)}[/tex]

    [tex]a(t)=-\omega^2 A\sin{(\omega t+\theta_0)}[/tex]

    Our initial variables:

    [tex]x(0)=-x_0[/tex]

    [tex]v(0)=\sqrt{2gl}[/tex]

    [tex]a(0)=mg[/tex]

    Solving for [tex]A[/tex] and [tex]\theta_0[/tex] provides us with the following:

    [tex]\tan{\theta_0}=\sqrt\tfrac{mg}{2kl}[/tex]

    [tex]v(t)=0[/tex]
    [tex]\omega t +\theta_0=\tfrac{\pi}{2}[/tex]
    [tex]t_{SHM}=\frac{\tfrac{\pi}{2}-\theta_0}{\omega}[/tex]

    Which are all known quantities.

    [tex]t_{total}=t_{freefall}+t_{SHM}[/tex]

    Now, I can't help but feel like there was something massive I've done wrong. :\ Could someone please look over my calculations?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 7, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    For the first question:
    OK.

    Double check that second term. (Look for a sign error.)

    For the second question: Your response looks good.

    For the third question: You're on the right track. Think in terms of the period of the SHM. At what time in the cycle did the jumper enter SHM? At what time does he reach momentary rest?
     
    Last edited: Aug 7, 2009
  4. Aug 7, 2009 #3
  5. Aug 7, 2009 #4
    Whoops, thanks for catching that, Doc Al. And thanks a lot for the link, Fightfish! Guess I won't need to bother you guys any more. :shy:
     
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