Question

Design a “bungee jump” apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while,with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 13 m long, and that the cordsstretch in the jump an additional 23 m for a jumper whose mass is 80 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit theground).

(a) At what instant is there the greatest tension in the cords? (How do you know?)

(b) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness ks for each of the two cords.

(c) What is the jumper's speed at this instant, when the tension is greatest in the cords?

(d) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.)

(e) What is the maximum acceleration |ay| = |dvy/dt| (in “g's”; acceleration divided by 9.8 m/s2) that the jumper experiences? (Note that |dpy/dt| = m|dvy/dt| if v issmall compared to c.)

(f) What approximations or simplifying assumptions did you have to make in your analysis that might not be adequately valid?

Answer 1

Wow, that's a very involved question. We'll assume thatthe cord obeys Hooke's Law but it only pulls the jumper towardsequilibrium, that is to say, the cord stretches but doesn'tcompress.

a) At first the cord is slack until 10 m have extended,at which point only constant gravity acts on the jumper (ignore air resistance). Once 10 m are out, we'll call thatthe equilibrium position.  Below equilibrium, theelastic force, which is variable acts on the jumper to bring himback to equilibrium. Once above equilibrium, only gravityacts again, and he undergoes some sort of harmonic motion, but notreally SHM.

b) The elastic force is greatest when he is furthestdownward from equilibrium, that is 20 m below. This is basedon Hooke's Law, as 'Fs' is proportional to distortion'x'.

c) At that instant, the speed is zero. I didn'thave enough space to draw all the pictures, but the velocity has tochange from downward to upward at some point, and that happens atthe very bottom, and the speed is zero at the instant itchanges direction

d) The momentum is changing at that instant, so dp / dtis nonzero. Given a constant mass, we can rewrite dp /dt as    m dv / dt. The velocity isindeed changing there, based on the argument from c. above.

e)   I was thinking maybe we're supposed towrite a position function but it's rather complicated as we havetwo kinds of Physics going on, the spring forceacting (SHM), and free fall. Instead, try energyconservation. Define the referene level as the equilposition. Make up a mass of 80 kg for the jumper.

Energy at top = Energy at bottom

KE + PE_g +PE_s = KE' +PE'_g + PE'_s

          m gh         = m gx + 2*1/2 kx²       ---> since there are 2 cords, k is for one ofthem

                                                                          each cord has a certain amount of PE

  

        m g (h -x)   = k x²

m g (h - x)  /x² =k               Theway it was set up, h > 0 and x< 0

m g (|h| +|x|)  / x² = k

∴ 80 kg * 9.8 (10 - (- 20 m)) / (20 m)² =k

                                             58.8N/m  = k

(f) Substitue this into Hooke'sLaw:     Fs = k x

                                                                =58.8 N/m * 20 m

                                                               = 1176 N  or 1.5 times hisweight

Interestingly, we get this same result for any mass ofthe jumper based on how the cord is designed to stretch.

ratio = Fs / mg

      

        = k x /mg

        = mg (|h| + |x|)  /x² * x /mg

        = (|h| + |x|) / x    = 30 m / 20 m = 1.5

(g) The max acceleration occurs when Fs ismaximized, which is again, at the bottom of the fall. Notice there are two cords pulling up, so we must double the Fscalculated above

a = ΣF / m = (2Fs - mg) / m

                  = (2Fs / m)   - g

                   = 2*1176N / 80 kg - 9.8  =  19.6 m/s²

The direction is upward

(h) Assumed no air resistance and that the elastic obeysHooke's Law