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 12
m long, and that the cords stretch in the jump an additional
19
m for a jumper whose mass is 120
kg, the heaviest adult you will allow to use your bungee jump
(heavier customers would hit the ground).
(a) It will help you a great deal in your analysis to make a
series of 5 simple diagrams,
like a comic strip, showing the platform, the jumper, and the two
cords at the following times in the fall and the
rebound:
1 while cords are slack (shown here as an example to get you
started)
2 when the two cords are just starting to stretch
3 when the two cords are half stretched
4 when the two cords are fully stretched
5 when the two cords are again half stretched, on the way
up
On each diagram, draw and label vectors representing the
forces acting on the jumper, and the jumper's velocity.
Make the relative lengths of the vectors reflect their relative
magnitudes.
(b) At what instant is there the greatest tension in the cords?
(How do you know?)
When the person
has fallen 12 m.At the top, when
the person has fallen 0
m. When the person
has fallen between 12 m and the bottom.At the
bottom, when the person has fallen 31 m.When the
person has fallen between 0 m and 12 m.
(c) What is the jumper's speed at this instant, when the tension is
greatest in the cords?
v
=
m/s
(d) Is the jumper's momentum changing at this instant or not? (That
is, is
dp_{y}/dt
nonzero or zero?)
Yes, the
jumper's momentum is changing.No, the jumper's
momentum is not changing.
(e) Which of the following statements is a valid basis for
answering part (d) correctly?
After a very
short time the momentum will be upward (and
nonzero).Since the
momentum is zero, the momentum isn't changing.If
the momentum weren't changing, the momentum would remain zero
forever.Since the net
force must be zero when the momentum is zero, and since
dp_{y}/dt is equal to the net force,
dp_{y}/dt must be zero.A very
short time ago the momentum was downward (and
nonzero).
Check to make sure that the magnitudes of the velocity and force
vectors shown in your diagram number 4 are consistent with your
analysis of parts (c), (d), and (e).
(f) Focus on this instant of greatest tension and, starting
from a fundamental principle
,
determine the spring stiffness
k_{s}
for each of the two cords.
k_{s}
=
N/m
(g) 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.)
F_{T}
=
N
(h) What is the maximum acceleration |a_{y}|
= |dv_{y}/dt|
(in "g's") that the jumper experiences? (Note that
|dp_{y}/dt|
=
m|dv_{y}/dt|
if
v
is small compared to
c.)
|a_{y}|
=
g's (acceleration in m/s^{2}
divided by 9.8 m/s^{2})
(i) What is the direction of this maximum
acceleration?
upwardno
direction, since the acceleration is
zero downward
(j) What approximations or simplifying assumptions did you have to
make in your analysis which might not be adequately valid? (Don't
check any approximations or simplifying assumptions which in fact
have negligible effects on your numerical results.)
Neglect air
resistance, despite fairly high speeds.Assume the
speeds are very small compared to the speed of
light.Assume that the
gravitational force hardly changes from the top of the jump to the
bottom.Assume tension
in cord proportional to stretch, even for the very large stretch
occurring here.