Composition Setup |
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A more complicated version of the Bungee Jump problem. |
(Photo courtesy Wikimedia Commons, uploaded by user Che010.)
Bungee cords designed to
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U.S.
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Military
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specifications
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(DoD
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standard
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MIL-C-5651D,
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available
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at
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)
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are
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characterized
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by
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a
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force
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constant
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times
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unstretched
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length
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in
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the
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range
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kL
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~
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800-1500 N. Jumpers using these cords intertwine three to five cords to make a thick rope that is strong enough to withstand the forces of the jump. Suppose that you are designing a bungee jump off of a bridge that is 30.0 m above the surface of a river running below. To get an idea for the maximum cord length, calculate the unstretched length of cord with kL = 4000 N (a system of five 800 N cords) that will result in a 118 kg jumper who leaves the bridge from rest ending their jump 2.0 m above the water's surface. (The 2.0 m builds in a safety margin for the height of the jumper, so you can neglect the height of the jumper in your calculation.) Since you are finding a maximum cord length, ignore any losses due to air resistance or dissipation in the cord. Ignore the mass of the rope.
Solution
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The jumper (treated as a |
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The system constituents interact via gravity, which contributes |
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We begin with an initial-state final-state diagram for this situation, along with corresponding energy bar graphs.
Initial State | Final State |
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As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model:
Latex |
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N. Suppose that you are designing a bungee jump off of a bridge that is 30.0 m above the surface of a river running below. To get an idea for the maximum cord length, calculate the unstretched length of cord with _kL_ = 800 N that will result in a 118 kg jumper who leaves the bridge from rest ending their jump 2.0 m above the water's surface. (The 2.0 m builds in a safety margin for the height of the jumper, so you can neglect the height in your calculation.) Since you are finding a maximum cord length, ignore any losses due to air resistance or dissipation in the cord. Ignore the mass of the rope. System: The jumper (treated as a [point particle]) plus the earth and the bungee cord. The system constituents interact via gravity, which contributes [gravitational potential energy], and via the restoring force of the cord, which contributes [elastic potential energy]. External influences are assumed negligible. Model: [Constant Mechanical Energy]. Approach: Shown above is an initial-state final-state diagram for this situation, along with corresponding energy bar graphs. As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model: {latex}\begin{large} \[ E_{\rm i} = mgh_{\rm i} = E_{\rm f} = mgh_{\rm f} + \frac{1}{2}k x_{f}^{2} \] \end{large}{latex} {info}Bungee cords provide a restoring force when stretched, but offer no resistance when |
Info |
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Bungee cords provide a restoring force when stretched, but offer no resistance when "compressed", since they fold like an ordinary rope. Thus, the initial spring energy is zero in this case. {info} |
This
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equation
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cannot
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be
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solved
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without
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further
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constraints,
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since
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we
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do
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not
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know
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k
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.
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The
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extra
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constraint
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that
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we
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have
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is
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given
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by
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the
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fact
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that
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the
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jumper
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has
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fallen
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a
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total
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of
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28.0
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m
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(descending
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from
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30.0
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m
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above
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the
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water
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down
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to
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2.0
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m
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above
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the
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water).
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This
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distance
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must
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be
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covered
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by
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the
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stretched
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cord.
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This
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gives
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us
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the
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contraint:
Latex |
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}\begin{large} \[ h_{\rm i} - h_{\rm f} = L + x_{f} \] \end{large}{latex} |
Solving
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this
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constraint
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for
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x
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f
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and
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substituting
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into
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the
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energy
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equation
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gives:
Latex |
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}\begin{large} \[ -2mg(\Delta h) = k((\Delta h)^{2} + 2 L \Delta h + L^{2}) \] \end{large}{latex} |
Where
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we
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are
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using Δh = hf - hi to simplify the expression.
We still have two unknowns, but we can resolve this by using the fact that kL is a constant for the rope. Thus, if we multiply both sides by L, we have:
Latex |
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Δ_h_ = _h_~f~ - _h_~i~ to simplify the expression. We still have two unknowns, but we can resolve this by using the fact that _kL_ is a constant for the rope. Thus, if we multiply both sides by _L_, we have: {latex}\begin{large} \[ -2mg(\Delta h)L = C((\Delta h)^{2} + 2L \Delta h + L^{2}) \] \end{large}{latex} |
where
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we
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have
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replaced
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the
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quantity
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kL
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by
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C
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(=
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800
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N)
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for
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clarity.
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With
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this
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substitution,
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it
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can
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be
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seen
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that
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we
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have
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a
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quadratic
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equation
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in
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L
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which
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can
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be
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solved
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to
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find:
Latex |
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}\begin{large} \[ L = \frac{-2(C+mg)\Delta h \pm \sqrt{4(C+mg)^{2} (\Delta h)^{2}-4C^{2}(\Delta h)^{2}}}{2C} = 13.3 \:{\rm m}\;{\rm or}\;58.9 \:{\rm m}\] \end{large} |
It is clear that the appropriate choice is 13.3 m.
Note |
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Does the other root have a physical meaning? |
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According to bungee jumping enthusiasts, a good rule of thumb is to expect that the maximum length reached by a bungee cord during a jump will be 210% of its unstretched length (assuming the cord is rated for the weight of the person jumping). Thus, a 10 m cord should stretch to 21 m during a jump. How does this rule of thumb compare to our calculated estimate? |
Follow Up – Checking Assumptions
In our constraint that:
Latex |
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\begin{large}\[ h_{i} - h_{f} = L + x_{f}{latex} \] \end{large} |
we neglected the fact that the jumper's center of mass might drop an extra meter or more (depending upon the point of attachment of the cord to the person). What effect would such an extra drop have on the final height?
Follow Up – Checking Assumptions
Suppose a 118 kg jumper foiled your calculations by leaping upward from the bridge with an initial speed of 2.25 m/s. Will the jumper hit the water, assuming the 13.3 m cord with kL = 4000 N that we found in Part A?
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