| Composition Setup |
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| Excerpt | ||
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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 U.S. Military specifications (DoD 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
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N.
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Jumpers
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using
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these
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cords
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intertwine
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three
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to
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five
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cords
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to
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make
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a
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thick
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rope
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that
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is
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strong
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enough
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to
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withstand
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the
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forces
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of
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the
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jump.
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Suppose
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that
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you
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are
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designing
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a
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bungee
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jump
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off
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of
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a
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bridge
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that
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is
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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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surface
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of
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a
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river
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running
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below.
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To
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get
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an
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idea
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for
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the
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maximum
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cord
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length,
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calculate
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the
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unstretched
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length
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of
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cord
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with
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kL
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=
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4000
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N
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(a
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system
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of
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five
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800
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N
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cords)
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that
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will
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result
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in
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a
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118
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kg
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jumper
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who
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leaves
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the
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bridge
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from
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rest
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ending
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their
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jump
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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's
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surface.
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(The
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2.0
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m
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builds
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in
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a
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safety
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margin
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for
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the
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height
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of
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the
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jumper,
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so
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you
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can
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neglect
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the
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height
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of the jumper in
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your
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calculation.)
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Since
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you
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are
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finding
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a
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maximum
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cord
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length,
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ignore
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any
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losses
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due
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to
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air
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resistance
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or
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dissipation
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in
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the
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cord.
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Ignore
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the
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mass
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of
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the
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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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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: We begin with an initial-state final-state diagram for this situation, along with corresponding energy bar graphs. {table}{tr}{td:valign=bottom}!Bungee Jump^bungee1a.png!{td}{td:valign=bottom}!Bungee Jump^bungee1b.png!{td}{tr} {tr}{th:align=center}Initial State{th}{th:align=center}Final State{th}{tr}{table} 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 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:
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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}{latex} |
It
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is
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clear
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that
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the
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appropriate
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choice
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is
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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:
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\begin{large}\[ h_{i} - h_{f} = L + x_{f} \] \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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| m.
h3. Part B
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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