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Bungee jumps involve elastic and gravitational potential energy. |
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(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
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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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50.0
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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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You
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have
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read
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that
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you
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should
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expect
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the
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cord
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to
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stretch
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(at
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peak
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extension)
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to
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about
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210%
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of
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its
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natural
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length.
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You
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have
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also
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read
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that
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you
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should
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use
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3
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cords
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together
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for
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jumpers
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with
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weights
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in
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the
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range
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100-150
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lbs,
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4
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cords
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for
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150-200
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lbs,
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and
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5
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cords
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for
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200-250
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lbs.
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Suppose
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you
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decide
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to
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use
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cords
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of
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length
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20
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m,
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which
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would
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seem
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to
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offer
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a
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safety
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zone
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of
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about
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8
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m
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(or
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really
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about
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6
...
m
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if
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the
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cord
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is
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attached
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at
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the
...
ankles).
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Part A
Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that kL = (4)(800
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N)
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=
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3200
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N.
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Since
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you
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are
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evaluating
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the
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safety
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factor,
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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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We begin with an initial-state final-state diagram and the corresponding energy bar diagrams:
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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 (treated as an object of infinite mass) and the bungee cord and the bridge (also treated as an object of infinite mass). The system constituents interact via gravity, which contributes [gravitational potential energy], and via the restoring force of the cord, which contributes [elastic potential energy|Hooke's Law#epe]. External influences are assumed negligible. Model: [Mechanical Energy and Non-Conservative Work]. Approach: We begin with an initial-state final-state picture and the corresponding energy bar diagrams: {table:border=1|cellspacing=0}{tr}{td:valign=bottom}!bungee1a.png!{td}{td:valign=bottom}!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} {note}Bungee cords provide a restoring force when stretched, but offer no resistance when |
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Bungee cords provide an elastic restoring force when stretched, but offer no resistance when "compressed", since they fold like an ordinary rope. Thus, the initial is zero in this case. {note} |
It
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seems
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that
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we
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have
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a
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problem
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here,
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because
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we
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do
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not
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know
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h
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f or
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x
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f.
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The
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easiest
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way
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to
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deal
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with
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this
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problem
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is
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to
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utilize
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our
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freedom
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to
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choose
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the
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zero
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point
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of
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the
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height
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axis.
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If
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we
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restructure
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our
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to
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place
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h
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=
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0
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m
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at
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the
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point
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where
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the
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cord
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is
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stretched
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to
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its
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natural
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length
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L
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(as
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shown below).
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Initial State | Final State |
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This redefinition of the origin of the height in our coordinate system greatly simplifies the equation describing the evolution of the jumper's mechanical energy:
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below) then we can rewrite our equation: {warning}Note that we are assuming here that the person is much shorter than the cord, and so we will treat the person as a point particle.{warning} {table:border=1|cellspacing=0}{tr}{td:valign=bottom}!bungee2a.png!{td}{td:valign=bottom}!bungee2b.png!{td}{tr} {tr}{th:align=center}Initial State{th}{th:align=center}Final State{th}{tr}{table} {latex}\begin{large} \[ mgL = - mgx_{f} + \frac{1}{2} k x_{f}^{2} \] \end{large}{latex} {note}You can also solve using the initial coordinate system. You would simply have to substitute _h_~f~ = 50 m - _L_ - _x_. After cancelling _mg_(50 m) from each side, you recover the same expression.{note} We now have a quadratic, which is solved to obtain: {latex} |
| Note |
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You can also solve using the initial coordinate system. You would simply have to substitute hf = 50 m - L - x. After cancelling mg(50 m) from each side, you recover the same expression. |
The discussion around the diagrammatic representation has left us with a quadratic equation in xf, which is solved to obtain:
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\begin{large} \[ x_{f} = \frac{mg \pm \sqrt{(mg)^{2} + 2 k mg L}}{k} \]\end{large}{latex}
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It
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is
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not
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sensible
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that
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we
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should
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find
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a
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negative
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value
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for
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x
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f,
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so
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we
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must
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select
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the
...
plus
...
sign,
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giving:
| Latex |
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}\begin{large} \[ x_{f} = \frac{mg}{k}\left(1+\sqrt{1+\frac{2kL}{mg}}\right) = 21.5\:{\rm m} \] \end{large}{latex} {tip}Does this bear out the original estimate that the cord should stretch to 210% of its initial length?{tip} {tip}Can you convince yourself by looking at the form of the symbolic answer that a 150 lb person attached to 3 cords and a 250 lb person attached to 5 cords would have the same _x_~f~ as the 200 lb person on 4 cords?{tip} h3. Part B Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords? |
| Tip |
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Does this bear out the original estimate that the cord should stretch to 210% of its initial length? |
| Tip |
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Can you convince yourself by looking at the form of the symbolic answer that a 150 lb person attached to 3 cords and a 250 lb person attached to 5 cords would have the same xf as the 200 lb person on 4 cords? |
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Part B
Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords?
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