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{excerpt:hidden=true}Bungee jumps involve elastic and gravitational potential energy. {excerpt} {table:align=right|cellspacing=0|cellpadding=1|border=1|frame=box|width=30%}{tr}{td:align=center|bgcolor=#F2F2F2}*[Examples from Energy]* {td}{tr}{tr}{td}
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!bungeejump.jpg|width=70%%%!
(Photo courtesy Wikimedia Commons, uploaded by user Che010.)

Bungee cords designed to U.S. Military specifications (DoD standard MIL-C-5651D, available at [http://dodssp.daps.dla.mil]) are characterized by a force constant times unstretched length in the range _kL_ ~ 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 50.0 m above the surface of a river running below. You have read that you should expect the cord to stretch (at peak extension) to about 210% of its natural length. You have also read that you should use 3 cords together for jumpers with weights in the range 100-150 lbs, 4 cords for 150-200 lbs, and 5 cords for 200-250 lbs. Suppose you decide to use cords of length 20 m, which would seem to offer a safety zone of about 8 m (or really about 6 m if the cord is attached at the ankles).

h3. Part A

Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that _kL_ = (4)(800 N) = 3200 N. Since you are evaluating the safety factor, ignore any losses due to air resistance or dissipation in the cord. Ignore the mass of the rope.

h4. Solution

*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).

*Interactions:* The system constituents interact via gravity, which contributes [gravitational potential energy|gravitation#negpegravitation (universal)#negpe], 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 variant.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 "compressed", since they fold like an ordinary rope. Thus, the initial spring energy is zero in this case. {note}
It seems that we have a problem here, because we do not know _h{_}{~}f~ or _x{_}{~}f~. The easiest way to deal with this problem is to utilize our freedom to _choose_ the zero point of the height axis. If we restructure our coordinate system to place _h_ = 0 m at the point where the cord is stretched to its natural length _L_ (as shown 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 variant.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}\begin{large} \[ x_{f} = \frac{mg \pm \sqrt{(mg)^{2} + 2 k mg L}}{k} \]\end{large}{latex}
It is not sensible that we should find a negative value for _x{_}{~}f~, so we must select the plus sign, giving:
{latex}\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}

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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?