Bungee Jumping: Unraveling the Physics Involved

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    Bungee Physics
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The discussion focuses on the physics of bungee jumping, particularly the calculations related to spring constants, gravitational potential energy, and the height a jumper rebounds after the fall. Participants analyze the relationship between spring energy and gravitational forces, noting that the jumper will not reach the original height due to energy losses from air resistance and the properties of the bungee cord. It is clarified that without these losses, the jumper could theoretically oscillate indefinitely or return to the starting height. The conversation emphasizes the importance of accounting for energy dissipation in real-world scenarios. Ultimately, the complexities of bungee dynamics highlight the interplay between physics principles and practical outcomes.
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we've been studying te physics involved in bungee jumping. We went to a bungee site, and recorded data, such as length of rope, weight of jumper, distance to fall, etc.

With the data we collected, I've been able to determine spring constant, equilibrium point, force exerted etc.

But I've been having trouble to work out the height the jumper will bounce back up to after reaching the lowest point.

Here's an example:
k= 50 N/m
x= 50 m (this includes the length of cord)
length of cord= 10m

so f=k(x-L)
=50 x 40
= 2000 N

and that his elastic potential energy at the bottom = E grav potential at top
m= 80kg

mgx=.5k(x-L)^2
Eep=40000 J

But using this how do i find the height the jumper will bounce to? I'm pretty sure he wouldn't reach the same height he fell from, due to gravity.
 
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Assuming they stop at the bottom of the bungee then they have no kinetic energy, so all the spring energy goes into the potential energy they have at the top of the bounce.
so 1/2 k x^2 = m g h
where x is the spring extention and h is the height (above the lowest point)
 
but wouldn't the height they return to be affected by gravity, therefore lowering the height he would return to?

1/2 k x^2 = m g h

h= 40000/ 800
= 50m

and this would not be correct.
 
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glugger said:
but wouldn't the height they return to be affected by gravity, therefore lowering the height he would return to?

Not affected by gravity. Affected by losses to air resistance and losses into the rope (it's a lossy spring)
 
berkeman said:
Not affected by gravity. Affected by losses to air resistance and losses into the rope (it's a lossy spring)

so without these losses, the rope would be in constant motion, and never stop?
 
glugger said:
so without these losses, the rope would be in constant motion, and never stop?

Not exactly. Without losses, the system is basically a mass on a spring. The mass oscillates up and down, and the spring goes with it.

http://en.wikipedia.org/wiki/Simple_harmonic_motion

.
 
yes, but it would never stop without these losses?
 
glugger said:
yes, but it would never stop without these losses?

I'm sure Berkeman didn't understand what you were driving at or some subtle point. Anyway, without energy loss in the cord or through air friction, you would oscillate forever, or if you wish, return to your starting point at take-off, with everything else being ideal.
 
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  • #10
Phrak said:
I'm sure Berkeman didn't understand what you were driving at or some subtle point. Anyway, without energy loss in the cord or through air friction, you would oscillate forever, or if you wish, return to your starting point at take-off, with everything else being ideal.
Yes... If there are no losses (no energy converted to heat) then the bungee jumper just keeps bouncing forever.

I think I'd vomit...
 
  • #11
ok thank you. that solved the problem...sorta.
 
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