The Physics Behind Bungee Jumping

[toosm:)ey]

Hey, this is my first time on the forum and I have to do a project about bungee jumping.

I need to talk about Hooke's Law, calculate The Conservation of Energy and the Conservation of Momentum...

I know that F=k*x, but I am still a little unsure on how to relate it to bungee jumping.

Assuming that there is no loss of energy due to friction, air resistance etc etc, I am confused on how the conservation of momentum and conservation of energy relate to bungee jumping. I would like to know how to calculate these topics for bungee jumping.

Of course, this is my homework, so I am not expecting you to do it for me, although it would be a great help!

Thankyou!

Smiley

 

[cookiemonster]

Before you jump, you're at a certain height above the ground. We can attribute to this height a certain amount of potential energy. When you jump off, this energy starts to be converted into kinetic energy, and you speed up while you fall. Then the bungee cord kicks in and starts transforming the kinetic energy and potential energy to elastic energy, which is stored in the cord and why it stretches. Once you stop, most (hopefully there'll still be SOME left so you don't smash into the ground) of the potential energy and all of the kinetic energy (hence, stopped) has been converted to elastic energy.

It's all about the energy! If you idealize your world enough, the calculations are pretty simple, too.

Does that help?

cookiemonster

 

[toosm:)ey]

Yes! Thankyou!

Potential = m*g*h (mass * gravity * height) ?
Kinetic = 0.5 * m * v^2 (mass * velocity^2) ?
Elastic potential = ?

 

[cookiemonster]

$$\textrm{Elastic Energy} = \frac{1}{2}kx^2$$
where k is an empirically determined constant and x is the displacement off of the rest position of the cord.

cookiemonster

 

[toosm:)ey]

How do I find k?

 

[cookiemonster]

I imagine the manufacturers of bungee cords are required to publish them. If not, you just test it by putting a few weights of known mass on the cord and measure how far it stretches. The graph of displacement off of rest position vs. mass will yield a straight line, the slope of which can be used to calculate the constant.

cookiemonster

 

[toosm:)ey]

Wicked! Thanks! You're a great help!

If I have any more questions, I'll come back, for now, I think I am good =)

Thanks again

 

[toosm:)ey]

Okay, I think I have things figured out so far. Could you check it for me please?

Height of bridge = 100m
Rope Length = 60m
Elongation = 50% (30m)
Mass = 75kg

To find k in (F=kx)

F=m*g F=(75kg)x(9.81m/s^2) F=735.75N

F=k*x 735.75N=k(30m) k=24.525

That good so far?

As the rope is 60m, does this mean that the equilibrium is at 60m? And because the rope does not compress, the equilibrium is also at 60m to 100m high?

The person is at his maximum speed at 60m? How do I find his maximum speed? I know speed=distance/time, but I do not know the time.

I'd like to go bungee jumping and see, but I don't have time for that!

~Thanks for all your help!

 

[cookiemonster]

Yes, your calculations are correct.

How are you getting your data? It's generally advisable to use many data points in order to minimize error, although it is quite possible to determine the constant from a single point, as you just demonstrated.

Oh, and you might want to stick the units on the constant.

cookiemonster

 

[toosm:)ey]

I made my data up... It's a hypothetical experiment. If there is no air resistance/drag, then he will be at terminal velocity at 60m?


The units for the constant are (N/m)?

 

[Chen]

No, the maximum (or minimum) velocity is achieved when the acceleration is zero. This is because the acceleration is the derivative of the velocity, and as you know f(x) has a maximum (or minimum, or "twist") when f'(x) = 0.

This happens when the force that the rope makes on the person is equivalent to the gravitational force, but in the opposite direction. As you probably realize, this is exactly the equilibrium point, 90 meters below the bridge. (Of course, in this case the person doesn't stay still at the equilibrium point, since he still has kinetic energy.)

 

[cookiemonster]

I think that now that we got this discussion of maximum velocity going on, we should more precisely define a few things.

Imagine the guy isn't dropping. Imagine that he's simply hanging from the rope. The guy will be resting when the force is zero (thank you, Captain Obvious), and this will happen when the elastic force is exactly equal and opposite the gravitational force, i.e. kx = mg, x = mg/k, where x is the displacement from the cord if there were zero mass on it. So there is a difference between the rest position of the cord and the equilibrium position of the cord. They're different by a factor of mg/k. Note that this shift has no bearing on the remaining calculations, i.e. F still equals -kX, where X is defined as (x - mg/k).

That being said, the maximum velocity will be as Chen said, at the equilibrium position. It's not difficult to see by writing the full differential equation (it's not as scary as it sounds) for the motion of the bungee jumper and then deriving the equations of motion.

There's probably one other thing to keep in mind. I might be wrong about this, but I've never seen many bungee cords that compressed. So the elastic force should be F = -kX for X >= mg/k and F = 0 for X < mg/k, at which point the jumper is in free fall and the rope just kind of loops up. It complicates things, but it's not very difficult to deal with.

And yes, the spring constant will have units N/m.

cookiemonster

 

[toosm:)ey]

Using

mass = 75kg
height, before = 200m, after = 140m

Energy before = Energy After

Potential + Kinetic = Potential + Kinetic
mgh + 0.5mv^2 = mgh + 0.5mv^2

(75)(9.81)(200) +(0.5)(75)(0^2) = (75)(9.81)(140) + (0.5)(75)(v^2)
147150J + 0 = 103005J + 37.5v^2
44145J = 37.5v^2
v^2 = 1177.2
v = 34.31m/s

This is the velocity at 140m (a 60m dive)... I think. Can anyone see a flaw in my calculations?

velocity = displacement / time
time = displacement / velocity
time = 60m / 34.31m/s
time = 1.75 s

The jumper is 40m above ground at 1.75 seconds.


Now for when the rope has elastic potential energy at maximum displacement:

Energy Before = Energy After
147150J = Potential + Kinetic + Elastic
147150J = mgh + 0.5mv^2 + 0.5kx^2
147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(k)(x^2)

We need the height, the constant in the rope, and the maximum displacement,x,

Using F=kx and F=mg to determine k.

mg=kx
(75)(9.81)=kx

The maximum elasticity of the rope is 50%, so a rope that is 60m will stretch a maximum of 30m, before it loses it's elasticity (Hooke's Law). So, x = 30m.

735.75=-k(30)
k=-24.525N/m

147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(-k)(30^2)
147150J = (735.75)(h) + 0 + -11036.25
735.75h =158186.25
h = 215m

However, how can it be 215m? it would be 15 metres above where it started... This is where I am confused.

 

[Chen]

Originally posted by toosm:)ey
velocity = displacement / time
time = displacement / velocity
time = 60m / 34.31m/s
time = 1.75 s

The jumper is 40m above ground at 1.75 seconds.

That would be the first mistake. Generally speaking:

$$\Delta V \Delta t = \Delta X$$

In our case, however, the person's velocity is not constant. So the equation above is not true for large time intervals (it can only be used to calculate the momentary displacement).

You need to use the more general formula of:

$$X = X_0 + V_0 t + \frac {1}{2} at^2$$

Luckily for you, the initial displacement is 0 and so is the initial velocity. So finding the time should be a problem. :)

 

[Chen]

Originally posted by toosm:)ey
We need the height, the constant in the rope, and the maximum displacement,x,

Using F=kx and F=mg to determine k.

mg=kx
(75)(9.81)=kx

The maximum elasticity of the rope is 50%, so a rope that is 60m will stretch a maximum of 30m, before it loses it's elasticity (Hooke's Law). So, x = 30m.

735.75=-k(30)
k=-24.525N/m

147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(-k)(30^2)
147150J = (735.75)(h) + 0 + -11036.25
735.75h =158186.25
h = 215m

However, how can it be 215m? it would be 15 metres above where it started... This is where I am confused.

If the maximum elasticity of the rope is 50%, I'm afraid our jumper will not live to see another day. Because only when the rope is elongated by 50%, it exerts enough force to cancel the weight of the person. Until that point, the gravitational force was still stronger than the rope's force. This means that only at 50% the person's acceleration is zero, but as I explained above this is also the point in which the person has the maximum velocity. The rope must elongate by more than 50%, or it would tear apart. (In this specific case)

The second mistake is that your equation implies that the person is resting at 50% elongation, which is not true as I said. What you need to do is express h as a function of x (or x as a function of h), so that your equation only has one variable. The answer for x is only the minimum elastic range of the rope, if you want your daredevil jumper to survive.

 

[cookiemonster]

What level of math can you do? Can you solve the differential equation

$$m\ddot{y} = -mg - (y + \frac{mg}{k})$$

?

cookiemonster

 

[toosm:)ey]

You need to use the more general formula of:

$$X = X_0 + V_0 t + \frac {1}{2} at^2$$

Luckily for you, the initial displacement is 0 and so is the initial velocity. So finding the time should be a problem. :)

I am assuming that a is the acceleration, v is velocity, and t is time. Without the acceleration or time, I can't find the velocity.

What level of math can you do? Can you solve the differential equation

$$m\ddot{y} = -mg - (y + \frac{mg}{k})$$

?

I am in High School Calculus. We have learned how to differentiate equations, but not those with so many variables.

If the maximum elasticity of the rope is 50%, I'm afraid our jumper will not live to see another day. Because only when the rope is elongated by 50%, it exerts enough force to cancel the weight of the person. Until that point, the gravitational force was still stronger than the rope's force. This means that only at 50% the person's acceleration is zero, but as I explained above this is also the point in which the person has the maximum velocity. The rope must elongate by more than 50%, or it would tear apart. (In this specific case)

Once the rope begins the elongate, the jumper loses velocity because his kinetic energy is transferred to elastic energy. His maximum velocity is at 140m, when there is no elastic potential energy. His maximum velocity is still 34.31m/s.

Energy Before = Energy After.

Susbsequent to a long night of thinking about it, my equation did not contain all the energies. Thermal (frictional) eenrgy was not included, which is huge because of the anture of bungee ropes. All those little stretchy things inside the rope rub against each other. In a rope of 60m, the thermal energy will be large? Of course, there are many other types of trace energies that are not counted for. But as they are so little, it would not alter my answer tremendously.

 

[toosm:)ey]

But I do not know how much energy is Elastic energy because I do not know the ropes k value.

 

[Chen]

Originally posted by toosm:)ey
Once the rope begins the elongate, the jumper loses velocity because his kinetic energy is transferred to elastic energy. His maximum velocity is at 140m, when there is no elastic potential energy. His maximum velocity is still 34.31m/s.

Energy Before = Energy After.

That is not precise. Once the rope begins to elongate, it does acquire elastic potential energy. But at first that energy comes at the expense of gravitational potential energy, not at the expense of kinetic energy. In fact, even when the elastic energy is growing, so does the kinetic energy until some point. That point is the equilibrium, in our case 50% elongation. Only then does the force of the rope begin to outbalance the weight of the person, which means his acceleration is now negative (he is decelerating). Only then does the kinetic energy begin to decrease.

His maximum velocity is not necessarily at the point where he has no elastic energy, it is at the point where he has the minimum of potential energy, which includes both elastic and gravitational energy. If you write down the equation of the person's potential energy (which is preserved, if we ignore friction and heat) you will get:

$$\Sigma E_p = E_{gra} + E_{els} = mgh + \frac{1}{2}kx^2$$

If you express h as a function of x you will get:

$$\Sigma E_p = mg(200 - x) + \frac{1}{2}kx^2$$

To find the minimal potential energy, get the derivative of the function and see where it is zero:

$${E'}_p = -mg + kx = 0$$
$$x = \frac{mg}{k}$$

This is the point where the potential energy of the diver is minimal, and therefore it is the point where his kinetic energy (and velocity) is maximal.

Do you understand or should I explain it a different way?

 

[toosm:)ey]

I understand... you're right

However, because of the lack of time, I don't have time to do all those calculations...

My website is at http://ca.geocities.com/physicsroxmysox/ , I am sure you can find lots of mistakes in my work, because you're all really intelligent at Physics!

Thankyou for all your help! I'll be sure to come back if I need any help

~Smiley

 

[cookiemonster]

Looks like you ran out of bandwidth. =\

cookiemonster

 

[physics student]

hey I'm also studing bungee jumping is there any other information that i could use to help with my assignment

 

[mathiasrask]

First of all I must excuse my English. I’m from Denmark.

Second, in all the calculations you have assumed that the elastic force is proportional with the extension, as a spring, which is not entirely true because, as #15 writes, there’s a limit to the elasticity.
At some point the graph gets steep as a vertical asymptote but also at small expansions the graph is not at all linear. This fact complicates things very much.

When you do bungee jumping, the amount of energy you have depends of the high. Therefore numeric size can be calculated as potential energy.
Epot = m*g*h
Which naturally is turned into kinetic energy and therefore determents the speed of get.

Because you don’t want hit the ground when you jump, the potential energy has to be annulled by the elastic energy.
Epot = Eela
The elastic energy is equal to the work of the elastic, which of cause defined as force multiplied with distance/expansion. The force of the elastic is not constant but is equal to the area under the graph, which is equal to the integral of the elastic function.
Eela = x0 * integral from 0 to x0 of f(x) dx

Hope its usefull!

 

[canadain_girl_0]

Wondering

Dear who ever,

I am a grade eleven student from Canada. I have to do an independent study for my math class. I was wishing to do my project on Bungee jumping and the physics and math behind it.I do not understand anything really, i found equations,ect. But i can not unerstand what all the letters stand for. if you can help me it would mean alot. thank you for your time

Sincerely
Jennifer MacArthur

you can either relay on here or you can email me at jenn_macarthur@hotmail.com

Thank you

 

[rcgldr]

Assuming bungees are made out of rubber similar to latex tubing, then F=k*x doesn't hold, it's a curve and the return curve has slightly less tension than the stretch curve (which is why rubber is good for reducing vibration). However for your homework problem I'm assuming you can use F=k*x. Also the max stretch is way more than 50%. Latex tubing can be pulled 350% (so total length = 450%) without "memory" effects. Other types of rubber can be pulled even more without "memory" effects.

Here's a link to a website with a graph of the stuff used for radio control gliders.

http://www.hollyday.com/rich/hd/sailplanes/rubberdata.htm

On a sidenote:

I use Latex tubing to supply a moving tension force to launch my radio control glider similar to a person launcing a kite by running. 60 feet of tubing that produces 24lbs of tension at 300% pull, and 27lbs at 350% pull (I vary this based on head wind), 210 feet of monofilament fishing line and a chute that pulls shut during launch but opens up to bring the line back downwind. It's a 4 1/4 lb glider with a 10' 2" wingspan. A video of a couple of launches:

jrartms.wmv

 

[canadain_girl_0]

thanks

Do you know anything about the full equation or what the bungee person would use? Or do you know someone able to help me with the part, or possible look at? But thank you ofr your help I kinda get it. I have not really studied anything in physics or the forces. But thank you for your help.

Jennifer MacArthur

 

[rcgldr]

This website below has some info, plus equations, so the original poster should not look at this web site.

However I'll summarize most of the information in this post. As I suspected, bungee's use latex rubber. For a normal (head down) bungee jump, the goal is to limit maximum tension to 3 times the jumpers weight (so 3 g's total force, including gravity). This is achieved with a 200% stretch, (maximum length of the bungee line is 300% the original length). With a 300% stretch, accomplished by using a smaller diameter chord for the same weight jumper, or the same diameter chord but a heavier jumper, the maximum tension is only lowered a bit to 2.7 g's. This fact eliminates the need for a large number of chord diameters at bungee sites (latex rubber produces 175 lbs of tension per square inch of area at 300% stretch).

To allow for even more variation, there is a static line (it doesn't stretch) used to connect the bungee chord to the platform. The static line can be adjusted vertically, to adjust the free fall distance before the bungee chord starts decelerating a jumper.

Although the actual operators of bungee sites may not be physics experts, they use tables generated by the experts to select chord diameter and static line length, based on the weight of the jumper and the desired maximum distance of the jump.

If the height of the bridge is 100m, then maximum bungee chord length for a 300% stretch would be 25m, and for a 200% stretch, 33m. In reality, the chord will be a bit shorter to provide margin for error (or some breakage of the latex strings forming the bungee).

bungee.htm

 

[canadain_girl_0]

thank you

Jeff,
Thank you very much. the website and it makes sense now.. i can not thanky uo enoug.h. ill write on here if i have anymore question :D Again thank you..
Jennifer MacArthur

 

[Ammar SH]

I have a similar question;
What makes a bungee cord safer than using a normal rope when jumping.?

I understand that its elasticity and Hooke's law explains that compression and expansions are created to transform energy which helps you make a safer jump but

What has impulse and change in momentum got to do with the bungee cord being better than the rope?

 

[Drakkith]

Ammar, for future reference, you should make a new post instead of posting in a thread that is 4 years old at the latest post. You can find the date and time in the top left above everyones name. But no biggie! It happens!

 

[Ammar SH]

Oh...sorry ...
I just joined and read the rules...
I wanted to avoid making a new post cause my title is the same

and thanks for the info :D

 

[Ammar SH]

I mean my question... is similar

 

[Drakkith]

It's ok. No harm done.

 

[Drakkith]

I have a similar question;
What makes a bungee cord safer than using a normal rope when jumping.?

I understand that its elasticity and Hooke's law explains that compression and expansions are created to transform energy which helps you make a safer jump but

What has impulse and change in momentum got to do with the bungee cord being better than the rope?

The issue with a rope vs a bungee cord is acceleration. In this case we refer to it as deceleration since it is negative acceleration. A normal rope isn't elastic. When it goes taut it doesn't stretch at all, it simply tries to stop your falling as quick as it can. This is very bad for you!

A similar experience happens in a car accident. The car collides with something and very quickly comes to a stop. Without safety features this can apply huge amount of G force to your body. What happens if your head suddenly weights 100 times normal? Broken neck. If we buckle up and have an airbag, then instead of you slamming into the steering wheel at 100g, you are gradually stopped over a much longer period of time, experiencing maybe 10g instead. Much better! (Just making up those numbers, I don't know what the actual amount is)

A bungee cord does the same thing. Instead of the rope, which stops your fall in about half a second, the bungee cord stretches, absorbing the energy of the fall over 5-10 seconds, making the deceleration much gentler and enabling you to survive the fall.

Acceleration = change in velocity, and velocity is one of the factors that determines momentum. So change in velocity = change in momentum = acceleration.
Impulse is simply a measure of how much momentum has changed. Accelerating quickly causes a larger impulse than accelerating slowly in the same amount of time.

 

[Ammar SH]

that was a good explanation

thanks for the answer and forum info..

one last thing(you might laugh but) how do I my own post lol
still can't figure it out...