modeling a bungee jump using differential equations a bungee jump using di erential equations eric...

Modeling aBungee Jump

usingDifferentialEquations

Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters

SolvingwithoutNumericalSolvers

Our model

NumericalSolvers

Findingequations forthe Model

Graphs of theJumper

Bibliography

Modeling a Bungee Jump using DifferentialEquations

Eric Moon, John Thompson

May 16, 2014

Eric Moon, John Thompson Modeling a Bungee Jump using Differential Equations



Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Purpose of this Model

Finding a model such as this is most beneficial to peoplewho are in control of bungee cords used in bungeejumping.

This model, shown later on, can be used to generategraphs and have the ability to tweak certain aspects of themodel to fit a specific desired motion.

We created a MATLAB GUI, with our model as thedriving equation behind it.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Assumptions made for the Model

This model only covers position with respect to time, anddoes not cover motion in two or three dimensions.

One assumption made is that the jumper is a point masswith the terminal velocity of a human.

Also, the Young’s modulus of the cord stays constant withwhatever strain is imposed on it. This means that it doesnot enter the plastic phase and keeps its propertiesthroughout the model.

The wind resistance imposed on the jumper is simply alinear relationship with velocity.

The mass of the jumper is 80 kg.

The length of the cord used in our model is is 30 meters,unless otherwise specified




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Hooke’s Law

Since we treat the cord as a spring, we apply Hooke’s law,F = kx

Also, because the force of the spring is opposite of theforce due to gravity, we use −kx .

k represents the stiffness of the cord, meaning that ahigher k-value will mean that the cord won’t stretch asmuch as another cord when the same force is applied.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Newton’s Second Law

The backbone of this model is Newton’s Second Law,∑F = ma, along with Hooke’s Law.

As stated earlier, this model is only looking at the positionwith respect to time, however by differentiating we findthe velocity and the acceleration.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Basis of the Model

We use Newton’s second law, F = ma, as well as Hooke’s Law,F = −kx , to get the sum of the forces. Using these we end upwith the equation∑

F = mg − βv − kx = ma

This equation is our starting point for finding a sufficient modelfor the motion of the bungee jumper.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Basis of Our Model

For our model, we found separate equations for the motionof the jumper whe in free fall, and when their motion isunder the influence of the cord. These will be discussed inmore detail later.

In order to use a numerical solver, such as MATLAB’spplane8 or ode45, we have to specify a few things, andmake it so it recognizes it as a second order equation.

Our final model is

x ′′ = g − β

mx ′ − k

m

and where we got the values indicated will be discussedshortly.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Wind Resistance

In order to find β, we have to see that we are only usingpart of the wind resistance expansion.

For example, the next term is a constant α that ismultiplied by the velocity squared, meaning that it hasmore of an influence when the jumper is going faster.

However, the calculations to get that constant is not asstraightforward as finding β, and finding the equations forthe motion including this extra term proved to beexceedingly difficult, so only the linear relationship will beused.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Finding β

β is the wind resistance constant and has a linearrelationship with the velocity.

To find it, we set the free fall equation, mg − βv = ma,equal to zero as acceleration tends towards zero.

simplifying this, we have

β =mg

v

, which is how we calculated our vlaue for β

Our value we found is 13.08




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Finding k

As said earlier, the value of k represents the stiffness ofthe cord.

Using our experimental data from our shock cord (anotherterm for bungee cord ), we found the Young’s modulus ofthe material of the cord.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Using this finding, and realizing that the expression forfinding the Young’s modulus,

Y =F/A

∆L/L

can be rewritten as:

F =YA

L∆L

We see that the value of k equals YA/L

Our calculated value is 2.73, which means that thisparticular cord is rather stretchy.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Solution without a numerical solver

Solution for free fall section.

x = C1 + C2e−βt/m +

mg

βt

Solutions for each of three cases for when the cord is pulling.Case 1: (b/m)2 − 4(k/m) > 0

x = C1e(−β/m+

√(β/m)2−4k/m)t/2 + C2e

(−β/m+√

(β/m)2−4k/m)t/2

+gm

k

Case 2: (β/m)2 − 4(k/m) = 0

x = (C1 + C2t)e−βt/2m +gm

k




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Case 3: (β/m)2 − 4(k/m) < 0

x = C1e−βt/2m cos t

√|(β/m)2 − 4(k/m)|

2+

C2e−βt/2m sin t

√|(β/m)2 − 4(k/m)|

2+

gm

k




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

For our model we assumed that the following characteristicswere true: k = 2.73,m = 80, L = 30, g = 9.81, and β = 13.08.We also assume that the bridge is 100m tall.For our model we found that it fell into case 3 when thebungee cord was stretched. We found that the solution was:

x(t) = 366.97e−.1635t + 60t − 396.97

x(t) = −316e−0.164t cos (0.166t) +−169e−0.164t sin (0.166t)

+ 288

The first equation is accurate for 0 <= t <= 2.67 and thesecond equation is accurate for t > 2.67




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

0 1 2 3 4 5−40

−20

0

20

40

60

80

100

t

x

Figure: x appears to be 0 when t = 2.67




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

0 20 40 60 80 100−100

0

100

200

300

400

t

x

Figure: Exact solution




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Using our Model in a Numerical Solver

Using a numerical solver, such as pplane8 or ode45, is aconveinent way to generate a graph of the motion.

Because the Cord only begins its effect after the jumperfalls past the un-stretched length of the cord, theexpression k(x >= 0), 0 otherwise, is used to account forthis.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

written as a planar system, the equations are:

x ′ = v

v ′ = g − β

mv − k

mx

plugging the values in we have stated we get the equation

x ′′ = 9.81− 0.1635x ′ − 0.03466x

The benefit of having this equation in this form, asopposed to the forms stated earlier, is that this form ismuch easier to visualize with a numerical solver.

Using our calculated values for the parameters, wegenerated this graph of the motion, velocity, and theacceleration of the jumper.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Graph of the Model

0 20 40 60 80 100−100

0

100

200

300

400x’’=9.81−0.1635x’−0.034125x

Time (s)

Pos

ition

(m

)/ V

eloc

ity (

m/s

)

PositionVelocity

Figure: Position and velocity graph of the jumper




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Of course, now that we have the model set up, we havethe ability to show what differences in the k-value, and thelength of the cord will have on the motion.

We can also display the acceleration as a function of timeas well, which gives insight as to how much accelerationthe jumper is experiencing.

Too much acceleration, and the jumper will not survive.




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

0 20 40 60 80 100−100

0

100

200

300

400x’’=9.81−0.1635x’−0.034125x

Time (s)

Pos

ition

(m

)/ V

eloc

ity (

m/s

)

PositionVelocityAcceleration

Figure: Inclusion of the Acceleration




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

0 20 40 60 80 100−50

0

50

100

150

200x’’=9.81−0.1635x’−0.034125x

Time (s)

Pos

ition

(m

)/ V

eloc

ity (

m/s

)


Figure: k=6, Length=50




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

0 20 40 60 80 100−50

0

50

100

150x’’=9.81−0.1635x’−0.034125x

Time (s)

Pos

ition

(m

)/ V

eloc

ity (

m/s

)


Figure: k=10, Length=50




Eric Moon,John

Thompson

Assumptions

Physics

Modeling themotion

Parameters


Our model

NumericalSolvers


Graphs of theJumper

Bibliography

Bibliography

Adkins, Samantha, Katy Arthur, Jeff Benefield, and BrianBoling. ”A Mathematical Representation of the Motion ofa Bungee Jumper.” Okstate.edu. Oklahoma StateUniversity, 14 Oct. 2002. Web. 20 Feb. 2014.

Polking, John, Albert Boggess, and David Arnold.Differential Equations. 2nd ed. Upper Saddle River:Pearson Education, 2005. Print.

Kelly, Nicole. ”Modeling a Bungee Jump.” Fvcc.edu.Flathead Valley Community College, 17 May 2013. Web.21 Feb. 2014.

Serway, Raymond A, John W. Jewett. Physics for Scientistsand Engineers. 9th ed. Cengage Learning, 2014. Print.


modeling a bungee jump using differential equations a bungee jump using di erential equations eric...

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