Bungee Cord Physics

1. Oct 2, 2014

Abid Rizvi

1. The problem statement, all variables and given/known data
A daredevil plans to bungee jump from a balloon 59.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 13.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.65 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

2. Relevant equations
Hookes Law = 1/2 kx^2
Potential Energy = mgx

3. The attempt at a solution
Okay so for starters I said when he falls out of the balloon he falls a distance x at which he is now not falling.
x = 59-13 = 46
Then I set out to find k for Hookes law. I said mg = ky where y is the 1.65 m. So
k = (mg)/y
I then said r is the actual length of the chord, and L is the length that the chord stretches. So
r + L = x
I then said the total potential energy of the devil is mgx, and the amount of work in opposition to him by the chord is 1/2KL^2 = 1/2 * (mgL^2)/y
I set this equal to mgx: 1/2 * (mgL^2)/y = mgx
= (L^2)/(2y) = x
= L = sqrt(2yx)
solving for L and subtracting it from x to get r (the actual length of the chord) I get 37.3 m. Which apparently is wrong. What am I missing? Thanks in advance

2. Oct 2, 2014

ehild

k=mg/y Is the spring constant of an 5 m long chord. What is K for an L m long chord of the same kind?

Also, the potential energy of a stretched spring depends on the change of length (r) instead of the unstretched length (L) as you wrote.

ehild

3. Oct 2, 2014

Abid Rizvi

Hi and thank you for responding
I did set L as the change of length and r as the unstretched length.

Also I just assumed K = mg/y. Does K change because of length?

4. Oct 2, 2014

BvU

Imagine a 0.5 m length of chord. Would it stretch to 2.15 m if the fellow hung from it ?

5. Oct 2, 2014

Abid Rizvi

Lol no I suppose not. Ok so K =mg/(stretched length) for a cord. I will try it with this.

6. Oct 2, 2014

Abid Rizvi

Ok so I put K = mg/L instead of mg/y for the chord, and I get L=2x which is obviously incorrect. I'm lost on what to do now...

7. Oct 2, 2014

haruspex

No, it is still the case that for a given piece of elastic K = mg/(length of stretch). Your problem is how to convert the K for one length of material to the K for a greater length. It's the innate property of the material that's constant. If a given load stretches a rope length L by x, by how much would it stretch a rope of twice the length?

8. Oct 2, 2014

ehild

So k=mg/1.65 for the 5 m long chord. Assuming it is r m long, what is the spring constant?

Think: 5 m chord stretches 1.65 m. You bind two 5 m chords together. How much does that 10 m chord stretch?
If the chord is r m long, how much does it stretch by the weight of the man?

ehild

9. Oct 2, 2014

Abid Rizvi

YES! I got it. Okay so from what I understood, if a 5 m chord stretches 1.65 m, a 10 m chord should stretch 3.3 m (=1.65*2). So to do the conversion I need to find the factor to multiply by and that factor is given by f = (r/5) where 5 is the length of the small cord. So mg/fy = K. Putting that back into the original stuff, I get 2.5L^2 = yx(x-L), and therefore L = 25.156 m. So to find the length of the cord r, x- l = r which is about 20.8 which the website accepts as correct. Thank you!

10. Oct 2, 2014

ehild

Well done! :)