Derive Equations for Bungee Jumper!

1. The problem statement, all variables and given/known data

[PLAIN] [Broken]

Derive the equations of the motion for the jumper. Consider only the vertical motion of the jumper. (g = 9.81 m/s^2)

2. Relevant equations

Straing = (Stated Above)

Strain Rate = v / lo = v/25

F = -kx (Hooke's Law)

3. The attempt at a solution

We aren't really given formula, but required to go out and seek and find them and find a way to solve the Question.

I know the Equation needs to be a piecewise equation.

First part will be for the free fall the bungee jumper will feel before hitting the 25 meter mark where the bungees spring force will be felt by the bungee jumper.

I think we can just use the simple equation:

d = u*t + 1/2 *a*t^2

Since d = l, u = 0

l = 1/2 * a * t^2 = 9.81/2 * t^2

So for 0 < l < 25 , l = 9.81/2 * t^2

For the second part of the equation I am a little stumped.

We do not need to take air resistance into account.

Thanks in advance, and any and all help will be greatly appreciated.

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Science Advisor
Homework Helper
Sheen91: Similar to what you computed, the bungee jumper velocity at the end of his first free fall will be vo = -(2*g*Lo)^0.5, where g = 9.81 m/s^2, and Lo = 25 m. You can get this from a kinematics equation.

Let us define the origin (or datum) of the x axis as the location where the bungee jumper will come to rest; and let the positive x axis point upward. Notice, the distance between this origin to the top end of the bungee cord is Lo + (m1*g/k) = 25 + (75*9.81/200) = 28.6788 m, where m1 = 75 kg, and k = 200 N/m. Let us define t = 0 s as the first instant the bungee cord starts to stretch; xo and vo are the bungee jumper position and velocity at t = 0 s.

Let's rename your cord viscous proportionality constant to c1 = 150 N*s. We can see that the cord coefficient of viscous damping in your problem is c = c1/Lo = (150 N*s)/(25 m) = 6.0 N*s/m.

Now look in any vector mechanics, dynamics, or structural dynamics text book for "damped free vibration of single degree-of-freedom (sdof) systems," and you will see that your equation of motion problem has already been solved, is listed therein, and is as follows.

x(t) = U*exp(-beta*t)*cos(wd*t - alpha),​

where beta = 0.5*c/m1,
wd = [(k/m1) - beta^2]^0.5,
U = (A^2 + xo^2)^0.5,
xo = m1*g/k = 75*9.81/200 = 3.678 750 m,
A = (vo + beta*xo)/wd, and
alpha = atan(A/xo).

However, the above x(t) equation is applicable only from t = 0 s, until approximately t = 2.2668 s, which is when x(t) = xo again, meaning the cord goes slack again. Whenever x(t) > xo, the bungee jumper is just a projectile in free fall. Putting these transitions together automatically would probably require some nontrivial programming.

By the way, whoever wrote the given problem statement image attached in post 1 (and I know it was not you) currently does not know how to write unit symbols correctly. Please see the last paragraph of post".
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