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Bungee Jumping Physics

Introduction
Recently a consultant of the Fair of Sciences sent me a message requesting the solution of a proposed subject, in college entrance exam, by FURG 2007. Here is the proposed subject:

"An adventurer plans to jump of the high of a bridge tied in an elastic (a radical sport known for 'bungee jumping') cable. The other extremity of the cable is tied in the bridge. In the beginning, the jumper's movement is a free fall. Starting from the point in that the cable is stretched out, the jumper begins to slow down until a certain position, where it stops. Of this moment in before, the cable begins to pull the jumper upward. This position, where the jumper inverts the fall sense, it marks his/her largest vertical displacement D regarding the bridge. Naturally that the height of the bridge should be larger than D. it Considers a jumper's of mass 80 kg hypothetical situation now using an elastic cable of 20 m of length. The elastic constant of the cable is 160 N/m. Calculate the value of D.
Observation: the mass of the cable can be despised in relation to the jumper's mass. For acceleration of the gravity, use the value 10 m / s²:
The) 20 m.  B) 25 m.  C) 40 m.  D) 36 m.  And) 10 m."

Being fallen back upon the search motors with the discriminador "bungee jumping" is had, as return, hundreds (or else, thousands) of places; that you/they say everything on the sport, less like him it works really --- and this was the objective of the proposed subject exactly. This article has as purpose to fill out this gap and, as it is Science, every suggestion / critic will be well arrival.

I will always maintain in view the solution of the proposed subject and the possible roads for you solve her. Like this, let us enhance the data of the subject:

the jumper's mass: m = 80 kg
natural length of the cable: L = 20 m
constant elastic of the cable: k = 160 N/m
local acceleration of the gravity: g = 10 m / s²
(to neglect effects of the air and mass of the cable)

We should determine until that distance D will go down the jumper, to count of the bridge.

First solution
In this first direction I will follow the technique of Newton's laws, verifying the active forces in the system. In a second solution we will use of the conservation of the energy.

Referencial and phases of the process
Let us adopt as reference axis y, in the vertical that contains the point of the jump and, in the bridge we adopt y = 0; the positive sense is 'down.' The "jump" presents two different phases.
In the first a free fall is verified (grandma = 0) because the cable still is not being requested; so that, for certain interval of time, in other words, while y < L, the only force agent in the man (therefore, to resultant) is his/her own weight; his/her acceleration is g = constant and his/her speed is that of an object in free fall.

Equacionamento of the first phase

Y < L....... F = P = m.g...... the = g...... v² = 2.g.y........ (1)

In this phase (1) ..... the = g > 0; v > 0..... as the and v have same sign (the vectors g and v are both vertical ones down, same sense of y), we have a movement evenly accelerated.

When the man's distance to the bridge, reaches y = L, passed for the second phase: the elastic cable begins being deformed (to stretch out) and raisin to apply a restoring force of the elastic (law of Hooke, f = -k.s) type in the man (f been opposed F). Like this, for y > L the cable will present deformation s = y - L, so that:

Y > L...... f = - k.s = - k. (y - L)...... (2)

The man is represented by the "material point", (he/she plans black) the gravitational (weight) force for the blue vectors and the elastic (f) force owed to the cable, for the red vectors. For 'to help' the presentation, the diagrams are 'sliding' for the right, for not happening overlap of vectors. Do notice that, until then, don't we know where it will happen the maximum deformation (D?).

Equacionamento of the second phase
While y > L, the new resulting force F' that acts in the man will be:

Y > L.......... F' = P - f = m.g - k. (y - L)........ (3)

This is a combination of forces that you/they act in opposed senses: a descending force of constant (mg) intensity and an ascending force continually growing, k. (y-L).
Starting from y = L, the force mg continues acting and it stays of constant value, but the contrary elastic force is going increasing of intensity, so that to resultant F' it is going decreasing of intensity, staying positive until a point B of coordinate yB, where, then, the resulting force is annulled (point of balance of the system; F' = 0; in the illustration above, blue vector = red vector, / /).
Formally:

y = yB......... 0 = m.g - k. (yB - L)...... yB = L + mg / k..... (4)

For the data of this matter subject it is had:

mg / k = 80.10/160 = 5 m;    yB = 20 + 5 = 25 m.

Note: If the man went going down prisoner slowly to the cable, but being held in a rope, he would be in balance in this point B, distant 25 m of the bridge. It could release the rope, because the cable applies him/her the force equilibrante f = - P.

Once the man goes by the position B (y > yB), his/her descent continues (once he/she still has vertical speed down), but, starting from y = yB, the elastic force becomes every time larger than the weight, so that to resultant he/she starts to have sense appearing upward.

Like this, being applied the Fundamental Beginning of the Dynamics (second law of Newton), it is had:

y > yB..... F' = mg - k (y-L) = m.a... the = g - (k/m) (y - L)...... (5) (law of acceleration of the movement for y > yB)

With y < L have the = g, the acceleration of the gravity; starting from y = L the acceleration decreases lineally with y because of the action of the cable (it forces restoring of the elastic type that follows the law of Hooke).
When L < y < yB (passage between L and B) the acceleration is positive (to resultant he/she has same sense that the axis; down); when y > yB (passage after B) is negative (resultant has sense contrary to the axis; upward).

Obviously the point B doesn't characterize the lowest point reached by the man. Observe that while y < yB, the acceleration has the same sign of the speed; like this the man is accelerating. When passing for B the acceleration it changes of sign, but the speed continues positive; this characterizes a retarded movement; the frenamento of the movement begins. In the point B the man has maximum speed!

We know that the acceleration is her derived her of the speed in relation to the time and, as we already learned, a flowed null corresponds to the maximum of the function. Like this, to obtain the maximum value of y we have two roads: (the) to integrate the function acceleration with regard to y to obtain the function of the speed in relation to y --- and soon afterwards to annul v (y) to obtain the yD; (b) to work with the definition of acceleration in function of the time and to try to obtain v (y).

Obtaining of the speed
As the answer, I think, to have urgency, I will just present the result of the integration and, to proceed, I will make the complementation, with integral calculation and without using integral calculation; just flowed and consequences.

result of the integration of the function (5):  v² = - (k/m) (y - L) ² + 2.g.y (it repairs well... does one of the portions of that result know?)

Equaling to zero, for y = yD and solving:

- (k/m) (yD - L) ² + 2gy = 0
yD = L + (mg / k) [1 + (1 + 2kL/mg) ½]..... reminding the (4) ... yB = L + mg / k..... be:
yD = yB + (yB2 - L2) ½.

For the subject:

yD = 25 + (25² + 20²) ½ = 25 + 225½ = 25 + 15 = 40 m

Complementations
(1) obtaining of the speed quadratic function of y, for integration:

The expression of the acceleration the = g - (k/m) (y - L) it is of the type: THE = THE - B².y, or be,

dv / dt = THE - B².y or, being derived once in relation to the time,

d²v / dt² = - B².v (he/she remembers that dy / dt = v)

This equation differential (typical of MHS), very known, has as solution, the function:

v = grandma.sen (B.t - ß)... (eq.01) or

sen² (B.t - ß) = (v / grandma) ²... (eq.02)

The (eq.01) it can be written: dy / dt = grandma.sen (B.t - ß) that, integrated supplies:

y = (- grandma / B).waistband (B.t - ß) + C or (y - C) ² = (- grandma / B) ².cos² (B.t - ß) or

(y - C) ² = (grandma / B) ².[1 - sen² (B.t - ß)]... (eq.03)

Being taken her/it (eq02) in the (eq.03), it is had:

(y - C) ² = (grandma / B) ².[1 - (v / grandma) ²] or

(y - C) ² / (grandma / B) ² = 1 - (v / grandma) ² or

(v / grandma) ² = 1 - (B / grandma) ².(Y - C) ²

Finally:

v² = vo² - B².(y - C) ² or, with our notations v² = 2.g.y - (k/m) (y - L) ²  

To confirm, we will derive that function in relation to the time:2.v.dv / dt = 2.B².(C - Y).dy / dt = 2.B².(C - Y).v or
THE = B².C - B².y, that is the original form of the acceleration, with THE = B².C.

(2) obtaining of the speed quadratic function of y, for derivations and differential:

Of Newton's second law it resulted:  the = g - (k/m) (y - L)

the = dv / dt... (introducing a change of variables, it comes:)

the = (dv / dy).(dy / dt)... (reminding that: v = dy / dt, is had:)

the = v dv / dy = ½.D (v²) / dy... (then:)

dv² / dy = 2.a = 2g - 2 (k/m) (y - L) or

Dv2/dy = 2g - 2 (k/m) (y - L) = d [2gy - (k/m) (y - L) 2] / dy... (and, finally:)

v2 = 2gy - (k/m) (y - L) 2  

Solution of the subject for the method of the energies

Let us indicate for d the experienced deformation for the cable from the position L to the position it exalts where it happens the inversion of the movement. The jumper's maximum removal, in relation to the bridge, will be: D = L + D.

The jumper's gravitational potential energy set out from the bridge to this extreme position, he/she turns into elastic potential energy of the cable; we wrote:

m.g. (L+d) = (1/2).k.d²2.m.g.L + 2.m.g.d - k.d² = 0.... dividing m.the.m. for k, and making 2mg/k = u is,

u.L + u.d - d² = 0 (quadratic equation)

d', d" = [- u ± square (u² +4uL)] / (-2) ... formula of Baskara; square = square root.

With the data of the subject, it is had: u = 2mg/k = 2.80.10/160 = 10 and L = 20.

The numeric roots, of the equation, are: d' = 20 and d" = - 10

Reinforcing: d is the maximum deformation of the cable; increasing L (length of the cable) the jumper's maximum distance is had to the bridge: D = L+d.

With the data of the subject, a solution is D' = 20+20 = 40 m and the other is D" = -10+20 = 10 m.

The second solution can be spared because the cable doesn't apply force in the jumper for y < 20 m.

About the Author

Mechanical Engineer of profession. Visit this at Geradores and see the Geradores Usados

What is a good science fair topic for a highschool physics class??

It doesnt really have to be a physics topic but it needs a independent and dependant variable.

I would suggest looking up science fairs on the Internet and finding something you like for your own project.

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