Finding General Formula for Bungee Jumping - Maths C Assignment

[Eldawg]

Good morning/evening fellow physics enthusiasts.

For my year 12 maths c class, we're doing an assignment on bungee Jumping (Task sheet can be found by googling "maths c bungee jumping" first link will be the relative one (pdf file)).

For this assignment, the first question asks you to find a general formula for the maximum depth reached in the jump in terms of the jumpers mass and the natural length of the rope (where air resistance in negligent).

The relevant information given on the specifications of the rope being used is that: when a mass of 75kg is applied to the rope, the rope will stretch to a length twice its' equilibrium length (length without stretching).

The main difficulty I've encountered in answering this question has been attempting to find a general formula for k (Hooke's law) in terms of l (the natural length of the rope) and/or the mass.

After countless hours considering this problem, my best attempt at determining k has been to substitute the following into my equation for the velocity of the jumper for the section of the fall where tension plays a role: x(extension beyond natural length of rope)=l when m=75 and v=0, since when the velocity reaches 0, the mass will have reached its' lowest point.

Where g=acceleration due to gravity (9.81ms^-2)

1/2(v)^2=gx - k(x)^2/(2m) + (2lg)^.5
0=gl - k(l)^2/(2*75) + (2lg)^.5

And so on and so forth until the equation is rearranged to give k

Please, tell me if I'm using a valid method to determine k, and if not, please give me some hints on how I can, I've already read two physicsforums posts, posted by someone else, about this maths c assignment which were no help.

 

[member 428835]

if i read you correctly you need Newtons second law; F=ma. now consider:

mx''=mg-k(x+L) where x is position, g is acceleration of gravity, L is length of spring (chord) extended after the body of mass is attached eg equilibrium length in your case, and k is the proportionality constant.

recognize "down" is now oriented positive direction for this diff eq to make sense. now if we are at equilibrium, it seems x=0 which implies mg=kL. i think now we may solve.

good luck

 

[Eldawg]

cheers

 

[Eldawg]

Hello, after re-reading your reply I've noticed what seem like some errors. You state that at equilibrium, which is when the position (x) is equal to the length of the rope (l), then x=0. x=0 at the beginning of the jump, when no forces are applied to the mass, if you're using x as position.

Also, the solution "mg-kl" cannot be right since F=ma, F=mg-kl (Resultant force) which means that mg-kl=ma. If mg=kl, and they act in opposite directions (which they do, since mg travels in the positive, down, direction and kl (Tension) travels in the negative, up, direction) then acceleration = 0. In other words, if the forces are of equal magnitude and opposite direction, the resultant force will equal 0, which means that 0=ma, also meaning that a=0. Acceleration can never equal 0 during the jump (unless terminal velocity is reached) since, after tension kicks in, the velocity begins to decrease from a positive value, to 0 (when the mass reaches the bottom). After the instant the velocity =0, it springs back up, traveling in the negative (upwards direction).

If velocity is going from positive, to 0, to negative, then acceleration is negative, and therefore not 0. If acceleration isn't 0, then mg cannot =kl.

Also, k is a constant determined by the springiness of the rope, if mg=kl, then k=mg/l. This is stating that the constant k is determined by the mass and length of the rope, when in actuality, the mass is independant of the k value, a rope has a k value before a mass is applied to it.

The only thing which could determine k (Hooke's constant) therefore could be specifications of what mass causes what extension, however, since the formula k=mg/l doesn't take into account the extension of the rope, even with the specifications of the stretching, k cannot be determined by this formula.

 

[member 428835]

please consider the following:

1) The force due to gravity will always act upon the object of course. This force is: mg

2) We are going to assume that Hooke’s Law will govern the force that the spring exerts on the object. This force will always be present as well and is: -k(l+x)
Hooke’s Law tells us that the force exerted by a spring will be the spring constant, k > 0, times the displacement of the spring from its natural length. For our set up the displacement from the spring's natural length is L + x and the minus sign is in there to make sure that the force always has the correct direction.

Let’s make sure that this force does what we expect it to. If the object is at rest in its equilibrium position the displacement is L and the force is simply Fs (force spring) = kL which will act in the upward position as it should since the spring has been stretched from its natural length.

If the spring has been stretched further down from the equilibrium position then will be positive and Fs will be negative acting to pull the object back up as it should be.

Next, if the object has been moved up past its equilibrium point, but not yet to its natural length then u will be negative, but still less than L and so L + x will be positive and once again Fs will be negative acting to pull the object up.

Finally, if the object has been moved upwards so that the spring is now compressed, then u will be negative and greater than L. Therefore, L + x will be negative and now Fs will be positive acting to push the object down.

So, it looks like this force will act as we expect that it should.
3) don't worry about damping or external forces, as we have none i can see
4) we now have the following equation we seek: mx''=mg-k(l+x)

Also, k is a constant determined by the springiness of the rope, if mg=kl, then k=mg/l. This is stating that the constant k is determined by the mass and length of the rope, when in actuality, the mass is independant of the k value, a rope has a k value before a mass is applied to it.

this is oxymoronic, as you state k=mg/l, which implies m,g,l are independent, and thus determine k. Now, when the object is at rest in its equilibrium position there are exactly two forces acting on the object, the force due to gravity and the force due to the spring. Also, since the object is at rest (i.e. not moving) these two forces must be canceling each other out. This means that we must have mg=kl

let me know if you still disagree

best of luck

 

[voko]

The coefficient k can be determined from the static condition that the rope extends twice its length under the mass of 75 kg at rest.

For the jump, consider the change in the potential energies of the jumper and the rope at the top and the bottom.

 

[Pkruse]

You might find it interesting to look into how the fall protection engineers design their equipment. Most have a simple formula based upon conservation on energy. Some also take onto account energy absorbed by the rope. Some rope is designed to absorb much energy.

 

[HallsofIvy]

The energy required to stretch a rope, with "spring constant" k, a distance x is $(1/2)kx^2$. Set that equal to the change in potential energy, mgh, in falling a distance h. (Note that x is the distance below the point where the rope begins to stretch, h is the total distance fallen.)

 

[Eldawg]

I see now, I'm sorry, my teacher had initially informed me that x=l for a mass of 75kg at the bottom the jump. I now see this is not true, you're right that mg=kx at v=0 when the forces are at equilibrium. The big confusion was that I was told not to consider any point after the spring begins to stretch back up in the negative direction, however, you need to consider that to find that mg eventually equals kx, and at this point, x=l. I was assuming that x=l for the instantaneous velocity = 0 down the bottom, the instant before the spring began to bounce back up.

Thanks tonnes for helping with this question, if I hadn't seen my mistakes I no doubt would have handed in a faulty assignment this Thursday, and changed my A to a C, in my most important year of school.

However, now with this information correct, I have managed to complete 3 out of 5 questions. The 4th however be-riddles me:

4. At present the model does not include air resistance. Discuss the changes that would have to be made to the model to include air resistance that is proportional to the velocity of the jumper. Discuss the strengths and limitations with the mathematics of this model.

I'm supposed to determine a constant of air resistance (which I have called b) from the information that Air resistance (which I have called J): is proportional to: velocity.

A friend of mine suggested substituting in small values of velocity, when velocity nears 0 to determine b.

My problem with this is I work with the equation a=g-bv, then I integrate with either respect with time or displacement.

To substitute in a value of velocity therefore, I must know the time or displacement at which that velocity occurred: if I already knew velocity at any time or displacement, I wouldn't need to figure it out in the first place.

The idea of substituting in small values of v, where air resistance is assumed to be negligent doesn't make any sense. How am I supposed to determine a constant which determines air resistance, when I'm assuming air resistance is negligent?

Any help would be muchly appreciated, as I'm running out of time on this one.

 

[member 428835]

glad to have helped. however;

I'm supposed to determine a constant of air resistance (which I have called b) from the information that Air resistance (which I have called J): is proportional to: velocity.

if i read you right b is the proportionality constant of velocity? if so, the new and improved model is:
mx''=mg-k(l+x)-bx'
ill leave you to deduce why we write -bx'

unfortunately this model does not account for external forces, f(t). adding other forces yields mx''=mg-k(l+x)-bx'+f(t) which is much more realistic than our current model.

as for solving the second order ode, remember x'(0)=0 and x(0)=-2L (We call the equilibrium position the position of the center of gravity for the object as it hangs on the spring with no movement)

 

[voko]

the new and improved model is:
mx''=mg-k(l+x)-bx'

This model applies to the part of the trajectory where the rope is under tension. Before that, the equation is mx'' = mg - bx'.