Max Acceleration of a Bungee Jumper

[harvellt]

Homework Statement

Consider a bungee jumper of mass 75kg, with a 9.00-m Cord tied to his ankles. The spring constant of the cord is 150 N/M. The jump of point is 9.00m. Set origin to where the cord becomes taunt.
At what point dose the jumper attain maximum speed?
What is the value of the max speed?
At what point dose the jumper attain maximum acceleration?
What is the maximum acceleration?

Homework Equations

x>0 x=mgx
x<0 x=mgx+ (1/2)kx²
(1/2)mv²

The Attempt at a Solution

I set second equation to zero took the derivative and solved to find the point x where the potential energy is lowest and then set the potential energy equal to the kinetic energy at that point and solved for v. I can also use the max kinetic energy to solve for the max stretch in the cord, but I am lost in how to solve for the acceleration. Any advice?

 

[quantumdude]

Are you aware that this is a problem in simple harmonic motion (SHM), and that with SHM we have $y(t)=y_m\cos(\omega t+\phi )$? If you have the displacement as a function of time, you also have velocity and acceleration as functions of time.

 

[harvellt]

The point of the problem is to solve it with the graph of the curve of potential energy.

 

[turin]

... mass 75kg, ... 9.00-m Cord tied to ... [it].
... spring constant of ... cord is 150 N/M.
... jump of point is 9.00m.
Set origin to where the cord becomes taunt.

At what point ... maximum speed?
What is ... the max speed?
At what point ... maximum acceleration?
What is the maximum acceleration?

... this is a problem in simple harmonic motion (SHM), ...

I don't think this is true, for two reasons. Firstly, there are two forces, and they are not both harmonic forces centered at the same point. Secondly, the restoring force is not symmetrical about the equilibrium point (is it?).

 

[turin]

x>0 x=mgx
x<0 x=mgx+ (1/2)kx²

Not quite. If your equations were true, then, in the region where x>0, g would not be an acceleration; it would be an inverse mass equal to 0.013 kg^-1, and, in the region where x<0, g would still be an inverse mass, although I don't know what it would be, k would be unitless, and x would be a fixed value. In other words, your equations don't make sense. However, that may just be a simple typo ...

I set second equation to zero ...

I don't know what this means. 0=0? x=0?

... took the derivative ...

You took the derivative of zero? What was your result?

... set the potential energy equal to the kinetic energy at that point ...

What is the phsysical interpretation of the condition that PE=KE at a point?

I can also use the max kinetic energy to solve for the max stretch in the cord, ...

Please show us how you would do that.

... I am lost in how to solve for the acceleration. Any advice?

Newton's second law.

 

[harvellt]

To solve for the lowest potential energy with respect to x I used
E=mgx+1/2kx²
DE/DX=mg+kx
0=(75kg)(9.81)+150x
x=-4.905
Then I find V at -4.905
with a($$\Delta$$x)=$$\Delta$$v²/2
v=15m/s
as long as x>0 then a=g but how do I find a through out the stretching of the cord?
mv²=kx²
So the max stretch in the cord is 10.6m.
I know it is small and I am going to smack my self in the face when I see it but how do I goto a from here?

 

[harvellt]

and yes sorry torin that was a typo I ment E=mgx+1/2kx²

 

[turin]

To solve for the lowest potential energy with respect to x I used
E=mgx+1/2kx²
DE/DX=mg+kx
0=(75kg)(9.81)+150x
x=-4.905

Yes, I agree with the procedure (given certain caveats) and the number. However, why did you do this?

Then I find V at -4.905

Why?

with a($$\Delta$$x)=$$\Delta$$v²/2
v=15m/s

Is that a so-called kinematical equation? Do you know the condition on $a$ when it is valid? Is that condition satisfied here?

as long as x>0 then a=g ...

I concur.

... how do I find a through out the stretching of the cord?

See my previous post.

mv²=kx²

This is not true, at least not the way you have written it. I'm not trying to get into your head but consider this observation: I think your understanding of physics may actually be suffering from the math that you have learned, because perhaps you think, subconsciously, that you can substitute the physics principle with raw mathematical power. It does not usually work this way. I am only saying this because I went through that myself, and I had to realize two difficult lessons: 1) my math wasn't as sufficient as it needed to be in order to do this, and 2) ultimately, this stuff is mathematically meaningless.

 

[harvellt]

Ok I see what you are saying when the spring is taunt acceleration begins to vary with respect to x as the cord stretches.
So to find max velocity at x=-4.9. If I move my origin I can use the equation.
mgh=(1/2)mv²
Where h=9+4.9 soooooo,
(75kg)(9.81m/s²)(13.0)=(1/2)(75kg)v²
v=14.9m/s
To answer your question of "why?" I did it because it is what the problem asked for.
I see what your saying about trying to mathematically muscle threw problems. This isn't my favorite problem I'm just trying to obsess over it because this is section in the book(same title as the post) that I feel like I am having some conceptual disconnect with. I think I am trying to push threw math I am not completely comfortable with yet and just throwing formulas at it.

 

[turin]

I will give you an example of what I mean by thinking through this without too much math (just very basic algebra).

- Above the origin (dangling bungee), what force(s) act on the jumper (including direction of force)? Using Newton's second law, write the acceleration of the jumper, symbolically, in terms of said force(s) and the mass of the jumper.

- Below the origin (taught bungee), what force(s) act on the jumper (including direction of force)? Using Newton's second law, write the acceleration of the jumper, symbolically, in terms of said force(s) and the mass of the jumper.

- Do you see an interesting/definitive comparison between the case above the origin and the case below the origin?

It may help to sketch a graph of the acceleration as a function of height. Once you figure out what's going on with the forces and Newton's second law here, this should be pretty easy to do. I will tell you that you should also consider energy in this problem, but don't ignore the utility of Newton's second law.