How Does a Bungee Jumper's Energy Change During a Jump?

[patrickmoloney]

Homework Statement

A bungee jumper of mass $m$ attaches a light elastic string to his foot, and the other end to railing of a bridge on which he stands. The string has natural length $l_0$ and modulus of elasticity $\lambda$. Write down the energy equation. use conservation of energy to find the maximum distance the jumper descends after stepping off the bridge.

Homework Equations

Gravitational potential energy = $mgh$
Elastic Potential energy = $\frac{1}{2}kx^2$
Kinetic energy = $\frac{1}{2}mv^2$

The Attempt at a Solution

I understand the principles of energy conversion during a bungee jump. So during the first interval, stepping of the bridge, the only force acting on the system(bungee jumper) is gravity(free fall). As the bungee jumper falls, the gravitational potential energy is converted into kinetic energy. When he reaches the point where the bungee cord begins to stretch, gravitational potential energy begins to be converted into the elastic potential energy of the cord. Eventually, all of the kinetic energy is also converted into potential energy of the cord, and he comes to rest for a short period of time.

I'm not really sure how to write this in mathematics. I can probably make thing easier by setting up a coordinate system with an origin. If I used the origin at $l_0$ below the bridge since that would 0 some terms in the conservation equation.

I found the relationship between the spring constant and elasticity, which is $$k = \frac{\lambda}{L}$$

The conservation of energy is $$PE_i + KE_i = PE_f + KE_f$$. I'm not really sure how to tie this all together. Do I have to model a differential equation or is it simpler than that?

I apprecitate any help,
Thanks

 

[PeroK]

Putting the origin at $l_0$ below the bridge is a good idea. What does this give you for the initial energy of the system?

Your description is fine and, as you suggest, you just need to translate that into the maths.

 

[BvU]

1. Make a drawing, clearly indicating you choice of coordinate system.
2. Convert $k = \frac{\lambda}{L}$ to that system. Oops, you need k only, so: done.
3. Choose the initial and final points in time for the energy balance -- naturally $t_i = 0$ is the initial point when he/she jumps and $t_f$ is at the lowest point. (the exercise isn't about $t$, but nevertheless.
4. Write the energy balance. What are the kinetic energies (which I guess you call $KE_i$ and $KE_f$ ?

 

[patrickmoloney]

I think I might have made some progress since the post.

Let $h$ be the distance the jumper falls
$PE_i=mgy=mgh$
$PE_f=mgy = mg(0) = 0$

The potential energy of the elastic string is $\frac{1}{2}kx^2$ where $x$ is the amount the string has stretched. At the bottom of the fall the string is stretched by an amount $(h-l_0)$. But the cord doesn't begin to stretch until he falls a distance $l_0$. Therefore the final elastic potential energy is $\frac{1}{2}k(h-l_0)^2 = \frac{\lambda}{2l_0}(h-l_0)^2$

 

[BvU]

What is your coordinate system ? You confuse me with an x and a y -- and perhaps yourself too (if not here then later on in your curriculum)

 

[patrickmoloney]

ok so I've changed it a little bit. The initial position at the top before the jump is a distance $h$ from the bottom when the jumper is at rest momentarily. Should I use a different variable for the length the cord stretches? I currently have it as $x$.

$l_0$ is the unstretched length of the string. Is there an easier way to coordinate this?

 

[patrickmoloney]

the initial kinetic energy is zero and the final kinetic energy zero, since he is at rest both of those times.

is the energy equation $$mgh = \frac{\lambda}{2l_0}(h-l_0)^2$$

 

[PeroK]

the initial kinetic energy is zero and the final kinetic energy zero, since he is at rest both of those times.

is the energy equation $$mgh = \frac{\lambda}{2l_0}(h-l_0)$$

How have you defined $h$? Is that a constant or a variable? Don't you think you need a square somewhere?

 

[patrickmoloney]

I defined $h$ to be the distance from the initial position to the final position at the bottom. It's a constant, the only variable of length that changes is the length of the cord, when it extends. I did forget to square the term $(h-l_0)$

 

[PeroK]

I defined $h$ to be the distance from the initial position to the final position at the bottom. It's a constant, the only variable of length that changes is the length of the cord, when it extends. I did forget to square the term $(h-l_0)$

Okay. so what next?

 

[patrickmoloney]

I need to find the maximum distance the jumper descends after steeping off the ledge. The maximum distance corresponds to the lowest point of the fall where the velocity is 0 (momentarily).

When he is standing on the top, he is at rest. Therefore his kinetic energy is zero.
When he is at the bottom, he is also at rest so the kinetic energy is also 0 (since the velocity is zero.)

That being said, we need to be careful because there is also potential energy in the string as it stretches.

Do I need to solve the differential equation $$mgh=\frac{\lambda}{2l_0}(h-l_0)^2$$

 

[PeroK]

I need to find the maximum distance the jumper descends after steeping off the ledge. The maximum distance corresponds to the lowest point of the fall where the velocity is 0 (momentarily).

When he is standing on the top, he is at rest. Therefore his kinetic energy is zero.
When he is at the bottom, he is also at rest so the kinetic energy is also 0 (since the velocity is zero.)

That being said, we need to be careful because there is also potential energy in the string as it stretches.

Do I need to solve the differential equation $$mgh=\frac{\lambda}{2l_0}(h-l_0)^2$$

Solving the equation sounds like a good idea!

 

[patrickmoloney]

$$\begin{align*} mgh &= \frac{\lambda}{2l_0}(h-l_0)^2 \\ mgh &= \frac{\lambda}{2l_0}(h^2-2hl_0+l_0^2) \\ mgh &= \frac{\lambda}{2l_0}h^2 - \lambda h +\frac{\lambda l_0}{2} \\ \end{align*}$$

Then I found this quadratic equation that needs to be solved using the quadratic formula
$$\begin{align*} \frac{\lambda}{2l_0}h^2 - h(mg+\lambda) + \frac{\lambda l_0}{2} = 0 \end{align*}$$

Is this correct so far? I ended up with

$$h = \frac{(mg+\lambda)l_0\pm l_0\sqrt{mg(mg+2\lambda)}}{\lambda}$$

This looks a little messy though so I'm not sure it's correct.

 

[haruspex]

$$\begin{align*} mgh &= \frac{\lambda}{2l_0}(h-l_0)^2 \\ mgh &= \frac{\lambda}{2l_0}(h^2-2hl_0+l_0^2) \\ mgh &= \frac{\lambda}{2l_0}h^2 - \lambda h +\frac{\lambda l_0}{2} \\ \end{align*}$$

Then I found this quadratic equation that needs to be solved using the quadratic formula
$$\begin{align*} \frac{\lambda}{2l_0}h^2 - h(mg+\lambda) + \frac{\lambda l_0}{2} = 0 \end{align*}$$

Is this correct so far? I ended up with

$$h = \frac{(mg+\lambda)l_0\pm l_0\sqrt{mg(mg+2\lambda)}}{\lambda}$$

This looks a little messy though so I'm not sure it's correct.

Looks right.

Do I need to solve the differential equation

It is not differential.

 

[patrickmoloney]

Thanks very much for the help.