Energy and Work Energy and Work Physics Help Forum 
Jul 7th 2008, 08:57 PM

#1  Junior Member
Join Date: Apr 2008 Location: Pangkor Island, Perak, Malaysia.
Posts: 28
 Bungee!!!
A 60kg bungee jumper jumps from a bridge.
She is tied to a bungee cord that is 12m long when unstretched and falls a total distance of 31m.
Calculate the maximum acceleration experienced by the jumper.
The answer given is
$\displaystyle 22ms^{2}$.
In my opinion:
she falling freely under gravitational acceleration when the cord unstretched.
When the cord is stretched, her acceleration should be less than gravitational acceleration.
Her acceleration during she is pulled upward by the cord I can't imagine.
After that, she is under free fall again.
If I ignore her acceleration during she is pulled upward by the cord, her maximum acceleration should
$\displaystyle a_{max}=9.81 ms^{2}$.
Therefore, is she experienced the maximum acceleration
$\displaystyle a_{max}=22 ms^{2}$.
during pulled by the cord?
What is the concept?
How to calculate?
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Electrical Engineering

 
Jul 8th 2008, 04:25 AM

#2  Forum Admin
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Originally Posted by SengNee A 60kg bungee jumper jumps from a bridge.
She is tied to a bungee cord that is 12m long when unstretched and falls a total distance of 31m.
Calculate the maximum acceleration experienced by the jumper.
The answer given is
$\displaystyle 22ms^{2}$.
In my opinion:
she falling freely under gravitational acceleration when the cord unstretched.
When the cord is stretched, her acceleration should be less than gravitational acceleration.
Her acceleration during she is pulled upward by the cord I can't imagine.
After that, she is under free fall again.
If I ignore her acceleration during she is pulled upward by the cord, her maximum acceleration should
$\displaystyle a_{max}=9.81 ms^{2}$.
Therefore, is she experienced the maximum acceleration
$\displaystyle a_{max}=22 ms^{2}$.
during pulled by the cord?
What is the concept?
How to calculate? 
The problem is written incorrectly. It should be asking for the maximum speed. For the record you are correct about the maximum acceleration.
There are a couple of ways to attack this problem. All of them depend on the concept that the motion of the diver is simple harmonic motion. The simplest is to model the motion, calling the origin the equilibrium position of the bungee (12 m), taking positive upward, and starting time from when the diver reaches the equilibrium position of the bungee.
$\displaystyle y = (31  12)~cos ( \omega t)$
Then
$\displaystyle y' = \omega A~sin(\omega t)$
We need $\displaystyle \omega$. Since this is a SHO and the bungee is essentially a spring, we know that
$\displaystyle \omega = \sqrt{\frac{k}{m}}$
From the condition that the object is (temporarily) stationary at the bottom of the motion we know that
$\displaystyle mg = k \Delta y$
$\displaystyle k = \frac{mg}{\Delta y} = \frac{mg}{31  12}$
You can do the rest noting that the maximum speed will take place when the diver is at the equilibrium position, in this case at the origin. (You can use the second derivative to derive that the maximum acceleration is at the endpoints of the motion and is g.)
Dan
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Jul 9th 2008, 12:40 AM

#3  Junior Member
Join Date: Apr 2008 Location: Pangkor Island, Perak, Malaysia.
Posts: 28

Originally Posted by topsquark The problem is written incorrectly. It should be asking for the maximum speed. For the record you are correct about the maximum acceleration.
There are a couple of ways to attack this problem. All of them depend on the concept that the motion of the diver is simple harmonic motion. The simplest is to model the motion, calling the origin the equilibrium position of the bungee (12 m), taking positive upward, and starting time from when the diver reaches the equilibrium position of the bungee.
$\displaystyle y = (31  12)~cos ( \omega t)$
Then
$\displaystyle y' = \omega A~sin(\omega t)$
We need $\displaystyle \omega$. Since this is a SHO and the bungee is essentially a spring, we know that
$\displaystyle \omega = \sqrt{\frac{k}{m}}$
From the condition that the object is (temporarily) stationary at the bottom of the motion we know that
$\displaystyle mg = k \Delta y$
$\displaystyle k = \frac{mg}{\Delta y} = \frac{mg}{31  12}$
You can do the rest noting that the maximum speed will take place when the diver is at the equilibrium position, in this case at the origin. (You can use the second derivative to derive that the maximum acceleration is at the endpoints of the motion and is g.)
Dan 
I see. I haven't go through simple harmonic motion yet, that why I can't do and the question has problem. My lecturer put this question in "work and energy".
Thank you very much.

 
Jul 9th 2008, 04:56 AM

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Posts: 2,862

Originally Posted by SengNee I see. I haven't go through simple harmonic motion yet, that why I can't do and the question has problem. My lecturer put this question in "work and energy".
Thank you very much. 
In that case, after you find the spring constant k, you can use the workenergy theorem. Setting the lowest point of the motion as the gravitational potential energy zero point (and positive upward) the mechanical energy of the system is entirely due to the spring potential energy. The diver is at his/her maximum speed when passing through the equilibrium position of the diver, so we get that
$\displaystyle \frac{1}{2}k(31  12)^2 = mg(31  12) + \frac{1}{2}mv^2$
Dan
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Jul 10th 2008, 12:18 AM

#5  Junior Member
Join Date: Apr 2008 Location: Pangkor Island, Perak, Malaysia.
Posts: 28

Originally Posted by topsquark In that case, after you find the spring constant k, you can use the workenergy theorem. Setting the lowest point of the motion as the gravitational potential energy zero point (and positive upward) the mechanical energy of the system is entirely due to the spring potential energy. The diver is at his/her maximum speed when passing through the equilibrium position of the diver, so we get that
$\displaystyle \frac{1}{2}k(31  12)^2 = mg(31  12) + \frac{1}{2}mv^2$
Dan 
By using conservation of energy, why I get the answer v=15.13m/s, but not v=2m/s ?
__________________
$\displaystyle {\color{blue}\text{Pangkor Island}}$ is a beautiful island.
Coming here for a $\displaystyle {\color{blue}\text{holiday}}$ is a wonderful decision.
(My grammar is not good, if there is any mistake, send a private message to correct me. Thanks.)
I study in Malaysia Science University(USM) now.
Electrical Engineering

 
Jul 10th 2008, 12:58 AM

#6  Junior Member
Join Date: May 2008
Posts: 29

Originally Posted by topsquark
From the condition that the object is (temporarily) stationary at the bottom of the motion we know that
$\displaystyle mg = k \Delta y$
$\displaystyle k = \frac{mg}{\Delta y} = \frac{mg}{31  12}$
You can do the rest noting that the maximum speed will take place when the diver is at the equilibrium position, in this case at the origin. (You can use the second derivative to derive that the maximum acceleration is at the endpoints of the motion and is g.)
Dan 
At its lowest point, we must have $\displaystyle k \Delta y > mg$, otherwise the spring won't have a net force upwards.
My unsure attempt:
At the top of the cliff, let the bungee jumper have gravitational potential $\displaystyle mg(31)$
At the bottom of motion, the bungee jumper will have elastic potential $\displaystyle \frac{1}{2}kA^2 = \frac{1}{2}k(3112)^2$
Equating, $\displaystyle mg(31) = \frac{1}{2}k(3112)^2$
$\displaystyle 62mg = k(19)^2$
$\displaystyle k = \frac{62mg}{361} = \frac{62 \times 60 \times 9.8}{361} = 100.99 N/m$
(LOL nevermind looks like you already found this)
So the maximum acceleration will be at the bottom and is $\displaystyle \frac{kA mg}{m} = 22.18ms^{2}$
Actually I have a problem with this question. If the amplitude of oscillation is 19m, and the unstretched spring has length 12m, then won't that mean that at the top the bungee jumper will be 7m above the cliff!!!? How can this be?
Last edited by DivideBy0; Jul 10th 2008 at 01:32 AM.

 
Jul 12th 2008, 06:35 AM

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Originally Posted by DivideBy0 At its lowest point, we must have $\displaystyle k \Delta y > mg$, otherwise the spring won't have a net force upwards.
My unsure attempt:
At the top of the cliff, let the bungee jumper have gravitational potential $\displaystyle mg(31)$
At the bottom of motion, the bungee jumper will have elastic potential $\displaystyle \frac{1}{2}kA^2 = \frac{1}{2}k(3112)^2$
Equating, $\displaystyle mg(31) = \frac{1}{2}k(3112)^2$
$\displaystyle 62mg = k(19)^2$
$\displaystyle k = \frac{62mg}{361} = \frac{62 \times 60 \times 9.8}{361} = 100.99 N/m$
(LOL nevermind looks like you already found this)
So the maximum acceleration will be at the bottom and is $\displaystyle \frac{kA mg}{m} = 22.18ms^{2}$
Actually I have a problem with this question. If the amplitude of oscillation is 19m, and the unstretched spring has length 12m, then won't that mean that at the top the bungee jumper will be 7m above the cliff!!!? How can this be? 
Heh. Nice job. I was too wrapped up in finding a value for k (which I did incorrectly) that I didn't even notice that my acceleration was 0 at the bottom of the motion.
Thanks for the catch!
Dan
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