Potential energy of rubber bands

In summary, the conversation discusses the calculation of potential energy in rubber bands, specifically in the context of a catapult. The equation includes variables such as the number of rubber bands, spring constant, unstretched length, mass of the ball, and moment of inertia of the arm. The conversation also mentions using a marshmallow and provides other data such as initial velocity, time, and maximum height. The mentor notes that the equation is unusual and suggests checking the moment of inertia and number of bands used. The mentor also provides a link for further information and explains the meaning of the variable "n".
  • #1
rasalzari
27
2
<Mentor's note: moved from a general forum and therefore no template>

upload_2017-4-12_4-28-28.png

This is the equation of potential energy in rubber bands, however, I don't know how to calculate the inertia of the arm of catapults or get the k constant. And what does n mean? Basically, i don't understand the whole equation.
The energy balance, where N is the number of rubber bands, k is the spring constant of a single rubber band, x0 is the unstretched length of a rubber band, m the mass of the ball, and I is the moment of inertia of the arm.
Can someone please show me an example of how to calculate this?
I used a marshmallow which is 0.007 kg, I stretched the catapults to 0.1 m, it has an angle of 35 degrees
range=0.793
initial velocity=2.88 m/s
time=0.337 s
max height=0.139 m
horizontal velocity=2.36 m/s
vertical velocity=1.65 m/s
Thats all i know and i hope it helps! please help me figure this out!
 
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  • #2
rasalzari said:
<Mentor's note: moved from a general forum and therefore no template>

View attachment 150616
This is the equation of potential energy in rubber bands, however, I don't know how to calculate the inertia of the arm of catapults or get the k constant. And what does n mean? Basically, i don't understand the whole equation.
The energy balance, where N is the number of rubber bands, k is the spring constant of a single rubber band, x0 is the unstretched length of a rubber band, m the mass of the ball, and I is the moment of inertia of the arm.
Can someone please show me an example of how to calculate this?
I used a marshmallow which is 0.007 kg, I stretched the catapults to 0.1 m, it has an angle of 35 degrees
range=0.793
initial velocity=2.88 m/s
time=0.337 s
max height=0.139 m
horizontal velocity=2.36 m/s
vertical velocity=1.65 m/s
Thats all i know and i hope it helps! please help me figure this out!
That equation is rather unusual. It implies the force needed to stretch the rubber rises as the square root of the extension, whereas the usual law says it rises linearly. Is it possible that x here refers not to the actual extension but to some other displacement within a mechanism supporting the rubber band? Is there a diagram?
 
  • #3
haruspex said:
That equation is rather unusual. It implies the force needed to stretch the rubber rises as the square root of the extension, whereas the usual law says it rises linearly. Is it possible that x here refers not to the actual extension but to some other displacement within a mechanism supporting the rubber band? Is there a diagram?
Yes here you go
upload_2017-4-12_6-3-53.png


And here is a link with further information
http://www.sciencebuddies.org/content/pdfs/projectideaskits/phys_p089/phys_p089_20131021.pdf
 
  • #4
Ok, I see that the text states that the rubber bands provided do have a nonlinear force/extension relationship.
rasalzari said:
I don't know how to calculate the inertia of the arm
It gives a formula in note 3 under Experimental Procedure.
However, I see that it defines I as the moment inertia about the mass centre of the arm. That is not what you want. The arm rotates about one end, not its mass centre. The text says to use I = 1.1 x 10-3kg*m². If that really is the MoI about the mass centre then you need to add ML2/4 where L is the length of the arm and M its mass. But it is possible they mean the MoI about one end is 1.1 x 10-3kg*m².
You can check this by measuring L and M and using:
MoI about mass centre = ML2/12
MoI about one end = ML2/12+ML2/4=ML2/3.
See which of those is closer to 1.1 x 10-3kg*m².
rasalzari said:
get the k constant
See note 5 under Experimental Procedure.
rasalzari said:
what does n mean?
As you wrote, N is the number of bands in parallel. I think it is up to you how many you attach.
 
  • #5
haruspex said:
Ok, I see that the text states that the rubber bands provided do have a nonlinear force/extension relationship.

It gives a formula in note 3 under Experimental Procedure.
However, I see that it defines I as the moment inertia about the mass centre of the arm. That is not what you want. The arm rotates about one end, not its mass centre. The text says to use I = 1.1 x 10-3kg*m². If that really is the MoI about the mass centre then you need to add ML2/4 where L is the length of the arm and M its mass. But it is possible they mean the MoI about one end is 1.1 x 10-3kg*m².
You can check this by measuring L and M and using:
MoI about mass centre = ML2/12
MoI about one end = ML2/12+ML2/4=ML2/3.
See which of those is closer to 1.1 x 10-3kg*m².

See note 5 under Experimental Procedure.

As you wrote, N is the number of bands in parallel. I think it is up to you how many you attach.
Ohh i see thank you so much for your reply, however it only applies for non-linear? but my graph is linear... is there any other way to get the potential energy for linear relationship?
 
  • #6
rasalzari said:
my graph is linear
What graph? What values are you plotting on the axes?
 
  • #7
haruspex said:
What graph? What values are you plotting on the axes?
upload_2017-4-12_7-23-32.png
 
  • #9
haruspex said:
Interesting.
What are the values of x2-x0 for each band?
Oh sorry I am a newbie in physics and math, could you explain what x2-x0 mean? :frown::frown::frown::frown::frown::oldfrown:
 
  • #10
rasalzari said:
Oh sorry I am a newbie in physics and math, could you explain what x2-x0 mean? :frown::frown::frown::frown::frown::oldfrown:
They appear in the equation you posted originally. They are defined in the link you provided. I have the feeling you have not read that page fully.
 
  • #11
haruspex said:
They appear in the equation you posted originally. They are defined in the link you provided. I have the feeling you have not read that page fully.
no, I did but I didn't cover much of physics so i couldn't understand that's why i asked in this website so i can learn more
 
  • #12
rasalzari said:
no, I did but I didn't cover much of physics so i couldn't understand that's why i asked in this website so i can learn more
This much is not a question of physics. What does it say x0, x1 and x2 represent?
 
  • #13
haruspex said:
This much is not a question of physics. What does it say x0, x1 and x2 represent?
I don't really know but at first, I thought it might mean initial velocity, is it?
 
  • #14
rasalzari said:
I don't really know but at first, I thought it might mean initial velocity, is it?
No, they're lengths of the elastic. As it says, x0 is the unstretched length. I thought it defined x1 and x2, but maybe I imagined it. x1 is the stretched length just before firing. At the end of firing (release of projectile) it may not have returned to its unstretched length, instead being length x2.
 
  • #15
haruspex said:
No, they're lengths of the elastic. As it says, x0 is the unstretched length. I thought it defined x1 and x2, but maybe I imagined it. x1 is the stretched length just before firing. At the end of firing (release of projectile) it may not have returned to its unstretched length, instead being length x2.
ohhh alright thank you sir! i will read more about the topic and if i have more questions ill ask :woot:
 
  • #16
In general, rubber is nonlinear and nonconservative. This means that
(1) force will not be proportional to the elongation;
(2) work put in will not be fully recovered when the stretched piece of rubber is released.
This makes energy calculations with rubber bands rather iffy.
 
  • #17
Dr.D said:
ber is nonlinear and nonconservative. This means that
(1) force will not be proportional to the elongation;
(2) work put in will not be fully recovered when the stretched piece of rubber is released.
To which I will add that rubber bands also display hysteresis; the work that is recovered depends on what has already been done to the band in terms of stretching and relaxing.
 

Related to Potential energy of rubber bands

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or condition. It is the energy that is stored within an object, which can be released and converted into other forms of energy.

2. How is potential energy of rubber bands related to their stretching?

The potential energy of a rubber band is directly related to its stretching. When a rubber band is stretched, it stores potential energy in the form of the tension force within the band. The more the rubber band is stretched, the more potential energy it has.

3. What factors affect the potential energy of a rubber band?

The potential energy of a rubber band is affected by several factors, such as the length and thickness of the band, the degree of stretching, and the type of material the band is made of. The greater the length, thickness, and stretching of the band, the greater its potential energy.

4. How does temperature affect the potential energy of rubber bands?

Temperature can affect the potential energy of rubber bands in two ways. First, at higher temperatures, the rubber band will have more kinetic energy, which will cause it to stretch further and have more potential energy. Second, at higher temperatures, the rubber band will also become more elastic, allowing it to store more potential energy when stretched.

5. Can potential energy of rubber bands be converted into other forms of energy?

Yes, potential energy of rubber bands can be converted into other forms of energy. When a stretched rubber band is released, its potential energy is converted into kinetic energy, causing the rubber band to snap back to its original shape. This snapping motion can be harnessed for various purposes, such as propelling objects or powering machines.

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