Motion of a non-linear pendulum with air resistance

  • #1

Homework Statement


I need to come up with an equation that would model the motion of a non-linear pendulum with air resistance. [/B]


Homework Equations


Fc=mgsintheta
Fdrag=(1/2)p(v^2)CA

The Attempt at a Solution



I started with mgsintheta-(1/2)p(v^2)CA=ma

After substituting v=r*omega and a=r*alpha, I get the following (look at the image)
i6bfv0


However, after graphing, I do not get a damped oscillating curve, because of the v^2 not changing signs. Can someone guide me? See second image:
i6bh0j
http://prntscr.com/i6bh0j[/B]

https://prnt.sc/i6bfv0
 

Answers and Replies

  • #2
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Sorry, we can't see the 1st image
 
  • #3
epenguin
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To know whether there is a problem, how many cycles did you follow it through?
 
  • #5
To know whether there is a problem, how many cycles did you follow it through?
I'm not quite sure what you mean by cycles. I set the domain on the graph to be [0, 2pi].
 
  • #6
epenguin
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I'm not quite sure what you mean by cycles. I set the domain on the graph to be [0, 2pi].
To be sure there is no damping try 20π or 100π?
 
  • #7
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It does make sense that your damping doesn't work if the damping force is always pointing in the same direction and not against v. Maybe you should try to replace v^2 by |v|v.
 
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  • #8
To be sure there is no damping try 20π or 100π?
I did, and there does not seem to be any. However, the amplitude does seem to change, but on average it remains the same.
 
  • #9
It does make sense that your damping doesn't work if the damping force is always pointing in the same direction and not against v. Maybe you should try to replace v^2 by |v|v.
Thanks, this does work. For some reason I didn't think of the absolute value as a solution to the v^2 problem. Now I get a graph that looks like the following. It does not pass through the x-axis like it should.
https://prnt.sc/i6cgpx
 
  • #10
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Thanks, this does work. For some reason I didn't think of the absolute value as a solution to the v^2 problem. Now I get a graph that looks like the following. It does not pass through the x-axis like it should.
https://prnt.sc/i6cgpx

Another sign error. When you're at positive theta, the force must point in the negative direction. So it should be -mgsin(theta).
 
  • #11
Another sign error. When you're at positive theta, the force must point in the negative direction. So it should be -mgsin(theta).
yes, I thought I already changed that. Thanks again.
 
  • #12
Another sign error. When you're at positive theta, the force must point in the negative direction. So it should be -mgsin(theta).
Did you look at the derivation of that equation I came up with? I just need someone to check if its correct (apart from the signs, and absolute values)
 
  • #13
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Did you look at the derivation of that equation I came up with? I just need someone to check if its correct (apart from the signs, and absolute values)

Seems fine to me.
 
  • #14
Ray Vickson
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Seems fine to me.

Here is what I get using Maple to solve the IVP
$$\theta^{''} = -\frac{g}{L} \sin(\theta) - \frac{pLcA}{2m} \theta' |\theta'|, \; \theta(0)=1.5, \theta'(0)=0$$
using your input parameters (but using notation ##w(t)## instead of ##\theta(t)##):
upload_2018-1-27_9-12-58.png
 

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  • #15
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Thread moved. Problems involving differential equations belong in the Calculus & Beyond section.
 

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