Motion of a non-linear pendulum with air resistance

[Omkar Vaidya]

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)

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:

http://prntscr.com/i6bh0j[/B]

https://prnt.sc/i6bfv0

 

[Nguyen Son]

Sorry, we can't see the 1st image

 

[epenguin]

To know whether there is a problem, how many cycles did you follow it through?

 

[Omkar Vaidya]

Sorry, we can't see the 1st image

There are two images. Here are the links:
http://prntscr.com/i6bh0j
https://prnt.sc/i6bfv0

 

[Omkar Vaidya]

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].

 

[epenguin]

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?

 

[Gigaz]

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.

 

[Omkar Vaidya]

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.

 

[Omkar Vaidya]

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

 

[Gigaz]

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).

 

[Omkar Vaidya]

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.

 

[Omkar Vaidya]

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)

 

[Gigaz]

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.

 

[Ray Vickson]

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 $\theta(t)$):

 

[Mark44]

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