Setting up Differential Equations of Motion


by NeutronStar
Tags: differential, equations, motion, setting
Clausius2
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#19
Nov6-04, 05:29 AM
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Quote Quote by NeutronStar
If the pendulum was a solid rod instead of a spring would it be correct then to write the Total Kinetic Energy as simply:

[tex]T=\frac{1}{2}M*(\dot{x}^2+\dot{y}^2)=\frac{1}{2}M*(r^2+\dot{\theta}^2)[/tex]

Or do you drop the r altogether since it's no longer a degree of freedom?
If the rod is solid and non elastic (K=0) then:

[tex] r(t)=R [/tex]

furthermore

[tex] \dot r=\ddot r=0[/tex]

All the terms that contain the above derivative are dropped, also the elastic terms. Check your new kinetic energy expression because that you posted is wrong.

EDIT: This is another advice, it's a traditional advice said by engineers to begineers in physics and mechanics:

Pay attention to physical DIMENSIONS of the expressions you write. For example:
[tex]T=\frac{1}{2}M*(\dot{x}^2+\dot{y}^2)=\frac{1}{2}M*(r^2+\dot{\theta}^2)[/tex]

It hasn't got dimensions of energy (Joules), so that it is wrong at first sight without thinking anymore. Checking dimensions will enhance you to be forewarned about errors.
dextercioby
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Nov6-04, 07:25 AM
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Quote Quote by Clausius2
If the rod is solid and non elastic (K=0) then:

[tex] r(t)=R [/tex]

furthermore

[tex] \dot r=\ddot r=0[/tex]

All the terms that contain the above derivative are dropped, also the elastic terms. Check your new kinetic energy expression because that you posted is wrong.

EDIT: This is another advice, it's a traditional advice said by engineers to begineers in physics and mechanics:

Pay attention to physical DIMENSIONS of the expressions you write. For example:
[tex]T=\frac{1}{2}M*(\dot{x}^2+\dot{y}^2)=\frac{1}{2}M*(r^2+\dot{\theta}^2)[/tex]

It hasn't got dimensions of energy (Joules), so that it is wrong at first sight without thinking anymore. Checking dimensions will enhance you to be forewarned about errors.

Your kinetic energy would be then
[tex] T = \frac{1}{2}Mr^2 \dot{\theta}^2 [/tex] which is just the usual kinetic energy for the rotation movement:
[tex] T = \frac{1}{2} I\omega^2 [/tex],identifying the momentum of inertia I with [tex] Mr^2 [/tex].

The dimensional issue is valid only in classical dynamics.At a quantm level,usually this is neglected,because [tex] \hbar [/tex] and [tex] c [/tex] are set to 1.
Clausius2
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Nov6-04, 08:13 AM
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Quote Quote by dextercioby
The dimensional issue is valid only in classical dynamics.At a quantm level,usually this is neglected,because [tex] \hbar [/tex] and [tex] c [/tex] are set to 1.
I have specified the word engineer.

As a nearly future engineer, I don't think I'll deal anytime with quantum mechanics. We are so classical.....
dextercioby
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Nov6-04, 09:22 AM
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Engineers in my home town are taught a physics course,though only for one semester (it used to be 2,and it still is,but in other universities),where a lotta quantum physics knowledge is inserted.For example,they study wave mechanics,quantum statistical physics and,of course,their applications to solid-state physics especially semiconductors.
I don't know the way thing stay in Madrid...I guess it's less theory...
NeutronStar
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Nov6-04, 12:43 PM
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Quote Quote by Clausius2
It hasn't got dimensions of energy (Joules), so that it is wrong at first sight without thinking anymore. Checking dimensions will enhance you to be forewarned about errors.
Yes. I'm embarrassed.

I've fallen into this very bad habit lately of just looking at things from a purely intuitive view instead of thinking of them in terms of basic mathematical definitions. That will always get a person in trouble.

Thanks for all your help. I hope you are teaching this stuff because you're a good teacher.
Clausius2
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Nov6-04, 03:07 PM
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Quote Quote by dextercioby
Engineers in my home town are taught a physics course,though only for one semester (it used to be 2,and it still is,but in other universities),where a lotta quantum physics knowledge is inserted.For example,they study wave mechanics,quantum statistical physics and,of course,their applications to solid-state physics especially semiconductors.
I don't know the way thing stay in Madrid...I guess it's less theory...
You're wrong. Here we haven't got much money (the goverment seems not to be very interested) to make experiments, so that our knowledge is theoretical almost completely.

I think it depends on what kind of engineering are you studying. For mechanical engineers, quantum mechanics is some relatively far away from their usual environment. For fluid-thermal engineers (like me) quantum mechanics are a question relatively far away from our studies. I had to take a course in Relativistic Dynamics and Nuclear Physics, that last as a part of a course in Nuclear Engineering. Those and the usual knowledge we all had at general physics and chemics about Schrodinger equation is all what I know about something similar to quantum mechanics (also I read two Hawkins books ).

You should know the knowledge need space and time, neither I haven't got much space available in my brain for more equations or I haven't got much time available to learn something out of the target of my studies.


Quote Quote by NeutronStar
I hope you are teaching this stuff because you're a good teacher.
Teaching??, I'm just studying my final engineering year! And in the case I would become a teacher, Lagrange Dynamics are a bit bored to teach .
NeutronStar
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Nov6-04, 04:51 PM
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Quote Quote by Clausius2
Teaching??, I'm just studying my final engineering year! And in the case I would become a teacher, Lagrange Dynamics are a bit bored to teach .
If you think that's boring you should try teaching computer technologies! Now we're talking BORING!

In comparison, I would much rather teach differential calculus, lagrangian dynamics, or better yet, modern physics, or particle physics. Any of those would be more exciting that the computer technologies that I've taught. My high point was the electronics classes where I at least got to teach basic electronics with hands-on projects in a lab. That was as close as I ever got to teaching physics.

Although, when I think of teaching Lagrangian Dynamics I think about teaching the calculus that goes with it. Typically it doesn't happen that way. The mathematics is supposed to be a prerequisite. I'd rather teach them concurrently in the same extended class myself, but educational institutions don't typically operate that way.
Clausius2
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Nov7-04, 06:20 AM
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Quote Quote by NeutronStar
If you think that's boring you should try teaching computer technologies! Now we're talking BORING!

In comparison, I would much rather teach differential calculus, lagrangian dynamics, or better yet, modern physics, or particle physics. Any of those would be more exciting that the computer technologies that I've taught. My high point was the electronics classes where I at least got to teach basic electronics with hands-on projects in a lab. That was as close as I ever got to teaching physics.

Although, when I think of teaching Lagrangian Dynamics I think about teaching the calculus that goes with it. Typically it doesn't happen that way. The mathematics is supposed to be a prerequisite. I'd rather teach them concurrently in the same extended class myself, but educational institutions don't typically operate that way.
So you mean you are a teacher now?.

If once I become a teacher, I would like to teach Fluid Mechanics, Heat Engineering or Combustion Theory. That's not bored at all!!
NeutronStar
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Nov7-04, 12:59 PM
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Quote Quote by Clausius2
So you mean you are a teacher now?.
I'm actually retired now. But I would like to teach again. But no more computer technology. I want to teach mathematics or physics. I'd also like to teach it on a more informal level. I just took a refresher course in Modern Physics and I would enjoy teaching that.

My professor didn't seem to know it very well actually. When she was describing the Twin Brother's Paradox she kept insisting that acceleration has nothing to do with it. She told me that this is Special Relativity and not General Relativity and that acceleration plays no role in the Twin Brother's Paradox. That's simply incorrect. The brother that undergoes the acceleration is going to be the younger brother when they meet up again! I mean, I knew that before I took the refresher course. Yet she is a Ph.D. teaching this course at the college level!

I could teach it better! I'm sure! Unfortunately I don't have a Ph.D. so I'll never be allowed to teach it.
If once I become a teacher, I would like to teach Fluid Mechanics, Heat Engineering or Combustion Theory. That's not bored at all!!
Sound boring to me. But if that's what you like go for it! I just think you'd be a good teacher in general because you seem to have a good conceptual grasp of the concepts and an ability to explain them concisely.
tomkeus
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Nov8-04, 01:51 PM
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Quote Quote by tomkeus
Sorry, something's gone wrong with LaTeX. I'll try to fix.
[tex]\ddot{\theta}=\frac{d^2\theta}{dt^2}=\frac{d\dot{\theta}}{dt}=\frac{d\d ot{\theta}}{d\theta}\frac{d\theta}{dt}=\frac{d\dot{\theta}}{dt}\dot{\th eta}=\frac{1}{2}\frac{d}{dt}(\dot{\theta}^2)[/tex]
Sorry guys, I was verry sleepy, I made a mistake. It should be:
[tex]\ddot{\theta}=\frac{d^2\theta}{dt^2}=\frac{d\dot{\theta}}{dt}=\frac{d\d ot{\theta}}{d\theta}\frac{d\theta}{dt}=\frac{d\dot{\theta}}{d\theta}\do t{\theta}=\frac{1}{2}\frac{d}{d\theta}(\dot{\theta}^2)[/tex].

The point of all this is that you get simple differential equation in form of [tex]a\frac{dy}{dx}+by+csinx=0[/tex]

No more late night surfing for me.
NeutronStar
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Nov8-04, 09:00 PM
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Quote Quote by tomkeus
Sorry guys, I was verry sleepy, I made a mistake. It should be:
[tex]\ddot{\theta}=\frac{d^2\theta}{dt^2}=\frac{d\dot{\theta}}{dt}=\frac{d\d ot{\theta}}{d\theta}\frac{d\theta}{dt}=\frac{d\dot{\theta}}{d\theta}\do t{\theta}=\frac{1}{2}\frac{d}{d\theta}(\dot{\theta}^2)[/tex]
I don't quite follow this step. Can you elaborate here?

[tex]\frac{d\dot{\theta}}{d\theta}\dot{\theta}=\frac{1}{2}\frac{d}{d\theta}( \dot{\theta}^2)[/tex]
dextercioby
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Nov9-04, 04:42 AM
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Quote Quote by tomkeus
Sorry guys, I was verry sleepy, I made a mistake. It should be:
[tex]\ddot{\theta}=\frac{d^2\theta}{dt^2}=\frac{d\dot{\theta}}{dt}=
\frac{d\dot{\theta}}{d\theta}\frac{d\theta}{dt}=\frac{d\dot{\theta}}{d\ theta}\dot{\theta}
=\frac{1}{2}\frac{d}{d\theta}(\dot{\theta}^2)[/tex].

It's totally bull,what u did here.
[tex] \dot{\theta} [/tex] is the time derivative of [tex] \theta [/tex],which is a function of time.Therefore,it's derivative is also a function of time,and not of its integral wrt to time...Got it??

In the worst scenario,it should have been
[tex]\ddot{\theta}=\frac{d^2\theta}{dt^2}=\frac{d\dot{\theta}}{dt}=
\frac{d\dot{\theta}}{d\theta}\frac{d\theta}{dt} =0 [/tex]
,because a function of one variable cannot depend of any other variable,including its integral wrt to that variable.Example
[tex] \frac{d\dot{\theta}}{d\theta} = \frac{1}{d\theta} (\frac{d\theta}{dt}) [/tex] ,which is a nonsense.
Remember to apply the chain rule for differentiation where u have implicit dependance of the variables,in this case time.
tomkeus
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#31
Nov10-04, 05:14 PM
P: 79
No ti is not totally bull, it is very standard trick you can se in lot of books. In Lagrangian formalism you dont take Lagrangian as a function of only t (at the bottom, everything in physics is function of time only, but in most cases, that is not the most productive way to look at things), but you take it as [tex]L(\dot{q_i}, q_i, t)[/tex] where [tex]\dot{q_i}[/tex], and [tex]q_i[/tex] are general speeds, and coordinates. Examine this step [tex]\frac{d\dot{\theta}}{d\theta}\frac{d\theta}{dt}=\frac{d\dot{\theta}}{d\ theta}\dot{\theta}[/tex]. In it you just replace [tex]\frac{d\theta}{dt}[/tex] with [tex]\dot{\theta}[/tex] because [tex]\dot{\theta}[/tex] is in Lagrangian formalism coordinate the same way [tex]\theta[/tex] is coordinate.
sach1tb
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#32
Jun15-07, 11:05 AM
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Quote Quote by NeutronStar View Post
Also can you explain what each of these terms represents from a Netwonian point of view?

[tex]r\ddot{\theta}+2\dot{r}\dot{\theta}+g\sin\theta=0[/tex]

I know that the last one comes from the gravitational force. But what are the other two terms? I'm thinking that they might be related to torque, or angular momentum but I can't quite figure out what's going on here.

I'd really like to fully understand the mathematical description of this system perfectly before I move on to to the next system in the book which is far more complex.
The first term is the centrifugal force, and the second one is Coriolis force. Feynman's lectures Vol 1. pages 19-8, 19-9 have an excellent (non-mathematical) explanation for Coriolis force.
"This other force is called Coriolis force, and it has a strange property that when we move something in a rotating system, it seems to be pushed sidewise. Like the centrifugal force, it is an apparent force. But if we live in a system that is rotating, and move something radially, we find that we must also push it sidewise to move it radially. This sidewise push which we have to exert is what turned our body around"


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