MAE5790-3 Overdamped bead on a rotating hoop
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- Опубликовано: 12 сен 2024
- Equations of motion. When can we neglect the second derivative? Dimensional analysis and scaling. A singular limit.
Reading: Strogatz, "Nonlinear Dynamics and Chaos", Section 3.5.
Professor Strogatz, thank you once again for a fantastic lecture on Overdamped bead on a rotating hoop. These lectures really emphasize stable and unstable systems in Nonlinear Dynamics and Chaos.
great professor!!!!!!!!!!!!!!!!!!!!!
+Yao Lu read the book as well) easy to learn
+Yao Lu i learn it for fun...
Wonderful lecture! Well done with the dimension units! It is a big idea, which I am always confused when I see it in fluid dynamics. Finally, I understand it!
big help!!! thank you from the bottom of my heart !! I wish I had you as my professor
Thank you Prof Strogatz, wonderful series.
In this video, the professor talks about the overdaped pendulum. At first, he derives the equation of motion from Newton's dynamics. Then he analyses the equation with a neglected term to get the bifurcation behavior of the system. In addition, dimension analysis is applied to analysis when a term can be neglected. Finally, he discussed why we can solve the 2-D system in the 1-D system.
it is not really a pendulum, the hoop is rotating at constant angular velocity, and the bead is constrained to the hoop.
dimensionless formulation its pherhaps the most powerful tool for applied math
Great lectures...At 7:52 the angle the tangent makes with the direction of mg is not phi but 90-phi else the component of force at 8:30 cannt be mg sin(phi).
He is a great teacher.
Thank you very much, Professor!
Using Lagrange's equations of the 2nd kind is a much better way to get to the governing equation of motion for this system. The question regarding the existence of a coriolis force is an understandable point of confusion since, when the bead slides along the wire, its distance from the fixed axis of rotation is changing and the product rho_dot times omega (which is the coriolis acceleration term) is non-zero
Since you are considering a frame rotating with the hoop, the relative velocity of the bead is phi_dot r in the tangential direction and the angular velocity of the rotating frame with respect to the fixed frame is Omega k, assuming k is the versor in the vertical direction that is common to both frames. Then I would say that the Coriolis acceleration is 2 Omega k X r phi_dot t where t is the tangential versor, so that the Coriolis acceleration is pointing outside the blackboard and it is balanced by the reaction force of the constraint. Am I getting something wrong?
@@carlosinigaglia2978 that seems correct
For HOMEWORK visit:
www.coursehero.com/sitemap/schools/11-Cornell-University/courses/8061175-MATH4210/
If you want assignment H/W for previous lectures:
1drv.ms/b/s!AjHH76aU3K1fgohKynY8BeF5Tk6P7A
Great Professor, beautiful series of lectures. Somehow it reminds me of the mathematical physics course by Professor Carl Bender. Similar stuff
The last 10 min is awesome, the idea was awesome but bit fast and abstractive, I guess a lot students were lost
I actually found this one to be rather uninformative. Too much time just wading through a morass of details instead of covering concepts. The material on the value of dimensionless analysis was good, though.
COUNT ME IN.
why at r > 0, the diagram points the arrows away from the node, instead of going away and going towards the node......sore of both ways.....is a question....care to explain....reference to lecture 2
Does this course have an official web page where I can download the course
material? I want to do the assigned homework!
Great lectures. Thanks for making these available.
At ~ min 31, in the diagram for γ > 1, I think the stable fixed points should be aligned vertically, not horizontally.
Scratch that (you can't have adjacent stable fixed points). Still, I'm definitely confused, because Φ as drawn, is confined to the right side of the hoop.
17:02 Why is there no Coriolis term? Because we already choose the rotational reference frame attached to the bead, the relative motion of the bead to the rotational reference frame is zero. Therefore there is no Coriolis force.
Brilliant.
@1:06:00 two timescales & two initial conditions.
Wow, Cornell students are fantastic!
Can we infer that the second order component is trivial at steady state and when approaching a constant angular velocity? The transient case is pretty straightforward and to be expected although the example was good.
I meant negligible
Hi
I would appreciate the help of someone who took the course
or has the material to provide me with the assignments or problem sets in this
course which are typically chosen from the textbook just problem numbers from
the textbook for each assignment.
Kind regards
What is the purpose of introducing T? I could not get after it was assumed that m goes to zero. I think the case of light weight bead is almost trivial.
The T is introduced to make the equation dimensionless. The idea is to set a standard for small using 1 instead of another reference mass, making it a more general case.
Why is the following true? Phi_double dot = 1/T^2 Phi ‘‘ 47:39
He assumed phi ' = d phi/ d tau. So in that sense phi " = d2 phi / d tau2
He did a chain rule involving phi and t. I get how T is dimensionless. Why does T seem constant.
Jamiiru Luttamaguzi T is constant but not dimensionless.
Tau is the comparison of the time respect to T, therefore is dimensionless.
@18:16 or thereabouts, wouldn't rho equal r |sin phi| instead of r sin phi?
this would probably make solving this system an even more ridiculous mess, but i was thinking that it makes perfect sense for the bead to swing on over to the opposite side it started on, because that is a possibility here, and if you're just writing the system such that phi is secretly |phi| (that is, direction doesn't matter, just how far up the hoop the bead goes) then i guess it kinda works out? Still feels weird to me.
@@sylviewrath2199 Yes, the bead can swing over. In that case, rho would be negative. The centrifugal force would also switch sign. rho = r sin phi captures all this.
38:21 LOL
This lecture seems awfully heavy on the details of a single problem and not very heavy on the topic itself.
He introduces the concepts in a lecture before. This is the application on a single problem.
these lectures are a bit disappointing - 90% of it is clever mathematical manipulations and only 10% is talking about the bigger picture!
I've probably took more math courses than you ever have before you reached puberty. No one said anything about reiterating anything continuously! the professor has a unique grasp on the big picture which students like you don't have a chance of figuring out on your own.
Thanks for your comment Martin. Can you please suggest some resources where I can get a better understanding of the big picture regarding these fundamental concepts?
@@sabyasachisen8767 it's been 3 years,I doubt he is gonna respond to you
@@siddharthpandya7763 no worries I've found resources. The best one I have read so far is Julio Ottino's book on mixing