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.
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!
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.
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).
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?
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
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.
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.
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.
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.
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.
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.
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
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?