Why does Kapitza's pendulum oscillate upside down?
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- Опубликовано: 29 дек 2023
- Using Lagrangian mechanics to understand how a pendulum driven vertically at high frequency and small amplitude can remain in stable equilibrium in an inverted configuration, and deriving the condition for stability.
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About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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Great refresher. I haven’t done much lagrangian mechanics in a year and I hate feeling like the topics are becoming more distant simply because of lack of use.
I know that feeling, I'm glad this helped!
Back with a epic video 😁
Amazing video!!!!😍
What an explanation😇. Got all my doubts clarified😁.
That's great to hear!
great video, I am a physics 2 student working on a research project and this was extremely helpful thank you!
I'm glad it was helpful!
Your analysis was really insightful! The idea of employing various time scales as an approach to solve a differential equation hadn't occurred to me previously. I'm curious whether a similar approach could be used to address the problem of a heavy spinning top. Perhaps one could consider three frequencies: one for the rotation around the symmetry axis, another for precession, and a third for nutation?
It's definitely not an obvious method for solving differential equations, which explains why there was a 40-year gap between the first experimental observation of this effect and the first solution to the equation of motion! I like your idea, perhaps it could work for certain spinning tops where the three frequencies you mention satisfy ω₁
I wonder what would happen if the pivot were mounted to the edge of a rotating disc in the x-y plane, so instead of going up and down, it would move in circles. It seems like that would make it possible to build a simple, fast, yet quieter physical demonstration. It might introduce additional equilibria at slow speeds, but probably not stable ones.
Indeed, would be interesting to investigate this some time - you'd just need an extra oscillating term in the expression for x. Perhaps it could be solved in a similar way but I'd need to work out the details.
Very interesting, many thanks !
Thanks for the nice video (still watching...). At around 19:00, you neglect multiple terms but some may not be negligible AFAIU. For example, the terms containing g may in fact be larger than the ones containing A omega^2 whenever cos(omega t) ~= 0. Similarly, the second derivative of theta_0 may be temporarily larger than that of delta for specific times. It seems that there is some averaging across time that is implicitly applied...
Thanks for watching. For ω²Acos(ωt) to be comparable with g, cos(ωt) needs to be infinitesimally small. For most of its cycle, cos(ωt) is not infinitesimal, and while it does reach zero at specific times, the proportion of time it spends being infinitesimal is negligible.
Sir I was wondering whether there is any other mean to solve the differential equation?
Also is there any constraints in this system?
I don't know of any other methods to solve it analytically. Apparently there was a gap of around 40 years between the effect first being observed experimentally and the equation of motion first being solved - if there was an easier way then presumably someone would have solved it sooner! You could view the fact that the pivot has displacement a(t) as a constraint, but here we take care of it automatically by building it into the Lagrangian, rather than using the Lagrange multiplier method that you might have seen in some of my other videos. We don't need Lagrange multipliers here because we don't care about the constraint force on the pivot.
@@DrBenYelverton ok sir
@@DrBenYelvertonHappy New Year sir , I was wondering what courses of physics you have studied in university and whether you have done any reasearch or not?
There are two ways to solve the equation with appropriate approximations:
(1) Building on the video, we could use the small-angle approximation on theta_0(t), reducing it to a simple harmonic equation, and get theta(t) = theta_0(t) + delta(t).
(2) We could use the small-angle approximation on theta(t) in the beginning, and obtain a solution in terms of Mathieu functions.
@@jonathan3372 thx
Thanks for the video! I like it very much. But it answers only the question "when?" we have an upside down oscillation and not "why?". Is it possible to define any virtual "force" that keeps pendulum vertically between those equilibrium points, similarly like Coriolis or centrifugal forces acts on a mass in rotating frames? As I understand, the first derivative of potential energy is a force, we can see that dU/dTheta(partial derivative) contains a sum, and we might assume that the second term is a gravitational force (-mglsin(Theta)) and the first one IS that force that I'm asking about: mw^2A^2*sin(Theta)cos(Theta)/2. Why it has this form? There's no answer in the video :(
There's an interesting looking paper called "Kapitza Pendulum: A Physically Transparent Simple Explanation" by E. Butikov - I haven't read through it myself yet but it may be worth a look if you're wanting to gain a more intuitive understanding!
nice but i was thinking of just using euler after the first equation in lagrange and just starting with a theta = 0.01 and t =0 and go on theta_ddot= etc... with some iterations. still cool
Numerical solutions are often a good starting point, but I always feel as though I've understood a system more fully after working through an analytical solution!
@@DrBenYelverton you are right, i'm getting lazy i guess :P
How can we compare the w^2A and g terms if the w^2A terms have a cos(wt) as well with them, since that will make it comparable with g at some points of time? I'm not able to justify this with separation of timescales because we assumed w is very large.
For ω²Acos(ωt) to be comparable with g, cos(ωt) needs to be infinitesimally small. For most of its cycle, cos(ωt) is not infinitesimal, and while it does reach zero at certain times, the proportion of time it spends being infinitesimal is negligible.
what if you considered the angle just smal and said goodbye to sin?
If you only care about small oscillations that should be fine, but the method here has the advantage of giving you the effective potential energy for all θ, not just θ ≪ 1.