Time derivatives in a rotating frame of reference
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- Опубликовано: 5 сен 2024
- Here's how to derive a very useful operator relation linking time derivatives in rotating and inertial frames. We'll use this relation to show how fictitious forces arise in rotating frames in my next video.
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.
My website: benyelverton.com
#physics #maths #math #mechanics #dynamics #fictitiousforces #coriolis #centrifugal #euler #forces #rotation #rotatingframes #inertialframes #calculus #vectors #differentiation #operators #particle #motion #circularmotion
Excellent, sir. Given your current number of subscribers, it's far below what you deserve. My wish is for your recognition to increase. 👌
Thanks for your kind words and support!
I finished Differential EQ and Linear Algebra a year ago, you've made it very simple to follow, good job 👍
Thanks, great to hear that it was a clear explanation. Back when I was a Physics undergraduate I never fully understood why dk'/dt = ω x k' and so on, so my hope is that this will help people who are now in a similar situation!
woow i found an underrated teacher i mean you can really go to any famous private institute or university and become their main teacher sir thanks for this effort for teaching here
Thanks for your kind words, I'm glad the videos have been helpful!
Thank you Ben. I can’t appreciate you more!
@@kienvo9072 Thanks for your support!
Amazing 🤩 . It's interesting and yr explaination is intuitive
Thanks for saying so!
Amazing. Absolutely amazing 👍👍👍
Thank you!
@DrBenYelverton thank u DrBen.I know it is a bit difficult to make a high quality intuitive video about elliptic functions ( integrals) & their applications in mechanics problems such as (not small angle pendulum or rigid body dynamic equations ) But in ur representation it would be great.Again thank u for ur great efforts 👌
This video was very explicative and interesting. I always tried to obtain the derivative relation you have shown at the beginning by using row, columns and matrices notation. Putting the symbols i', j,' k' like you did makes everything easier to understand. Thank you.
Thanks for watching and I'm glad it was helpful!
Excellent. In almost one slide the whole explanation. I really enjoyed watching your video.
Good to hear, thanks for watching!
Excellent!!! Simple and direct to the point.
Thank you for your excellent explanations and stating details in derivation .
🙏🙏🙏🌹🌹🌹
Thanks for watching, I'm glad it helped!
Thanks for making videos 🎆
Thank you for watching!
thanks. simplified explaination
Great video and excellent explanation!
Thank you!
Thanks a lot for the video...
You're welcome, I'm glad it was helpful!
Hello, thank you for the very clear video. I wanted to ask: at minute 5:24 we make: d/dt (A i' + B j' + C k'). But why this derivative follows the common rules of derivation of product as it was a product between scalar functions? It is the derivative of a linear combination of scalars and vectors..
Thanks for watching. Imagine writing out the equation in component form, so that you now have three separate equations in terms of the x, y and z components of all the vectors. Maybe the easiest way to see how to do this is just to write the vectors in column vector form. Each of the three individual equations now only involves scalars and you can differentiate as usual. After differentiating, write your three equations as a single equation again by switching back to vector notation and you'll arrive at the result in the video.
@@DrBenYelverton, if I understood properly, you mean this:
d/dt (A i' + B j' + C k') = d/dt ( A(t) * [a(t), b(t), c(t)] + B(t) * [d(t), e(t), f(t)] + C(t) * [g(t), h(t), i(t)] )
Where a(t), b(t), ..., i(t) are the components of versors i', j', k', which are variable with time if examined in the fixed reference frame. And then you go with product rule, is it correct?
Yes, that's exactly right - it basically comes down to viewing your vector equation as three scalar equations, so you can then proceed with all the rules you know for scalars.
Or you could calculate the christoffel symbols in this new frame and find the covariant time derivative... Jk this was great, but it is an interesting idea to use the covariant derivative to verify this
Sounds like an interesting exercise - though probably not something I'd be able to do right now as I haven't used Christoffel symbols since studying GR around 7 years ago! Thanks for watching.
Nice
Hello Dr. Yelverton, thank you so much for a lucid description on this topic. However I have one source of confusion which is bothering me (as I have been experiencing with many other videos addressing this topic as well). Referring to the second term of the last expression at 17:30 min, when you cross omega with the vector A, the latter has to be expressed in/converted to its time varying form in the inertial reference frame S, by applying coordinate transformation to the corresponding prime components in the rotating frame S', right? You cannot directly cross omega with the stationary prime components? If we make things a bit simpler by making the vertical axes of the two frames coincident, then the components in the S domain will be sums of components in the prime domain multiplied with Cos(wt), Sin(wt), etc. ? In the general 3D case shown this will be more involved with, e.g., qaternion rotations? Thank you for any advice to clarify my thinking.
Hello and thanks for watching! With the ω x A term, what matters is that both vectors are expressed in the same basis. This could either be the basis of S (i, j, k) or S’ (i’, j’, k’) - whichever is more useful for the problem at hand. If A were constant in S’ and you wanted the time derivative in S, then you would indeed need to transform the components of A into frame S using a rotation matrix before taking the cross product with ω expressed in S. Perhaps the most common usage of the relation derived in the video, though, is to find the equation of motion of a particle in frame S’ - so here A is a position vector which is varying with time in S’ and we want to know how exactly it’s going to evolve over time. In this case it’s more useful to keep A in the S’ basis and express ω in that same basis. I think the key point to remember is that the vector ω x A exists independently of any basis, but if we want to write this vector out in component form then the components of ω and A must be expressed in the same basis. Hope this helps a little, there are some more videos later in my rotating frames playlist which might also be useful: ruclips.net/p/PLTntRzhyShSEugh4o6AARrJRS-PxxqjiI
@@DrBenYelverton Thank you so much! Yes, I continued to be tormented by the question since I posted the comment🙂. And came to the realization that I was actually conflating between "which frame a vector is with reference to" and "which coordinate system the vector may be expressed in" (as you mention above). My question had to do with the latter. I was able to make peace with it by assuming that both systems used a common Z axis, and freezing time at a multiple of 2pi/omega. Then the basis vectors would coincide in both systems, and all equations would go through smooth. In general, as you mentioned, A|S and A|S' are consistently oriented at any instant of time, no matter the bases, but if the actual cross product is to be computed then a common set of coordinate bases should be used (mapped if necessary).
Thank you also for sharing the full playlist; I will certainly be watching the subsequent videos in the series.
That's the first time I've seen how dihat/dt=w x ihat derivation
Thanks man💌
Thanks for watching!
@@DrBenYelverton Sir, I'm requesting you to Upload Bsc Physics videos please if you can. It really helps me understand.
@@AlterEgo-v7o Will do! I have many more undergraduate-level Physics videos planned.
@@DrBenYelverton that's great 😊
Wow ;!