Relative motion (with rotating axes) Summary

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Virtually Passed

Virtually Passed

Күн бұрын

Пікірлер: 33
@celesefernal2490
@celesefernal2490 5 жыл бұрын
Thanks this actually explained it so much better than my lecturer who, btw is a professor.
@atmikgoswami1289
@atmikgoswami1289 2 жыл бұрын
Thanks man! Excellent video, clearly explained. 👏 Made my concept clear.
@minmax3160
@minmax3160 3 жыл бұрын
11mins of utube video explains the concept better than my 2 hour lecture, y do i pay tuition bruh
@sengulseydaylmaz2637
@sengulseydaylmaz2637 2 жыл бұрын
You are amazing teacher
@virtually_passed
@virtually_passed 2 жыл бұрын
Thanks :)
@Scott21
@Scott21 5 жыл бұрын
It's a nice video, I just wish you talked about the coriolas acceleration.
@virtually_passed
@virtually_passed 5 жыл бұрын
Thanks for the feedback :) Maybe when I get some time I might make some videos on the Coriolis effect.
@mathschampion4377
@mathschampion4377 3 жыл бұрын
Is the acceleration is same in rotating and fixed frame
@mwont
@mwont 3 жыл бұрын
Amazing explanation. Thank you
@kedarkattimani2839
@kedarkattimani2839 4 жыл бұрын
this was really helpful in the sorts of understanding this topic intuitively and it all makes sense, but my prof has asked to understand and learn the derivation of these formulas. Could you please make a video on it? I am using RC Hibbeler Statics and Dynamics 14th Ed Thanks
@muhamadabdulaziz1319
@muhamadabdulaziz1319 5 жыл бұрын
Very helpfull thankss👍👍
@vitorpereira1579
@vitorpereira1579 7 жыл бұрын
thanks man, your videos are helping me a lot, keep it up. From brazil :D
@virtually_passed
@virtually_passed 7 жыл бұрын
Glad you liked it! From Australia :D
@rifat6049
@rifat6049 3 жыл бұрын
why is it necessary to imagine a point P? why can't i just say point A?
@virtually_passed
@virtually_passed 3 жыл бұрын
Great question. There's a very subtle difference. At any particular instant of time, Point P is defined to have the same position as point A, but point P is defined to be attached to the plate. Now imagine at this instant, the plate was rotating around point B. This means point P will be on a circular path and so it must have a velocity tangent to that circle. This is unlike point A. Point A is free to move on top of the plate, so even if the plate rotates purely around B, the velocity of A could be in an entirely different direction. I think this example will help: kzbin.info/www/bejne/aoHMf3yai5qhZpI&ab_channel=VirtuallyPassed
@josvangorp5618
@josvangorp5618 3 жыл бұрын
good video!!
@N0fear43
@N0fear43 7 жыл бұрын
Man, I dont remember any of my dynamics. I've never had to use this luckily. Glad you're doing these videos tho
@virtually_passed
@virtually_passed 7 жыл бұрын
Thanks for the kind words :)
@notSavant
@notSavant 6 жыл бұрын
this is perfect! Thanks Matthew :)
@virtually_passed
@virtually_passed 6 жыл бұрын
Glad you liked it :)
@thonnhomsnit6352
@thonnhomsnit6352 5 жыл бұрын
Thank you very much. but i have 2 question. 1.If I cancle an "a rel" and coriolis acceleration. it will look like relative motion using translating axes. Is the term " w x (w x r) " similair to " (-w^2)r " like translating axes. can I use -w^2 r in rotating axes. 2. How can i know which problem use translating axes or rotating axes.
@virtually_passed
@virtually_passed 5 жыл бұрын
Hey mate, you've got some good questions here. 1) w x (w x r) is pretty much the exact same as (-w^2)r. The only difference being that the first equation is a vector, whereas the second is a scalar. But if you remember that the direction of the normal acceleration term (-w^2)r always acts towards the center of rotation, then it's the same. 2) It depends on how you define your system. Usually you want to define a rotating reference axis (eg define x axis of a rotating bar so that x is always measured down the bar). Some example questions will help build some intuition for you: kzbin.info/www/bejne/aoHMf3yai5qhZpI Hope that helps :)
@thonnhomsnit6352
@thonnhomsnit6352 5 жыл бұрын
Virtually Passed Awww.I got it!!! Thank you very much.
@TheBrandoboy
@TheBrandoboy 6 жыл бұрын
I think you may be missing an term on your Vp/b I think it should be w x (w x r_a/b)
@virtually_passed
@virtually_passed 6 жыл бұрын
Hey mate, thanks for the comment, but I think the video is right. Here is a proof: kzbin.info/www/bejne/apOwlYenZsmCoaM
@kedarkattimani2839
@kedarkattimani2839 4 жыл бұрын
Thanks a lot!!
@giabao576
@giabao576 3 жыл бұрын
very helpful!
@hamedhilal7113
@hamedhilal7113 3 жыл бұрын
fantastic!
@jeremysquires4472
@jeremysquires4472 4 жыл бұрын
i think your just confusing me more incorporating P instead of just sticking with the A's and B's. My dynamics book makes no mention of P. Just sayin...
@GaryTugan
@GaryTugan 5 жыл бұрын
Love these series of vids! Thanks for making them :)) Apologies, but I found two significant corrections. The first: I Found at time 5:14 you describe .... Vrel as = to Velocity of A/B... Velocity of A relative to B. But this is NOT correct. Vrel is Vel of A/P, remembering the P is fixed to the plate. And therefore, will have only ONE direction away from where point p used to be. If you look at all the Relative veloctity "cancellations" it makes it perfectly clear... Va = V b + Vp/b + Va/p. which can be thought of when seeing the equation without the V's... A = B + P/B + A/P.... The B's "cancel", as well as the P's, leaving A = A. QED :) Likewise, the acceleration of A involves the coriollis effect AND the Arel term, but the sum of these two = V a/p. Understanding this concept and connection was HUGE in my understanding of these equations. So basically.... Aa = Ab + Ap/b + Aa/p, Where: 1) Ap/b expands exactly as you described...two terms that explain the 2 circular components that are the normal and tangential components of that acceleration of p from the perspective of B. And... 2) Aa/p = w x Vrel + Arel NOTE! This is where the video at time 10:44 makes the 2nd mistake. When you look at 2) above, you see that Aa/p has a normal and tangential description. Just like Ap/b did. But in our case with Aa/p, Arel is the tangential part, and the corriolis effect term (w x Vrel) is the normal part! Why... remember that Aa/p describes the motion (acceleration) "away from p", and therefore these two terms ARE the normal and tangential components. The corriolis effect IS the normal acceleration term (normal to the path of a when it is coincident with p). I hope this makes sense! When you see it with this understanding, then the corriolis effect terms is not magically just a leftover or just "there". It makes sense.
@virtually_passed
@virtually_passed 5 жыл бұрын
Hey Gary, first I just want to say I absolutely love your attention to detail! I'm not sure if you've noticed, but I've actually got two proof videos behind each of the formulas I've presented here - one for velocity and one for accleration: kzbin.info/www/bejne/apOwlYenZsmCoaM kzbin.info/www/bejne/m3rWnKqeiNSWns0 Please have a look at these videos and let me know if you still think there is a mistake.
@mollypete4270
@mollypete4270 5 жыл бұрын
lovely.
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