# 25:36 , If reference line is above , we should see how much deeper the mass has gone . But formula used for potential is considering the height from ground rather than depth from the reference level taken here. Is it justifiable ?
Potential should be -mglcos(theta) according to your reference point in case of pendulum beacuse in the term mgh , h is the height from reference point.
Sir pls EMT ke bhi PYQs krwa do JEST ke with concepts pls pls its my humble request 🙇🙇🙇🙇 i m weak in that subject and your way of teaching is too good 👏👏 i m hoping that by ur concept i will be able to solve questions from that subject in Jest exam pls sir
Thankyou so much sir❤ Sir I have a doubt. While writing potential height should be taken from refrence line right. In 29.59 in the bead based system reference is central line. So potential wil be -mgrcos € right?. I am confused with that.V=mgr(1-cos €) if take refrence at bottom of hoop no? .please clarify sir
In spring + pendulum case, there should be coupling of velocity components of spring and pendulum, then we will get net velocity
Ohh yes, you are right. Thanks to point out this.
What extra will be added here??
Helpful lecture 😊
Very good consept
Great job sir ...waiting for the next video lecture
It was really one great watch!!!! Thank you so much sir.
# 25:36 , If reference line is above , we should see how much deeper the mass has gone . But formula used for potential is considering the height from ground rather than depth from the reference level taken here. Is it justifiable ?
Good job 👍 ❤️ thank you sir ❤️
Best way of teaching... Oragnised
Thank you
Great lectures
This is very helpful
Sir thank you so much for these JEST series . Waiting for your next video.
Potential should be -mglcos(theta) according to your reference point in case of pendulum beacuse in the term mgh , h is the height from reference point.
27:02 Kinetic energy of sphere 7/10 mv^2 hoga na?
Yes.
In second case , to whom we should differentiate to get eqns of motion , both x as well as theta is there...?
thank you sir. I am preparing for JEST exam. Your lectures are very helpful. Hoping for more to clarify my concept before exam.
This vedio is very helpful for me... Thank u so much sir
9.50 if you assume reference as ceiling in which pendulum is attached
Then how the potential of block is zero ?
It should be _LMg ?
Choosing reference levels doesn't effect EL equation. You can consider that.
Thanku sir
Thank you ❤
Sir pls EMT ke bhi PYQs krwa do JEST ke with concepts pls pls its my humble request 🙇🙇🙇🙇 i m weak in that subject and your way of teaching is too good 👏👏 i m hoping that by ur concept i will be able to solve questions from that subject in Jest exam pls sir
Thank you sir :)
Your lectures are admirably organised and well explained. Keep up the good work.
sir,
In the spring. Pendulum system (12.50) why don't you include potential energy due to the spring?
Please clarify sir..
Oh yes, you are correct. Please add that, I forgot it.
OK sir
All important topics explain sir
Sir when will be the next video please made asap , it's really helpful
Dada operator, commutator ya average of L_x / S_x / E ka video lecture kar do plz. We have one week left. Thank you!! Namaskaram!!
Sir 3:39 me for cylindrical coordinate me T k expression me last wala term ź^2 hoga na??
suno net ke form fill ho gye h kya?
@@hdphysics4682 not yet started
@priya yes.
Sir please complete atleast classical mechanics course regarding jest before the exam it helps in getting the basics clear
Sir add some video lectures which are helpful for jest exam... Please 🙏 sir ....upload video in every 2days of interval
Wonderful sir👍
Although this lecture was a bit too fast😁
Next csir exam classes explain sir
I think of this after JEST exam.
Sir newtonian mechanics k question kraye pls
Thankyou so much sir❤
Sir I have a doubt. While writing potential height should be taken from refrence line right. In 29.59 in the bead based system reference is central line. So potential wil be -mgrcos € right?. I am confused with that.V=mgr(1-cos €) if take refrence at bottom of hoop no? .please clarify sir
Actually, it doesn't matter to chose reference line as r is constant. So you get same EL equation of motion.
@@Physframe okay sir
Sir please upload videos on more important topics for jest...your videos are really helpful
Bead on a rotating wire problem has done wrongly....r should be taken as rsin(theta) as the bead is moving in a circle of radius rsin(theta)