Common Misconception in angular momentum conservation | NMS Sir
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- Опубліковано 4 жов 2023
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I learnt most of the beautiful thing from nms sir..... Attitude ❤
Sir if I am correct on the reasoning I gave in the previous video then I think the same type of thing is happening here also as when we conserve angular momentum about point B on rod we get the correct answer even though the method is wrong like in the previous video when we ignored pseudo force even though exact form of equation is different the concept is same when we write integral(torque .dt) for whole system about point B there is no external real torque about point B so the only torque is the torque due to pseudo force so angular impulse(which is entirely due to torque due to pseudo force) = Iw + mvcmcross(r) now here once again the mistake done by most is writing Vcm in ground frame even though it has to be written in frame of point be so if after collision velocity of COM is Vcm(in frame of ground then MVcmbR = M(Vcm-vB)R and coincidentally that MVBR contribution exactly cancels out with angular impluse (the same type of cancellation that happened in that oly problem )
Kisi swal ki aukad nhi jo apse na ho ❤️
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Saleem sir explained the same thing during this topic in rotation . At 5:41 shayad wahan M1Lsquare/12 whomega aayega
Finnally, my classmate...
Sir please collision ka pura revision karaye
👍🏻
Saleem Sir also explained the same thing. 4:57
1st comment... I'm present Sir 😊 lot's of love from Nalanda, Bihar ❤
Ye baad me omega ka direction left ki taraf kese ho gya?? Ball collide krne k baad rod toh right ki taraf jaa rha hai
Sir hum toh apki t shirt se bhi seekh rahe hai 😅❤❤❤
awesome
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Suppose a mass A is in circular motion of center O suddenly it's radius changed to half then according to angular momentum conservation it's velocity will be dubbled
Then we can say it's kinetic energy has increased
Where this extra energy comes from
sir we could have also done by centre of rod frame because in that frame if we consider the system of ball and rod there is equal and opposite impulse so there torque will cancel out and coam will be valid and i beleive we will still get the same result. please reply if my concept is correct or not?
Still we have to consider relative velocity if point is on rod but if the point is in line of com we will be able to write eqn
Most of the teachers are not explaining this type of concept.
Nms sir ❤🔥🔥💥
Then why are you with them hame toh bata dia tha ak par
found the angular impulse myself after listening to your explanation. now what is on my mind is that, is this video a hint to solve the previous challenge about the olympiad problem where the torque equation was failing. so the fundamental question boils down to whether the instantaneous centre of rotation is a point on the rigid extension of the body or the general space point. I feel that it should be a point on the rigid extension of the body. Please clarify this point sir. I'm a bit confused regarding the instantaneous axis of rotation...
Very close my dear good
Think more you will reach at correct answer
I think for a point to be some type of "rigid extension" of a body it has to satisfy the original definition of a rigid body that is that point should be at a constant distance from every point on the rigid body which in the falling rod case the IAOR is not at a constant distance from the endpoints
Mind blowing very nice
Bro ICR can be anything actually because Just take two examples for this
1. A rigid body where velocity of one point is right wards and another point which is below the first point has a velocity towards left, So I hope u know ICR will lie on the body
2. A rod is sliding on a two walls that are at right angles to each other, here ICR is a point in space
I hope u got this now u can take as many examples as can and one more thing if the velocities are of two points are both in the right direction then ICR actually depends on value of that x(position of ICR of CM) whether it's positive or negative and if u can't do it this way just brute force the velocities of any points u are given and want in vector form and use the vector eqn rather than even using ICR (but ICR is quite interesting if u ask me) for eg. v(1)=v(2)+w(omega)×r(12)(position vector of point 1wrt2)
Ap har ek video banaie na sir