Thorough Discussion of Olympiad 2017 Physics Problem with multiple Solutions | NMS Sir
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- Опубліковано 30 вер 2024
- Motional emf, faradey law, instantaneous axis of rotation, IAOR energy consideration torque work energy theorem concept of pseudo force is discussed.
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Sir we neglected self inductance may be that's the reason we didn't consider pseudo force torque. Because if we consider pseudo force torque the ans will be exact.
the reason not to account pseudo torque maybe of the below two reasons:-
1.The pseudo torque and torque due to BIL cancel out each other.
2. We are dealing with IAOR , it is a point which we can consider to be at rest at a moment, so no need to consider any pseudo torque !
There are two points one is icr and other is the point on rigid body which is coinciding with icr . We cant apply torque =i(alpha) about icr as it is not rigid body we can apply wrto the second one only
Atleast discuss 10 questions sir 😊❤
Accelarations given at 19:55 are those of CENTRODE, not IAOR
Sir collision ka pura revision karwa dijiye please
Thank you very much for bringing up such a question sir...in fact a series on rotational motion would be highly appreciated by us. I tried this problem for so many hours. I am a dropper studying in an offline coaching,; I have already wasted too much time on this question and I am obsessed with it. I tried to write the power as Torque dot Omega and then obtained torque. From that, it was clear that there was a mistake in writing the net pseudo-torque. I am seeing this method of finding the acceleration of the instantaneous axis by differentiating its position twice for the first time. Our sir would make us discover the acceleration of the instantaneous axis by a different method. In that, the acceleration of the axis comes out to be the sum of the accelerations of the ends of the rod(now that I think about it, it's literally the same thing). So the acceleration of the axis is definitely correct. Now I went to the basic definitions of torque and angular momentum(assuming the velocities and accelerations of each point with respect to the instantaneous axis and summing them up vectorially). The equations that appear by assuming the velocity of the axis as v and its acceleration as a turns out to be simply
net real torque = Iaxis.alpha + (omega cross L) + M(r of cm wrt axis cross a of axis)
here omega cross L is zero as it is planar motion, now M(r of cm wrt axis cross a of axis becomes the pseudo torque when taken to the other side. Here is the spot I am stuck, anything I try I am not able to prove that it is wrong in any way. I wrote that equation from the fundamentals itself but still, it is not clear. So I tried to check by solving some basic problems where the instantaneous axis is accelerating. The famous irodov problem where there is a cylinder rolling down and is attached to a string connected to the edge of a pulley is the one I chose to analyse. In that, I assumed the centre of mass to be falling with an acceleration of a and the cylinder to be rotating with angular acceleration alpha. Now I wrote the position vector of the IAOR assuming the point of contact of the wall and the string to be the origin. I differentiated it twice and found that the acceleration of the IAOR is 'a', the same as the centre of mass of the body. Now I went back to my lecture notes where our sir made us solve this problem and guess what!! There we solved that question about IAOR but there is no damn pseudo torque. Now don't get me wrong, our sir made solve some good questions where the IAOR is accelerating. Now I am totally confused, when do we consider the pseudo-torque and when do we not?? Please give us a hint on how to find the mistake in this problem, sir. Give us one more chance to try to find the mistake, after that please give us a general idea about this topic. Thank you very much again sir.
sir in 4:34 shouldn't moment of inertia be about icor rather than com???. please reply rotation is one of my weak topic
Moi is about icor only
Sir this question was very good and the solutions provided are very beautiful. But, I am thinking about the 3rd solution and that small mistake in the concept..
sir, i got the right result using pseudo force but i had a different approach, i first wrote the coordinate of com of rod and then the coordinate of iaor and found that it is just double of com coordinate, so after double differentiation of coordinates to get accn of com of rod and iaor we also get accn of iaor is double that of com and is in same direction that means that the actual force on com will just become opposite in sign in frame of iaor and actually thats how we should have calculated torque about iaor. we got the right result in the 2nd method where you didnt account the pseudo force because we get same torque in both the case just the sign is opposite as the force becomes opposite in frame of iaor so it was just a coincidence.
thankyou for reading it.
Sir please continue the previous session
It will continue
20:06 why equate with omega × I × alpha...??
Thats wrong i feel
Another mistake he might have done is write accleration as a whole of i cap and j cap.
We will then break angular momentum into x and y and treat them indivisually..
Also the rod has mass m
It whould only be I × Alpha..... as far as i know
Torque by pseudo force will be 0 as force direction is parallel to the direction of position vector wrt instantaneous point of rotation
Sir i think the reason for this is the following:
Sir we can very easily derive that dL/dt = Torque external - M(R - r0)cross(accn of IAOR) from basic Definition of to torque and angular momentum where R is position vector of centre of mass and r0 is position vector of IAOR and if we write angular momentum about IAOR Iw + m(R-r0)cross(Vcm) and from my calculation d/dt(m(R-r0)Vcm ) = ml²(alpha)(cos²theta - sin²theta) and M(R - r0)cross(accn of IAOR) = ml²(alpha)(cos²theta - sin²theta) these contribution exactly cancel out !!! And we get I(alpha) = torque external
(Note: we can not directly apply I(alpha) = torque external as it derived when we assume that R - r0 vector is constant this is the same way parallel axis theorm is derived assuming R - r0 vector is constant but here it is not constant so one has to go to basics it was a coincidence that those extra terms of angular momentum derivative and torque exactly cancel out
Why do you think that (R - r0) is not constant ? Isn't IAOR defined as the shifting of origin i.e. r(cm) = C + r(IAOR) where C is a constant.
@@koq_utx you could yourself calculate the R vector and r0 vector in terms of angle theta and you would see that it's not constant and I don't think IAOR is defined by the definition you gave
@@chayanaggarwal3431 Yes I did the magnitude is always constant.
sqrt( (Lcos(theta) - Lcos(theta)/2)^2 + (Lsin(theta) - Lcos(theta)/2)^2 ) = L/2
@@koq_utx magnitude is constant but is the vector constant??
@@chayanaggarwal3431 Yeah vector is not constant its rotating from 0 to Pi/2
Pls tell us where is the mistake . It will be very helpful for us plz plz plzzz😢
Sir I think he should also equate fnet= macm and t=Ialpha the rod is moving in front also
Istantaneous axis or rotation(IOR) is considered to be a point which can be assumed to be momentarily at rest .Thus while writing any torque equation we don’t need to consider pseudo force .this is the absolute beauty of ior otherwise we could just use any random point to write our torque equations
Sir please please please ishi tarikr ke super detailed aur indepth wale discussions aur laeye , kyuki aisa to hame paid coaching me bhi nahi milta
Awesome
Sir acceralation tangential as well as radial hmm lenge
Sir , is it like the iaor is rotation so something like corrolis acceleration be added
More questions like this for class 11
Iaor not moving , it is on resr wrt to this ioar whole body is rotating
No nikita , iaor is moving it has acceleration
Goat❤❤❤
Maza aa gya guru ji ❤🙏🙏🙏
Iar is not on body. Pseudo force is wrt point lying ON body. Therefore whenever we use iar in standard rolling body also, we don't do pseudo force. Iar is a point very close to that point of body, but its not on body
Iar is always on body either on the part you can see or On the part you can not see. ( THODI BADI BAAT HAI PER SACH HAI)
iar normally x , y me change ho raha hai.. to uska position vector i aur j me likhna hi fundamental mistake hai... kyoki iar fixed tarike se thodi na change ho raha... aur accleration impulsive to uska derivative lene par iar ka accln nahi aayega