IAR means every point on body is instantaneously purely ROTATING about that axis. And for pure Rotation , by definition of V= w × r , w is along axis (IAR)
Making a topper AIR 1 is not a true teacher, A true teacher is who pushes a average student to top AIR. Thank You Sir I firmly believe that this channel aims at the same.
I can't believe it sir! I've just managed to solve that 2016 problem on my own at the very first time after watching this video! It's actually a breath-taking moment for an average student like me to be able to solve it
That question was not out of syllabus by any aspects. I know you might have watched the use of concepts like inertia tensor, but it is useless. You can easily solve it without the use of inertia tensor. Just litetally follow the same steps as in this question to get 2 values of omega, one along negative x and other at an angle theta to the vertical. Boom, that's it... (just make sure your concepts of com and other mechanics are strong, as that is a necessity...)
And we all know that the d part, which is actually the only correct answer to the question could be find out more easily than JEE Mains questions fr...😂
Fantastic explanation sir, it's a tragedy that I found your channel only recently and have begun exploring the jewels scattered around, with only 16 days left for Advanced, but I would recommend your channel to anyone who is interested in Physics of any level. This is and will continue to serve as a goldmine for current and future pupils of the subject of Physics.
Sir, you explain so well........ This one penetrated deep down my head, now I genuinely wish for a question to be asked in JEE-Adv. that tests these rich concepts!
I solved the question after my teacher owner of channel" INDIAN SCHOOL OF PHYSICS:-Nitin Sachan sir" explained the concept......... But to be honest then also ur explanation was out of the world ... Do check out some of most intriguing problem from the above mentioned channel... Thank you so much sir
Hi sir, the solution was great, but I propose a little solution to this (just to make things shorter). The superposing of the motions is the key. We can use the equations of circular motion to find out the angular velocity about the k direction. 10cm/s = 5√3 cm × ω Solve to get ω = 2/√3k̂ rad/s. Next we find the angular velocity about the -i direction by fulfilling the condition for rolling. 10cm/s = 5cm × ω' , Solve to get ω' = -2î . Now we have both the omega, we can find the resultant. ω(net) = -2î + 2/√3k̂ Find the magnitude to be 4/√3 which is 2.3 rad/s. This method makes it just a little shorter by not using the IAR. This is valid also for the 2016 Rolling question. Idk why people call it very difficult, if you use this same method, you'll again get both of the ω values, except that one wll be along î and other along an angle to the k̂. It is all about observation, some even called out that it was out of syllabus (using inertia tensor and what not)...the best thing about rolling questions is that you can easily visualize, and if your mechanics is strong, then it is easy to do any question...
Never had this level of explanation. Felt really blissful to having discovered such dedicated teacher of Physics .Thanks a lot sir for this level of education in physics that too for free
Awesome. I truly appreciate ur hardwork for us and clear such Explanation. I had done this earlier but after seeing back answer I got troubled .thanks so much for these video
Thank you 😀👍. I am happy to hear that. My sole aim is to make difficult Concepts understandable and reach out to greater number of students. Please kindly share this with your telegram and whatsapp groups and request them to check out the channel. It will help my channel grow 😊
@@PHYSICSSIRJEE your explanation is SO NICE, SO CLEAR , CONCEPTUAL ,and you explain tougher concepts such that it doesnt look tough at all....THANK YOU VERY MUCH SIR🙏🙏🙏
@@PHYSICSSIRJEEsir I'm new to ur course.. Please guide me if taking ur dropper batch really worth it.. I know ur teaching is fabulous but please guide
Amazing explanation sir ...though I was also able to solve the problem on my own also but the IAR thing was new and net omega will be along that I never realized this thing ....thanks for your help 😀😎🙏
Thankyou very much sir by this viedo I have been having doubt and I thought that there will be many mysteries in this problem thank you very much sir your concepts r very nice supper.......thank you once again
Sir if the Q was like a disc free to rotate in 2 planes (2nd plane is like a cap hinged at 2 diametrical ends) is attached with rod and rotated as in Q then because of angular acceleration the disc would also get into slant position ?
sir, in case of NOT PURE ROLLING ,consider a disc moving with some velocity and having some womega ,on rough surface , now it has slipping (either forward or backward) ,now if we write TORQUE ABOUT GROUND POINT just below contact point (since it is not accelerating point ,we dont have psuedo force on C.O.M) ,since friction,normal,mg all passes through it,so we have NET TORQUE IS ZERO ,SO ALPHA IS ZERO.. BUT if we write about C.O.M we have friction contributing towards torque ,so there exists some alpha for rigid body. BUT WE USUALLY WRITE TORQUE ABOUT GROUND POINT in some other cases ,AND it gives same as c.o.m, here why it is happening like that?? please KINDLY clear my LONG LASTING DOUBT SIR 🙏🙏🙏
I FIRST SAW 2016 ONE QUESTION THEN I SOLVED THIS AND GUESS WHAT...I WAS ABLE TO SOLVE PART 1 WITH ONE MINUTE...I REMEBERED THAT BODY IS SYMMETRICAL THATS WHY LO WILL BE PARALLEL TO WO.....FOR BOTH OMEGAS I SOLVED AND THEN APPLIED VECTOR SUM
Great explanation👍 Also, isn't Irodov 1.146 an upgraded version of this problem in which we have to see the dynamics also ? Although I could solve 1.146, would be great if you make a video on that too
Sir could you please confirm whether if in the translating frame of COM , whole cone would seem to be rotating about the axis parallel to IAOR and passing through COM and with magnitude as W(net) ??
I could not figure out the IAR in my first try So I tried to find w in two parts first W¹=v/r cos a. = 2v/√3 r W²=v/r =v/r And taking the resultant give the same ans as in the book I went for this in fort try as I could find the angle of the resultant w
@@PHYSICSSIRJEE sir could you please also confirm whether if in the frame of COM , whole cone would seem to be rotating with axis parallel to IAOR and passing through COM and with magnitude as W(net)
sir I get your concept and solution and thanku so much... but I have a doubt that in rotation starting we learnt that angular velocity of a rigid body is same about any point .... I use this concept and simply find omega of C wrt O and got 2/√3 ... what is the mistake in this statement or concept please clarify, I dont want to live with any misconception it hurts a lot💔
Hello sir , In this question if we try to find out the angular acceleration by writing the torque and then equating to I×(alpha) ......... Then the angular momentum of the disc would be ( Icm×Ws+Io×Wo) differentiating this would give us torque . Then now to equate this to I(alpha) to find (alpha) .....in this what is the value of I to be taken to find alpha...... My idea is that , an axis perpendicular to plane of paper passing through apex of the cone
Sir in the beginning of the video you mentioned about instantaneous axis of rotation. And so if the we drop the perpendicular from C to IAR, it's length will be r cos alpha and and we know the velocity of C. So using v = omega * r we can directly find omega as we know the velocity of C. Sir am I correct in this reasoning or am I missing something?
I realise that in the derivation, the point moves in a circle with radius r and has a velocity tangential to the circle. And I think this is not the case here as the velocity of C is not tangential. Sir is that case?
Sir if the cone is having a angular acceleration then the axis of rotation is providing a torque. So sir to calculate the torque, about which axis should we take the moment of inertia? Hope you will surely reply.
@@PHYSICSSIRJEE yes I know , magnitude was constant,, but I m telling a general expression of rate change of any vector which is rotating and changing magnitude,,, can I write above expression , I superimposed the two, may be I m incorrect, pls verify once .
Because, IAR means body is instantaneously purely rotating about this Axis , so net Omega by definition should be along that . For further reference : you can ask your teacher or read Morin or watch the playlist videos on rotation in this channel. 👍🏼🙂
Janardhan Sir, Great video as usual and great explanation. But just a doubt...in one of the videos very similar to this in INSP(th-cam.com/video/xCxBAIEUK84/w-d-xo.html), Nithin sir had taught us d(r)/dt=w*theta. I completely understood the reason why you wrote d(r)/dt=w*r but I really didnt understand how that equation is different from this and why 2 different equations
My video talks about cross product. The direction of w cross r comes out to be in tangential direction only. I was about to reply the same in the mail you sent and I saw your comment here
Also , please refer to my rotating vectors video, w cross r and theta cap mean the same direction. Also remember the difference between dr bar/dt and dr cap/ dt
Sir could you tell me the difference between dr bar/dt and dr cap/ dt. This is what I understood as the difference: dr bar/dt gives cross product and is used when a POINT is rotating, in our case the point along IAR. Also I watched the special video on rotation of vectors which is well explained... Sir but dr cap/ dt which gives us the result written by Nithin sir is used when we consider a LINE(vector omega) is rotating Sir Im I correct in my understanding?...please correct me if I'm wrong. Thank you in advance.
![[JEE ADVANCED] ANGULAR VELOCITY & ANGULAR ACCELERATION OF A ROLLING CONE [ SCHOOL PHYSICS PROBLEMS]](/img/n.gif)
sir is there any reason why net omega must be along iar or should we accept it as a fact
IAR means every point on body is instantaneously purely ROTATING about that axis. And for pure Rotation , by definition of V= w × r , w is along axis (IAR)
v = w cross r, if v=0 on IAOR then w and r vector must have 0 degree angle or 180 degree.
w is net angular velocity.
this is Physics not chemistry for which you have to accept few things as a fact.
Making a topper AIR 1 is not a true teacher,
A true teacher is who pushes a average student to top AIR.
Thank You Sir I firmly believe that this channel aims at the same.
I can't believe it sir! I've just managed to solve that 2016 problem on my own at the very first time after watching this video! It's actually a breath-taking moment for an average student like me to be able to solve it
Its not a tough question its a out of syllabus question thats why it is hyped
@@Ayush-mg6xw no its nt out of syllabus, just the diagram makes it look tough
@@Ayush-mg6xwno bro it was really tough
That question was not out of syllabus by any aspects. I know you might have watched the use of concepts like inertia tensor, but it is useless. You can easily solve it without the use of inertia tensor. Just litetally follow the same steps as in this question to get 2 values of omega, one along negative x and other at an angle theta to the vertical. Boom, that's it... (just make sure your concepts of com and other mechanics are strong, as that is a necessity...)
And we all know that the d part, which is actually the only correct answer to the question could be find out more easily than JEE Mains questions fr...😂
My Rotation is getting stronger day by day. Thanks a lot sir. 😀
Fantastic explanation sir, it's a tragedy that I found your channel only recently and have begun exploring the jewels scattered around, with only 16 days left for Advanced, but I would recommend your channel to anyone who is interested in Physics of any level. This is and will continue to serve as a goldmine for current and future pupils of the subject of Physics.
Sir, you explain so well........
This one penetrated deep down my head, now I genuinely wish for a question to be asked in JEE-Adv. that tests these rich concepts!
Yes sir, this question was a concept rather than a question
We get to apply the concepts we learn in this question in all other questions like this
Thank you sir for taking up a very good concept and expalining it in a systematic manner..
Thank you 😀, please kindly share it 🙏
I solved the question after my teacher owner of channel" INDIAN SCHOOL OF PHYSICS:-Nitin Sachan sir" explained the concept.........
But to be honest then also ur explanation was out of the world ...
Do check out some of most intriguing problem from the above mentioned channel...
Thank you so much sir
Mind -boggling solution to the one of the difficult question
Awesome content 🔥🔥
Beautiful.
I love the way u take up concepts involved in the question properly before proceeding with the solution
Concept 2 cleared my previous doubts of rotation 💥🔥
Hi sir, the solution was great, but I propose a little solution to this (just to make things shorter). The superposing of the motions is the key. We can use the equations of circular motion to find out the angular velocity about the k direction.
10cm/s = 5√3 cm × ω
Solve to get ω = 2/√3k̂ rad/s.
Next we find the angular velocity about the -i direction by fulfilling the condition for rolling.
10cm/s = 5cm × ω' ,
Solve to get ω' = -2î .
Now we have both the omega, we can find the resultant.
ω(net) = -2î + 2/√3k̂
Find the magnitude to be 4/√3 which is 2.3 rad/s.
This method makes it just a little shorter by not using the IAR. This is valid also for the 2016 Rolling question. Idk why people call it very difficult, if you use this same method, you'll again get both of the ω values, except that one wll be along î and other along an angle to the k̂. It is all about observation, some even called out that it was out of syllabus (using inertia tensor and what not)...the best thing about rolling questions is that you can easily visualize, and if your mechanics is strong, then it is easy to do any question...
Never had this level of explanation. Felt really blissful to having discovered such dedicated teacher of Physics .Thanks a lot sir for this level of education in physics that too for free
Thank you for the kind words 🙂🙏
Applications of concept required 2 was really a great insight that we were bestowed with and whole other content was perfect sir.
Thanks sir
Awesome. I truly appreciate ur hardwork for us and clear such Explanation. I had done this earlier but after seeing back answer I got troubled .thanks so much for these video
I am an average student in physics but this video cleared my basic concepts thoroughly. Thank you sir
Thank you 😀👍. I am happy to hear that. My sole aim is to make difficult Concepts understandable and reach out to greater number of students. Please kindly share this with your telegram and whatsapp groups and request them to check out the channel. It will help my channel grow 😊
Sure
@@PHYSICSSIRJEE your explanation is SO NICE, SO CLEAR , CONCEPTUAL ,and you explain tougher concepts such that it doesnt look tough at all....THANK YOU VERY MUCH SIR🙏🙏🙏
@@PHYSICSSIRJEEsir I'm new to ur course.. Please guide me if taking ur dropper batch really worth it.. I know ur teaching is fabulous but please guide
Amazing explanation sir ...though I was also able to solve the problem on my own also but the IAR thing was new and net omega will be along that I never realized this thing ....thanks for your help 😀😎🙏
Thank you 👍😊
Ok
Best video in TH-cam
Love you sir you teaching & understanding development.
You are great 🙏.
so easy to understand! thankyou sir
Thankyou very much sir by this viedo I have been having doubt and I thought that there will be many mysteries in this problem thank you very much sir your concepts r very nice supper.......thank you once again
I am glad you liked it, please share it with your friends who have a similar struggle 🙂
Wow! Beautifully explained. Thank u very much
Simplest and the best solution of this problem. 👌
Sir if the Q was like a disc free to rotate in 2 planes (2nd plane is like a cap hinged at 2 diametrical ends) is attached with rod and rotated as in Q then because of angular acceleration the disc would also get into slant position ?
Thank you very much sir🥳🥳🥳🥳
Revised it
WOWWWWWWWW BEAUTIFUL EXPLAINATION SIR .
Great question with great explanation
Excellent explanation. THANKS A LOT.
Very crucial problem for clarity in rotation
Simply brilliant sir... although I am teacher but learning like student.
🔥Excellent explanation 🔥
🔥Excellent question🔥
🔥Excellent Teacher🔥
Wow sir, just great explanation, thanks.
Such a Neat and Clean explaination.
Than you so much sir.🙏❤️
Thank you 😊, keep watching , sharing and promoting if possible 🙏
@@PHYSICSSIRJEE Yes sir, in my peer group every one watch your videos❤️You are the Best!!
@@kishan_kashyap21 okay
@@kishan_kashyap21 konsa peer grp bro
@@kishan_kashyap21 okay prince of priya🤣
# 52 For my revision ( Outstanding Explanation!!👌👌🙌❤)
All concepts are clear thank you sir
Wow aaj real meaning of Omega samajh Aya...
Great Great 👏👏👏Great
Awesomely explained sir
thank you ❤
I want a similar channel for mathematics 😭😭😭
jee maths unplugged
how do you make it look so simple?...Thanks a lot sir..
Sir, can you explain the para 31-32 in 2012 p-ll jee advanced sir.. Pls sir🙏
Thank you 🙏janardhan sir.
Amazing sir
sir, in case of NOT PURE ROLLING ,consider a disc moving with some velocity and having some womega ,on rough surface , now it has slipping (either forward or backward) ,now if we write TORQUE ABOUT GROUND POINT just below contact point (since it is not accelerating point ,we dont have psuedo force on C.O.M) ,since friction,normal,mg all passes through it,so we have NET TORQUE IS ZERO ,SO ALPHA IS ZERO..
BUT if we write about C.O.M we have friction contributing towards torque ,so there exists some alpha for rigid body.
BUT WE USUALLY WRITE TORQUE ABOUT GROUND POINT in some other cases ,AND it gives same as c.o.m, here why it is happening like that??
please KINDLY clear my LONG LASTING DOUBT SIR 🙏🙏🙏
Wow so easily solved great soln
Sir at @10.25 it is cosec(theta)
I FIRST SAW 2016 ONE QUESTION THEN I SOLVED THIS AND GUESS WHAT...I WAS ABLE TO SOLVE PART 1 WITH ONE MINUTE...I REMEBERED THAT BODY IS SYMMETRICAL THATS WHY LO WILL BE PARALLEL TO WO.....FOR BOTH OMEGAS I SOLVED AND THEN APPLIED VECTOR SUM
sir you solved my two problems with one vid ..one problem of irodov 1.55 2nd part and 1.57
crystal clear !!
thanks a lot sir !!
Yes , that was the idea and also to follow it up for 2016 JEE advanced rolling question. Please watch that too 🙂
I think so this is the priniciple which is used in gyroscopes
Amazing video sir ❤️
Excellent problem sir thank you
THANKYOU SIR REALLY A GREAT EXPLANATION .MAZAA HI AA GYA THANK YOU SIR ,THANK UUUUU SO MUCHHHH
I am glad you liked it 😊
Simply beautiful 😍
Thank you.🙂. If possible, please kindly share the content with relevant telegram and whatsapp groups 🙏
Thank u sir 🙏🙏👌👌
thanks sir, my coaching didn't taught me that
Awesome explanation
THANK YOU SIR!!
revised 22/12/2021
Sir another question of the same type 1.146 is also in irodov .. please do a explaination video of it also
Feeling confidence at rotation 🙂
Sir how can you say that choosen axis is instaneous axis of rotation
Thank you is not at all sufficient. 👏👏
Great explanation👍
Also, isn't Irodov 1.146 an upgraded version of this problem in which we have to see the dynamics also ? Although I could solve 1.146, would be great if you make a video on that too
😄😄trying different addressing s at start across the videos...👍
😁
Sir could you please confirm whether if in the translating frame of COM , whole cone would seem to be rotating about the axis parallel to IAOR and passing through COM and with magnitude as W(net) ??
Instantaneously , yes👍
@@PHYSICSSIRJEE Thanks sir all my concepts got cleared now 🙏
@@PHYSICSSIRJEE Sir if possible please also make video on irodov 1.267 problem
Excellent sir👍🔥
sir please upload more questions from irodov
Very nice video sir
I could not figure out the IAR in my first try
So I tried to find w in two parts first
W¹=v/r cos a. = 2v/√3 r
W²=v/r =v/r
And taking the resultant give the same ans as in the book
I went for this in fort try as I could find the angle of the resultant w
Gajab🔥
sir can you please make a video on 2017 jee adv question where a chain is rotated in a hand without slipping and explain why it is made a bonus
Ok sure 🙂👍
Perfect
Thank you very much sir
Sir could we have calculated kinetic energy with help of net omega if we know the moment of inertia about that side of cone as (I.W(net)^2)/2 ??
Yes , definitely 👍. You can actually in reverse use this idea to find I about slant height
@@PHYSICSSIRJEE but sir for that we need to know KE , so is there any other method of finding so ??
@@PHYSICSSIRJEE sir could you please also confirm whether if in the frame of COM , whole cone would seem to be rotating with axis parallel to IAOR and passing through COM and with magnitude as W(net)
Sir kindly can you please confirm it ?
sir I get your concept and solution and thanku so much... but I have a doubt that in rotation starting we learnt that angular velocity of a rigid body is same about any point .... I use this concept and simply find omega of C wrt O and got 2/√3 ...
what is the mistake in this statement or concept please clarify, I dont want to live with any misconception it hurts a lot💔
Thank you sir
Thank you sirJee
Hello sir ,
In this question if we try to find out the angular acceleration by writing the torque and then equating to I×(alpha) ......... Then the angular momentum of the disc would be ( Icm×Ws+Io×Wo) differentiating this would give us torque . Then now to equate this to I(alpha) to find (alpha) .....in this what is the value of I to be taken to find alpha...... My idea is that , an axis perpendicular to plane of paper passing through apex of the cone
My ans didn't match up by doing with the same method in case of "disc orbiting around an axis by rolling " .... Plz reply sir ..
L = I w is not valid when L and w are not parallel. Please read the corresponding topic from Resnick, krane book
Sir I've a doubt in a pathfinder Ques from electrostatics
Sir in the beginning of the video you mentioned about instantaneous axis of rotation. And so if the we drop the perpendicular from C to IAR, it's length will be r cos alpha and and we know the velocity of C. So using v = omega * r we can directly find omega as we know the velocity of C.
Sir am I correct in this reasoning or am I missing something?
V = rw is wrong. It's also answered in old comments
Sir can you please explain how it is incorrect. It may help clear my concepts
@@rohanshah6178 try recollecting , under what circumstances did you DERIVE v =rw . Is that applicable here ?
I realise that in the derivation, the point moves in a circle with radius r and has a velocity tangential to the circle. And I think this is not the case here as the velocity of C is not tangential. Sir is that case?
@@rohanshah6178 yes 👍🏼
Sir if the cone is having a angular acceleration then the axis of rotation is providing a torque. So sir to calculate the torque, about which axis should we take the moment of inertia?
Hope you will surely reply.
Torque is provided by couple of N and Mg and not by axis ( which is assumed to be smooth)
Sir there must also be an inherent angular velocity in the base of cone??? If i am bot wrong
All aspects of motion are already covered in the angular velocities assumed in the solution.
Time stamp of your doubt please?
SIR U R OP
The net rate of change of any vector can be written as
A bar dot = omega cross A + |A bar| dot
???pls verify
It was mentioned for constant magnitude vector in the question ( spl case) . Please go through the video again carefully
@@PHYSICSSIRJEE yes I know , magnitude was constant,, but I m telling a general expression of rate change of any vector which is rotating and changing magnitude,,, can I write above expression , I superimposed the two, may be I m incorrect, pls verify once .
@@krishsharma162 above expression, in general, is correct
Sir , voh capacitor wala que, how can I send you?
Sir for angular acceleration what's the problem in writing dΩ/dt = Ωω(orbital)
Net Omega is not rotating in a plane. In RHS of your equation instead of product it should be cross product
@@PHYSICSSIRJEE Thanks sir 🙂🙏
Sir why net omega along IAR?
Because, IAR means body is instantaneously purely rotating about this Axis , so net Omega by definition should be along that . For further reference : you can ask your teacher or read Morin or watch the playlist videos on rotation in this channel. 👍🏼🙂
Jee advanced questions made from irodov book.
😁👍👍👍
Janardhan Sir, Great video as usual and great explanation. But just a doubt...in one of the videos very similar to this in INSP(th-cam.com/video/xCxBAIEUK84/w-d-xo.html), Nithin sir had taught us d(r)/dt=w*theta. I completely understood the reason why you wrote d(r)/dt=w*r but I really didnt understand how that equation is different from this and why 2 different equations
My video talks about cross product. The direction of w cross r comes out to be in tangential direction only. I was about to reply the same in the mail you sent and I saw your comment here
Also , please refer to my rotating vectors video, w cross r and theta cap mean the same direction.
Also remember the difference between dr bar/dt and dr cap/ dt
Sir could you tell me the difference between dr bar/dt and dr cap/ dt.
This is what I understood as the difference:
dr bar/dt gives cross product and is used when a POINT is rotating, in our case the point along IAR. Also I watched the special video on rotation of vectors which is well explained...
Sir but dr cap/ dt which gives us the result written by Nithin sir is used when we consider a LINE(vector omega) is rotating
Sir Im I correct in my understanding?...please correct me if I'm wrong. Thank you in advance.
Opp
Sir i hope you can be eligible for monitisation please do it sir you will get upto 1,64,235.7988 ruppes
Awesome explaination sir