I have not seen anyone take as much effort as you do for each and every upload. It is truly impressive. Thank you for showing us the intricacies and beauty of physics instead of trying to make it into an 'easy' washed-down version, compressing information to be consumed in one shot.
I dont know Sir but there s surely something in your voice I was feeling stressed for IOQP after watching your video Im feeling quite relaxed .Although I have never met you in real life till now but I consider you a true mentor Sir. Love you 3000 Sir🙂🙂
We can easily calculate accelerations of COM by assuming geometric center has acceleration 'a' toward left and we have calculated alpha of ring... So acceleration of COM is in x direction is a-R/2 omega square and for pure rolling a=alphaR so we will get a'x'=0 and in y direction acceleration is alphaR that is g/2
Alternative : The cm of the system makes a circular motion about geometrical center of ring. so in order to find acceleration of CM we can just take its distance from ICR as r then Ax will be (alpha)r cos(thetha) - w^2(R/2) and Ay=(alpha)r sin (thetha)
Even if the questions level is of olympiad your explaination made it in one go...thanks a lot sir.....please keep posting videos like these.....really helpful sir❤❤
6:02 here is the angular acceleration w.r.t ICR or ground ??.......to be more precise , is α and omega independent of frame ?? And sir, in CM frame will there always be a centripetal force on all particles ( in any case, in general ) ? Lastly sir, what is the significance of fs = 0 ..... can u plz explain it sir... Sir plz don't get irritated by too many questions...I have many doubts in Rotation and this question strikes to all of them...such is the beauty of this problem....plz help me sir by clearing my doubts...
Alpha and Omega are not just same for any point on body but also same as seen from any "TRANSLATING" frame Wrt CM or any material point of a body in general planar motion , the other points are in circular motion . So we observe centripetal acceleration Fnet = mass of system x acm . In horizontal direction as acm is zero and f is only possible horizontal force , it has to be zero 👍🙂
12:50 sie can't we directly write accn of com from R√5/2 (alpha) because of icr and then split it into x and y components ? I didtn clearly understand why we need to change the frame pls help
janrdhan sir i have a doubt sir can u plz clarify,sir why exactly is the equation torque= Ialpha always written wrt to the com or icor why isnt the equation not written with respect to any other point
@@ponnapu It can be written about anypoint just that you need to consider the torque due to pseudo force for those cases, but if you are written Torque about COM Pseudo force passes through it so its effect on torque is 0, same for ICOR. So for the convenience to solve we take Torque about COM .Hope it clears your doubt
siir please please say, why cant we just sit on the icr and do acceleration analysis (instead of doing it wrt com).... if we are doing wrt icr we can directly find acceleration of com if we just know the distance bw icr and com right because wrt icr is it pure rotation??and then add Rw^2 which is the acceleration of icr (11:01)
I may be completely wrong, but at 5:05, shouldn't the torque be (mg+mg)*(R/2)? As the CM of the system is at R/2 from the geometric centre of the ring? Which would then give alpha = g/4R
You are forgetting to add a pseudo force (as explained in the video) at the position of CM (in yellow colour). In your method too, you will get the pseudo force at CM as 2mg , so another 2mg*R/2 gets added to your calculation , giving the same answer as my method
sir i found alpha by sitting on the centre of ring as beacuse the pseudo force applied will pass through centre itself.ended up with the same result.(4:48) and sir analysed the whole thing from centre of ring itself and found all results.
Thank you sir so much we are so lucky and glad to get a teacher like you thank you sir from my heart . Sir for revision before ioq I am doing david Morin , Purcell books mentioned by you sir is it enough for revision , please suggest .
Sir what if we take a point on ground just below ICR, will the body rotate about that point too? Also sir if we consider ICR and the point on ground just below it, the angular momentum will be same as Vcm is same wrt to both(as both points have 0 velocity, and using L=mvrsinθ+I₀ω we get same L) , but in torque about ICR we will have an extra pseudo force term which we don't have about point on ground. Why this contradiction?
Sir here the ring was moving with velocity v and had an acceleration a which we calculated as g/2. If consider only the rotating motion of the ring, cm has a speed in y direction which is equal to v/2 due to which it will be having a centripetal acceleration towards the centre of ring and if we see the translatory motion, the ring already has an acceleration a away from the center of the ring. On superposing the two, the resultant comes out to be zero, which gives the friction force acting on the ring to be zero. similarly the acceleration of cm in y direction is equal to r(alpha)/2 which is g/4 and it will give the normal reaction as 3mg/2. This method was a bit shorter than yours. Sir can you please tell that this method is correct or not? (revised 10/1/2022)
Hello, sir just wanted to know whether the icr is in a rotating frame or a linear frame of reference in this particular question , and thank you for your help
sir i have a general dbt suppose a ring of mass M radius R placed on the smooth surface and A Force "F" Applied on its centre then body does not rotate but i can write torque of this force about top point then about top point body experience a net angular accleration and sir ek pseudo force ka torque likhna padega but this pseudo force = (a(com) -alpha.R) M then net torque also is there
@@princekumar8285 you are thinking in wrong manner. Why will you write angular acceleration about center when you said you are sitting on top most point. W.R.T top point net force ( including pseudo ) is zero . So Torque = 0 implies alpha = 0. Case closed here
Good evening sir , I have a few queries. Alternative approaches : 1. Since there is no slipping at the bottom most point (ie.no tendency of relative motion) no friction will act therefore any horizontal acceleration will not be present. 2. Can we try to find the normal reaction by decomposing the rolling into pure rotation and pure translation? Then we can write the equations separately for the two cases. I couldn't do it, I am probably missing something. A doubt : www.sarthaks.com/344430/the-figure-shown-below-masses-blocks-and-respectively-the-force-constants-springs-and-are Why haven't we balanced the forces 👆 in this problem but used energy conservation? Has it something to do with "releases from rest" ? Thanks a lot for your time and effort 🙏
ANSWERS TO ALTERNATIVE APPROACHES : 1) just because initial velocity is zero , doesnt mean friction is zero . It can have static friction . here it became zero 2)No we cant forces are not superposable unlike KE and angular momentum . please read David morin for these basics , if you are curious
Sir one more thing in question 1.269 there is again a doubt that the point C is in translating frame or in a rotating frame of reference as it is still point but rotating about oo' axis ' being attached to the rod
15:00, we can also solve this question by considering the fbds of the two bodies separately. I solved this question in this way and it took me lesser time .
Sir can we solve it about the geometric centre? We write the torque equation about it as ( mgR-fR)=(mr^2+mr^2)a/r..... and we can also write the force equation....as ring is experiencing a frictional force along accelration ... f= ma......but then a =g/3....But actual answer is g/2...where am i wrong sir please help
REALLY LOVE YOUR Olympiad series sir♥️♥️ sir I am having confusion in the fact that if the velocity of the circle were more than √gR would it lift up the lowermost point. N=0? So no friction?
@@harshiljain2368 acceleration is not zero but upwards. Please watch it again 👍🙂 Normal reaction is drawn on entire FBD of body. Acceleration of bottom point is different from acceleration of CM If you want to draw fbd of only BOTTOM PARTICLE then there will be contact forces from its neighbours in its fbd
Wow excellent concept sir..same question was also given to us in our coaching class and I solved it😊😊in 2nd step I immediately went to frame of geometric centre and in one step it followed that ax=g/2-rw^2/2 =0 and ay= Ralpha /2 = g/4 ..
sir on more thing how did you consider that in( RESOLVED-06! A COMMON MISCONCEPTION! PSEUDO FORCE IN A ROTATING FRAME PHYSICS CHALLENGE)this video point p is on rotating frame as he also is a point hence, he should be also translating. And how do you differentiate between a point and a rotating frame observer
"observer" word or "rotating frame" attached to the axis statement should be mentioned in the question ( eg , read the IRODOV question wording of 1.268 till the end ).... So, if no mention of above words is made then "points" are considered and they just TRANSLATE
@@PHYSICSSIRJEE Sir one more thing in question 1.269 there is again a doubt that the point C is in translating frame or in rotating frame of reference as it is still point but rotating about oo' axis '
@@PHYSICSSIRJEE sir i love physics but because i have been studying physics lately too much i have not been able to focus on chemistry , hence not able to score good in mock tests for iit jee . can i enroll in any of your courses for studyiny physics .
If the particle's acceleration is a , - 2m a force should be put at CM... Only the tangential component will be responsible for pseudo torque, radial one will pass through it
@@mamtamittal3051 points don't rotate as mentioned earlier. 'Point' is different from "frame"... CM or ICR or point mass you asked are all equivalent to 'translating frames' (mentioned and clarified earlier too)
Sir in the same question if the particle is given charge q and and a constant electric field is applied and we have to find the displacement of particle if ring is rotated by by Theta .What would be it's approach?
Displacement is directly dependent on theta you gave . Write S = S of CM + S wrt CM . S of CM is R(theta) horizontally for pure rolling S wrt CM is 2Rsin(theta/2) making angle 90- theta/2 with horizontal Vectorially add the two Presence of electric field only changes the time taken but not the displacement
Sir is it allowed to shift like that , I mean u have shifted normal and frictional force from centre of mass to icr, or is it the same way Ans sir as we are calculating about that axis why did we neglect normal and friction?
Sir how can we deduce that the alpha you took about geometric centre ...remained alpha only while in the centre of mass frame and instantaneous centre of rotation frame ?????sir please reply its my conceptual doubt
Sir in that clockwise diagram the centre of mass itself has circular motion and the lowest point also is in circular motion then wrt com the bottom most point should be moving in straight path ?
Sir Why you not write acceleration of com directly by using ICOR then adding acceleration of ICOR to acceleration of com to get the acceleration of com wrt ground. SIR PLEASE REPLY I AM VERY CONFUSED ABOUT IT
At 4:23 , I checked , we calculated acceleration of ICR and used in next slide to calculate torque about ICR. Why does this require acceleration of centre of mass?
@@hemantgupta1257 at 7:23 acceleration of centre of mass was calculated in a different manner to clear elaborate concepts . Are you referring to this calculation , at this timestamp? You may then , alternatively use acceleration from ICR frame as you suggested .
@@PHYSICSSIRJEE sir it is my general query that can we write acceleration of any point of the Object wrt ICR if it is accelerated also and if yes then do we have to add the acceleration of ICOR to that point to get the acceleration of that point (any general point) wrt to ground
Sir at 8:02 can we analyse the acceleration components of com wrt to ICR only , I mean first we assing pseudo acceleration of g downwards and then write r(alpha) and w²r where r this time would be root5R/2 . Can we do this?
sir pathfinder solve karne ka realization abhi hua hai lekin sir uske mcq bhi nahi jaamte and agar jame to bhi time lagta hai sir kya karu sir advance ke pahle ,sir plz guide
I just made it for discussing icr . But similar questions have appeared in Olympiads before. In IRODOV a similar problem with work-energy idea is there
![[JEE ADVANCED] MOMENT OF INERTIA OF A HEMISPHERE | OBLIQUE AXIS | SCHOOL PHYSICS SIR JEE QUICKIES](http://i.ytimg.com/vi/7r9bZum2688/mqdefault.jpg)
![[JEE ADVANCED] MOMENT OF INERTIA OF A HEMISPHERE | OBLIQUE AXIS | SCHOOL PHYSICS SIR JEE QUICKIES](/img/tr.png)
![RESOLVED-22 [PHYSICS JEE ADVANCED] BERNOULLI'S EQUATION FOR TWO DIFFERENT STREAMLINES | VORTICITY |](http://i.ytimg.com/vi/ge9c1Ax8W1E/mqdefault.jpg)
I have not seen anyone take as much effort as you do for each and every upload. It is truly impressive. Thank you for showing us the intricacies and beauty of physics instead of trying to make it into an 'easy' washed-down version, compressing information to be consumed in one shot.
I dont know Sir but there s surely something in your voice I was feeling stressed for IOQP after watching your video Im feeling quite relaxed .Although I have never met you in real life till now but I consider you a true mentor Sir. Love you 3000 Sir🙂🙂
Kind words 🙏 . Felt really good 😊
You will ace through it.. Don't worry
Hard work pays off 😃😃
Best of luck !!
going into cm frame and then coming out of cm frame to comment on acceleration is such a genius move 🤩
This one is really special sir. A lot of concepts in a single video. Will have to watch it 2-3 times and contemplate, to digest all the things.
nice problem i have done this problem already i always prefer com frame to avoid pseudo torque thnx for giving insight of pseudo torque method
I am jee aspirant watch for advanced concepts from jee adv point of view.....just rocking
.sirji ...amazing we all love u ...and support u
We can easily calculate accelerations of COM by assuming geometric center has acceleration 'a' toward left and we have calculated alpha of ring... So acceleration of COM is in x direction is a-R/2 omega square and for pure rolling a=alphaR so we will get a'x'=0 and in y direction acceleration is alphaR that is g/2
I did it same way
can you please explain a bit
1:44 why we can't take icr attached to ground instead of ring .
The body is in rotation
about both the points
For a_x and a_y of centre of mass, if we analyze with respect to centre of ring, then we can get their values directly.
Alternative :
The cm of the system makes a circular motion about geometrical center of ring. so in order to find acceleration of CM we can just take its distance from ICR as r then Ax will be (alpha)r cos(thetha) - w^2(R/2) and Ay=(alpha)r sin (thetha)
Thank u sir....
Comment box me sb apne bhai log hi chhai hai....
Sahi kaha....sab jane pehchane ha....
Your efforts r praiseful. I m gonna tell all my frnds about ur channel.
Even if the questions level is of olympiad your explaination made it in one go...thanks a lot sir.....please keep posting videos like these.....really helpful sir❤❤
Your patience is ultimate sir. Thank you for multiplying my power sir.
such a great explaination sir,thank you very much
6:02
here is the angular acceleration w.r.t ICR or ground ??.......to be more precise , is α and omega independent of frame ??
And sir, in CM frame will there always be a centripetal force on all particles ( in any case, in general ) ?
Lastly sir, what is the significance of fs = 0 ..... can u plz explain it sir...
Sir plz don't get irritated by too many questions...I have many doubts in Rotation and this question strikes to all of them...such is the beauty of this problem....plz help me sir by clearing my doubts...
Alpha and Omega are not just same for any point on body but also same as seen from any "TRANSLATING" frame
Wrt CM or any material point of a body in general planar motion , the other points are in circular motion . So we observe centripetal acceleration
Fnet = mass of system x acm . In horizontal direction as acm is zero and f is only possible horizontal force , it has to be zero 👍🙂
@@PHYSICSSIRJEE Thank u Janardhan Sir....thank u very much SIR JEE
12:50 sie can't we directly write accn of com from R√5/2 (alpha) because of icr and then split it into x and y components ? I didtn clearly understand why we need to change the frame pls help
Thank you sir very good problem
Same question in FIITJEE review package 😀
Sir,your efforts are genuine
Excellent keep doing it's helping extremely !!
Was struggling with this concept ( now it's crystal clear )
Thank you sooo much sir 🙏
i cant believe how many qs i did wrong because i never considered pseudo torque. Thank you sir
thats a rookie mistake most of the teachers do too...
probably one of the best physics problems on this channel loved it sir
janrdhan sir i have a doubt sir can u plz clarify,sir why exactly is the equation torque= Ialpha always written wrt to the com or icor why isnt the equation not written with respect to any other point
sir plz reply when you see this message
@@ponnapu It can be written about anypoint just that you need to consider the torque due to pseudo force for those cases, but if you are written Torque about COM Pseudo force passes through it so its effect on torque is 0, same for ICOR. So for the convenience to solve we take Torque about COM .Hope it clears your doubt
Truly amazing sirji .. the amount of effort making this video is a lot ... i thoroughly understood the concept thanks a lot sir👍
Best Channel 🔥❤️
1:10 isn’t it the centre of mass of system you are asking for
Yeah , actually I wanted to calculate both the CM and center of ring accelerations at 5:20 onwards
@@PHYSICSSIRJEE ok sir...tq
Thanks sir, this situation came in fiitjee aits
I wrote all the equations in the ground frame because I am always cautious not to use non inertial frame. Got the same answer
Can u send ur solution
thank u very much sir...Was waiting for u to upload some ioqp problems since few days
siir please please say, why cant we just sit on the icr and do acceleration analysis (instead of doing it wrt com).... if we are doing wrt icr we can directly find acceleration of com if we just know the distance bw icr and com right because wrt icr is it pure rotation??and then add Rw^2 which is the acceleration of icr
(11:01)
yes we can do it that way also. I also checked by doing it wrt icor and got the same answer.
Thank you sir 🙌🙌🙌
This one is just nice problem with the best explaination 🙌
Thank you sir ❤️❤️
Thanks sir !
Thankyou sir 🙂🙂🙏❤🙌😇🔥
I may be completely wrong, but at 5:05, shouldn't the torque be (mg+mg)*(R/2)? As the CM of the system is at R/2 from the geometric centre of the ring? Which would then give alpha = g/4R
You are forgetting to add a pseudo force (as explained in the video) at the position of CM (in yellow colour). In your method too, you will get the pseudo force at CM as 2mg , so another 2mg*R/2 gets added to your calculation , giving the same answer as my method
Thanks, I was doing it from the ground frame and forgot to add the cross product r_cm x (m a_P), point P being the ICR. Now I get the same answer.
@@dr.bogdan24 Thanks for taking your time and commenting... Please keep visiting the channel 🙂🙏... All the best 👍🏼
9:34 how can we say that it will be pure rotation from cm frame
It's pure Rotation from any point on body . CM is one of them ... Basic idea of Omega being same from every point on a rigid body
This one was truely amazing, thank you so much sie🙏
sir i found alpha by sitting on the centre of ring as beacuse the pseudo force applied will pass through centre itself.ended up with the same result.(4:48) and sir analysed the whole thing from centre of ring itself and found all results.
Thank you sir so much we are so lucky and glad to get a teacher like you thank you sir from my heart . Sir for revision before ioq I am doing david Morin , Purcell books mentioned by you sir is it enough for revision , please suggest .
It should provide you enough confidence 🙂👍
Bro, in which class you are?
uttam bose just revise this channel asap don’t do books atm
Sir what if we take a point on ground just below ICR, will the body rotate about that point too?
Also sir if we consider ICR and the point on ground just below it, the angular momentum will be same as Vcm is same wrt to both(as both points have 0 velocity, and using L=mvrsinθ+I₀ω we get same L) , but in torque about ICR we will have an extra pseudo force term which we don't have about point on ground. Why this contradiction?
☹️interesting
Sir here the ring was moving with velocity v and had an acceleration a which we calculated as g/2. If consider only the rotating motion of the ring, cm has a speed in y direction which is equal to v/2 due to which it will be having a centripetal acceleration towards the centre of ring and if we see the translatory motion, the ring already has an acceleration a away from the center of the ring. On superposing the two, the resultant comes out to be zero, which gives the friction force acting on the ring to be zero.
similarly the acceleration of cm in y direction is equal to r(alpha)/2 which is g/4 and it will give the normal reaction as 3mg/2.
This method was a bit shorter than yours. Sir can you please tell that this method is correct or not?
(revised 10/1/2022)
Super problem
holy sh** this is awesome
Hello, sir just wanted to know whether the icr is in a rotating frame or a linear frame of reference in this particular question , and thank you for your help
Any 'point' related frame like ICR or CM are always 'translating frames' ... In Newtonian physics 'points don't rotate'🙂
@@PHYSICSSIRJEE sir is the particle of mass m in a rotating frame of reference
@@PHYSICSSIRJEE And sir can you also answer why the body always rotates about the center of mass.
@@mamtamittal3051 least moment of inertia about CM axis (parallel axis theorem)
Sir But yahan pai @14:30 2mg/4 nhi sirf mg/4 aayega .Rest All is perfect THankyou sir
We are using FBD of total 2m mass . So acm of 2m has to be written. Right?
@@PHYSICSSIRJEE oh yes actually i forgot to take com as of ring also
Really helpful ☺️❤️
Sir what is the reason behind observation of circular motion of any point wrt any other point on a rigid body?
Because distance between two points is ‘Fixed’ and that’s locus of a circle
and sir is the w(omega) about any point the same?
@@PratyushKumar-d9f for a planar ( laminar ) body in a general planar motion , yes
@@PHYSICSSIRJEE sir understood
thank you very much sir
sir i have a general dbt suppose a ring of mass M radius R placed on the smooth surface and A Force "F" Applied on its centre then body does not rotate but i can write torque of this force about top point then about top point body experience a net angular accleration and sir ek pseudo force ka torque likhna padega but this pseudo force = (a(com) -alpha.R) M then net torque also is there
Torque is zero about top point . Pseudo force cancels F and passes through center
@@PHYSICSSIRJEE but sir agar centre ke respect me ang accle likhe(top point ka ) toh F/M-alpha×R then pseudo force not equal to "F"
@@princekumar8285 you are thinking in wrong manner. Why will you write angular acceleration about center when you said you are sitting on top most point. W.R.T top point net force ( including pseudo ) is zero . So Torque = 0 implies alpha = 0. Case closed here
@@PHYSICSSIRJEE thank you sir now clear 😍😍
😊😊😊 nice video sir.
calculating a(com) is also easy from the IAOR right?
Good evening sir , I have a few queries.
Alternative approaches :
1. Since there is no slipping at the bottom most point (ie.no tendency of relative motion) no friction will act therefore any horizontal acceleration will not be present.
2. Can we try to find the normal reaction by decomposing the rolling into pure rotation and pure translation? Then we can write the equations separately for the two cases. I couldn't do it, I am probably missing something.
A doubt : www.sarthaks.com/344430/the-figure-shown-below-masses-blocks-and-respectively-the-force-constants-springs-and-are
Why haven't we balanced the forces 👆 in this problem but used energy conservation? Has it something to do with "releases from rest" ?
Thanks a lot for your time and effort 🙏
ANSWERS TO ALTERNATIVE APPROACHES :
1) just because initial velocity is zero , doesnt mean friction is zero . It can have static friction . here it became zero
2)No we cant forces are not superposable unlike KE and angular momentum . please read David morin for these basics , if you are curious
answer to the other DOUBT :
In the problem you posted " max extension" condition requires v=0 and not a=0 , thats why you cant use F=0 condition
Thank you sir for the correction and the book reference, awesome book by the way 👍👍
Sir one more thing in question 1.269 there is again a doubt that the point C is in translating frame or in a rotating frame of reference as it is still point but rotating about oo' axis ' being attached to the rod
C is Cm then translating frame
@@PHYSICSSIRJEE th-cam.com/video/Ej8LFGXiwus/w-d-xo.html here he has mentioned something else
Sir I tried solving the question in the frame of geometric center but I am not able to get the answere can suggest where I might have done mistake
Take a snapshot of your hand written Solution. Upload it to your Google drive. Make the link settings public and post it here as reply
sir can you please explain why won't there be pseudo force when we write torque equation about centre of mass?
15:00, we can also solve this question by considering the fbds of the two bodies separately. I solved this question in this way and it took me lesser time .
Sir can we solve it about the geometric centre? We write the torque equation about it as ( mgR-fR)=(mr^2+mr^2)a/r..... and we can also write the force equation....as ring is experiencing a frictional force along accelration ... f= ma......but then a =g/3....But actual answer is g/2...where am i wrong sir please help
REALLY LOVE YOUR Olympiad series sir♥️♥️ sir I am having confusion in the fact that if the velocity of the circle were more than √gR would it lift up the lowermost point. N=0? So no friction?
Yes. You can find that velocity too
@@PHYSICSSIRJEE sir but if the acceleration of the bottom point is zero, why would there be a normal?
@@harshiljain2368 acceleration is not zero but upwards. Please watch it again 👍🙂
Normal reaction is drawn on entire FBD of body. Acceleration of bottom point is different from acceleration of CM
If you want to draw fbd of only BOTTOM PARTICLE then there will be contact forces from its neighbours in its fbd
Thanks sir jee
Wow excellent concept sir..same question was also given to us in our coaching class and I solved it😊😊in 2nd step I immediately went to frame of geometric centre and in one step it followed that ax=g/2-rw^2/2 =0 and ay= Ralpha /2 = g/4 ..
sir on more thing how did you consider that in( RESOLVED-06! A COMMON MISCONCEPTION! PSEUDO FORCE IN A ROTATING FRAME PHYSICS CHALLENGE)this video point p is on rotating frame as he also is a point hence, he should be also translating. And how do you differentiate between a point and a rotating frame observer
"observer" word or "rotating frame" attached to the axis statement should be mentioned in the question ( eg , read the IRODOV question wording of 1.268 till the end )....
So, if no mention of above words is made then "points" are considered and they just TRANSLATE
@@PHYSICSSIRJEE Sir one more thing in question 1.269 there is again a doubt that the point C is in translating frame or in rotating frame of reference as it is still point but rotating about oo' axis '
Sir can we also do this by applying a centrifugal force on the point mass about geometric centre because I got all the ans correct by that method
True 👍 , good method 🙂✅
Thanks sir for the encouragement I feel very confident after ur comment
@@shreyyansh4098 by doing it could you find normal reaction because in that method I didn’t get
Thank uu sir !!
Sir is it okay to study the motion from centre of the ring
As the equations become simple .
Yes, all methods are welcome 🙂. Try it and see for yourself. In different problems, different methods become easier
@@PHYSICSSIRJEE sir i love physics but because i have been studying physics lately too much i have not been able to focus on chemistry , hence not able to score good in mock tests for iit jee . can i enroll in any of your courses for studyiny physics .
Hello sir if i was sitting on the particle of mass m what all pseudo torques i have to consider.
If the particle's acceleration is a , - 2m a force should be put at CM... Only the tangential component will be responsible for pseudo torque, radial one will pass through it
@@PHYSICSSIRJEE sir is it possible that the Euler pseudo force of r alpha / 2 will also be there as it is in rotating frame of reference
@@mamtamittal3051 points don't rotate as mentioned earlier. 'Point' is different from "frame"... CM or ICR or point mass you asked are all equivalent to 'translating frames' (mentioned and clarified earlier too)
So sir how do we identify which thing is in a rotating frame of reference and what is a “point”
Sir IAR rest pr hota at that instant tho ye Rw² ka acc kaha se dediye
Rest but has acceleration
Sir in the same question if the particle is given charge q and and a constant electric field is applied and we have to find the displacement of particle if ring is rotated by by Theta .What would be it's approach?
Displacement is directly dependent on theta you gave . Write S = S of CM + S wrt CM .
S of CM is R(theta) horizontally for pure rolling
S wrt CM is 2Rsin(theta/2) making angle 90- theta/2 with horizontal
Vectorially add the two
Presence of electric field only changes the time taken but not the displacement
Sir is it allowed to shift like that , I mean u have shifted normal and frictional force from centre of mass to icr, or is it the same way
Ans sir as we are calculating about that axis why did we neglect normal and friction?
Normal reaction is not acting at CM
Sir how can we deduce that the alpha you took about geometric centre ...remained alpha only while in the centre of mass frame and instantaneous centre of rotation frame ?????sir please reply its my conceptual doubt
Please kindly refer to my reply to "@pop Paul" below in comments section
@@PHYSICSSIRJEE thank u very much sir ...I was repeatedly wachting the video again to catch the concept 😅
Sir in that clockwise diagram the centre of mass itself has circular motion and the lowest point also is in circular motion then wrt com the bottom most point should be moving in straight path ?
Time stamp at the place of doubt? Bottom point or bottom material point?
@@PHYSICSSIRJEE 12:18 the point bottom most material point thro which IAR passes
@@PHYSICSSIRJEE I am unable to digest how wrt any material point other points are in circular motion
@@PHYSICSSIRJEE sorry to disturb sir but after watching the video again and thinking I understood everything 😍
Sir Why you not write acceleration of com directly by using ICOR then adding acceleration of ICOR to acceleration of com to get the acceleration of com wrt ground. SIR PLEASE REPLY I AM VERY CONFUSED ABOUT IT
Where? Time stamp of your doubt?
@@PHYSICSSIRJEE 4.23min why you not write acceleration of com directly from ICR
At 4:23 , I checked , we calculated acceleration of ICR and used in next slide to calculate torque about ICR. Why does this require acceleration of centre of mass?
@@hemantgupta1257 at 7:23 acceleration of centre of mass was calculated in a different manner to clear elaborate concepts . Are you referring to this calculation , at this timestamp? You may then , alternatively use acceleration from ICR frame as you suggested .
@@PHYSICSSIRJEE sir it is my general query that can we write acceleration of any point of the Object wrt ICR if it is accelerated also and if yes then do we have to add the acceleration of ICOR to that point to get the acceleration of that point (any general point) wrt to ground
1st like, 1st comment
Tnx sir
Sir can’t we write torque equation from ground frame and sir why did u choose the bottom most point and why not any other point
6:05 see from here
Sir a similar problem is in IPhO 2014
Sir at 8:02 can we analyse the acceleration components of com wrt to ICR only , I mean first we assing pseudo acceleration of g downwards and then write r(alpha) and w²r where r this time would be root5R/2 . Can we do this?
Yes , you can 👍
🤩🤩🤩🤩
sir pathfinder solve karne ka realization abhi hua hai lekin sir uske mcq bhi nahi jaamte and agar jame to bhi time lagta hai sir kya karu sir advance ke pahle ,sir plz guide
They take time but try to divide time from now to July 3 , try maybe 5 problems a day. You will learn a lot for JEE advanced each MCQ
@@PHYSICSSIRJEE thank you sir i will definitely try to do it
From which book or exam this question is?? May I know sir
I just made it for discussing icr . But similar questions have appeared in Olympiads before. In IRODOV a similar problem with work-energy idea is there
What books to follow for jee advance?? Plz tell me.. I will be humbled for ur help..
@@lumenphysics8608 NCERT , HCV , RESNICK ( partially), Anurag Mishra ( all volumes , especially solved examples) , IRODOV , Pathfinder MCQs , NCERT EXEMPLAR , FIITJEE and Allen AITS
@@PHYSICSSIRJEE thank you very much for your timely help
🙏🙏🙏🙌🙌🙌🥰👏❤❤🥳🥳🥳😇😇😇 thankyouuu sir
🔥
Bhai prep kaisa hein mujhe to INPhO ke q zyada easy laag rahe hein, NSEP lag raha nehi hoga mera to utna syllabus bhi khatam nehi hein😭
@@studentstudy162 bhai vaise question hamne kie hue hai nse me gande wale question aate hai usme speed matter krti hai
@@studentstudy162 kvpy me kitne bne
Sir bless me for ioqa and ioqc tmrw😀
Sir one doubt didnt prepare modern physics till now
Which book to do now😓
Last minute! H C verma solved examples.👍
As per the blessings : " May the force be with you" 🙏🙂
Thank you very much sir
Thank you very much sir
Thank you very much sir