JEE ADVANCED 2016 PROBLEM On ROTATION | TOUGHEST ROLLING PROB | ALL CONCEPTS EXPLAINED😍 Read Desc👇

LEARN ALL FINE NUANCES OF ANGULAR MOMENTUM THROUGH THIS PROBLEM😊. IT'S A JOY RIDE YOU CAN'T AFFORD TO MISS. THE OPTION D IS CALCULATED USING THE ASSUMPTION OF CM FRAME TAKEN AS ROTATING WITH Wo TO PROVE IT WRONG. IF WE TAKE ACTUAL CALCULATION, IT COMES OUT TO BE EVEN MORE GREATER THAN 76.06🙂
Q : Two thin circular discs of mass m and 4m, having radii of a and 2a , respectively , are rigidly fixed by a massless , rigid rod of length through their centers . This assembly is laid on a firm and flat surface , and set rolling without slipping on the surface so that the angular speed about the axis of the rod is . The angular momentum of the entire assembly about the point ‘O’ is (see the figure) . Which of the following statements(s) is(are) true ?
(A) The center of mass of the assembly rotates about the z-axis with an angular speed
(B) The magnitude of angular momentum of center of mass of the assembly about the point O is
(C) The magnitude of angular momentum of the assembly about its center of mass is
(D) The magnitude of the z-component of is
Errata : In the last step of Lo , made a small product error of 24/25 😋🙏 at 21:12 . Please note it doesn't change the final key/ Concept of the question
HERE'S LINK TO ROLLING CONE IRODOV 1-57 CHALLENGE
• IRODOV SELECT SOLUTION...
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[@amirthya]

Sir, I think in L_0 there might be a product error, seems to be missing a factor of 24/25.

[@PHYSICSSIRJEE]

Yeah , in the last step while multiplying . Added in the" errata" of the description 🙂🙏

[@muskaanyadav29]

THIS QUESTION BECAME EASY AFTER THAT IRODOV QUESTION. REALLY ENJOYED IT SIR

[@AdityaRaj04281]

Which one??

[@MrAnonymous91156]

​@@AdityaRaj04281lol you here😂

[@kashyaptandel5212]

which one?

[@ommakadia3815]

this is the epitome of quality content that one can get for jee adv phy on youtube... 🔥🔥 great video sir .. as always

[@yatharthgupta6468]

Wow ...I first time understood this question...so beautifully explained

[@varun1702]

The best part of ur Videos is that after we attempt we could see where we've made mistake on - Concepts Required🤩

[@kalashtrivedi4327]

Every question gangsta until you arrive sir🔥

[@wisdomkeytraining700]

Extraordinary.....just take a bow.....

[@pawanbhardwaj3153]

This channel is the best thing that has happened to me . Huge respect sir!! . I also teach physics and learning some new concepts from every video you post .

[@PHYSICSSIRJEE]

That's a heartening response coming from a teacher himself . Thank you 🙂🙏

[@Arnav-vt6xm]

​@@PHYSICSSIRJEEsir can you tell me that how have you taken W(spin)sin=wo

[@aaditya_63377]

Sir I am in Class 11th, I yesterday Found your channel and Saw this Rotation Problem , The Explaination Was just awesome Sir , Thanks Sir for Such a Great Explaination 👍👏😊

[@otakurequirements]

Hey you were in class 11th right? Did you clear jee ? If yes then what rank in both exams and which college?

[@sunritroykarmakar4406]

Wow..understood all ,easily the best question of jee

[@aritrabiswas4236]

Thank you sir for this wonderful video . Lots of doubts and confusions cleared . Plz keep uploading more such videos

[@sumithasuresh4597]

Oops...I was so wrong in that case. I didn't think carefully about the influence of ω₀ over the L𝒸. However, I've fully understood the point you wanted to acknowledge, and I visualize the contribution of ω₀ as a result of motion of CM of discs into the plane (leftward wrt CM if we were to place a guy sitting @ CM and facing towards motionless point ɪ as marked @18:27 even though the rod would rotate). Thankyou sir (Edit: The paper-setting panel is fairly correct in giving double key wrt the exam because expecting from we students to transcend past the known concepts and apply this degree of logic IN EXAM HALL is illogical)

[@shyamsundar4665]

You surely have no idea how many students are following you sir. There are number of corporate tutors out there who demand a lot but return you very less. But here is an educator who gives you what he finds in physics very prospective accessible in english to all students across the country.

[@arshitharajkumar3485]

Hats off to you Sir...you are doing a great job... Feeling grateful to be here....words cannot explain my gratitude to you.... Thank you Sir...

[@atmanirbharofficialindiaon2789]

First educational video with 0 dislikes

[@shivgurjar9602]

You are very clever sir, sir why you have not joined KZbin early, we needed you,

[@shyamsundar4665]

Don't have the authentic knack for solving a problem like this from first principles in the first attempt yet. But surely very comprehensible and satisfying.

[@AdityaRaj04281]

Finally aaj smjh aaya😇😇

[@gameraspirant]

Thank you so much sir i understood it completely...

[@aritrabiswas4236]

Sir , SHOULDN'T WE CONSIDER THE ANGULAR MOMENTUM CONTRIBUTION OF THE SYSTEM ABOUT CM CAUSED BY ROTATION OF OBJECT ABOUT CM IN CALCULATING L(z) ?

[@prathamrana8916]

Thanks sir really good ❤

[@physicsacademy3858]

Very informative Sir, Sir can u also elaborate the soln to us without using IAR? It will give us one more method to think.

[@PHYSICSSIRJEE]

Without IAR , you should use moment of inertia tensor analysis. You should also have an idea of "product of inertia" Concept ( out of JEE advanced syllabus)

[@physicsacademy3858]

Sir, IAR is used for A part, which is kinematics based. I was thinking it may be done without the knowledge of Moment of inertia (of higher level) as it is based on kinematics.

[@PHYSICSSIRJEE]

@@physicsacademy3858 yes, for only part A , you can👍🙂

[@aritrabiswas4236]

A better approach is to calculate angular momentum for individual discs and then add up . The difference between the value calculated this way and value calculated with respect to CM is the angular momentum of the system about CM along L(O) .

[@shashwatgupta4613]

Vv v helpful But what do u mean by ' cult' in thumbnail??

[@PHYSICSSIRJEE]

Alternative meaning for the word "cult" is used to describe something that has achieved a status of being popular and fashionable. No " rotational mechanics" discussion among Physics enthusiasts is complete without discussing this question 🙂

[@simplephysics104]

Really enjoyed sir throughout this solution.

[@equalcell1555]

The best Solution for this question

[@sohammandal3511]

L(s) is the angular momentum of the system about COM due to JUST w(s) and L(o) is the total angular momentum of COM about O how does adding them give me the net angular momentum of the SYSTEM about O, cuz the "about COM" expression is incomplete rt?? help me plz also is there any book/place where i can get all these theories ( like "net w lies along IAR", "L(p)=L(COM) + L(of P about COM)" ) altogether?? thnx

[@radewoosh3356]

Sir either the questions wasn't clear or you didn't say it, but Ls isn't the angular momentum of the system about the centre of mass due to omega s vector. Instead, Ls is the angular momentum of the system about the axis joining the centre of mass and point O due to omega s vector. This is because we know that angular momentum can be defined both, about an axis, and about a point. But here they ask us to calculate it just about the centre of mass but you have calculated it about axis passing through centre of mass and point O. Please tell me if I have made a mistake sir.

[@PHYSICSSIRJEE]

What you have said is correct . That's what I have said in the video too. About CM , you have to take both Omega s and Omega o . That's why they gave double key , not to penalise Students for a Concept beyond JEE. Please watch carefully the video again during my Calculation for Ls 👍

[@meenudubey9476]

Sir still in doubt same as what he said, even if we consider both omega s and omega o for calculation of angular momentum about CM But as per the calculation you did why axis was used through CM, it should had been point naah(as L,about point p=L of cm about point p +L of system about point cm(not an axis through it)). I think sir in this way here L about o is very tough to calculate as due to calculation about CM point

[@ANURAGGUPTA-fe4ek]

Sir in d part where you have calculated Lz =Locos(theta)-Ls sin(theta) you have taken Lo to be angular momentum of center of mass ,which is fundamentally wrong ,instead you should take vector sum of individual angular momentum of both disc i.e Lo= L1 +L2 (total angular momentum is sum of individual angular momentum)then Lo=(ml^2+4m(2l)^2)×(wa/l) Whose magnitude is different to what you have calculated ... This is given in hcv book eq 10.12 Please reply and correct me if i am wrong

[@PHYSICSSIRJEE]

Please read the first paragraph of "description" below the video and let me know if you still have a doubt , I will answer 👍. As explained in the video at the end about why option C was given (considering CM as ROTATING frame which it is not), option D was done using that idea. The value for Lz comes out to be more

[@ANURAGGUPTA-fe4ek]

@@PHYSICSSIRJEE Sir , lets first talk why Option C was given as double key , the professor who prepared this option made a language mistake please read carefully what option C is "THE MAGNITUDE OF ANGULAR MOMENTUM OF THE ASSEMBLY ABOUT ITS CENTER OF MASS IS 17ma^2w/2 " now here comes the mistake WHERE IT IS GIVEN THAT AXIS MUST GO THROUGH THE CENTER OF BOTH DISC AND PASS THROUGH COM, IT COULD BE ANY AXIS THROUGH COM ,IT CAN BE PERPENDICULAR TO PLANE OF FIGURE AND PASSING THROUGH COM, due to this ambiguity this option was made double key ,since language was ill defined. And for option D, calculating Lo you must consider sum of individual Angular momentum through O not angular momentum of COM. please take pen and paper solve the solve problem i am giving to you to understand - TAKE 2 MASS m and 4m connnect it with masless rigid rod of length l then connect m with massless rod of length l to rotating frame with constant angular velocity w, now find angular momentum of COM wrt frame and angular momentum of whole system wrt to frame . You will both answer to be different this is the mistake you have made while calculating Lo.. Please reply with a video if possible to my query . I will be grateful to you

[@PHYSICSSIRJEE]

@@ANURAGGUPTA-fe4ek Thanks again for taking time and pointing out the issue🙏🙂. This comment section is becoming difficult to exchange ideas. Let me try with Option C first. Regarding option C again, I don't think there is any language mistake in the Question paper . "The angular momentum about CM " is perfect to ask. CM is like point mass moving around z axis in a circular motion . Lo in my Calculation is defined as "Angular momentum of CM about O" So Lo is calculated ,like we do in case of a point conical Pendulum of mass 5m connected to point O, which is what I did . Yes, I made a TYPO (Calculation) error in hurry in last step in red of Lo at 21:06 (pointed out by my student, I have pinned that comment) which I acknowledged long back in ERRATA (2nd paragraph) in description . There is cos theta missing in last step of substitution . Nothing wrong in Concept I believe. But I am ready to see your view also on this 👍 Once ,this issue is resolved , we can move on to discuss my opinion about ACTUAL Calculation of Lz. You may communicate to me at [e-mail] (we can share pen-paper working like you said) , so that it's easier to resolve the issue further . Looking forward towards your email...

[@ANURAGGUPTA-fe4ek]

@@PHYSICSSIRJEE thank you very much sir jee for your reply now it is clear when you have explicitly mentioned that "while calculating Lo you have considered ANGULAR MOMENTUM OF COM , I was calculating angular momentum of whole system about Point O therefore I was getting higher value for Lo. Thanks a lot sir jee

[@PHYSICSSIRJEE]

@@ANURAGGUPTA-fe4ek you are welcome🙂 , I was very happy when I got a comment on this video after a long time . It was a healthy discussion and would help others here when they go through this comments. Please , Keep following the channel 🙏🙂 and if possible keep commenting 👍

[@sreenivasarao7041]

Explained it very clearly sir tq

[@praneethvarma1364]

Sir here( Lof body=Lof cm about point p+L about cm) now Lof cmabout point p is Lo and now L about cm also consists of both Lo andLs now here we are getting Lo vector2 times in the L of body but while calculting L component about z axis we considered Lo only once

[@jitendrapandey1085]

Yeh...same mine doubt...i have confirmed it with nitin sachan sir...there is a mistake

[@krishna-ng8ge]

Sir other than this book any other book is there which is frequently available about that concept (which you mentioned like resnick, Krane)

[@PHYSICSSIRJEE]

Slightly higher level : Morin

[@krishna-ng8ge]

@@PHYSICSSIRJEE thanks sir

[@vardaandua3585]

loved the explanation

[@radewoosh3356]

Sir if I use formula 3) which you showed at 6:55 and use it to calculate angular momentum of the "asymmetric" setup then I get two terms which are M * v * r (L of CM about P) and moment of inertia * omega ( L about CM). The first term is in the same direction as L0 as shown by you but the second term is in the direction of omega. When we take resultant then the resultant doesn't point in the direction of L0 as you said. Pls tell me my mistake sir.

[@PHYSICSSIRJEE]

You are correct 🙂but when and where in the video did I say 🤔that net angular momentum points in Lo direction. I don't think so. Net L Vector Calculation will be a complicated mess and that's why it was avoided in the Calculations in the video

[@nitinjuyal7739]

superb explanation ..easy to understand ..thank you sir

[@heshe1260]

Sir while analyzing the orbital motion, are we considering the system as two rigid discs or are we taking them as point masses?

[@PHYSICSSIRJEE]

Rigid discs

[@physicsgalaxy642]

it

[@l1mbo69]

Sir this is not directly related to what was asked in the options but I still wanted to ask- The orbital angular momentum (L_o) constantly changes direction as the assembly revolves, though it's magnitude remains constant. So, there must be a torque always perpendicular to it. My question is, what force is applying the torque?

[@amiyancandol4499]

Friction and normal force

[@ks2372]

superb explanationation sir

[@utkarshkumar8883]

one word......beautiful problem

[@nagamalaraju3182]

Sir you are super sir everyday I learn new things by watching your channel

[@aparnaarora5184]

i found a channel who does same thing for maths! kzbin.info/www/bejne/oZ6uaJ2qm9Smh9E

[@jatinjindal9135]

in the very last question, while calculating L_z , you considered L_o (ang. momentum of cm ) and L_s (angular momentum about com. due to w_s) but why you did not considered (angular momuntem about com. due to w_o) ?????

[@vol230]

thankyou sir .. concepts are fully cleared now :))

[@vaishaliagrawal9190]

Am not very comfortable with IAOR ... Is there any alternate way to understand the addition of the 2 omegas ?

[@PHYSICSSIRJEE]

Without that , I am afraid it would be difficult to solve

[@Arawn_0001]

Its more advance form is in david morin 🙂

[@SandeepKumar-nn6pr]

🔴 Wonderful communication. Despite you haven't used animation to impress the subscribers, yet your way of explanation is impressive. A minor error in calculation was ignorable. 🔴JEE paper setter must have asked to find angular momentum in CM frame while looking along the axle joining the disc.

[@PHYSICSSIRJEE]

Please kindly check the errata in description added at that time along with the pinned comment above . Thank you 🙂

[@SandeepKumar-nn6pr]

@@PHYSICSSIRJEE Thanks for a quick reply. I withdraw my comment.

[@PHYSICSSIRJEE]

@@SandeepKumar-nn6pr No need , you could have kept your comment. It's ok 🙂. It's always welcome when someone is watching the video till the end. Please keep visiting the channel and commenting... Thank you 👍🏼

[@m.sanjitreddy]

Sir L of the entire assembly about point O is equal to L of COM about O + L of that system about COM. But we don’t have the exact value for , L of system about COM,then how can we calculate L of system about point O?

[@krushnanshjain8712]

Did you get the solution of your problem? I too have the same doubt.

[@user-ny8di2ow5w]

Sir i am a 10th grade student and i solved this question in 2 min Idk I have done this by tukka or it is true but i felt i have don't solved this amd by my imagimation i have corrected this This for me was an real life imagination.

[@srisaisubramanyamdavanam9912]

Sir pls make a video of jee advance problem of 2017 from rotational mechanics

[@dhruvchaurasiya545]

Amazing explaination sir ❤️😍

[@rupajha9941]

Why angular momentum is not in Omega direction?

[@physicsera7366]

Nice explanation But At 13:14 there is a Correction.. kindly check!! L(about com) or Ls (u r right we should consider both angular vel. i.e Wo & Ws) But L(about com) or Ls should be less than (17mwa^2)/2.. because angle between Wo & Ws is obtuse.. i may be wrong but kindly check

[@mr.indianfactor9316]

Sir can you plz provide all questions In PDF form so that I can try all questions before JEE advanced 2023. In every video I'm learning something new . Thanks sir

[@nishantlamba.4463]

Is there any ambiguity in jee advanced 2021 physics paper 1 question 12 of optics in which IIT gave multiple answer key as (B,C,D) or (C,D)

[@atharvagarwal6412]

Amazing video sir. I have followed your channel's in depth analysis of concepts of rotational motion. Is there any book that you can suggest to read a bit further into this? For eg. you stated that ω and L may not be in the same direction for asymmetric bodies, which sounds very counter intuitive to me. Any way for me to dive deeper without straying too far away from jee syllabus?

[@PHYSICSSIRJEE]

You may try , Resnick halliday Krane for beginners or David Morin for slightly advanced version of your doubt

[@as.if_0077]

Very awesomely explained sir thank you

[@ks2372]

ok sir

[@compettishun7360]

Ok

[@devashishthakkar6723]

Tysm my dear sirjee

[@dishakayande3997]

watching for 5th time and now completely understood thanks sir jee 💓

[@aparnaarora5184]

i found a channel who does same thing for maths! kzbin.info/www/bejne/oZ6uaJ2qm9Smh9E

[@dishakayande3997]

@@aparnaarora5184 thank u very much

[@student-gr1gi]

sir please please resolve this doubt of mine. L vector of body= L vector of CM + L vector of body about CM. So here, L vector of body=L_o vector (which is L of cm) + L vector about CM. Now ,as you said, L vector about CM will not just be L_s vector and orbital omega will also contribute to it. So how can the net L vector of body be equal to L-o vector + L_s vector ( as you have taken while calculating z component of L) . Are we not missing the contribution of w_o in L about cm? Plz help me sir , or any student of sir's!🙏 edit: oh oh! I just noticed sir has clarified it in the description! You are a really great teacher sir!

[@Algoner]

Sir I think I can say ROCK and ROLL... Is the precursor for this question Just for first option

[@PHYSICSSIRJEE]

Yes 🙂

[@kevinadesara356]

I still don't understand one thing of when to use L = Iw and when not to .. please suggest me a video on this topic if u have already made..

[@boomlight437]

Glory to Janardan Sir, ♾️♾️

[@prakhargarg4166]

10:10 W0 can't also be calculated by applying the condition of pure rolling simply.

[@umeshiyengar5614]

hats off sir.

[@yashagrawal8592]

Sir if we were to choose any point on the body other than center of mass then will the direction of angular momentum of the body about that point will be parallel to the IAOR ??

[@PHYSICSSIRJEE]

No rule like that

[@yashagrawal8592]

@@PHYSICSSIRJEE Sir can you please clarify : when talking about Centre of Mass in this question , whether it means a rest point at the location same as center of mass or itself the centre of mass (in its frame)

[@PHYSICSSIRJEE]

@@yashagrawal8592 point moving along with position , velocity and acceleration of CM . It is the "CM"

[@yashagrawal8592]

@@PHYSICSSIRJEE So sir angular momentum about CM should be parallel to IAOR as body seems to be rotating about that axis with magnitude as I.w(net.) ?

[@PHYSICSSIRJEE]

@@yashagrawal8592 I think you are confused between angular momentum about a point ( CM ) and about an axis passing through CM

[@iamajayop]

Sir where can i get recorded lectures of urs.. Iam sure if someone like u teaches me i can score well.. I qualifies kvpy sa and ntse lv 2 but now i dont have any physics teacher to help me in jee .. Ihope u will help me

[@ShailendraYadav-eg6ev]

Sir while calculating angular momentum about centre of mass after calculating due to spin, when we go for calculating contribution of orbital motion for angular momentum, be observe that arm length of Central mass from centre of mass will be zero so angular momentum due to orbital motion will always be zero whatever velocity of centre of mass maybe? Please clarify this to me.

[@shivasudhan165]

Sir so L about CM means the spin angular momentum? That becomes 0 if CM is a rotating frame right? So even in both translating and rotating frames the L of CM can’t be eliminated 15:50 What is that angular moment of system about given in that formula

[@thingnamsunitase6106]

Late to this channel🙂

[@avneesh690]

great problem sir

[@rajvikramtiwari837]

thank u sir

[@praneethreddykoyagura4020]

sir, we needed total angular momemtum for Lz which is not calculated (which requires angular momentum about centre of mass). how could you calculate Lz by just taking spin angular momentum in place of angular momentum about com

[@PHYSICSSIRJEE]

Yes ,👍The Calculation was reduced to eliminate option D and actual Calculation for Option D comes out to be much greater than 76.06 as mentioned in the description below the video. The Calculation would require moment of inertia of system about CM (hor axis) making it tedious

[@praneethreddykoyagura4020]

@@PHYSICSSIRJEE thank u sir

[@physicsfun2022]

Sir , L of body should have 3 terms as you have explained L-cm, (one term) . L about CM should have two terms Spin and orbital that's why third option is wrong, But you have written only one term Lo which you have written should be L-cm (first term)?

[@r4music220]

for option c can i just write Icm * ( w(s) - w(o) sin theta), taking component of w along the axis about which it is rotating?

[@krishna-ng8ge]

Sir is morin better or Pathfinder better for jee advanced in your view,we follow that..

[@PHYSICSSIRJEE]

PATHFINDER objective is a must and problems from build your understanding if you have time ( JEE 2022)

[@sripathigajula8210]

AMAZING LONG VIDEO SIR BUT THERE IS LINK IN LINK OF THIS video I HAVE SEEN SIR YOU ARE MOVING CROSOR REPTEADLY FOR NEXT PAGE .. SHORTCUTS: NEXT PAGE :RIGHT ARROW , DOWN ARROW , PAGE DOWN ZOOM IN :CTRL + + ZOOM OUT CTRL + - LAST PAGE :LEFT ARROW , UP ARROW , PAGE UP HOPE THIS HELPED YOU SIR

[@PHYSICSSIRJEE]

Thank you for the help🙏, I will train myself in these shortcuts to become better 👍🙂

[@lemmeexplain743]

Sir I have understood the whole concept but sir now I came up with a doubt in jee adv 2021 question 18 paper-1 in which they asked angular momentum of system about point hinge point O iN WHICH a Rod is connected to a disc.... In that question rod was rotating with a angular velocity 1(ohm) and disc is rotating about its own axis with angular velocity 4(ohm ). I am getting answer by doing simple L)rod + L)disc where I write L)disc as MrV)cm + I(4ohm) where I is moment of inertia of disc about centre of mass and V)cm is velocity of CM but the the doubt is why we don't write angular velocity as 5(ohm), why we write it as 4(ohm) about CM of disc . CM is doing transalatory motion so in its frame angular velocity of rod should be 5ohm not 4ohm that I learnt from this video. Sir please reply and please clear this doubt ... If you reply then I will post this Video On my whatsapp status and I will send you 50 rupees from my pocket money....and this is what I can do as thanking you pls reply

[@PHYSICSSIRJEE]

I have already done a question on wheel and axle angular momentum in the rotation playlist. It was coincidentally done much before the actual JEE advanced 2021 paper. Students claimed it was similar. Please kindly refer to that video first. It would be easier to understand, I guess. You can comment there

[@cricvillager4889]

Sir w not mere khyl se angle thetha par nhi hoga with vertical

[@JitubhaiyaPhysics]

Nice sir

[@pawanbhardwaj3153]

Sir Thanks for explaining in such a lucid manner. I have a doubt regarding ICOR in case of pure rolling . Suppose a body is pure rolling with acceleration ‘a’ and angular acceleration alpha. Since ICOR is a point in space which is momentarily at rest and may be accelerated. But in case of pure rolling, would ICOR be bottom most point of the body or the top most point of the ground or the point in space at the intersection of both body’s bottom most point and ground top most point . And what would be the acceleration of that ICOR ? Please reply i am having this doubt for more than 2 years and none could clear it.

[@PHYSICSSIRJEE]

It's a part on the body or material extension ( imaginary) on the body with instantaneous zero speed. Please read first few paragraphs of this 👇 en.m.wikipedia.org/wiki/Instant_centre_of_rotation

[@pawanbhardwaj3153]

@@PHYSICSSIRJEE It means in case of body pure rolling on a stationary surface acceleration of ICOR is same as that of bottom most particle i.e acceleration in upward direction towards centre of body But sir ICOR always remain below the centre of the body and it moves on a straight line like centre does than should not its acceleration be the acceleration of centre of body ?

[@PHYSICSSIRJEE]

@@pawanbhardwaj3153 ICR is instantaneously the bottom point , shares it's acceleration. You don't follow it and say it moves in straight line. New instant , new point , apply concept again . You don't follow it's trajectory and write it's properties . It's an instantaneous concept

[@sasi8391]

Sir why resultant angular momentum and resultant angular velocity don't coincided please say sir

[@PHYSICSSIRJEE]

In general , it need not . Eg : conical pendulum

[@kushagra_Panjwani]

Sir , Pls tell why Vcm is not equal to aw in case of tilted disc

[@PHYSICSSIRJEE]

I have already replied in your previous comment. You are considering a special case situation of a rolling vertical disc and applying it's equation to this case. Vcm helps CM here to complete a circular motion of certain perimeter . Does the point on the rim cover same distance due to spin in the same period ? Also the bottom point has two motions one due to orbital motion around z axis and other due to spin . These two should cancel. The speed due to orbital motion of bottom Point is not Vcm but more than that as it's distance from z axis is greater than center because disc is tilted. So Vcm is not aw

[@yashagrawal8592]

Sir is angular momentum about centre of mass always I.w where I is instantenous axis passing through centre of mass and w is the angular velocity of the body or it is only true for symmetrical bodies ?

[@PHYSICSSIRJEE]

Symmetrical only

[@yashagrawal8592]

@@PHYSICSSIRJEE Yes sir ,so we also can't say in this question that angular momentum about centre of mass is vector sum of I1.w1 + I2w2 (orbital + spin).

[@PHYSICSSIRJEE]

@@yashagrawal8592 we can do that for L about axis passing through CM and not L about CM, like in your previous comment. Also you have to consider two different spins in this question , as mentioned in errata in description below the video

[@yashagrawal8592]

@@PHYSICSSIRJEE Ok sir , So calculating angular momentum about Centre of mass (option c ) would be very difficult ?

[@yashagrawal8592]

@@PHYSICSSIRJEE Also sir could you confirm whether that in description you mentioned that real answer of Iz part would be even more than 76.06 ,then the extra term in calculation will be I.Wₒ ? Where I is the moment of intertia parallel to z axis and passing through centre of mass.

[@nimitjain8163]

My doubt may seem stupid. i am not able to understand why we didnt add the angular momentum *about* centre of mass to angular momentum of centre of mass while calculating Lo at 18:42 . if we consider orbital motion only, then wouldn't the system be having angular velocity w0 about cm so why did we ignore it? kindly clear this doubt sir.

[@zacksnyder8903]

Sir I have seen certain other solutions where they calculate omega at angle theta to z axis and then take component along z. How is that solution related to yours .

[@PHYSICSSIRJEE]

I have explained in this video at the start ,why taking Omega at an angle theta is incorrect

[@ROHITKUMAR-hi4cu]

Sir o have a doubt that while calculating moment of inertia then why we have not considered rod moment of inertia ml2/12

[@ved2705]

Sir , at 13:11 you said that the L about CM is due to both w(o) and w(s) , then if we calculate L of assembly about O, we add L of cm wrt O and L about CM which gives L(o) +L(s) + L due to w(0) discussed at 13:11, but in calculation of L(z) at 26:01 why did we only consider L(o) and L(s)??

[@PHYSICSSIRJEE]

To disprove option D for exam purpose. Same was mentioned in the description below

[@ved2705]

@@PHYSICSSIRJEE sir I am a bit confused regarding calculation of L(z) , I have also seen the video on this question by Ambarish sir ( kzbin.info/www/bejne/q6OvZJeGrNt6ebs ) , in which at 11:30 sir just added l(o) and l(s) in the direction of z -axis for the calculation of L(z) but as per our discussion L about o is L(o) + L(s) +L due to w(o) which doesn't matches , what is wrong here?

[@sachinbeniwal8481]

👍👍👍👍👍👍

[@Algoner]

Sir can u bring a viedo by explannig direction of angular momentum iam very much confused

[@PHYSICSSIRJEE]

Watch this and the previous videos in the description Completely

[@Algoner]

Sir I saw but direction angular momentum is not mentioned .....u said to see from resnic haliday

[@PHYSICSSIRJEE]

@@Algoneryes, RESNICK, Halliday, Krane . Ok, I understood your issue. Let me plan a video on it in future 👍

[@Algoner]

@@PHYSICSSIRJEE thank you very much sir🙏🙏🙏🙏🙏

[@arunayarramreddy4235]

Sir,for caluculating Iz in lat part,I have a doubt,sir if we follow I of cm + I about cm rule ,then we must take both w0 and ws right? What I felt was that you only took ws in your soln effectivly..plz explait sir.

[@PHYSICSSIRJEE]

Yes , you are right🙂👌 . I continued the Calculation from the assumption of cm rotating frame obs (with wo)from previous option to prove Lz option wrong. Yes ,in Correct Calculation , the value will be even more than 76.06 due to involvement of wo. I have added this explanation in the description because calculating it in full becomes very cumbersome. Now , I have added this explanation to my pinned comment too🙂

[@amirthya]

PHYSICS SIR JEE - IIT JEE and OLYMPIAD - JANARDHAN Further on this note, it seems to me that the calculation of angular momentum about CM arising due to w_0 is computationally involved because (1) w_0 is not alone any principal axis creating the need for the whole momentum of inertia tensor, and (2) using a direct integral approach of r cross product v, there is lack of any symmetry due to as omega_0 is not in the plane of the wheels

[@amirthya]

In this spirit if a full calculation of angular momentum is needed, for calculating the angular momentum about the CM, we might consider resolving the angular momentum about CM (vectorial sum of w_s and w_0) into components parallel and perpendicular to the axle. Then the spin part will come from effectively w_s - w_0 sin(th) and the part due to w_0 cos(th) can be found again using the L of Cm plus L about CM for each of the disks

[@PHYSICSSIRJEE]

@@amirthya absolutely correct 👍 🙂. That's what I implied in the description . It's exactly how I would have gone ahead

[@unclasswithsuryasir5448]

@@amirthya Excellent dude..... If possible can you please attach the calculations. I would be very gald🙏

[@kushagra_Panjwani]

Dear SIR, For the system to roll purely, velocity of centre of small disc should be Ws*a . But by taking perpendicular distance of centre of small disc from iar and multiplying omega the result is different? Pls clarify this.

[@PHYSICSSIRJEE]

For pure rolling , bottom most point should be at rest . The condition becomes your first line in comment only if disc were vertical. So , here Vcenter is not a* Ws

[@kushagra_Panjwani]

@@PHYSICSSIRJEE Sir, Can u please give me a little feel that how that point is at rest in this situation. According to my visualisation if Ws*a in isn't being balanced by vcm than shouldn't the point slip?

[@PHYSICSSIRJEE]

I have already replied in your previous comment. You are considering a special case situation of a rolling vertical disc and applying it's equation to this case. Vcm helps CM here to complete a circular motion of certain perimeter . Does the point on the rim cover same distance due to spin in the same period ? Also the bottom point has two motions one due to orbital motion around z axis and other due to spin . These two should cancel. The speed due to orbital motion of bottom Point is not Vcm but more than that as it's distance from z axis is greater than center because disc is tilted. So Vcm is not aw

[@kushagra_Panjwani]

@@PHYSICSSIRJEE Thank you sir,for helping 😊😊.

[@pratyushkumar9194]

Sir there is a mistake while calculating L° . U should be multiplying another 24/25 to √24/5 . Pls show the calculation also sir ,why did you skip it

[@PHYSICSSIRJEE]

Check the errata in the description or in the pinned comment written at that time

[@pratyushkumar9194]

@@PHYSICSSIRJEE sir so the entire multiplying factor should be (81)(24)(√24)/125?

[@pratyushkumar9194]

@@PHYSICSSIRJEE am I correct ? Pls tell sir

[@Srinu19595]

Sir can you please suggest standard books for physics

[@bikki7524]

Ncert

[@Algoner]

Nice it's so exciting 😊 iam waiting

[@huzaifaabedeen7119]

Please suggest me a KZbin channel for chemistry for JEE advanced

[@krishna-ng8ge]

Sir wrt to point O both the discs are in translating frame and also COM so wrt to COM there will be no orbital angular momentum because relative angular velocity (orbital) is zero wrt to COM, this is my doubt sir....so please give reply

[@PHYSICSSIRJEE]

COM is a translating frame as explained in the previous video. About O also , discs are not translating as you mentioned in first line. Please watch the previous video mentioned and your basics will be clearer to follow this video. 🙂

[@VishnuKumar-yr5ky]

Sir is it the same case with jee advance 2021 angular momentum integer type question

[@hemantgupta1257]

sir your explanation is great but sir i think the answer that you find for ANGULAR VELOCITY is wrong and it sholud be 5W/24 according to my sir if you want to check then check at the channel rohit sir physics quantum AND SIR YOU CAN ALSO CHECK IT FROM CHANNEL AAKASH EDUCATION that sir has also given the answer for angular velocity is 5w/24.

[@PHYSICSSIRJEE]

He ( Rohit sir )has contacted me through another person ( Nilesh kulkarni ) to verify his video and I have told him already why his answer is wrong. You may find that comment below if he has not deleted it . 5w/24 is wrong answer not because they have given it , but because it's physically wrong. If it's given in any other channel also , it's ok , it's wrong. They are using a common misconception v=rw at bottom point , which is not valid for rolling disc which is slant🙂👍🏼

[@PHYSICSSIRJEE]

As expected that comment here , therefore my replies , after accepting that 5w/24 have been deleted for reasons known to them . You may refer to a similar question from comments of user "sour Diesel" and my replies why w/5 is correct and 5w/24 is wrong Here's snapshot link of his comment and his acceptance for why 5w/24 is wrong 👇. My reply , you can find is same as for sour Diesel's query drive.google.com/file/d/1FJOmEsh399MpItpUMuKxMJgHL-9UgHj0/view?usp=drivesdk

[@PHYSICSSIRJEE]

I hope you won't delete your comment here 😁

[@PHYSICSSIRJEE]

@@hemantgupta1257 please read my above comment or replies to "sour Diesel" comment below . In short : writing v=rw is wrong for a slant rolling disc. It's correct only for vertical straight disc rolling. That's a common misconception among students who apply the formula without understanding the meaning of rolling

[@hemantgupta1257]

@@PHYSICSSIRJEE ok sir thax for reply

[@mamtamittal3051]

Sir I think that you might have taken Wo wrong in Z direction , because as the discs are tilted then the W of the disc while doing circular motion must be at an angle of q form z axis . If i am wrong pls correct me sir

[@PHYSICSSIRJEE]

Wo means orbital angular velocity. This is perpendicular to principal plane of rotation of any point on the object (considered without spin). Take for eg CM itself. What's it's (principal) plane of circular motion as it "Orbits" around z axis... horizontal and parallel to XY plane, right? So the Wo is perpendicular to this plane and hence in z axis direction... please, kindly refer to the precursor irodov problem mentioned in this video at the start. It will help you understand it better. Also, for the further knowledge of "PRINCIPAL planes of rotation" ( not required for JEE in detail ), you may refer to "DAVID MORIN" book,,, ALL the best :)

[@mamtamittal3051]

@@PHYSICSSIRJEE Hello sir I didn't take the help of iaor but instead I used the formula for v=wr for pure rolling, hence when I first took orbital angular velocity wholly in the z-direction and then wrote Wz cross Rcom vector = W(spin angular velocity) cross(A)it didn't come right and when I took it at an angle q same as the discs are inclined (the orbital angular velocity direction ) it came correct that wz=w/5, hence I thought that since discs are tilted his might be the reason for it to be tilted, Moreover could you pls tell the page number of David morin for the principal axis.

[@PHYSICSSIRJEE]

@@mamtamittal3051As the disc is tilted, v=rw is wrong for tilted disc if you are using v Cm and w for w spin here. AND I don't remember the page number of Morin where I read it, you may search about the principal plane of rotation in any standard book's back index or Google the same. Also , you may mail the context of the above latest message of yours along with a pic to [e-mail]. It will be helpful to understand your context with the help of a diagram which KZbin doesn't allow. 🙂 Also, I am busy with my courses till April end on my app, I will surely note your email and we can discuss your solution then. All the best and keep visiting the Channel 🙏

[@mamtamittal3051]

@@PHYSICSSIRJEE Thank you sir I will send a reply, l have been watching your videos for a long time , and have followed other channels for physics also , Could you pls guide me how can I learn the concepts for Ipho while doing my prep for iit jee

[@ved2705]

@PHYSICS SIR JEE - JANARDHAN - IIT JEE and Olympiad sir I am a bit confused regarding calculation of L(z) , I have also seen the video on this question by Ambarish sir ( kzbin.info/www/bejne/q6OvZJeGrNt6ebs ) , in which at 11:30 sir just added l(o) and l(s) in the direction of z -axis for the calculation of L(z) but as per our discussion L about o is L(o) + L(s) +L due to w(o) which doesn't matches , what is wrong here?

[@krushnanshjain8712]

Did you get the solution for your doubt? I too have the same doubt.

[@ved2705]

@@krushnanshjain8712 not yet

[@hello3267]

Basic concept here is to add the ang momentum caused by the omega in z direction, Janardhan sir's methodology is to add that ang momentum by separately taking moi of system abt com thru a vertical axis while ambarish sir is accounting for that ang momentum due to w(k^) by changing the coordinate axis and taking component of omega in principal axis, so the Ls here isn't equal to Ls calculated by ambarish sir

[@SAURABHANANDLOVE]

There are still very confusing points in this question, which is arising due to orbital motion of body about Z axis because many teachers are considering different plane of center of mass...hence there is slight error.... The whole confusion is regarding any point on the body...let us take point Q....what will be the absolute velocity of point Q or velocity of point Q wrt point P ( both situation is same as point P is at rest)...from Janardan Sir's & Nitin sir's method velocity of point Q will be "24/25 aw" from Z axis also and fom IAOR axis also .....Now velocity of Point P with respect to Z axis is "aw" and with respect to point Q is "-24/25 aw"...how this will satisfy pure rolling motion because it will not add to give zero velocity of point P..... 1) Sir's case - Here Sir has considered plane of center of mass as perpendicular to orbital motion about Z axis..... If about Z axis we take velocity of point P, it is "w/5" (angular velocity about Z axis) multiplied by distance of point P from Z axis "5a" which will give "aw"....(where w is spin agular velocity)....but since point P is in pure rolling motion so it should have net zero velocity...so if we calculate velocity of point P with respect to point Q, by taking all correct vector dimensions of spin agular velocity "w" and distance of point P from point Q and then taking cross product of both, we will get "- aw" and that makes sense....Thus due to pure rolling motion we get net velocity of point P as zero.....Now since point P and Q lie on same rigid body and also point P is at rest, so velocity of point Q in absolute sense will also be "aw".... But this velocity of point Q which is "aw" ...will not come as "aw" from IAOR as calculated by Sir and Nitin sir....both of them are showing net angular velocity which is along IAOR as "w cos θ"....If we find the velocity of point Q either from IAOR or from Z axis it will come same as "24/25 aw "but not equal to "aw", which is slightly less than "aw"..... from Sir's method, if absolute velocity of Q is "24/25 aw" then velocity of point Q with respect to point P will also be same as "24/25 aw" because point P is in rest...then velocity of P wrt Q will be negative of that i.e; " -24/25 aw" ....velocity of P from orbital motion about Z axis is "aw"....if we superimpose velocity of point P due to "orbital motion and spin angular motion" then it will not be zero and hence not satisfy pure rolling motion condition.... 2) Other teacher's case- If we take plane of center of mass perpendicular to Lo axis, then there will be no such ambiguity.....angular velocity of that plane will be parallel to Lo axis , let that angular velocity be "Zeta" which is w/√24 , which is inclined with θ angle from +Z axis.......then net angular velocity of the body along IAOR will not be ( w cos θ) ..but net angular velocity along IAOR will be (w cosθ + zeta sinθ).....In this approach, no slight error arises and pure rolling motion condition at point P is satisfied ..... I am attaching a video of IPS Ganesh Kumar Sir's method.....Janardan Sir, Nitin Sir, Ganesh Sir...all are scholars of physics ....only for this particular question I am sharing this video....this video may not satisfy that assymetrical axis concept (from Resnick ) which Janardan Sir has mentioned...but with this video's approach pure rolling condition is satisfied.... kzbin.info/www/bejne/bqDUcn6uoNqFbc0 Respected Janardan Sir...I am a big fan of your teaching and knowledge....kindly share your thoughts over my area of concern in this particular question.....

[@PHYSICSSIRJEE]

Hi, I am highly impressed by the time you took to write down your thoughts in so much detail. But I must confess, that I didn't read the entire comment properly owing to my busy schedule with the Crash course on my 'physics sir jee' app. It will be finished by March end (40 days from posting of this comment) .I will surely go through each and every detail of this comment and respond. But YT comments section would be a difficult medium to communicate without any diagrams available. You may , if possible, mail me the same issue with the diagram to [e-mail], we will definitely connect 🙂. PS: It may not be related to your comment above... But the pure rolling condition when the disc is inclined like here is NOT Vcm = r Ws , Ws is spin angular speed of disc... The entire issue with angular speed being a derived quantity is greatly discussed in "David Morin" especially the first few problems in "many Omega " situations. Let me connect with you through the email or phone and go through the above discussion along with the diagram by March end 🙂🙏