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 👍👏😊
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.
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.
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 .
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)
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)
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.
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 🙂
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) .
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) ?????
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
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.
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 👍
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
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.
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
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
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
@@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
@@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 physicssirjee@gmail.com (we can share pen-paper working like you said) , so that it's easier to resolve the issue further . Looking forward towards your email...
@@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
@@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 👍
🔴 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.
@@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 👍🏼
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?
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.
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
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?
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
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?
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
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
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 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)
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.
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
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
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.
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
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
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 sir I am a bit confused regarding calculation of L(z) , I have also seen the video on this question by Ambarish sir ( ua-cam.com/video/urL3dPwvGCY/v-deo.html ) , 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?
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 .
20:05 Sir when we find Vcm using the centre of mass formula, Vcm=(m1v1+m2v2)/m1+m2 We get it as 9aw/5 which is same as (9lwsin(theta))/(24)^1/2 but is different from what you have got Can you please identify the error?
@@theintegratedguy9528 if w in your equation is spin w... then you are using a wrong assumption v = rw for rolling... It's an inclined disc. Please try to understand why we use v=rw for a vertical disc. The real meaning of the expression and not the formula. You will then understand why it's not applicable here for discs which are inclined
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!
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.
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🙂
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
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
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
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
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.
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
@@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?
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
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 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).
@@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
@@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.
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
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 :)
@@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.
@@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 physicssirjee@gmail.com. It will be helpful to understand your context with the help of a diagram which UA-cam 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 🙏
@@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
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)?
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
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. 🙂
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
can we take component of angular velocity like that sir, it seems highly non intuitive and doubtful, like you took for z axis i.e. w sin theta directly
@@PHYSICSSIRJEE sir then why can't we use v=wr to calculate the velocity of the center of mass , as v=w/5*24a/5=24wa/25, which is not the velocity of the disc???? also sir by your logic if rod is rotating about an vertical axis and it is inclined at an angle theta to the vertical then we can calculate its angular velocity of the rod about its own axis(the axis along the length of the rod) by taking components although it is not even rotating about that axis i recently opened a thread related to this question on physics forum.com, sharing its link if you could contribute to a solution some people are building on it www.physicsforums.com/threads/existential-dilemma-on-angular-velocity-of-a-complex-rigid-body.1003792/
@@gourvi6339 very long doubt and not properly worded . Responding to First para of your doubt only : " center of mass of small disc" or "CM of system" ?? can you be very clear whose v did you attempt to calculate ? V= rw is not a standard formula and should be used very carefully . w vector is a derived vector of convenience ( not a fundamental vector like angular momentum) . I request you to go through David Morin text " introduction to classical mechanics" chapter 8 section 8.9 , problems 2( many omegas) , problem- 3(rolling cone) and n sequence with its multiple solutions . Also go through preceding text on understanding omega in that chapter. Once you are done with these , please revisit the first two paragraphs of your doubt above. Then we can discuss further once you have clarify over the diagram I drew for inclined rod rotating.👍🏼🙂
@@PHYSICSSIRJEE on the inclined rod rotating, I was commenting that if we take components of angular velocity like you told about in this video along the axis which lies along the length of the rod, then we do get a w but it can be easily seen that the rod is not rotating at all about that axis???
Sir, I think in L_0 there might be a product error, seems to be missing a factor of 24/25.
Yeah , in the last step while multiplying . Added in the" errata" of the description 🙂🙏
this is the epitome of quality content that one can get for jee adv phy on youtube... 🔥🔥 great video sir .. as always
Wow ...I first time understood this question...so beautifully explained
The best part of ur Videos is that after we attempt we could see where we've made mistake on - Concepts Required🤩
Every question gangsta until you arrive sir🔥
THIS QUESTION BECAME EASY AFTER THAT IRODOV QUESTION. REALLY ENJOYED IT SIR
Which one??
@@AdityaRaj04281lol you here😂
which one?
Extraordinary.....just take a bow.....
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 👍👏😊
Hey you were in class 11th right?
Did you clear jee ? If yes then what rank in both exams and which college?
First educational video with 0 dislikes
Real teacher who cares about concept and carrier of students
Handsofff to your service 🎉
Thank you so much for helping to students like us sir🌟
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.
You are very clever sir, sir why you have not joined UA-cam early, we needed you,
Wow..understood all ,easily the best question of jee
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.
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 .
That's a heartening response coming from a teacher himself . Thank you 🙂🙏
@@PHYSICSSIRJEEsir can you tell me that how have you taken W(spin)sin=wo
Thank you sir for this wonderful video . Lots of doubts and confusions cleared . Plz keep uploading more such videos
The best Solution for this question
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...
Finally aaj smjh aaya😇😇
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)
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) ?
Very informative Sir,
Sir can u also elaborate the soln to us without using IAR? It will give us one more method to think.
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)
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.
@@physicsacademy3858 yes, for only part A , you can👍🙂
Vv v helpful
But what do u mean by ' cult' in thumbnail??
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 🙂
Really enjoyed sir throughout this solution.
one word......beautiful problem
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) .
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) ?????
Sir other than this book any other book is there which is frequently available about that concept (which you mentioned like resnick, Krane)
Slightly higher level : Morin
@@PHYSICSSIRJEE thanks sir
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
Thank you so much sir i understood it completely...
Glory to Janardan Sir, ♾️♾️
Thanks sir really good ❤
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.
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 👍
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
loved the explanation
Explained it very clearly sir tq
superb explanation ..easy to understand ..thank you sir
superb explanationation sir
Sir while analyzing the orbital motion, are we considering the system as two rigid discs or are we taking them as point masses?
Rigid discs
it
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.
thankyou sir .. concepts are fully cleared now :))
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
Yeh...same mine doubt...i have confirmed it with nitin sachan sir...there is a mistake
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
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
@@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
@@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 physicssirjee@gmail.com (we can share pen-paper working like you said) , so that it's easier to resolve the issue further .
Looking forward towards your email...
@@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
@@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 👍
🔴 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.
Please kindly check the errata in description added at that time along with the pinned comment above . Thank you 🙂
@@PHYSICSSIRJEE Thanks for a quick reply. I withdraw my comment.
@@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 👍🏼
Am not very comfortable with IAOR ... Is there any alternate way to understand the addition of the 2 omegas ?
Without that , I am afraid it would be difficult to solve
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?
Friction and normal force
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.
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
Why angular momentum is not in Omega direction?
Sir you are super sir everyday I learn new things by watching your channel
i found a channel who does same thing for maths! ua-cam.com/video/kmK7jtfosQo/v-deo.html
Sir pls make a video of jee advance problem of 2017 from rotational mechanics
Amazing explaination sir ❤️😍
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?
Did you get the solution of your problem? I too have the same doubt.
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)
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
watching for 5th time and now completely understood thanks sir jee 💓
i found a channel who does same thing for maths! ua-cam.com/video/kmK7jtfosQo/v-deo.html
@@aparnaarora5184 thank u very much
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?
You may try , Resnick halliday Krane for beginners or David Morin for slightly advanced version of your doubt
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
Sir @17:17 wha will be the other angular momentum wrt COM OTHER THAN spin angular momentum wrt COM ?
Very awesomely explained sir
thank you
ok sir
Ok
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..
Its more advance form is in david morin 🙂
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
Sir I think I can say ROCK and ROLL... Is the precursor for this question Just for first option
Yes 🙂
Sir is morin better or Pathfinder better for jee advanced in your view,we follow that..
PATHFINDER objective is a must and problems from build your understanding if you have time ( JEE 2022)
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 ??
No rule like that
@@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)
@@yashagrawal8592 point moving along with position , velocity and acceleration of CM . It is the "CM"
@@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.) ?
@@yashagrawal8592 I think you are confused between angular momentum about a point ( CM ) and about an axis passing through CM
hats off sir.
10:10
W0 can't also be calculated by applying the condition of pure rolling simply.
Tysm my dear sirjee
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?
Sir why resultant angular momentum and resultant angular velocity don't coincided please say sir
In general , it need not . Eg : conical pendulum
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.
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
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
@@PHYSICSSIRJEE thank u sir
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.
Sir ,
Pls tell why Vcm is not equal to aw in case of tilted disc
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
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
Thank you for the help🙏, I will train myself in these shortcuts to become better 👍🙂
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)??
To disprove option D for exam purpose. Same was mentioned in the description below
@@PHYSICSSIRJEE sir I am a bit confused regarding calculation of L(z) , I have also seen the video on this question by Ambarish sir ( ua-cam.com/video/urL3dPwvGCY/v-deo.html ) , 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?
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 .
I have explained in this video at the start ,why taking Omega at an angle theta is incorrect
20:05
Sir when we find Vcm using the centre of mass formula,
Vcm=(m1v1+m2v2)/m1+m2
We get it as 9aw/5 which is same as (9lwsin(theta))/(24)^1/2 but is different from what you have got
Can you please identify the error?
Are you referring to the small calculation error mentioned in the pinned comment and errata in the description?
No Sir
It’s about the velocity of COM
@@theintegratedguy9528 Vcm calculation at left top corner screen of 21:08 is correct. It should be same in all methods of calculation
Vcm=(aw*m+4m*2aw)/5m =9aw/5…
@@theintegratedguy9528 if w in your equation is spin w... then you are using a wrong assumption v = rw for rolling... It's an inclined disc. Please try to understand why we use v=rw for a vertical disc. The real meaning of the expression and not the formula. You will then understand why it's not applicable here for discs which are inclined
Sir w not mere khyl se angle thetha par nhi hoga with vertical
Please suggest me a UA-cam channel for chemistry for JEE advanced
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!
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.
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🙂
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
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
@@amirthya absolutely correct 👍 🙂. That's what I implied in the description . It's exactly how I would have gone ahead
@@amirthya
Excellent dude..... If possible can you please attach the calculations. I would be very gald🙏
Sir is it the same case with jee advance 2021 angular momentum integer type question
great problem sir
Sir can you please suggest standard books for physics
Ncert
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
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
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.
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
@@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?
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
@@PHYSICSSIRJEE Thank you sir,for helping 😊😊.
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 ?
Symmetrical only
@@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).
@@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
@@PHYSICSSIRJEE Ok sir , So calculating angular momentum about Centre of mass (option c ) would be very difficult ?
@@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.
Sir can u bring a viedo by explannig direction of angular momentum iam very much confused
Watch this and the previous videos in the description Completely
Sir I saw but direction angular momentum is not mentioned .....u said to see from resnic haliday
@@Algoneryes, RESNICK, Halliday, Krane . Ok, I understood your issue. Let me plan a video on it in future 👍
@@PHYSICSSIRJEE thank you very much sir🙏🙏🙏🙏🙏
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
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 :)
@@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.
@@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 physicssirjee@gmail.com. It will be helpful to understand your context with the help of a diagram which UA-cam 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 🙏
@@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
Sir o have a doubt that while calculating moment of inertia then why we have not considered rod moment of inertia ml2/12
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)?
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
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. 🙂
I scored +2 in this question by jugaad in my mock test
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
Late to this channel🙂
Sir can you plz solve L about CM and give the correct ans. You just eliminated it by saying its wrong option but then what is the correct ans?
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
Check the errata in the description or in the pinned comment written at that time
@@PHYSICSSIRJEE sir so the entire multiplying factor should be (81)(24)(√24)/125?
@@PHYSICSSIRJEE am I correct ? Pls tell sir
thank u sir
Nice it's so exciting 😊 iam waiting
can we take component of angular velocity like that sir, it seems highly non intuitive and doubtful, like you took for z axis i.e. w sin theta directly
Once you define a physical quantity as vector , it should obey vector laws . Taking components while performing addition is one of them
@@PHYSICSSIRJEE sir then why can't we use v=wr to calculate the velocity of the center of mass , as v=w/5*24a/5=24wa/25, which is not the velocity of the disc????
also sir by your logic if rod is rotating about an vertical axis and it is inclined at an angle theta to the vertical then we can calculate its angular velocity of the rod about its own axis(the axis along the length of the rod) by taking components although it is not even rotating about that axis
i recently opened a thread related to this question on physics forum.com, sharing its link if you could contribute to a solution some people are building on it
www.physicsforums.com/threads/existential-dilemma-on-angular-velocity-of-a-complex-rigid-body.1003792/
@@gourvi6339 very long doubt and not properly worded . Responding to First para of your doubt only : " center of mass of small disc" or "CM of system" ?? can you be very clear whose v did you attempt to calculate ? V= rw is not a standard formula and should be used very carefully . w vector is a derived vector of convenience ( not a fundamental vector like angular momentum) . I request you to go through David Morin text " introduction to classical mechanics" chapter 8 section 8.9 , problems 2( many omegas) , problem- 3(rolling cone) and n sequence with its multiple solutions . Also go through preceding text on understanding omega in that chapter. Once you are done with these , please revisit the first two paragraphs of your doubt above. Then we can discuss further once you have clarify over the diagram I drew for inclined rod rotating.👍🏼🙂
@@gourvi6339 I am talking about this Book , you will find pdf of it on net 👇
scholar.harvard.edu/david-morin/classical-mechanics
@@PHYSICSSIRJEE on the inclined rod rotating, I was commenting that if we take components of angular velocity like you told about in this video along the axis which lies along the length of the rod, then we do get a w but it can be easily seen that the rod is not rotating at all about that axis???