Centre Of Mass 11 || Trick For COM of Remaining Part || When Mass is Removed IIT JEE MAIN / NEET ||

[@PhysicsWallah]

Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App. Download the App from Google Playstore ( bit.ly/2SHIPW6 ) Physicswallah Instagram Handle : instagram.com/physicswallah/ Physicswallah Facebook Page: facebook.com/physicswallah Physicswallah Twitter Account : twitter.com/PhysicswallahAP?s=20 Physicswallah App on Google Play Store : bit.ly/2SHIPW6 Physicswallah Website: physicswallahalakhpandey.com/ LAKSHYA Batch(2020-21) Join the Batch on Physicswallah App bit.ly/2SHIPW6 Registration Open!!!! What will you get in the Lakshya Batch? 1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey. 2)Live Classes and recorded Video Lectures (New, different from those on KZbin) 3)PDF Notes of each class. 4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level. 5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture. 6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021) 7)In case you missed a live class, you can see its recording. 8)You can view the videos any number of times. 9)Each chapter will be discussed in detail with all concepts and numericals 10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams. ****Test Series for XI & XII**** We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year. The test series follows very logical sequence of Basic to Advance questions.& Evaluation of Test and Solution to all the questions at the end of the test. 11 chap 7 || System of Particles - Centre of Mass 01 || Introduction Of COM for IIT JEE / NEET || kzbin.info/www/bejne/d5yYpoZpd516oJI 11 chap 7 | Centre of Mass 02 | COM of Continuous Bodies | COM of Semicircular Ring ,Disc,Triangle | kzbin.info/www/bejne/aKGwdaWBgs6Jh5I 11 chap 7 | Centre of Mass 03 | COM of Continuous Bodies | COM of Hollow and Solid Hemisphere , Cone kzbin.info/www/bejne/i4vbYZKkhJd5atU Class 11 chapter 7 System Of Particles | Centre of Mass 04 | Motion of Centre Of Mass IIT JEE / NEET kzbin.info/www/bejne/fpvHqnesnbB7Z7s Class 11 Chapter 7 System Of Particles | Centre Of Mass 05 | Conservation Of Linear Momentum IIT JEE kzbin.info/www/bejne/iGbCpJmse82Wq6s Class 11 Chapter 7 | Centre Of Mass 06 | Conservation of Momentum in Bomb (Shell ) Explosion IIT JEE kzbin.info/www/bejne/sJWzfXSNoqeBb7c Centre Of Mass 07 || Collision Series 01 || Elastic Collisions in 1 -D || IIT JEE MAINS / NEET | kzbin.info/www/bejne/lXfFooytos6Mras Centre Of Mass 08 || Collision Series 02 || Elastic Collision in Two Dimension IIT JEE / NEET || kzbin.info/www/bejne/lZ2VeZKAnZxgr5I Centre Of Mass 09 || Collision Series 03 || Inelastic Collisions IIT JEE / NEET || kzbin.info/www/bejne/bpDFY4hte7hoY8k COM 10 | Collision Series 04 | Coefficient Of Restitution | Elastic and Inelastic Collisions IIT JEE kzbin.info/www/bejne/jpm3f4Vmbq2fjJY Centre Of Mass 11 || Collision Series 05 | Oblique Collision | Elastic Inelastic Collision JEE /NEET kzbin.info/www/bejne/Zl7EqmCue7eIgbM Centre Of Mass 11 || Trick For COM of Remaining Part || When Mass is Removed IIT JEE MAIN / NEET || kzbin.info/www/bejne/Y4urgWCnjaZ5qqM Class 11 chapter 7 | Systems Of Particles and Rotational Motion | Rotational Motion 01: Introduction kzbin.info/www/bejne/rIStZ2l3ea-XaNE Class 11 chapter 7 | Rotatational Motion 02 || Torque - Moment Of Force - Turning Effect Of Force | kzbin.info/www/bejne/jF7YkIWAjplqosU Class 11 chapter 7 | Rotational Motion 03 | Rotational Equilibrium IIT JEE / NEET | Torque Problem | kzbin.info/www/bejne/qXOZi2adp7CrgLM Class 11 chapter 7 || Rotational Motion 04 || Moment Of Inertia - Introduction || kzbin.info/www/bejne/fpSxmHiOqr6Hac0 Rotational Motion 05 | Moment Of Inertia Of Continous Bodies - Rod , Ring ,Disc, Cylinder,Triangle kzbin.info/www/bejne/aJfaYqVqj5uZZtU Rotational Motion 06 || Moment Of Inertia Of Sphere and Cone || MOI of solid Sphere JEE MAINS /NEET kzbin.info/www/bejne/fHLNgn-YoquAgbc Rotational Motion 07 || Perpendicular and Parallel Axis Theorem Moment Of Inertia JEE MAINS / NEET kzbin.info/www/bejne/q4bbfGCOZdijrdE Rotational Motion 08 | Best Numericals of Rotational Motion and Rigid Body Dynamics JEE MAINS /NEET kzbin.info/www/bejne/nYGqgJSBrpybaNE Rotational Motion 09 | Hinge Forces | Rigid Body Dynamics JEE MAINS / NEET | Rotational Numericals kzbin.info/www/bejne/a3POYqSEftlgZ8k Rotational Motion 10 || Kinetic Energy of a Rotating Body | Work Done By Torque IIT JEE MAINS / NEET kzbin.info/www/bejne/n2HLZnpsqMmSq6s Rotational Motion 11 || Angular Momentum IIT JEE MAINS / NEET || Angular Momentum of Rotating Body kzbin.info/www/bejne/qF7PYYVmeNamqpo Rotational Motion 12 || Conservation Of Angular Momentum || Angular Momentum IIT JEE MAINS / NEET kzbin.info/www/bejne/bqbXhnmgg8aleqs Rotational Motion 13 | Rolling Series 01 | Combined Translation + Rotational Motion |IIT JEE / NEET kzbin.info/www/bejne/gKfVcqWcmcSSepI Rotational Motion 14 | Rolling Series 2 |MOI , KE and L expression for Translation + Rotation Motion kzbin.info/www/bejne/iqjOY5qKn62JfKc Rotational Motion 15 | Rolling Series 3 | PURE ROLLING JEE MAINS / NEET | Uniform PURE ROLLING kzbin.info/www/bejne/n4WrYZ1-ltWtqpY Rotational Motion 16 | Rolling Series 4 | Forces in PURE Rolling | Pure Rolling IIT JEE MAINS / NEET kzbin.info/www/bejne/aGSkXmhqn898fbc Rotational Motion 17 | Pure Rolling on Inclined Plane IIT JEE MAINS / NEET | Rolling Series 5 kzbin.info/www/bejne/gqqcgKyDZbSHbtU

[@srinivasuparsa2829]

VASU

[@ritikmishra5709]

Sir last question ka answer ( r/2) h

[@kimhana2406]

@sidharth Verma same ans here.....but is it confirmed that its right?

[@subhradeepsardar786]

@@kimhana2406 I got the same answer

[@Indian_898]

Plz koi btayega moment of area kon sai lecture main hai

[@PhysicsWallah]

Kindly Poll for next chapter in physics and Chemistry starting 21st November. Poll going in community tab of our channel

[@achutanandshukla7238]

Thermodynamics

[@vivekyadavbholu7077]

Shm

[@dgroup5731]

Shm

[@rahulkarepurkar1148]

Goc

[@kushagraupadhayay]

Shm

[@alana_jaykar12426]

Sir, the answer to the HW question is:- -2r/3(4-π) Thank you sir for teaching us❤️

[@prashantundrikar7690]

Sunday ho ya monday Roj dekho Physics Wallah ke naye funde. Salute to you sir☺️☺️😀😀

[@sanjaybhargav6058]

Bete moj kardi

[@meralakshya1050]

Waah

[@asimtiwary991]

Aur Bhai iit se pass out hua yaa nit se? 😂

[@NikolaiNova]

​@@asimtiwary991tu bta😂

[@AjVerma-jj2sn]

👏👏😂

[@nirmalmistry867]

helpful lecture, Last question A¹ = 2(r^2) A² = PI × (r^2)/2 X¹ = -r/2 X² = -4r/3PI C.M. = -2r/(3(4-PI))

[@harleendhillon6760]

👍🏻

[@epikherolol8189]

But will sigma be same for both as one is a rectangle and other is disc??

[@masud_writes]

​@@epikherolol8189 yes. Same material na.

[@sleepymemes2133]

@@epikherolol8189 ermm wht tda sigmaa

[@srijitachakraborty4980]

27:22 The answer will be -2r/3(4-π).

[@maparesshparthasarathi9921]

How?

[@hellsowner8513]

My answer for the last question is coming out to be ... 2r/3(4-pi)... Rectangle COM: (-r/2,0).. Semicircle COM : (-4r/3 pi) ... since they are to the left of origin!

[@payal2905]

I also got same ans:-)

[@rajgupta12a60]

Correct😉😉😉😉😉. Answer. C.O.M. of remaining portion =2r/3(4-Π).

[@Medvibes-bh6oj]

Same answer 😄😄

[@yedgaming6577]

I also got the same :)

[@Chinnu-je7hr]

Whatis the final answer

[@s62soujatyapahari75]

COM of rectangle is at (-r/2,0).....taking origin according to sir and coordinate of semicircular cut portion is (-4r/3π,0)........the area of rectangle A1 = 2r*r=2r^2 ......... and area of semicirclular cut portion A2 = πr^2/2 applying these two area and the the coordinates of x axis in the equation..........Xcom = (A1*x1 - A2*x2 )/A1 -A2........ the answer is coming ...... -2r/3(4-π)

[@Messi-m1k]

Correct hain👍

[@techguru6851]

29 videos it was really useful.. thank you so much sir❤️

[@Zartan-XR]

Bro variable mass system kaha hai...

[@techguru6851]

@@Zartan-XR mtlb

[@techguru6851]

@@Zartan-XR sir ne questions me kr waye h

[@Zartan-XR]

gubahi bahi.. Oblique collision ke baad na variable mass system hai... Ase kuch external force thrust force equal to m dv/dt....ye sir padhaye hai kiya bahi

[@Zartan-XR]

@@techguru6851 [phone] my no.... Bro thoda help chaiye call karna...

[@labakumardebnath9682]

Sir,I am from Tripura. I have no words to appreciate your way of teaching.প্রণাম নেবেন স্যার ।🙏🙏

[@arindombabu9844]

দেবনাথ কেমন আছো

[@SanjayKumar-mu9qo]

I'm from Tripura too , panisagar

[@arindombabu9844]

I'm from bengal

[@docsujan]

Tripura ta ki bangla vasha chole naki

[@docsujan]

Arindom tor wb er kothey bari?

[@AbhijitDas-tt4os]

Sirf padhne se koi student nahi hota and koi padhane se teacher nahi banta...... Teacher kaa real meaning app pe suit karta hein sir... A lots of love from Assam ❤

[@Ray52163]

I am starting watching your videos in lockdown.i have not take admission in any institute. I am dependent only on your videos and mujhe ye kehne mein kuch hestitation ny ho rhi ki aapki di gyi concept har institute ke barabar ya us se bhi achhi hi hogi......love u sir .and thanks to u very very much😊😊😊

[@trickswithhrs]

💯%%

[@t4thakkali993]

Did u crack😌

[@suyashnegi8344]

Bata hua clear kuch??? Bata na himmat hai to??

[@sheero231]

@@suyashnegi8344 😁

[@MUSICWITHMEGH]

@@suyashnegi8344 😆

[@pranav3835]

Sir iska answer (2r / 12-3pie)

[@mamta_jha]

Tumne rectangle ka com kaise nikala

[@rhythmmotwani6710]

Ekdam sahi hai dost I also got that

[@mamta_jha]

@@rhythmmotwani6710 Haan mujhe bhi smjh AA gya baad mein

[@pranav3835]

Thanks bro answer confirm krne ke liye

[@tanmaysukhija4933]

It's -ve bro

[@ShivaniSharma-vs1ys]

Sir you are doing great job for us I respect you with bottom of heart Thank you so much sir 😊😊😊😊

[@AdityaYadav-jd8ly]

27:23 Answer of hw question will be -2r/3(4-π).

[@vikramkhatri6657]

Physics wallah is great. If you agree hit like

[@prajwal6495]

@physicswallah Sir always be happy😁😀😊☺❤ Aapki smile 😊 Dekh ke mere chehre Par smile aa jaati hai We all make our parents and @you proud Thank you so much 👍

[@reshmavesuwala555]

Sir, your methods & logics are superb.. You are the one who is making physics easier & easier for us.. Thank u so much

[@ashutoshsharma6353]

Sir abhi kuch chapter ke kuch topics Bache Hain (1) kinematics-projectile on an inclined plane (2) Newton's laws of motion- weldge constraint (3) center of Mass - spring questions (4) states of matter -ediometry (5) ionic equilibrium- acid base titration

[@aayushjha3243]

Yes Sir Waiting For those Lectures

[@yashwnanaap5106]

Its made

[@ushayadav2387]

Ye complete lecture hai kya iss chap ke

[@Prashantkaim2]

Jindagi tonsabki change hoti h per alakhsir ki to bahut jyada change ho gayi hai 😀😀

[@musicophile1012]

Ans of the HW is 2r/3(4-π)...thank you sir for this good video...pehle to yeh topic mujhe samajh nehi ati bt after watching this video I never forget the whole concepts..🙂

[@ankitgaming9662]

I got same

[@Ruhal123-l9i]

Hlo plz reply

[@mdshoyebakhtar4595]

(-) hoga pehle

[@Spatial-mm6et]

up 2r nahi 8r hoga?

[@Gzrigel]

The answer to the last question is 2r/3(4-pi). Thank You Sir. The entire topic of Centre Of Mass including Conservation of momentum and collision is finally done. I am extremely happy today . Thanks a lot Sir.

[@rishabyadav181]

No 4r/3pi

[@nilmonichatterjee897]

@@rishabyadav181 wrong

[@rishi6587]

@@rishabyadav181 glt hai vai

[@gulzarkanizquadri8205]

Mra answer -4r/3(4-π) arha hai...????

[@raunakbatra848]

​@@gulzarkanizquadri8205no I think answer is -2r/3(4-π)

[@vivekgopakumar4497]

0:53 bahuth pyara he sir🤩🤩🥰

[@sujaychakravarty160]

The answer is coming -2r/3(4-π). Thank you sir for the concept. 😊

[@harry--ajourney...2440]

Sir you are Anand Kumar for us . When I saw super 30 I was imagining you as Anand Kumar sir

[@AnandKumar-dn6ju]

I am ANAND KUMAR😎😎

[@mohitchauhan7669]

😂😂😂

[@srishtiyadav7963]

@@AnandKumar-dn6ju 🤣🤣🤣🤣🤣

[@nitinsandhu7368]

He is super 3 million

[@devendrapratap9463]

@@AnandKumar-dn6ju 😆🤣

[@Nabija.]

A1=2r^2, x1= r/2 ,A2= πr^2/2, x2=4r/3π now calculate🤗🤗 the answer is 2r/3(4-π).

[@ghufranaamber1950]

Yes,I also getting the same

[@levix9886]

Yeah I got the same too

[@nilakanthakar1627]

Thanks bro

[@beenagautam8518]

Minus me aega

[@Noone-xh1ld]

​@@beenagautam8518nahi dono center mass ke cordinate mines me aae ge

[@mohityadav6836]

Area of Rectangle= Lb(in this question 2rXr) and com in exist -r/2 from origin.semi circle area =pia r square/2 and com -4r/3pia after calculating,the answer is coming -2r/3(4-pia),

[@anvigupta2331]

Correct... And in literal terms, it is 0.775r

[@vikas-kumar727]

-ve hoga 2r/3(4-π)

[@aarnachauhan6277]

@@vikas-kumar727it is negative with reference to the origin sir has taken

[@vikas-kumar727]

@@aarnachauhan6277 Ya 👍

[@pralhadsonkamble7634]

thanks bade bhaiya mera galat aa rha tha

[@Jems-uo9ni]

Chahe kitne v naye teachers aa jaye digital bords le aye but alakh sir ki baat hi kuch or h love you sir ❤️

[@TheSpyFardin]

27:21 My answer of the last question!!🇧🇩 The coordinates of the com with respect to the point 'O' the center of the semicircular disk is (-0.776R, 0) Xcom={-2R÷(12-3π)}≈-0.776R Ycom=0 🇧🇩

[@sohamgohil1722]

First, my ans came wrong. Then, after cross-checking with others, i got it correct which is -2r/(3(4-π)). Thank You Alakh Sir. 😀

[@rishi6587]

Negative ?? Vai apne origin right hand side se liya hai ??

[@shutosh2006]

Mera to 4r/(12 - 3π) aa raha hai bro

[@harvirsingh3762]

2r/3(4-pi)

[@LostZoro798]

​@@shutosh2006 same

[@pranjalgupta6292]

-4R/(12-3π)

[@shauryamishra8101]

Answer of the last question is -2r/3(4-π) on putting the value of π =(22/7)we get -7r/9 ... Thank you sir for your great effort towards us.. I am watching this video in February 2022.. ❤️❤️

[@aadibeautyparlour896]

No the answer is -r/3(4-π)

[@sarojthakur4561]

tu top karega well done

[@KHAMA_10]

@@aadibeautyparlour896 -2r/3(4-π) will be the answer....

[@psykon7148]

Once a legend start with this🥺😭op lecture

[@dikshajain3767]

Superb lectures by u sir...sare concepts ek dum clear ho jate h..thanku a lot for supporting us..👏👍

[@thakurjigupta1093]

answer for the last question Xcom is -2r/3(4-pi)

[@r.grecipes8048]

Shi h....mera bhi ye hi h..

[@AmritPalSingh-io5ts]

yep same

[@AmritPalSingh-io5ts]

Guys this is the correct ans

[@Joshishield123]

Right brother

[@devkumar9889]

R/3(4-π) from bottom centre of circle

[@Ssbofficial2005]

Still evergreen and unmatchable🔥🔥❤️‍🔥❤️‍🔥♥️💯

[@AnjaliSKyadav]

Coordinate of COM of rectangular body..(r/2,0) Coordinates of COM of semicircular disc.... (0,4r/3π) Therefore the Ycom of remaining will be 0 and Xcom will be 2r/3(4-π)

[@azmiumar9947]

✓ 26:39 Solution= •Ycom= 0 •Xcom= = {(2r×r)(-r/2) - (πr^2/2)(-4r/3π)} ÷ {(2r^2) - (πr^2/2)} On solving above eq. you get__ = -2r/3(4-π) Then, Position of COM {-2r/3(4-π) , 0}

[@kaichettykalpana3362]

Area is only πr^2 know why πr^2/2

[@VictoR_Daada]

Sir ji Jo aap lectures ki starting mai lecture samajh aane ki surety dete that's build up our confidence.....God bless u 😎

[@infinitus1233]

we have studied this topic in kota but we haven't been told the CONCEPT OF NEGATIVE MASS. , They just told the formula and how to apply and believe me or not knowing this thing helps a lot ❣️😊

[@CRiCket_TV_09]

Bhai tum kota me padhte ho?

[@infinitus1233]

@@CRiCket_TV_09 haa bro

[@CRiCket_TV_09]

Bhai sir ka content kota ki tarah hi hai?

[@infinitus1233]

@@CRiCket_TV_09 dekho bhai , mai apne experience k basis pe bata skta hu baaki yaha pe kayi coaching h aur har coaching me alag batch k alag teachers , so they may differ in their opinion 1) Sir kuch kuch cheezo ka prove dete h jo ki kota me kuch batches me nii diye jaate bcoz those are not relevant 2) sir jitna content dete h lagbhag utna hi hum b padhte h kisi kisi chapter me kbhi thoda zada vgerah b padhaa dete h hm logo ko 3) kota is not just for the study purpose , kota me aane k baad tum bohot kuch seekhte ho bcoz u r physically present there , the classroom experience is entirely different, tum class me padho to aisa feel aata h jaise saari theory , formulae yaad ho gaye h aur agr ghar ja k ek baar revise krlo tb to sach me yaad ho jata h 4) agar online padhna h tb I would not prefer kota bcoz pata nii q bhale hi teachers yaha k excellent h aur hame bohot saari resources provide ki jaati h but still for phy and chem theory , agar koi doubt ho to mai ALAKH SIR ka video dekhta hu (not the new ones but these old videos jisme sir hi padhaate the ) aur maths k liye unacademy k Sameer sir ka video , online me ye log bohot hi sahi padhaate h Agar offline padhna chaahte ho to kota se better jgh I don't think kahi milegi

[@DivyanandSharma0809]

I study in prayagraj in regional coaching and this concept of negative mass was told there 😊

[@jaikirankumari8339]

Bhagwaan aapko hamesha khush rkhe sir, kyuki aap hai to humlog hai sir ! Namste guru g! 🙏🙏🙏🙏🙏

[@mohitrawat7762]

Last question answer is 2r/3(4-π) those who are not getting it take com of rectangle as -r/2 and com of semicircle -4r/3π

[@harshulsingh5876]

Yeah I did

[@pikachugamer2652]

bro plz help me for last question how to solve

[@pikachugamer2652]

bro i will got and thanks for give clue in above -r/2

[@ketan6182]

It's - 2r/3(4- pi) , just a minor correction from his answer. If you took centre of the semi circle as origin.

[@shyantanroychowdhury8899]

I did a calulation mistake and then getting frustrated that why isn't my answer matching xd

[@lonerashu6152]

-2r/3(4-pie) . Ans of h/w q

[@Surviving17]

Answer for last question is : -2r /3(4-π) ..thank you legend for this wonderful explanation 🥺❤️🥰

[@shivanipriya3418]

Same

[@crispo2346]

Same same didi🙂🙂🙂🙂

[@crispo2346]

Same same didi🙂🙂🙂

[@babybear9998]

@@harshitcharanagrawal4570 mera bhi yahi aa raha

[@harshitcharanagrawal4570]

@@babybear9998 are wo main galat solve kar rha tha, ab sahi answer aa gya hai

[@akshatthakur7402]

answer -2r/3(4-π)

[@alana_jaykar12426]

Sir you have won the heart of many people and there blessings are always with you..

[@afridajinnat5408]

Sir Physics Wallah hamare liye bahut pyara hai.

[@aenameena961]

The correct answer is.... - 2r/ 12 -3π ( -ve coz .. The COM of both rectangular nd semi circular disc lies on (-ve X-Axis ).

[@basudevmallick7658]

only -r ki -2r Check karke dekho toh?? 😅

[@aenameena961]

It's -2r .

[@sahil_j6918]

--2r nahi -r hai

[@mokaifansari2018]

PLZ FULL SOLUTION

[@a_man_jaiswal]

_0.8r for exact

[@anjalipramodtembhare7769]

00:51 bohot pyara hai sir 🤩🔥

[@homephone711]

COM = 2R/ 3(4-π) ; R Thank u sir for clearing our concepts 🙏🙏

[@shutosh2006]

Mera to 4r/(12 - 3π) aa raha hai bro

[@anshtayal1992]

@@shutosh2006 beta galat h ji aapka

[@mishra.__.shivamm]

Answer of the last question or hw question is -2r/[3(4-π)] ❤️❤️❤️❤️

[@sawanghosh5756]

Is there any similar concept to calculate the COM of remaining portions in 3-D bodies? So that we can easily find out hollow spheres or cones etc??

[@gr8gamer_]

I think it's formula should be V1x1-V2x2/V1-V2

[@TathagataMondal-w8y]

-2r/(12-3pi), thank you so much sir, my concepts of rotational motion are clear now

[@bireshprasad5166]

the answer to the homework question is: 2r/3(4-pie)

[@mr.buttowski3340]

Explain krdo bro

[@azmiumar9947]

Wrong

[@azmiumar9947]

✓ 26:39 Solution= •Ycom= 0 •Xcom= = {(2r×r)(-r/2) - (πr^2/2)(-4r/3π)} ÷ {(2r^2) - (πr^2/2)} On solving above eq. you get__ = -2r/3(4-π) Then, Position of COM {-2r/3(4-π) , 0}

[@rajshekhar308]

Today is 26 Nov, After 6 days the video will be uploaded 5 years ago, We love old videos not the new videos (I also don't know why I do not see the new videos ) I thik that ....I in this video sir's hard work is clearly seen (t-shirt is sweating ) and there are no ads and..even sir by themselves say that if you know the concept than do not waste your time and sir are looking very simple and sobour . Thanku soooooo much sir 🥺🥺🖤🖤 Wish you a happy new year buddies bb byee

[@akashsharma-pk3fk]

Great teacher, big respect for you

[@justsathwik941]

my teacher exolained in a complicated way which i didnt understand.. HATS OFF for this man.. now im able to understand it very easily

[@biswarupmajumdar]

Last question A¹ = 2(r^2) A² = PI × (r^2)/2 X¹ = -r/2 X² = -4r/3PI C.M. = -2r/(3(4-PI))

[@akshayyadav6713]

X^2= R - 4R/3pie

[@rohanaf1914]

Why u take radius r by 2 ? Of semicircle

[@ubkofficial9117]

Ans of hw question after putting value of pi is 0.77r or 2r/3(4-π)

[@ECE_Pratyush]

Answer of last question- 2r/3(4-pi)....sir U R great....please sir make chemistry videos for all topics...only U can make....

[@anshtayal1992]

correct

[@anshtayal1992]

I mean mera bhi same aa rha h

[@anshtayal1992]

vaise sir ne padhaya h to sahi hi hoga

[@dharmendrakumarverma3839]

Sir ans for last question is ___ *2r/3(4-π)* ...

[@rkswain7115]

Sir u r the best . What a Great teacher!!!

[@satyanarayaniitkgp3484]

Best physics Channel on KZbin

[@sarijitsaren2128]

Those who are unable to solve-- Take (-r/2) as the COM of rectangle and (-4r/3π) as the COM of semicircle from origin.

[@manoharpatel763]

Thank

[@chetanagaikawad1345]

Thanks

[@SKY16022]

🤔🤔🤔

[@SKY16022]

Thanks..

[@ashutoshpaliwal8517]

Tu usse tere answer kya aaya h bo bta

[@priyanka...464]

Thank you so much sir 😊🙌

[@J_M_V0704]

We want a movie upon ur concept sir... PHYSICS WALLAH - WHERE JOY AND KNOWLEDGE MEETS

[@youthubesrts]

Aa gai

[@youthubesrts]

15 dec ko release v ho gai

[@theultimatecoder3433]

and it happened-

[@neetilamba2671]

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[@RahulKumar-st9ys]

HW question : 2r / 3(4- π)

[@snehabrahamane8109]

My answer is also same

[@RahulKumar-st9ys]

@@snehabrahamane8109 congrats

[@PhysicsloverPW]

Explain me

[@RahulKumar-st9ys]

@@PhysicsloverPW see bro , Use the formula - A¹r¹-A²r² / A¹-A² Here A¹ = Area of Rectangle= L×B = r X 2r = 2r² A²= Area of semicircle = πr² /2 r¹ = Distance of COM of reactangle from our choosen origin .( As for a rectangular plane , com will be at its center ) - distance from the choosen origin = r/2 r²= distance of Com of semicircular disc from chosen origin ( for a semicircle the com from center is 4r/ 3π on its perpendicular axis , as derived in the previous videos ) Putting all the values in the formula , and calculating carefully , you will get the answer as 2r/ [ 3(4-π)]

[@PhysicsloverPW]

@@RahulKumar-st9ys thanks ☺️

[@arunjakhel5877]

Thank you so much sir . Aaj tak kisi ne itna acha explanation nhi kiya.....

[@eshapaul5692]

#Homework -2r/3(4-π)

[@satyamtripathi7736]

👍👍

[@chinu2322]

Finally I did it

[@lakshyaagrawal6679]

Best understanding...sir..thank you..for your kind effort...

[@sanjana7223]

Bahut bahut bahut...... Pyara😘😘😘 Channel hai sir physics wallah

[@Sahil-gb1jh]

Sir answer of homework question is -2r / (3)(4-π)

[@vanshikabhagat7480]

Could you please tell me how because I am getting 4 in place of 2

[@Chetan-fv1mo]

I also get 4 in place of 2

[@panu07-Thala]

Alakh Sir supremacy ❤

[@omrajput9112]

Thank you sir for this lecture I was not able to solve questions Now I am confident to solve this kind of questions. You are god gifted 🙏🙏🙏🙏

[@ajaysharma2399]

Ans of h.w......is...2r/3(4-pie)...

[@pranavbadhe5318]

Sir the HW Answer is :--- 2r/3(4-pie)

[@GOAT1906gfg]

Esa padhate he sir ki maan na ho tab bhi admi dekh hi leta he❤

[@wajidsaeed4771]

Answer to hw question : 2r/8-pi÷3(4-pi),0

[@abhijitshirwal]

Sir 14:44, even I got (-9,0)☺

[@abhijitshirwal]

Thoda tough tha

[@bindiyajain8672]

The answer for the last question is Ycom = -2r/3(4-π) or = 2r/3(π-4) To remove negative sign took minus out from denominator and cancel

[@milan7130]

COM of Rectangle from the given axis ( r/2,0) Value rkho [ 2r/3(4-π) , 0 ]

[@dhirajrai8471]

Congratulation 3M subscriber.... 🎊 🎊

[@henajabeen5608]

Timestamps Explanation and formula for COM of Remaining Portion with example of Disc - 1:00 Q1 - 7:15 Q2 - 12:58 Q3 of Annular Semi Circular Disc - 19:14 HW Qn - 25:36 Answer of Homework Qn is (-2r)/[3(4-pi)]

[@mahak2858]

Same answer 🙌

[@ishankhadge1558]

The answer is (-7/9)r. Put the value of pi as 22/7.

[@malingamohit3870]

Sir My Dream to meet you one day....... You'r teaching way sooo easy nd smart........thanks you vry much😘😘😘

[@unnikrishnanthaliyil9590]

Thanks......... sir.........🙏 Answer =2r/3(4-pie).

[@anuragsinha6780]

The remaining area can be cla culated by subtracting the area of semi circular portion from rectangular portion, ⟹remaining area = area of rectangle - area of semi circular portion, now,let us define σ= mass per unit area = 2r 2 M ​ , where M= mass of rectangle of area 2r 2 mass of semicircular portion = σ×πr 2 /2 = 4 Mπ ​ now centre of mass of remaining portion = m 1 ​ −m 2 ​ m 1 ​ r 1 ​ −m 2 ​ r 2 ​ ​ , where m 1 ​ = mass of rectangle and m 2 ​ = mass of semi circular portion r 1 ​ and r 2 ​ are position of centre of mass of rectangle and semi circular portion with respect to O r 1 ​ =r/2 and r 2 ​ = 3π 4r ​ centre of mass of remaining portion= M−Mπ/4 Mr/2−(Mπ/4)×(4r/3π) ​ = 3(4−π) 2r ​

[@yalagapurisathwickrao5981]

Hi

[@arpitwasnik]

How did you wrote that

[@PradipDas-vv4qt]

Excellent Guru of Modern Science

[@karanveersinghdeora5085]

Sir growing some beard then talking about "chaku " might fear some but he is a legend with a sword in his hand fighting for us against physics

[@hemanandadas1010]

dfcgbhjnmk,

[@harikrishna6021]

The correct answer is 4r /12-3pie and also area of rectangular is 2r square and

[@yogeshkumar778]

Answer of the last question: {-2r/ [3(4-π)] }

[@subhradeepsardar786]

Yeah I got the same answer....

[@souvikmondal5276]

Yeah I got it

[@navneetmishra5533]

Same answer

[@rudrapratapsingh1187]

👍🏻

[@aishwaryapatel7393]

Rectangle ka COM ka lena hai??

[@fatmazohra7956]

answer of homework question is negative of 2r/3{4-pi}

[@Travelwithraj5]

H.w ka Answer ke hai=-4r/(12-3r)

[@Santosh_843_]

Finally l completed all the classes in this series 😅😂

[@vikramkhatri6657]

You are Einstein for us

[@hrishikeshbharadwaj6807]

Last question Answer 👉 position of COM is (-2r/12-3pie ,0)

[@ASAlan-sh8kf]

It's wrong if the value is in minus than com is outside the rectangle .I got the value in positive

[@sensei680]

​@@ASAlan-sh8kf 😑😑 see where is origin

[@easthercules]

Thanks a lot sir I got stucked in this kind kind of problem but now i can easily solve them

[@sneharathore4454]

Thankyou sir ❤apne tough se tough concept and questions ko ek dum easy😎 (halwa) bana diya 🔥🔥🔥 ab kabhi bhi c.o.m and collision ke questions ko dekh dar nhi lagega 🤙🤙🤙 again thank you❤ very very much sir. Ans of the H. W question is (-2r/12-3π) ✌✌✌

[@ADVERSE04]

This video was utterly required sir thank you

[@ramprakashgupta4498]

Answer is 2r/3(4-π)