Solution Problem #4 Pendulum Difficult

[@bios546]

AWESOME HAT! AWESOME JACKET!! AWESOME PERSON!!! AWESOME AWESOME AWESOME!!!!

[@Usrbinry4n]

Wow! This was a fun problem, I goofed up in a few places. Thank you for posting the solution and I hope you continue to post problems.

[@bartubmaiden]

Yay! got it right! :):):) Mr. Lewin, you are the most inspiring physics teacher hands down. Your never-ending enthusiasm not only makes your lectures and problems extra-extra-ordinary, but also adds up to the enthusiasm of the students themselves! It's a privilege to live in our era (the era of technology) and be able to access such content as yours from the other ends of the world. Physics works indeed!

[@lecturesbywalterlewin.they9259]

smile

[@debjyotibanerjee6382]

You have a great sense of fashion sir... Not only learning from you but also seeing you gives us great joy and hope...

[@lecturesbywalterlewin.they9259]

:)

[@aeroscience9834]

I think it's much easier to calculate the torque by saying it's the total weight Mg acting through the center of mass which is L/2 from the end and thus L/2-x from P. Thus T=(L/2-x)Mgb. Much quicker.

[@icystrangers5482]

+Bob Le Blanc That kind of proves that you were told in high school, not taught. If you had been taught, you would surely also be aware of the method used in the video. It is conceptually simpler.

[@lecturesbywalterlewin.they9259]

Aeroscience It's Perfect to use ONLY MG at the CM.In my opinion it's more transparent to show students TWO torques (one CW and one CCW) instead of one torque, but one is PERFECT! of course.

[@lecturesbywalterlewin.they9259]

Bob Le Blanc - It's Perfect to use ONLY MG at the CM.In my opinion it's more transparent to show students TWO torques (one CW and one CCW) instead of one torque, but one is PERFECT! of course.

[@vishaltailor03]

sir, that was very very amazing explanation. I never seen such precious description. salute to your effort. you are truly innovative institution.🙏🙏🙏

[@lecturesbywalterlewin.they9259]

:)

[@thales7951]

I really enjoyed solving this problem. If you could post more "tough" problems like this, I would really enjoy it. Thanks professor!

[@AdityaSingh-ks3zc]

Prof. Lewin, the admiration to be like you and to teach with empirical facts is a simply desire that i have ...

[@lecturesbywalterlewin.they9259]

>>>>> I have asked people to define why a thing sinks in water, why a thing floats or remains somewhere between and of course the answer is density and buoyancy. when I ask about the same thing in the atmosphere people always say gravity.>>> Helium filled balloons rise in the Earth atmosphere for the VERY same reason that wood rises in water becos of *density and buoyancy*. There is no difference.

[@esa062]

That is very interesting. You wouldn't expect such a seemingly simple system to be counterintuitive. I was pretty sure that my assumption about the interval increasing with X all the way would be correct. But that's why math beats intuition.

[@debayanchandra2700]

Sir, this time you remind me my grad days....we had a experiment on Kater's pendulum. Where we have to find the CM of an uneven weighted rod by calculating the time periods w.r.t. both ends of the rod. Being a lazy fellow I find out a simple method.... I used to suspended the rod horizontally about my 2 fingers to understand where might be the possible CM. Thank you sir for this video....it explains a lot what I did unknowingly. Take care of your health. 🙏

[@bobmarley9905]

could u post a link to the problem?

[@trialen]

A most interesting problem. Please can we have more questions like this.

[@ganymedkallisto5561]

Hello, to get the torque that is acting on the pendulum you dont need to devide the pendulum into two peaces. The gravitational force still acts at the center of mass on the rod. And you named the distance between point P and the center of mass "d". Then the torque of course is simply: M*g*sin(b)*d. Or in terms of x, since d = L/2 - x: M*g*sin(b)*(L/2 - x). Or after the small angle approximation and factoring out 1/2 you get: M*g/2 *(L - 2x)*b which is the same than in your solution. It's that easy. Best Regards

[@odysseashlap]

Thanks for a mind blowing problem. To say the truth i thought it would be even harder by listening to your scary advice to quit if we didnt solve the previous one but it wasnt the case. However i especially enjoyed when you gave an entirely new dimension to the problem by examining all the possible T for each x/L . Thanks for this chance to go our minds one step further and next time i promise i will be more eager to explore all the nice details of the problem even if they are not part of the question.

[@meeharbin4205]

Back to your lectures it is

[@KcMsPik7278]

Sir, I have no words to express my gratitude towards you and your inexplicable devotion to work! I mean seriously waking up for the whole damn night just to teach us at this age! I think I can not control my tears, sir! Just like others, you have always been my motivator as always. But I can proudly say now you have always been and now being an octogenarian you still are my super motivator towards everything. Hats off! :)

[@lecturesbywalterlewin.they9259]

thanks for your very kind words

[@kankagoswami4267]

Respected sir, The problem was Wonderfully presented! I am very sorry that I missed the correct solution by making a small mistake. I took the counterclockwise torque incorrectly. Instead of (L-x)/2 mistakenly wrote (L/2 - x) :-P So landed miles apart from the correct answer. "Recheck, recheck and again recheck your calculation" this was the takeaway today. Thank you, professor, your methodology of explaining and validation is unique and unfortunately less practiced! The way you teach defines you as not only a great teacher but also a great artist! I awe each and every neat presentation you deliver with style (in classrooms and also on the other videos) and Want to become a teacher like you! May you have a healthy life, sir. Thank you.

[@lecturesbywalterlewin.they9259]

thank for your kind words

[@AcaPele007]

Excellent lecturer always make every solution looks easy. Thank you very much for this explanation, Sir.

[@zenapsgas]

Thank you! Understood everything, and I like your way of presenting.

[@kushagramisra4429]

Feeling happy! found my faults, THANK YOU sir.

[@25Aditya25]

Today was IIT Jee Advanced 2017.

[@manishprasad1298]

How was the paper ?

[@manishprasad1298]

It is going to be my turn next year 😀

[@25Aditya25]

I didnt wrote the exam but saw a few solution videos.The difficulty seems to be medium

[@25Aditya25]

Best of luck.Unlike me may you get in☺

[@manishprasad1298]

Well Thank you 😊

[@assadij]

To stabilize the image, you can use KZbin's video editor. It has a very nice feature that works very well. Thanks for the video :)

[@lecturesbywalterlewin.they9259]

how do I use it?

[@assadij]

I'll try to explain as simple as possible, but may not be very straight forward, like your lectures :-) 1- Click on "My Channel" on the left menu 2- Click on "Videos" tab under the channel art picture 3- Click on the video that you wish to edit 4- Under the video, click on the wound which is the "Enhancements Icon" 5- Click on the "Stabilize" button on top-right 6- to see the effect, you can drag the vertical line in the middle of the screen to left to see more 7- click the "Save" button on the top-right 8- Click the "publish" button on the right side of the video image

[@assadij]

I'll put a video together for you. That is much easier I think. Because there are too many steps.

[@lecturesbywalterlewin.they9259]

that would be cool! Thank you KEEP INMIND, your instructions will have to be "idiot-proof" If you skip one or if one is ambiguous then I will be lost.

[@lecturesbywalterlewin.they9259]

I followed your video instructions when I click "save" It is written on the thumbnail "edit in progress" but that NEVER changes. Thus I am stuck. I cannot get past that point.

[@alwysrite]

very interesting. thanks for sharing your knowledge to the world even after retiring from MIT : ) and you are still young at 81 !!

[@lecturesbywalterlewin.they9259]

:)

[@SamuelLopez-ql9ki]

Great video, nicely explained! I understand it better now!

[@rutvijgholap8647]

dear prof. thank you for sharing this problem with us. looking forward for new questions cheers!!

[@akshat-kharbanda]

i hope we will still be friends

[@rohitsanjay1]

For crying out loud, I was telling you that it was x=L/3 and you kept telling me that it was incorrect. Amazing.

[@lecturesbywalterlewin.they9259]

That is NOT true Your first attempt must have been WRONG thus I wrote INCORRECT If your 2nd (or 3rd of 4th) attempt was correct I did not post it. I NEVER told you that it was wrong. This must have happened to many many viewers.

[@padraiggluck2980]

At 12:58 you say the clockwise torque is positive and the counterclockwise torque is negative. It’s actually the other way around. I believe that the sign difference in the final differential equation affects the period T. Also, you meant to divide out the M in the torque when you did the calculation for T at about 15:30.

[@onlyfun7765]

i don't know that the rod will have two torque. i just put the torque from the center of mass. i thought, the weight of the rod Is just 'appear' in the center of mass (if i consider the rod has homogeneous mass). but suprisingly, my answer Is correct (and the equation period at the point P Is also same). thank you, Professor

[@Dan474834]

Walter, do you listen to classical music? I'm just curious.

[@lecturesbywalterlewin.they9259]

:)

[@manassrivastava7448]

Sir thanks for posting the solution. You always keep me curious by your lectures and these problems. Sir today is really a very special day As you posted this solution this day and today is also My birthday. Thank You Once Again For The Solution

[@lecturesbywalterlewin.they9259]

Heppeee BD

[@manassrivastava7448]

Thank You

[@Noulis92]

there is no oscillation when x=L/2. In order to have one you need to have an equilibrium point and a force/torque of restoration. Every single position of the stick is an equilibrium point (Στ=0) so there is no way that you can have a torque of restoration (it's direction is always towards the equilibrium point). But you also need something for the oscillation to begin. As I said, you can't have a swing width and a single initial velocity will make tha stick to rotate, in which case we can talk about an other period :)

[@lecturesbywalterlewin.they9259]

rotational freq about the center of mass is not a resonance freq. ANY omega will do. There is NO equilibrium.

[@KevinS47]

I absolutely LOVE how he is still teaching Physics ! Very very nice! There is nothing that I love more than Physics in my life, but I still have a long way to go before considering myself to be an actual Physicist. I just have one non related question: in one of Lewin's lectures of Electromagnetism, where he derived Maxwell's equations, I was wondering if that is part of the master degree programme at MIT, or is considered to be simple enough to be included in the bachelor ? Thanks again for constantly uploading !!!

[@lecturesbywalterlewin.they9259]

Maxwell's eqs cannot be derived! The course is for freshman at MIT. This is their first encounter at MIT with E&M.

[@neevparikh685]

Lectures by Walter Lewin. They will make you ♥ Physics. why can't Maxwell's equations be derived?

[@lecturesbywalterlewin.they9259]

Newton's 4 Laws cannot be derived either. Coulomb's law can also not be derived.

[@KevinS47]

Lectures by Walter Lewin. They will make you ♥ Physics. Again thanks so much for answering (I had already ask you the same on the actual video, and already got an answer) but the reason for which Maxwell's equations cannot be derived i don't know. I am currently studying the basics of classic Physics at IB (which is an international school) so not yet at University level. I spend quite a lot of time watching your lectures, but I do not always understand, as my level is just to low for now. Thanks again !

[@neevparikh685]

Interestingly, I am in IB as well, have my 11th finals this week. I, too am curious as to why can't Maxwell's Equations, along with Newton's Laws and Coloumb's Law, cannot be derived.

[@chotirawee]

looking forward for your next problem

[@lecturesbywalterlewin.they9259]

It's a math problem - It's very very very easy. But it's ONLY meant as a warmup for the problem that will follow which is much much more difficult.

[@arvildasgupta5938]

sir you are truly dedicated I really love this thing about you You just dont have any idea how much motivated i feel whenever i see you Its a request whenever you are coming to India do let me know I will surely meet you please sir

[@lecturesbywalterlewin.they9259]

:)

[@arvildasgupta5938]

Lectures by Walter Lewin. They will make you ♥ Physics. Sir you will inform us when you are coming to India right ?

[@raulcastiblancocastiblanco8798]

thanks teacher Walter, it was a very interesting way to study a phisyc pendulum....thanks for share your knowledge, for me is very usefull ( forgive me bad English)

[@paulcfo]

You need to get a chalk board in your house Dr. Lewin :)

[@kallewirsch2263]

I definitly support that. Just the amazing dotted lines would justify that. It is so good to see that Mr. Lewin is well and still has fun in educating.

[@karimsabbah1116]

Hi Dr. Walter, I have a question that is related to this problem and many more similar ones. I know how inertia is calculated for different geometries. For the case of a rigid rod It's (1/12)mL^2 (after integrating)at the centroid and by parallel axis theorem is (1/3)mL^2. Here comes the question. If I only knew the inertia at the end((1/3)mL^2) and I want to know it's inertia at (L/4), where L is the length. I cannot just subtract (L/4)^2*m by parallel axis theorem. The Math actually works out to where I have to subtract 3*(L/4)^2*m. In order to get an equivalent quantity as if I went from the center and just added (L/4)^2*m. I know the math works out and that should be sufficient but I would like some intuition. Thanks for your time.

[@lecturesbywalterlewin.they9259]

Parallel axis theorem is not always very intuitive but it can be proven.

[@j.r.graham6117]

great video very clear

[@lecturesbywalterlewin.they9259]

:)

[@shengwencheng8040]

Great

[@HenkJanBakker]

Question as to a matter of 'physics protocol' in general with this solution as an example; Does plotting the answer count as proof (provided the basis is correct of course)? In this case for there being (no more than) 2 points? Or is it just visual aid showing why? I can see reasons for both being valid but I bet there is

[@lecturesbywalterlewin.they9259]

Of course plots are OK if they accurately represent the physics (thus math). But they will at best tell you that x= approx L/3.

[@Hobbit183]

I tried using the parallel axis theorem to find the moment of inertia for the new axis ( I=(ml^2)/12 +m(l/2-x)^2) then i used it in the period equation for a physical pendulum and set the periods equal so that after diving by 2 pi and squareing both sides: ( (ml^2)/12 +m(l/2-x)^2)/(mg(l/2-x))=2l/3g And simplified: x(x-l/3)=0 It was fun

[@lecturesbywalterlewin.they9259]

:)

[@nickjaramillo6321]

you're an amazing human being.. I wish you would cone teach physics at my University, Midwestern State. I had the pleasure of meeting Dr. Michio Kaku, i would love to meet you sir.. best regards Nicholas jaramiilo

[@lecturesbywalterlewin.they9259]

If you are in the Cambridge, MA USA area, we can meet.

[@BroadeningHorizonsos]

I LOVE YOUR HAT

[@MilanKarakas]

Sorry Professor, but my math is rusty. On your video with the question I just put answer: "X=1/3L", but I did not make any video, nor any math proof or calculations. Also, I found old video named "The Professor - Center of Percussion" from Professor Julius Sumner Miller - where he simply explains that period of oscillation of the physical pendulum is the same if pivot point is at one end, and if the pivot point is at 1/3 of the length of the rod. So, instead calculating, I just simply wrote the answer, which is probably correct (X=1/3L), but without any proof...

[@lecturesbywalterlewin.they9259]

I asked for x larger than 0 and smaller than L/2. The answer to my problem is x=L/3. Of course you can turn the rod 180 degrees. But that adds no more physics.

[@MilanKarakas]

Yes, I know.... I think that X=L/3 is equal X=1/3L ?

[@lecturesbywalterlewin.they9259]

>>X=1/3L>> you mean (1/3)*L which is not 1/3L

[@MilanKarakas]

Oh... yes... my bad.

[@bobmarley9905]

Mr. Walter Lewin why did you graph x/L vs. T (specifically the x/L part confuses me)?

[@lucav2732]

Thanks for your videos! :)

[@shaktirajdaudra7285]

sir, what would be your next problem? I am excited about it!

[@lecturesbywalterlewin.they9259]

it is embarrassingly simple math problem but it is meant ONLY as an introduction to the next problem after that and that one is much much more difficult.

[@icystrangers5482]

ha ha! > 86400 seconds... reminds me of when you crushed the egg with your hand. I think that you can also calculate T by saying that the entire mass M acts through the centre of mass, which is at a distance of L/2 - x from the pivot point, giving a torque of Mg(L/2 -x) * b

[@lecturesbywalterlewin.they9259]

>>>ha ha! > 86400 seconds... reminds me of when you crushed the egg with your hand.>>> YESSSS there was a parallel

[@odysseashlap]

yes i did exactly the same

[@kanadkrishanusengupta2578]

Sir my ans was correct which I command on your previous video.Sir you didn't reply me.When I watch this I realize my ans was correct.Thank you sir.You are my favorite physics teacher.

[@SoumilSahu]

+Kanad Krishanu Sengupta he didn't point out ANY correct answers, since it's no fun for others if they can see what the right answer it is.

[@lecturesbywalterlewin.they9259]

:)

[@bios546]

the x/L goes upto 1... and there is a point of discontinuity at x/L=0.5 there are two more points... 2L/3 and L where the same result is obtained... so, 4 points exist

[@lecturesbywalterlewin.they9259]

I ONLY asked what x is between 0 and L/2. As that contains all the physics. Of course you can turn the rod over 180 degrees.

[@thales7951]

Hey professor, you already answered that before, but I cant find the answer anymore. What physics books do you recommend to study?

[@hiranmayroy5910]

which artist inspired the design of the hat that you have on your head prof.?Is it from the cubism movement,or the impressionist?

[@lecturesbywalterlewin.they9259]

Piet Mondrian is the inventor of Neo Classicism which changed the world in 1917-1930. I suggest you google him to learn more bout him and about "De Stijl" Movement of which he was the godfather. "De Stijl" also had a great impact on architecture. Google "Gerrit Rietveld". I own 4 pieces of furniture by Rietveld among them the famous red and blue chair. www.google.com/#q=rietveld+red+and+blue+chair&safe=active&tbm=shop Also take a look at www.google.com/search?q=mondrian+neoclassicism&safe=active&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwi0wJDY9YzUAhWCOiYKHYmmC3gQsAQIJg&biw=1200&bih=613 Show less

[@ShashankShekhar-de4ld]

hi Sir hope you are doing great

[@fjeldesc]

Great demostration #sensei #WalterLewin

[@beertje601]

If all teachers where like Walter the world be a very different place , Love the video's even though I only understand 20% of them , but I keep trying :) .

[@mraliamer798]

you are very very great man

[@lecturesbywalterlewin.they9259]

:)

[@mraliamer798]

you make me change my life to study phzics you make me love phzics

[@lecturesbywalterlewin.they9259]

SUPER -- soon you will probably also know how to spell the word "physics".

[@ShashankShekhar-de4ld]

I have a question According to Gauss' law flux is equal to electic field dot area but I think this is definition of rate not flux because flux is defined per unit area per unit time. Rate =flux*Area Please help me out Wish you great day thanks

[@lecturesbywalterlewin.they9259]

use google

[@mathematicalworld7337]

professor, how can I calculate the magnetic field of closed loop of wire in a shape of symmetrical six cornered star and current I flowing through it.

[@lecturesbywalterlewin.they9259]

watch my 8.02 lectures and also some relevant videos in my 8.02 Help Sessions.

[@rohandevil]

Nabaneet Sharma I did it the same way today

[@dadaimiza]

❤️

[@BloomwinxSuper1]

Loving the hat!

[@abhigyansingh890]

A uniform solid cylinder with pure rolling along horizontal plane collide with a plane inclined at some angle with the horizontal so what must be the value of angle for the cylinder to stop immediately after impact? ???

[@lecturesbywalterlewin.they9259]

To stop in 0 seconds?

[@abhigyansingh890]

yes

[@abhigyansingh890]

this problem is troubling me a lot.

[@lecturesbywalterlewin.they9259]

to stop in time 0 is never possible. If the inclined plane is at 90 degrees, the object will stop in a collision times which may be several microsec it may then bounce back. If the collision is elastic it will have lost KE when it bounces back. Any angle

[@abhigyansingh890]

I thought the same but when I checked the answer it was 120 degree. Maybe we can change the question by saying in the least time instead of 0 seconds

[@ernestschoenmakers8181]

Ik heb de pittige slingerbeweging opgelost met de Euler-Lagrange vergelijkingen wat berust op het Principe v.d. kleinste (stationaire) werking. L = T - U ; d/dt(dL/dv) - dL/db = 0 waarbij v = db/dt , L is Lagrange en T - U is kinetische min potentiéle energie en b is de draaihoek.

[@lecturesbywalterlewin.they9259]

dat is prima VERY ADVANCED! Vond je het juiste antwoord? x=L/3

[@ernestschoenmakers8181]

Jazeker want in feite is de E-L vgl een stuk gemakkelijker te gebruiken dan Newton's vgl. voor ingewikkelde vraagstukken.

[@lecturesbywalterlewin.they9259]

Ja dat is juist. De meeste van mijn viewers hebben nog nooit met de E-L vgl gewerkt. Ik heb het daarom niet eens genoemd.

[@ernestschoenmakers8181]

Ja, Natuurkunde is zo interessant en breed, we hebben er in ons leven bewust of onbewust mee te maken. Ik vind het nog altijd frappant dat als ik in een auto zit en ik rem dat de wagen vóór mij van mij af versnelt terwijl ie met constante snelheid rijdt. Dit is relativiteit van beweging en in dit stelsel moeten Newton's wetten aangepast worden omdat ze alleen gelden voor inertiaal bewegende waarnemers. Ik ben van plan om een lerarenopleiding Natuurkunde te gaan volgen in België.

[@lecturesbywalterlewin.they9259]

:)

[@cecilialindagatonaoliveira882]

thanks

[@chaitanyagiri76]

sir after how many days you upload problems and their solutions..?

[@lecturesbywalterlewin.they9259]

I am posting problems about twice per month. I post the solutions typically 5-6 days after I post a problem. I posted the L/3 solution yesterday.

[@chaitanyagiri76]

👍

[@pranjalhazarika7361]

sir,can you post video related to capacitance and resistor problems ....I am facing problem while solving such problems...please help me sir

[@lecturesbywalterlewin.they9259]

use my 8.02 Help Session videos - they re on this channel.

[@pranjalhazarika7361]

thank you ,sir

[@fugacityxdunk6782]

how were you at maths during college?

[@lecturesbywalterlewin.they9259]

this is a well-kept secret

[@baldpolnareff7224]

Yesss! My second attempt was correct :)

[@bios546]

Sir, there is more to that graph...

[@rudrakshsharma6537]

Is this real walter lewin???

[@lecturesbywalterlewin.they9259]

NO this is my twin brother.!

[@rohitbhosle6521]

Rudraksh Sharma lol😆FUNNY

[@thiernosall7779]

I want Professor Walter Lewin to explain the inadequacy of Newton's theory in explaining the advance of the perihelion of the planet Mercury. Thank you Professor and have a good day

[@lecturesbywalterlewin.they9259]

I suggest you ask Einstein - he KNEW (GR!!)

[@thiernosall7779]

This is not bad. Very FUNNY, professor Walter Lewin.

[@thiernosall7779]

I will ask mister Einstein if I see him, of course. FUNNY

[@SourabhBhat]

I laughed hard at the end of the video. :D

[@spacetime3121]

professor i said answer was 2/3 L but u rejected it. [2/3 L is same as 1/3 L from other side of rod ] P.S. ---> I also said either x = 0 or either x = 2L/3

[@lecturesbywalterlewin.they9259]

I asked for the x values between x=0 and L/2 answer: x=0 and x=L/3 sorry

[@spacetime3121]

cmon professor its the same thing. anyways, its ok

[@lecturesbywalterlewin.they9259]

there are 4 locations if we accept x between 0 and L. x=0, x=L/3, x=2L/3 and x=L.

[@tejasgajra2731]

I was right😎

[@ShashankShekhar-de4ld]

gravitational force is always attractive but elctro magnetic force is both attractive and repulsive why

[@lecturesbywalterlewin.they9259]

good question. Use google. gravity is due to GR photons are the carriers of electromagnetism

[@anubhabbiswas7767]

love the hat!!

[@lecturesbywalterlewin.they9259]

:)

[@anubhabbiswas7767]

i got it right!

[@avirani510]

I solved by comparing the given time period by T=2π √I/√mgl I was unable to solve it plzz help

[@AMotoVlogger]

So professor, do you have a conjecture as to why 21% was the slowest period? Perhaps because when kids turn 21 then tend to 'slump..' in their newfound addiction to the bottle? 😂

[@lecturesbywalterlewin.they9259]

of course, that's it

[@sanwer2996]

💝💝💝💝💝💝💝💝💝💝💝💝

[@isidremasmagre2020]

I wonder what could be the misterious secret that you kept so well to find the minimum of a function... xD

[@salehzaidan1220]

Isidre Mas Magre find the derivative of T with respect to x and set it equal to 0. You'll get about 0.21L

[@lecturesbywalterlewin.they9259]

:)

[@isidremasmagre2020]

I knew that but I found funny he couldn't tell

[@hariprasadjayaprakashan9625]

Prof. you are looking like a sailor.. Have a mouth piece and we call u "Popeye"

[@psobocik]

? Why You then describe my amendment 1/3x as incorrect?

[@lecturesbywalterlewin.they9259]

I NEVER DID Your first attempt must have been WRONG thus I wrote INCORRECT If your 2nd was wrong I must have written INCORRECT again, but if your 3rd attempt was correct I did not post it as I do not post correct answers. I NEVER told you that it was wrong. This must have happened to many many viewers.

[@psobocik]

Anyway, thank You for this physics reminder ;) Best Regards!

[@jalaljo2136]

there is a missing "M" in the differential eq… .

[@lecturesbywalterlewin.they9259]

oooops sorry this is called a slip of the pen

[@jalaljo2136]

Lectures by Walter Lewin. They will make you ♥ Physics. thank you Sir for this videos… 🙂

[@lecturesbywalterlewin.they9259]

:)

[@ubantu100]

Yay! I got it right :)

[@howtosport425]

I basically got it right

[@Deepanshu72]

I did it within 35 seconds

[@MrIndomit]

I would never solved it myself

[@rachydamine9458]

I got the answer correct that's why it got deleted a I guess :)

[@lecturesbywalterlewin.they9259]

YES

[@rachydamine9458]

+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for the reply professor

[@benefactor4309]

sir I have a humble request, I have created a very good question in Newtonian mechanics, kindly give ye email address I will mail it to you its a very good problem, kindly have a look and if it's of a good level kindly make a video of its solution, it will be an honour sir.

[@lecturesbywalterlewin.they9259]

send the question to me here on my channel. No one will be able to see it as I will then not make it public

[@benefactor4309]

is the problem worth yr attention

[@lecturesbywalterlewin.they9259]

it's very easy but it is also unrealistic. You will have to state that you start the tangential acc when the object is hanging vertically down with zero speed. Then you apply a constant tangential acc (not realistic) . I have already recorded a problem with similar physics but which is more practical and very realistic.

[@benefactor4309]

sir did you go through the problem is the approach of question correct