8.01x - Lect 13 - Potential Energy, Derive Simple Harmonic Motion using Energy

[@nailyarakhmetova2003]

Dear Walter Lewin, Your teaching is excellent and your lectures are truly understandable. Being from Kazakhstan 🇰🇿, I find your approach especially valuable. Thank you for making physics so accessible and enjoyable.✨

[@alexeialayo951]

Hello, I am from Cuba. When I was in preparatory I used to compete in physics national challenges, and a dear friend taught me to derive Simple Harmony Motion equations from energy conservation equations. That was amazing, together with the cosine of theta approximations. That was amazing. Many years ago I see it again. Your lectures are amazing. Thanks very much.

[@ゾカリクゾ]

I think I have a great force towards these lectures, enough to accelerate myself against the direction of the force of sleep (It's 3 AM and I'm watching this!!!)

[@spiralgaming8940]

😂😂😂 great

[@kinglogic1729]

Me too!

[@thienthanhtranoan6723]

8.01x Lecture 12 *THANKS TO WALTER LEWIN. WE CAN LEARN PHYSICS FROM THE BEST TEACHER, BUT EVERYTHING IS FREE (I’M GRATEFUL FOR HIM)* *ENJOY!* *SOME MAIN IDEAS AND THE TIMELINE* 0:00 Gravitational Potential +graph/ diagram (focus intension) (Close to the Earth) + 2:49 do not close to the earth The gravitational force always opposite To the increase of Gravitational Potential 7:06 spring force (+plot)=> restoring force 10:26 => we define the force from the potential (you must know the physics Meaning of this equation + some example 15:55 focus intension potential diagram => equilibrium (Stable/unstable) 19:55 simple harmonic motion on a potential well (by a concept of conservation of energy) 25:25 motion on a track (example)- amazing approximation ~small angle approximation (SHO) ~pendulum Walter Lewin: *“POPPELEPEE!”* 38:00 KEY QUESTION: “gravity is really the only force that does work, is that true?” 40:17 mind-boggling demonstrations + experiment (such amazing) 47:00 experiment with a track (huge difference)

[@marcuswada877]

UAU!!! What a lecture!!! Now let's rewatch and take notes because the first time is to be savored. Thanks from Brazil.

[@devinwindham9656]

I am ill and depressed right now, but your lectures are keeping me going!

[@scienceandphilo]

Best wishes to you.

[@ronakrawat70]

What was your age 2 years ago

[@herzikkimolog]

is it have to something to do with rolling motion of the ball? As I remember that ball makes both rotational motion AND translational motion so it has two types of kinetic energy (1/2mv^2 + 1/2Iw^2), unlike the first oscillator which only had kinetic energy from translational motion. Thus, some of the ball's potential energy will be consumed for its rolling and it will lose tangential speed by time

[@Higgs000Boson]

I think so too...

[@carultch]

You are correct. It is that not all of the kinetic energy of the ball is in pure translation, that slows it down. The fact that it rolls gives it another mode of storing its energy (rotational KE), that is directly linked to its speed due to being in pure roll. A frictionless slider would move like a simple pendulum on that track, but a rolling ball needs to have its rotational dynamics considered to predict its period. A solid tungsten ball inside a plastic shell, would be a way to get it much closer to translational-only mechanics, by concentrating as much of the mass as close to the axis of rotation as you can..

[@AyanHussain05]

I also think that 🤔

[@ernestschoenmakers8181]

@@carultch Yeah agree but there must be static friction for the ball to roll.

[@jaihind6472]

sir you are the only one who shows why approximations work so well.That is why you make us love physics

[@vivekpathak5717]

The understanding and excitement I get, from watching your videos is as good as reading Sir Richard Feynman's book : Lecture on physics .........THANKS professor.

[@lecturesbywalterlewin.they9259]

Thanks Vivek for your kind words

[@webdevacademy3496]

I am also reading the same book currently

[@TheSamuelWells]

Is it because the ball has angular momentum? It takes energy to get the ball rolling and to reverse the roll to go back and forth so the period of oscillation is slower. The air track weight isn't spinning/rolling, so there's no angular momentum.

[@lecturesbywalterlewin.they9259]

+TheSamuelWells Yes that is correct. No need to use the words "angular momentum". The rolling ball has rotational KE (0.5*I*omega^2) + 0.5*mv^2 (here v is the tangential speed aolng the track) . A sliding object has only KE=0.5*mv^2. Therefore the average "speed" along the track will be lower for the rolling ball than for a sliding object.

[@johannsebastianbach3411]

+TheSamuelWells Exactly what I thought yay :D

[@NSBeverything]

thus, larger the diameter of the ball slower will be avg speed on track?

[@vaidikgupta6b967]

So if the circular track was placed on a frictionless track then would we had got the right result ?

[@chakshupunj5943]

I also thought about it and now got the confirmation!!!!!! Yeah!!!! :)

[@neetachoudhary3980]

Thanks to god to give such a wonderful people like u sir. #walter lewin the great. LOVE FROM INDIA🇮🇳🇮🇳 😇😇🌷🌷❤❤

[@dreamjourneys2703]

I love you so much because you made me love physics more and more. I just wanna get reply. It’ll be a great honour for me. Love from Pakistan.

[@sebbyteh9203]

the ball is rolling on the track, it has an additional rotational kinetic energy (0.5 I omega^2), period is higher than normal sliding ball

[@lecturesbywalterlewin.they9259]

correct

[@mylesbishop1240]

Teh Yong Lip it is a very rare thing for Mr. Lewin to say "correct" well done.

[@md65000]

No s**t -- he's obviously not mentioning that because they haven't reached that part of the course yet.

[@parthkatke6706]

I thought the energy is wasted in the noise it's producing 🤦😂

[@paulproofmath323]

@@parthkatke6706 Me too Parth. I thought part of the energy is wasted as sound.

[@piyushbhardwaj9950]

Could Someone Please explain this? at 34:00 I did not understand the taking of derivative of theta wrt time, and how did the chain rule come into iT?

[@dimal7439]

Let say v=dθ/dt so we have v^2 and we want the derivative so d(v^2)/dt = (d(v^2)/dv)*dv/dt which is 2*v*dv/dt so is 2*(dθ/dt)*(d^2θ/dt^2) for the time derivative of θ^2 you will get 2*θ*(dθ/dt) I didn't understand your first question

[@obayev]

Big thank you for such a great explanations and demonstrations which give me incredible insight!

[@lecturesbywalterlewin.they9259]

My pleasure!

[@sreelathareddymanda4318]

It’s because in second case the ball also has rotational kinetic energy.time period increases by root of 7/5 times

[@RAHULaphysicsfan]

can u explain it more please....

[@RAHULaphysicsfan]

but T=2pi ×roit under l/g

[@manjusingh4620]

Yup.... U r right the difference is becouse of the rotational kinetic energy of the ball

[@Dr10Jeeps]

The master at work!

[@debasishdutta9661]

Sir today on this 5th of September we are celebrating Teachers' day in India. Wish you a very happy teachers' day.

[@lecturesbywalterlewin.they9259]

thank u kindly

[@kathandesai5896]

5:13 when we go A to B' , why gravity is doing positive work i.e. in the opposite direction ? Isn't it should be in our direction and therefore, we have to work less?

[@sudarshann7194]

For last question. "The normal force is doing work ". He already gave a big clue in middle of lecture itself. I love his lectures more than any MOOCs available on internet. Sir plz reply me if you see this . ill be very happy😃

[@sudarshann7194]

danm me , i was wrong!!!!!!

[@vaanivijay6552]

So the ball is taking more time because normal force is doing work on it?

[@airgunner2177]

This ball while rolling down is spinned up. Rolling up - spinning down. Non-zero mass to consume an energy.

[@LearnWithFardin]

50:50 how about the rotation of that ball? It must have some rotational kinetic energy supplied by the initial potential energy! 🤔🤔🤔🤔 🇧🇩

[@arvildasgupta5938]

by the way I am sitting in my university and since no classes are going on I am utilising the time by watching your video

[@b.b.k.k.5578]

could you please provide me the coefficient of kinetic friction of the material on which the solid spherical ball is oscillating?

[@lecturesbywalterlewin.they9259]

I do not know. My guess is 0.1

[@yatindersingh_]

To those looking for the answer to 51:00, The case 2 ball rolls, which means simple KE for translational motion isn't enough and the ME equation changes to ME = KE(rolling) + KE(tranlational) + PE. This breaks the SHM approximation we did earlier.

[@lecturesbywalterlewin.they9259]

yup

[@thomasnoonan2039]

Is it possible to represent the S.H.O. equation at 35:12 in terms of x instead of theta? If you did, how could you?

[@lecturesbywalterlewin.they9259]

sure, try it

[@vijayr1485]

I've seen that graph at 5:31 drawn the same from earth radius to infinity, but differently from 0 to earth radius. Is it equally true to draw a line with a negative slope from 0 until it meets the 1/r part of the curve?

[@lecturesbywalterlewin.they9259]

Near Earth mgh works for large distances you have no choice you MUST choose the 1/r relation. Choose the signs any way you want to AS LONG AS the gravitational potential INCREASES is LARGER fir B then for A if B is farther away from the Earth surface than A.

[@mehmetalivat]

50:38 answer; difference is rolling kinetic energy

[@mortezaoskuei9738]

one more question, in mintue3:00 you say we can put u=0 where you pleased, but it is nonintuitive a little, I think if we put u=0 on the top of a mountain, it could not defend against your fall, could it? so can we put u=0 everywhere we pleased?

[@lecturesbywalterlewin.they9259]

question unclear. when een apple falls from a tree, at the moment that it starts falling its speed is zero.

[@mortezaoskuei9738]

@@lecturesbywalterlewin.they9259 on top of a building, we can put u=0? so why an apple will fall if we put u=0 on top of a tree? how can we put energy minus? what is the meaning of minus for energy as a scalar variable?

[@VickysTuition]

Result at 24:01 was already derived at 11:51 when potential energy was differentiated for force

[@michael73931]

Professor, at around 23:00 and 34:00, what was your intuition for making the variables go on the L.H.S and then taking the time derivative? It looks like magic to me.

[@lecturesbywalterlewin.they9259]

23:00 Total energy = KE + PE 34:00 ME is mechanical energy = KE + PE If any of that is magic for you I advise you to start 8.01 with lecture 1 and work you way up from there.

[@sagarrathore5400]

IT'S THE AMPLITUDE! The amplitude of the larger arced SHO is much smaller as compared to the amplitude of the ball on the smaller arc in the second experiment. I hope I'm right. It was fun joining this lecture professor. Luckily I showered a couple hours ago :)

[@carultch]

It's not the amplitude. It takes at least a 22 degree amplitude of a pendulum to achieve a 1% error in the period. Hyperphysics has a large amplitude pendulum period calculator, that implements the infinite series to more terms to calculate this. It's the fact that the ball rolls, while the air track slider is in pure translation, that explains it. The kinetic energy of the ball is more than just 1/2*m*v^2. It is 1/2*m*v^2 + 1/2*I*omega^2, where omega = v/r, r is the ball's radius, and I is the moment of inertia of the ball about its central axis. Put it all together, and you end up with KE_total = 1/2*v^2 *(m + I/r^2). Apply this equation to the procedure of finding the period of oscillation from this lecture and get the following. Note that big R is the track radius: T = 2*pi*sqrt(((m + I/r^2)*R) / (m*g)) Set I = K*m*r^2, where K is the factor in front of m*r^2 in moment of inertia, and get: T = 2*pi*sqrt((1+K)*R / g) Set K=0 for the case of pure translation, and get: T = 2*pi*sqrt(R / g) Set K = 2/5 for the case of a solid uniform sphere and get: T = 2*pi*sqrt(7/5 * R/g) With the given data, we predict T=2.2 seconds, for the ball rolling, which is consistent with the experiment.

[@affandanish2594]

@@carultch when you asked your dad for fork on dinner.

[@mt22956]

​@@carultch that was great explanation .

[@SoumilSahu]

Here's my guess: Is it because the formula you derived doesn't take rotational kinetic energy into account?(It worked in the first case because there wasn't any rolling) Also, some energy was lost in the form of sound.

[@lecturesbywalterlewin.they9259]

YES! during the trip back and force, PE is being used for translation and rotation. Thus the ave speed along the track is slowed down by the rotational KE. But at the extremes when the ball stands still it should be back at the same height above the track (assuming insignificant friction and air drag).

[@priyanshugupta7904]

Soumil Sahu u r wrong

[@akhileshsharma4977]

@@lecturesbywalterlewin.they9259 You are right

[@aliciaisabelcastrorejon4090]

​@@lecturesbywalterlewin.they9259😮😮

[@ゾカリクゾ]

34:21 don't try to outsmart him buddy...

[@lecturesbywalterlewin.they9259]

:)

[@sflm9277]

@@lecturesbywalterlewin.they9259 You *did* point at the exponent of R when calculating the derivative, not at the exponent of theta... that may have triggered the student's reaction.

[@bornPhysics-t5z]

To achieve the correct result, you would need to incorporate the sphere's moment of inertia.

[@sweetysingh7113]

at 2:08 you said gravity is doing positive work (when move object from A-B') sir please tell me why ?i know question is trivial and sorry for this.But i really want to know the answer because gravity always attract objects then how it is doing positive work

[@lecturesbywalterlewin.they9259]

+sweety singh When I lift an object over distance h, I have to do + work +mgh because my force and my displacement are in the same direction. The gravitational force is down. Thus gravity is doing negative work as F_grav and the displacement are in opposite direction.

[@sweetysingh7113]

+Lectures by Walter Lewin. They will make you ♥ Physics. thank you sir i get the point

[@dr.fourier9943]

For the rolling ball, T = 2*π*sqrt(7R/5g). Plug the numbers in! Proff. Walter Lewin is right. PHYSICS WORK.

[@paulproofmath323]

How did you get that formula, please?

[@dimal7439]

@@paulproofmath323 When you do the calculations the kinetic energy for a rolling body is not (m*v^2)/2 but you have to add to this the kinetic energy from the rotation which is (I*w^2)/2 so the kinetic energy is KE= (m*v^2)/2+(I*w^2)/2 and this will give the difference

[@pravitsacademy3673]

It is going slower because we haven't inculcated the rotational energy of the ball whereas the block does not exhibit any rotation.

[@CM-Gram]

Hello Mr Walter, i hope your doing very well! i have been commenting on your videos for 4 months now, i dont know if you remember me but i sure do remember you :) i followed your lectures in 8.01 last semester and i achieved a very good grade , "A" actually and i thank you for that :) my course was pretty aligned with your lectures , how ever this semester i cant really get a grip on which of your lectures should i watch this semester we are studying simple harmonic motion and waves, like waves in a string attached to a wall. where can i find these lectures at your channel? by the way the lectures of this semester are based on are chapter 14 15 16 in Serway reference

[@lecturesbywalterlewin.they9259]

GREAT!

[@Jason-ke4jf]

Look for 8.03x.

[@animals42life8]

At 30:00 why there's only linear kinetic energy and gravitational potential energy? What about angular kinetic energy? Thanks in advanced.

[@lecturesbywalterlewin.they9259]

good question. Do this problem kzbin.info/www/bejne/eqq5p2qNr9x7bKM

[@yash29210]

At 29:48, you say: "independent of theta"........I understand that gravity is conservative force so it does not depend on which path do we choose......so what did you mean by "independent of theta"?

[@lecturesbywalterlewin.they9259]

I listened to it a few times. What I meant is that the eq (conservation of energy) always holds regardless of the value of theta (release angle - see picture on black board). Of course, the larger theta (at release) the larger is the total energy. E_tot= PE + KE (and it is conserved) which is always true for conservative fields regardless of how large E_tot is thus in our case regardless of the value of theta. If a non-conservative force like friction is involved, then E_tot is not conserved.

[@pabloastoreca8726]

what's the direction of the force in the plot (10:20 20:20) when x=0? Couldn't it exist there?

[@lecturesbywalterlewin.they9259]

SHM, when x=0 the force is zero. F=-kx.

[@aakashchakraborty3673]

A very odd result pop out professor because in harmonic motion how can the velocity can be zero as u have considers it while deriving the equation of harmonic motion from mechanical energy ????????????

[@ayush_study_circle]

Sir what could be the ways to minimize Damping in the motion of ball ??

[@prakharbhalla9461]

at 8:42 sir you wrote FWL = +kx but since we are doing dot product of f and dx which are at zero degree angle to each other so it would be magnitude of f * magnitude of dx only. so there is no meaning of plus sign with force. is it right?

[@lecturesbywalterlewin.they9259]

I watched from 8:30 - 8:50. I cannot add to the clarity. Maybe you should watch it again.

[@ismailhaggag1859]

8:50 why the work equal 1/2*k*x^2 not just k*x^2 since work between the two points will equal the force*distance (force=kx , distance=x) , Thank u a lot best physics prof ever Walter Lewin

[@lecturesbywalterlewin.they9259]

MATH! integral of kxdx

[@ismailhaggag1859]

why integral is needed in this case Thanks for ur time prof

[@lecturesbywalterlewin.they9259]

watch my lectures in which I cover "energy and work" this in great detail.

[@ismailhaggag1859]

Lectures by Walter Lewin. They will make you ♥ Physics. Alright professor thank u for ur attention

[@vineethkumar.v8647]

@@ismailhaggag1859 why integral needed.................simple force is changing with the position x. if force is const: , then work is area under the rectangle of side x and that const force but here the force change linearly with x and we have to cut that rectangle diagonally gives us a right angled triangle of area 0.5bh (force *x).hope you understand ...no big deal....suggest you 3blue 1brown channel ...invent calculus yourself.....

[@shaktikashyap6]

What is the reason of difference in observed value of Time Period of last demonstration? Small angle approximation?

[@lecturesbywalterlewin.they9259]

In one of my Bi-weekly Physics Problems I do this propblem - try to find it

[@backyard282]

It's amazing how the problem of sliding across a track is the same as that of a pendulum. I wouldn't think of it at first.

[@unknownguy8979]

For which combination of students were you teaching sir ??

[@exactempire7854]

Sir, I don't understand the last expirement why does the ball go slower if not because friction?

[@lecturesbywalterlewin.they9259]

friction in pure rolling condition is negligible.

[@exactempire7854]

@@lecturesbywalterlewin.they9259 so why then does the ball slow down?

[@subbiahmarimurugan1039]

1:01... how the gravitational force (F=mg ) is negative but gravitational potential energy (U=mgh ) is positive? Then how does the the force and potential energy of prof lewin differs?pls reply to my doubt,,,,..

[@lecturesbywalterlewin.they9259]

watch my lectures I cover this in detail

[@esmealabnoof5278]

.Hello,I am from Sudan, because in swinging pendulum suspend with wire and in rolling ball not suspend.

[@lecturesbywalterlewin.they9259]

explanation incomplete. Be more precise.

[@lecturesbywalterlewin.they9259]

what is your age?

[@esmealabnoof5278]

I am 38 years from day I born until now,but I am about 4 to 5 years in physics . I have BSC IN PHYSICS AND MATHEMATICS,but in our country's it means nothing.I have story if you dont mind ([e-mail])

[@yashwanthsreech5170]

Sir the last brain teaser that u gave is because the angular amplitude is more in second case I think anyway it’s an honor learning from ur lectures I am a student from Sri chaitanya vijayawada thx for ur videos. That are kept on KZbin

[@carultch]

It has to do with the fact that the ball is not in pure translation, but is also rolling. There is more to its kinetic energy than just 1/2*m*v^^2.

[@khandkertiashazad6858]

Dear prof, at 43:33, shouldn't your reaction time stay the same (+-0.1) regardless of whether you are pushing the timer after 3 oscillations or just 1?

[@andrei-lucianserb1771]

Khandker Tiash Azad the reaction time stays the same, but the error goes down. This is because the first error is the error on 1 oscillation, caused by the pressing of the button at the beginning of the single oscillation and at the end of the same single oscillation. In the case of 3 oscillations the button is pressed at the beginning of the first and at the end of the third, so there is no error at the end of the first, start of the second, end of the second and start of the third.

[@sanyam340]

Sir, at 7:30 - 8:20 you hold the object with equal and opposite force of the spring force. Sir I do not understand that than how the motion would be possible. Hoping for your early response sir.

[@lecturesbywalterlewin.they9259]

I watched and cannot add to the clarity of my lecture - sorry

[@sanyam340]

But you can explain sir.

[@lecturesbywalterlewin.they9259]

I cannot explain it any better than I did in my lecture.

[@sanyam340]

Also sir, what is the direction of spring force when the object is returning for its maximum amplitude position to zero amplitude position.

[@lecturesbywalterlewin.they9259]

Fspring=-kx Thus if x is positive Fspring is in the negative x direction if x is negative Fspring is in the positive x direction.

[@surendrakverma555]

Excellent lecture Sir. Thanks and Regards 🙏🙏🙏🙏🙏🙏🙏🙏

[@lecturesbywalterlewin.they9259]

Most welcome

[@sainte5]

studying for my exam with these videos!Hope I succeed!

[@giannisr.7733]

Are there any courses about movements with objects with mass that is constantly changing? For example water droplets gaining mass with a constant rate as they are falling, generally dm/dt≠0

[@lecturesbywalterlewin.they9259]

watch my lecture on rockets

[@giannisr.7733]

@@lecturesbywalterlewin.they9259 thanks a lot, you truly are a lifesaver

[@mortezaoskuei9738]

so after all, what will be the behavior of gravitational force behind the R? on the other hand, inside the earth, near the earth nuclei, what will be experienced?

[@lecturesbywalterlewin.they9259]

both questions are unclear.

[@mortezaoskuei9738]

@@lecturesbywalterlewin.they9259 what is the potential energy inside the earth? in the radiuses under the R?

[@lecturesbywalterlewin.they9259]

@@mortezaoskuei9738 it is one of my bi-weekly physics problems. Try it

[@mortezaoskuei9738]

@@lecturesbywalterlewin.they9259 thank you for writing Sir Walter

[@sheriefatalla7391]

Prof Walter, I think that the fundamental difference between the first and second demo is; the first one was almost straight but the second one was curved. so gravitation force in the second demo is opposing the ball every time when it goes higher and direct it to the lowest point"stable eqilibrium".thnx

[@lecturesbywalterlewin.they9259]

that explanation is ncorrect

[@sheriefatalla7391]

Prof Walter please give us a note since all explanations on that question are not correct.thnx

[@sheriefatalla7391]

Air drag!

[@lecturesbywalterlewin.they9259]

NO not air drag. I leave it up to you to find the solution. It is VERY basic.

[@sheriefatalla7391]

is it something with (h)?

[@adsalah7254]

there is some thing not clear . in curve 4:18 you said that the EP will decreases when the R decreases suppose we arrive at R=0 the center of the earth logicaly we must have EP=0 similar as point x=0 in the spring when EP = 0 but the EP in R=0 still have a value even it's very small and it is not logical the curve must show the Zero in R=0 because if we have a value of EP at R=0 that mean there is work done from no thing

[@lecturesbywalterlewin.they9259]

You should watch this again. The curve I drew is NEGATIVE and it only holds for r>R. R is the radius of the Earth. Notice I erased the curve for r

[@lecturesbywalterlewin.they9259]

Using the concept that PE is zero at infinity, we can calculate the PE at the center of Earth. It's even LOWER than the PE at the surface for r=R. for r>R the mass is always the Earth mass. But when you go inside the Earth the mass decreases. If the mass density were constant (rho), then the mass within radius r is rho*(4/3)pi*r^3. It's easy now to calculate what the PE is at r=0. Give it a try!!! I decided to do this for you. The work you have to do to move a mass m from the center of the Earth to the surface (assuming a constant mass density rho of 5.7*10^3 kg/m^3) is the integral from r=0 => R of (4/3)m*pi*rho*G*rdr which is (4/6)pi*R^2*rho*m*G which is about 3.27*10^7*m Joules. The PE curve inside the Earth is a parabola. Thus the PE of mass m at the center of the Earth is 3.27*10^7*m LOWER than the PE at the Earth surface as I calculated in class. It is -mMG/R = -6.28*m*10^7 J. Thus at the center of the Earth the PE of mass m is about - 9.5*m*10^7 Joules (and ZERO at infinity). Keep in mind that what counts is NOT the zero point of PE but differences in PE between 2 points. The most sensible thing to do is then to choose zero at infinity.

[@deepshikhasingh4965]

But Sir what would be the physical meaning of having some potential energy at the center of the Earth , as then when we place an object at the center of the Earth it should move in the direction of the force ? Though I do understand that when we are calculating for PE , we are considering the work we have to do against the increasingly lesser gravity , so by the definition of PE we get a value for it at the center of the Earth.

[@rogerfederer1622]

Who is here during the lock down

[@BPHSadayappanAlagappan]

Still in lock down

[@samarthsai9530]

Hello dear Sir, I wanted to ask you that whether or not I need to buy book(s) or these lectures are enough for a good understanding of physics. Thank you for these lectures.

[@lecturesbywalterlewin.they9259]

you decide

[@samarthsai9530]

Lectures by Walter Lewin. They will make you ♥ Physics. Yes sir.I got your point.

[@DM-sl9hp]

Samarth, I would suggest taking notes on the lectures without pausing. Then go back and recopy those notes neatly and make sure you can fill in any details that you didn't have time to write down during the viewing. If you can't, then watch again and if that fails, check a book. Do all that note taking by hand. I know it may sound silly, but it really does help with learning. It also wouldn't hurt to find some sample problems to work out on your own. Good luck. Physics is a fascinating subject.

[@aakashchakraborty3673]

In last experiment some energy is converted into rotational energy and heat energy in friction. Actually I noticed it before u even asked the question and also keep on shouting time period will get larger professor ur experiment will fail 😆😆😆😆😆😆. Hahahaha nice lecture professor really love u and ur lectures

[@farhananjum932]

Sir where can I get your 'For the love of Physics' book?

[@danielmelchert7945]

Why is the period of the rolling ball larger than predicted? -> My guess: Because the diameter of the ball must be added to the length of the virtual string. The ball touches the slide only at one point at the bottom, so this point becomes the ball's center of mass and that makes the effective length of the virtual pendulum.

[@lecturesbywalterlewin.they9259]

+Daniel Melchert Daniel, your explanation is not correct. Try again.

[@ultradudexd]

So why is the period of the ball larger than the prediction? Other than friction and gravity there are no other forces acting on the ball.

[@lecturesbywalterlewin.they9259]

tachyon Tachyon you should be able to figure that out yourself. Get back to me in a week.

[@vasantbarve4651]

The angle of oscillation in first case is smaller compared to second so the assumption cos(theta) = 1-theta^2/2 is not valid. It is a case of pendulum with large angle of oscillation. Period for this case is larger.

[@lecturesbywalterlewin.they9259]

+Vasant Barve You explanation is incorrect.

[@yash29210]

At 4:41, Why the curve didn't exist between r=0 and r=R???

[@lecturesbywalterlewin.they9259]

THINKKKKK

[@yash29210]

Is it because essentially we cannot go r

[@yash29210]

I can understand that mathematically it will work but practically couldn't understand so I thought this.......

[@lecturesbywalterlewin.they9259]

OF COURSE we can below the surface of the Earth. But that's not the reason why -mMg/r (M is the mass of Earth) does not work for r

[@yash29210]

there will be no attractive force left between 'M' and 'm' since the force between 'M' and 'm' is g*M*m/(x^2) where 'x' is the distance between 'M' and 'm' and when we are r

[@soniathongam2927]

Sir, I could watch some of the lecture video of classical mechanics. Please let it available again. 🙏

[@5dsgoyo]

Is the reason that since the rolling ball has a rotational inertia to overcome, it needs some of its potential energy to be converted to rotational energy as well as the kinetic energy to move back and forth

[@lecturesbywalterlewin.they9259]

martin robledo Yes of course. A sliding object on a frictionless track with the radius of my demo would have a period that I "predicted". But the ball is NOT sliding, it's rolling. It has rotational KE and that will slow down the translational motion.

[@somnathpal7444]

+Lectures by Walter Lewin. They will make you ♥ Physics. it means that friction is taking part to make the ball to rotate

[@lecturesbywalterlewin.they9259]

the friction is doing no work - the ball is not sliding - it's pure roll.

[@rupanshuremember4541]

what are the topics in mathematics you recommend are must for a physicist........Any book you recommend in maths and physics both....oh it goes without saying feynman lectures and of course for the love of physics....thanks

[@ReDDeViLW1nS]

If the friction is insignificant, why does the ball gets so much lower after 10 oscillations ?

[@lecturesbywalterlewin.they9259]

friction lowers the amplitude but would have a near negligible effect on the frequency

[@anvarva9377]

Dear Prof. Walter Lewin, Wonderful lectures, glad to see you are replying to all relevant questions. There is a similarity in equations of motion for pendulum and spring. Can we consider motion of a pendulum as a combined motion of two identical perpendicular springs (different Xmax and Ymax) ?

[@anvarva9377]

Ok, in Y direction is it like modulas of SHM ( I mean only in one direction ) ?

[@lecturesbywalterlewin.they9259]

The motion of a mass at the end of a spring is simple harmonic (SH) A*sin(wt). The motion of a pendulum of length L is also SH. For small angles of alpha_max: alpha(t)=alpha_max*sin[sqrt(g/L)*t]. x(t)=L*sin{alpha_max*sin[sqrt(g/L)*t]} and y(t)=L -- L*cos{alpha_max*sin[sqrt(g/L)*t]}. x(t) is to good approximation a SHM (for small alpha_max), but y(t) is not. That is obvious even without my equations. During one complete oscillation v(x) is zero twice but v(y) is zero 4 times.

[@anvarva9377]

Thank you for the detailed explanation. :)

[@anvarva9377]

+Lectures by Walter Lewin. They will make you ♥ Physics. Does that mean y(t) is not SHM at all or SHM with a period half that of x(t) and sin() term with it if x(t) has cos()

[@lecturesbywalterlewin.they9259]

I gave you all the ammunition to answer this now by yourself - do not be lazy

[@rainerwahnsinn3262]

Is generally a force always in the opposite direction of increasing potential energy? And what relation have "conservative forces" with the direction of increasing potential energy?

[@lecturesbywalterlewin.they9259]

1. Yes! F = -- div V 2. F=qE F and E are vectors. In case of a Resistor, the electric potential decreases in the direction of the current I, thus in the direction of E

[@carultch]

F = -grad V is the equation, rather than F= -div V. You can't take a divergence of a scalar function. You can use the del operator on a scalar function and the only operation that becomes is the gradient. The gradient is analogous to scalar multiplication, divergence is analogous to the dot product, and curl is analogous to the cross product. The del operator is a vector of differential operators that then gets applied to the following function. In pure math, we may have a vector field capital F, and a related scalar field lowercase f. When F = grad f, with no negative sign, we call f the potential function of vector field F. This term is inspired by the physics application of a potential function, being the concept of potential energy. In Physics, it becomes necessary to introduce the negative sign in the relationship between potential energy and force, so that positive work is done by the force when potential energy decreases.

[@carultch]

A vector field is conservative, when it is the gradient of a scalar field. If no such scalar field exists, the vector field is non-conservative. You can use the curl operation to check for whether a vector field is conservative, and if it is, there is a series of integrations to do to find that field. You'll find contradictory results in the process of integration if you attempt to calculate the potential function of a non-conservative field. In the forces / potential energy application of vector calculus, the fact that a force is conservative mean it is useful to define the concept of potential energy. It means that work done by the force is independent of path, and the potential energy can be used as a shortcut for evaluating work done by the force.

[@glaucofernandes3270]

There's something with the angular momentum of thhe ball in the last experiment?

[@sebastianrada4107]

Professor I have seen many of your lectures and I have understood a lot of physics thanks to you. Nevertheless there is a mystery that doesn't let me sleep at night and that is Why do you put different things like donuts or bananas in your shirts?

[@namanjain9672]

Professor centripetal acceleration also works on the pendulum so why didn't we take into that consideration and why just tension.

[@lecturesbywalterlewin.they9259]

we take not only tension into account. we use gravity and tension. when the pendulum is at its lowest point, the centripetal force at that moment will be the difference between the tension and mg. If we use the torque=I*alpha (small angles) the centr force is automatically taken into account as it is the Tension - mg.

[@namanjain9672]

@@lecturesbywalterlewin.they9259 Thank you sir

[@newagarwalcomputech2952]

@@namanjain9672 j

[@carultch]

@@namanjain9672Centripetal acceleration is by definition, perpendicular to velocity. As a result, the dot product of the force causing centripetal acceleration and displacement is zero, and thus does zero net work. Only the components of forces tangential to the path (i.e. parallel to velocity) will do work on an object.

[@billchu2]

Given that the period must be longer due to rolling kinetic energy (by my solution by sqrt(7/5)), I assume the fact that the amplitude on the small track is getting smaller must be due to friction?

[@lecturesbywalterlewin.they9259]

+William Chu That is not correct

[@billchu2]

OK. :-) Can you point me in the right direction? I incorporated the rolling kinetic energy into the differential equation you wrote for theta -- assumed rolling without slipping -- and it looks like the same SHO equation except g/R is replaced by 5g/7R (assumes solid sphere). I'm assuming the decaying amplitude over time must be due to some damping. Hints? Btw, these a lectures and help videos are a godsend!

[@lecturesbywalterlewin.they9259]

+William Chu A decay in amplitude always occurs. That's the result of friction. But that has a near zero effect on the period of the oscillation. The question I asked class is why is the period of this rolling object smaller than what I calculated for a sliding object.

[@billchu2]

Yes. Thank you. I understand why the period is longer. That is purely due to having to add kinetic energy of rotation (0.5*I*omega^2) and assuming rolling without slipping. I was curious as to why the amplitude was decaying. This is due to friction as you mentioned above.

[@lecturesbywalterlewin.they9259]

+William Chu amplitude would decay both in sliding object and rolling object. There is always air drag and also some friction in the track.

[@eggonwalterlewinsshirt1071]

Cliff hanger resolved: neglecting rotational KE in the equation

[@aayushkhandelwal4912]

How can some students look sleepy with type of teacher and education system . We never get this type of facility . To see every topics practical.

[@gemeitgoel5104]

What's the mean of angular frequency in oscillation?? Angular frequency - unit - rad sec^(-1) Means particle who is in periodic motion move 1 rad in one second in reference Circle. I'm right or no.

[@gemeitgoel5104]

My question is that SHM can be liner or something more but we use angular frequency for SHM who relates motion to a circular motion. And can we use the term angular frequency for liner Harmonic motion? English is weak.😔

[@lecturesbywalterlewin.they9259]

if an object oscillates in SHM with period T, then the ang freq ω= 2π/T. If T changes in time you can calculate the mean value of ω in a given time interval.

[@lecturesbywalterlewin.they9259]

A 1 D SHM with period T (e.g., an object oscillating on a spring) has an ang freq ω= 2π/T.

[@carultch]

@@gemeitgoel5104 Simple harmonic motion means that you can describe the oscillation as a sinusoidal function of time (i.e. sine or cosine function). The input to the sine function expects radians as the unit. Angular frequency is the parameter that translates the time into corresponding units of radians, each representing a two-pi'th of a cycle. The complete cycle is 2*pi radians. The angular frequency doesn't physically manifest into any angles you can measure on a protractor in the real situation it describes. It is a way to translate time into the abstract angular domain, that the sine function expects as an input. The general equation for a sine wave is: y = A*sin(omega*x - phi) + y0 A is amplitude omega is the angular frequency, equal to 2*pi* the full cycle frequency. phi is the phase offset, that specifies how many radians through the sine wave the cycle starts y0 is the equilibrium position

[@bidishachakraborty7723]

Is it because as theta decreases wrt time implies omega decreases as a result time period increases n this happens because the mg cos theta component increases wrt time if we consider each event at the start of each oscillation as discrete event n the value of amplitude is changing at the starting of each cycle

[@lecturesbywalterlewin.they9259]

Bidisha Chakraborty No this is not correct. What is the difference between a swinging pendulum and this rolling ball?

[@vijayr1485]

Is it a difficult job to include rotational kinetic energy and calculate a new equation for the SHO. It would add an extra term in the final solution, correct?

[@lecturesbywalterlewin.they9259]

Rot KE can be taken into account, but we didn't. That's why the period is so large.

[@vijayr1485]

Thank you sir!

[@vijayr1485]

I calculated the period with Rot KE: quite a fun exercise: SHO is: (theta double dot) + (mgR / (mR^2 + I))(theta) = 0, therefore, T = 2pi*sqrt((mR^2 + I) / mgR). For the spherical ball case, moment of inertia of a sphere = 2/5 * MR^2: Period = 2pi*sqrt(7R/5g). Both are independent of the mass as expected. This gives us a prediction period of approx. 2.19 seconds +- 0.07 seconds. Experimental period = 2.27 += 0.1 seconds Physics Works!

[@lecturesbywalterlewin.they9259]

GREAT thanks - good agreement!

[@vijayr1485]

:)

[@abdulazizfaisal607]

though the expression for sliding case is not valid for the case of rolling so the experimental result doesnot agree with the calculated value, but what is the reason for continuous reduction in amplitude of the ball?? by the conservation of mechanical energy a ball which roll down from some given height does go back to d same provided there is not a loss of M Energy, which in the demonstration is surely too low to explain the large amplitude reduction.

[@lecturesbywalterlewin.they9259]

reduction of amplitude is due to friction and air drag.

[@abdulazizfaisal607]

upto my knowledge friction due to ramp will support and cause the rotation instead of dissipation. and air drag may not be huge enough to cause such fast reduction of motion. plz comment

[@lecturesbywalterlewin.they9259]

The motion of a rolling object can be close to pure roll but never for 100%.The frictional force acting on the rolling (partially sliding) steel ball will remove energy. Air drag alone cannot explain the reduction in amplitude of the oscillations. Every rolling steel object on a perfectly horizontal plane will come to a halt. For the same reason.

[@shdowdragon3]

Hey, Are their any lectures to do with light here? e.g Refraction, Snell's law.

[@lecturesbywalterlewin.they9259]

+Shadow Yes there are. Look under 8.02x

[@carl6167]

The derivation of Energy approach to the pendulum is way more precise than the one using Newton's Second Law, since you are using a degree two approximation. Right ?

[@lecturesbywalterlewin.they9259]

*No not right.* Any method you use will lead to a SHM ONLY for small angles.

[@lecturesbywalterlewin.they9259]

In Lect 13 I could have derived the SHM period of a *sliding* object on the circular track in less than 5 minutes by using torque=I*alpha. However, at that time I had not introduced torques yet. Thus I decided to go the route of conservation of energy. It took 10?15? minutes? but it was useful in that the students saw for the first time that SHM can always be approached that way.

[@carl6167]

Lectures by Walter Lewin. They will make you ♥ Physics Ooops, by taking the derivative you get back to degree 1 for the sinus (sin(x)=x). I apologize. Thank you very much, professor.

[@kashishvidhate4490]

It might be the air drag in the second condition .. it might also be in the first one but the shape of the first object would have handled it and reduced it .. (due to its less surface area in motions direction)... The second was a sphere so more area was exposed while motion and air would have made it lil slow ... Plz clear if I'm wrong .. 😅

[@dimal7439]

You are wrong, first the air drag it will be very small so it can't be it, the reason is that the sphere rotates so it has kinetic energy due to its rotation K=(I*ω^2)/2 which you have to add in the initial relation and if you do it, you will find that the T is 2,19 instead of 1,85 which is very close with the experimental one

[@NYKYADU]

sir picture a block kept on a plate ,plate is on the floor (friction only btw block and plate) block is attached to the ceiling via an elastic string (string is vertical at the moment) now the plate is moved horizontally till the block starts sliding to calculate work done by friction I know I can use change in mechanical energy=work done by friction now can I calculate it as integral of F*ds where both I can write it down as a function of angle formed by string with the vertical? I know it's a big problem but if you could just help me thank you

[@lecturesbywalterlewin.they9259]

make first a free-body diagram. Keep an eye on magnitudes and directions. Work done by a GIVEN force is always the integral of F dot dl. DOT product!

[@NYKYADU]

Lectures by Walter Lewin. They will make you ♥ Physics. sir is there any way I can send you the picture of the problem so I can ask you more clear questions?.. please

[@lecturesbywalterlewin.they9259]

people send me pictures on this channel. They create a website with the picture.

[@NYKYADU]

Lectures by Walter Lewin. They will make you ♥ Physics. 😀😀😀 anyways thank you sir

[@NYKYADU]

Lectures by Walter Lewin. They will make you ♥ Physics. tinypic.com/r/106ezpx/9 question number 1.141

[@TheNC100]

Professor, I have a doubt. Since you estimate your reaction time to be 0.1 seconds, you write it as the uncertainty, because you admit you may be off by 0.1 seconds at any measurement of time. Well, the doubt is the following: to measure the period of the oscillation shouldn't you consider an uncertainty of 0.1 second both at the start and at the end of the timing, resulting in a total uncertainty of 0.2 seconds? Thanks for the attention

[@lecturesbywalterlewin.they9259]

my *total reaction time including start and stop* was 0.1 sec in 1999. In 2011 it was closer to 0.2 sec. Thus that's the uncertainty in my measurements.

[@TheNC100]

Lectures by Walter Lewin. They will make you ♥ Physics. Thank you professor

[@sweetysingh7113]

sir one more question you have written at 4:05 gravitational force = -GMm/r^2 and U = -GMm/r since both have negative sign then why you said that force is always opposite to gravitational PE. SORRY SIR I M taking your precious time

[@lecturesbywalterlewin.they9259]

+sweety singh U = -GMm/r is defined this way as I explain. Gravitational force = -GMm/r^2 Here you have a choice. The force on one of the two masses is -- then the force on the other is +.

[@trilokimaurya1479]

Sir please tell why there is a huge error in the end?? u didn't tell i'm not getting the answer

[@lecturesbywalterlewin.they9259]

kzbin.info/www/bejne/eqq5p2qNr9x7bKM

[@kinglogic1729]

You should have used Lagrangian mechanics

[@gemeitgoel5104]

Please one lecture on damped and forced oscillation and ressonance.

[@lecturesbywalterlewin.they9259]

they are already on this channel under Playlist 8.03

[@jimmyday656]

Imagine being in a rocket and hearing the good doctor exclaim "What Happened".....pucker time.

[@nerd3685]

Why does the force always have to oppose the direction of increasing potential energy?

[@lecturesbywalterlewin.they9259]

climb the stairs. Your PE increases, but the gravitational force on you is in the opposite direction. watch my lectures or use google

[@nerd3685]

Lectures by Walter Lewin. They will make you ♥ Physics. Thanks, Sir..

[@MrAyangan]

How do we define metastable equilibrium and saddle points?

[@lecturesbywalterlewin.they9259]

+Ayan Gangopadhyay I suggest you Google "metastable equilibrium" and also "saddle points".

[@OlayRivera]

the BIG difference is to use an arc of only 20cm in the calculations. nobody checked that it would start 20cm from the equilibrium point but the object actually started way more than 40cm from the equilibrium point.

[@lecturesbywalterlewin.they9259]

No that's not the reason

[@OlayRivera]

but just by the look of it the object was not started at 20cm from rhe equilibrium point.

[@lecturesbywalterlewin.they9259]

that's not the reason for the larger period. It's WAYYYY more fundamental.

[@judicatoraldaris7604]

So professor what is that difference exactly

[@lecturesbywalterlewin.they9259]

calculate it!!!

[@arvildasgupta5938]

sir we could have written g as GM/r(square) then it would have become a function of r(square) then we will not get a st.line or would we consider g as constant sorry sir but could you please help me to clear this concept .

[@lecturesbywalterlewin.they9259]

question unclear - how many minutes into the lecture?

[@arvildasgupta5938]

i am extreeeemely sorry sir extreeeemely sorry , i never thought that i will ask you such a silly question please please please please forgive me i never thought i will make such a silly question please sir sorry

[@spiralgaming8940]

@@arvildasgupta5938 no question is a silly question