*8.01x Lect 12* *Conclusion* *“It’s a difficult lectures, but it’s so amazing”* *Have fun!* 0:09 *drag force / resistive forces* Equation (viscous term C1>Temperature/ pressure term C2, 4:30 the terminal velocity *2 regimes* 6:30 *critical velocity* 1.Vvcrit 9:40 *conclusion* 11:00 syrup experiment + interesting diagram *important - must-watched* See more here: ruclips.net/video/K389Fw2U5sk/видео.html ??? *Key question* 1.How long does it take for the terminal speed to reached? In the syrup) 28:25 in air *Assignment 4* just do it 35:00 Experiment with a big ball +graph (pebble) ??? *Key Question* 2. how the air drag influence the trajectory (assignment)? 3. Last question 48:50 *8.01x Lect 12* *Thanks to our teacher, Walter Lewin.* *Physics is so Beautiful.*
Brilliant lecture. Things are getting harder. Needed more than three hours to have everything figured out. or so I believe/hope. The pause button, ah, such a great thing!
Another masterpiece of a lecture! Beautiful approach using just c1 and c2 instead of the more difficult to understand viscosity for the linear dependence on v and friction coefficient for the square dependence on v of the drag. Love it!
My guess is that it takes a shorter time to go from O to P than the one it takes to go from P to S. The reason is considering only the y-component , the ball starts with speed V0y up and decelerates with g + something (more than g) until it reaches 0, while on the way back it accelerates with g - something (less than g). Besides the speed might even stop at a terminal value, so if V0y is big enough, it's clear that the average speed from O to P is bigger than the one from P to S.
I would argue that, although the average deceleration between O and P is somewhere above g and the average acceleration between P and S is definitely somewhere below g, the answer is not as clear-cut as it seems, since dVop is also larger than dVps. So it may well be possible for Top to be smaller/larger/equal to Tps depending on the values of C2 and Vy0. Please correct me if i am wrong!
@@mejente1 if there is no air drag than we get a symmetric parabola and the times are identical: so no dependence of Vy0 on the time difference between going up and down.
I believe they will be the same. The vertical component, which is the only part that matters vis á vis time, is entirely controlled by gravity once the object has its initial vertical upward velocity. There are no outside VERTICAL forces of any kind except for air drag, which at any given velocity will be the same going UP as it is at the same point going DOWN.
@@jwills8606 You're not entirely wrong... but it is gravity (accelleration) not velocity that is constant for both up and down directions. As Tarik has said above this accelleration (or decelleration) is acted on asymmetrically, causing the difference in time taken. You are right that the resistive force is proportional to velocity, but the velocity is determined by the accelleration (and time) and so it is in turn affected by the (asymmetrical) resistive force.
Goodness, Prof. Lewin is fantastic. He explains topics so elaborately yet so easy to understand. Me, a eleven year able to understand what he says is nothing but proof that Prof. Lewin is really good at explaining Physics
It's a pretty interesting feeling to watching your aging over the course of lecture series. The education that I've got from your lectures is not only bounded with physics sir.
After hours of suffering with my attempts to calculate air drag, I remembered to check your lectures. This easily solved my problem. As a high school student, I really appreciate that there are quality lectures online to learn from. Thank you
Sir, u are the finest creation of GOD because you are someone who inspires millions all over the world and creating an impact, as it is said that "SUCCESS" is all about making a difference and you have done it in millions of student's life. Richard Feynman sir is Richard Feynman but walter lewin is Walter lewin. You have brought very big change in the we think physics as just a subject that is compulsary if we take science not only in me but millions of people all over the world. We may not get a chance to meet you because by that u may be alive or not but yes till I die u will always be within our hearts. Thank you, PIYUSH CHOUDHARY. LOVE FROM INDIA🇮🇳🇮🇳🇮🇳.
Thank you very much! Great lesson and it is so brilliant to have an experiment for each topic ! This physics exam I have tomorrow in Computer Engineering is really been tough and altogether very boring, until I have discovered your channel prof. Lewin and I started really to understand what I was up to while demonstrating a thing or solving a problem
7:30 how we get to know that regime 1 is where when velcity v is much much less than critical velocity causes viscous term dominating over the pressure term? sir pls reply...
Navneet Mishra Look, at the pressure term the velocity is squared. If vvcrit, the v squared will be extremely high, so the pressure term will be high in comparison with the viscous term. I hope I helped
Awesome!! Finally i can SEE THROUGH THR EQUATIONS because this topic was just a big amount of unclear equations for me thank you professor walter lewin 🙂
You are amazing teacher Dr. Lewin. Thank you for these lectures they helped me a lot. I am a new Physics teacher and I would like to tell me how you prepare the lecture and thank you in advance.
Sir, I have some trouble understanding why it takes less time for the object too travel from O to P, than from P to S, Because when the object is traveling from P to S, it is accelerating due too gravity and the pathlength is shorter, wich makes me think the object encounters less air molecules, which reduces the airdrag. So what am i not getting here ?
Hi professor. At 7:30, you drew conclusions about dominant terms in the drag force expression for a sphere (regime 1 & 2). I'm not entirely sure how you drew those conclusions, I had to take ratios of term 1 & 2 to convince myself of it, why wasnt there more elaborate math to prove the approximations? Was it intendent to be obvious?
no it is not obvious. I decided to give the constants and leave it with that. My graduate student (Dave Pooley) used these constants and I show his results.
Hello sir. My old math teacher showed you in my class I thought you were really awesome. Now I am planning of watching all of your lectures. I do have a question on this one. What is C1 and C2 specifically and how do you calculate them. I’ve watched the lecture over and over again but can’t wrap my head around it.
I would be extremely grateful for a specific path -> of topics -> to study -> in order to eventually understand this lecture easier, where each subsequent topic builds on the one before it. Thank you.
Hello professor. I have started watching your lectures with parallel reading from Ohanian's book (2nd edition - 1989). In the pdf file you suggest reading pages 142 to 146 and page 190 from the book, the topics of which seems to be unrelated to the topic of the lecture.
Hello :) If we have a vertical shot towards the ground and the initial velocity is not 0 and we take into account the air resistance, could the terminal velociry be less than the initial or is it always greater?
I think it will take less time to go from O to P than to go from P to S...........because from P to S the deacceleration is greater than the acceleration from P to S...... Is it correct?
answer is correct - reasoning a bit strange >>>>the deacceleration is greater than the acceleration from P to S>>>> Correct reasoning: on the way up both gravity and air drag forces are downward, on the way down the net force down is the grav force minus the airdrag. Thus the slow down in speed in vertical direction is smaller on the way down than on the way up. Perhaps this is what you meant.
Sir, if I may ask, what exactly is the phenomenon at 19:27 called, which gives rise to the air bubble at the surface? I ask because I was having difficulty searching for it and finding any pertaining formulae.
physics is not about equations, physics is about concepts. Math is the language of Physics. To answer your question. it's the result of surface tension. Notice that at first the ball bearing had difficulties the break through the surface tension.
Eternal Greetings to you sir and please accept my pranam 🙏🙏🙏 at the beginning of this letter. I beg to introduce myself as a grade 11 student from India. The topics mentioned below are not covered in the 3 audios . 1.Relative velocity (1D and 2D) 2.Constraint relations 3.Few parts of Friction (1.Two block system on a rough/Smooth horizontal surface 2.Minimum angle at which a pushing force should be applied 3.Minimum aangle at which a pushing force should be applied) I would be obliged if you recommend any lectures on you tube so that I can learn the topics mentioned above. Any advice you give me, will be highly valued by me.Moreover,above all I want to thank you for the lectures you upload for the greatest benefit of mine and others. Your most debtful subscriber and student Sudeb Saha.
Sir, the assignments refer to some problems in specific pages, would you kindly tell the name of the book whose page numbers are being mentioned in the assignments of lectures 1 - 12? Thank You.
Professor Lewin great lecture. May I ask if the Terminal speed of a falling man from the sky ist around 120 Mph, how did man a who jumped from the Stratosphere manages to go with 1300 kmh. It was a word record in 2012 I think.
at 130,000 ft altitude the density of our atmosphere is is only 0.3% of what it is at sea level. Thus the air drag is negligibly small in the early part of the fall. By the time he reaches the ground his terminal velocity is about 120 mph.
I think that, by energy considerations,from O to P the mechanical state loses some K energy by air drag, and then even more for P to S. So, E(O,P) > E(P,S); an object with less energy is going to take more time to travel the same trayectory - taking into account only the y component. Thanks prof. Lewin, sorry if my english is bad. P.S. But i dont completely understand how the energy is conserved as whole... is correct to say that the mass in his trayectory is making the atoms in the medium vibrate? Therefore transfering his kinetic energy?
prof Lewin in my book they use another equation for drag force: D=1/2 C ρ A v^2 , is it the general form of the drag force for object of any shape?, in the lecture, F=C2r^2v^2 was used for spherical objects only (3:18)
I cant find any of the equations mentioned in this lecture on the internet either. All i see is the equation R=1/2pCAv^2 for drag and terminal velocity is mg=1/2pCAv^2. I cant study any follow up material if i cant find it elsewhere so ive been forced to learn it that way.
Your smiley face doesnt really help anyone. The original poster had a good question, why didnt you want to answer it? I think you really need some constructive criticism what with all the people who suck up to you on a daily basis
Walter, tell me what is the result of the integral(v_t(1 - e^((-k/m)t)) where v_t is terminal velocity and constant. K is the constant of proportionality in the equation of air drag at low speeds f =kv. And m is mass and constant. It should be : v_t(t + (m/k)e^((-k/m)t). Right? In my book, it says that it is : v_t(t -(m/k)(1-e^((-k/m)t). It includes a -(k/m)v_t. Is that right? Why?
Hi Walter, I have a question about the parabola, when we compare two parabolas with and without air drag or the resistence of a liquid, can I say that the highest point in Y and the further point in X are proporcional, just like P point and S point in 47: 28 ? Thank you, professor!
@@lecturesbywalterlewin.they9259 oh, I see now. But just for sure: We have two trajectories, one with air drag(1) and other one without air drag(2), let me put like this: trajectory 1 as T1 and trajectory 2 as T2, now, imagine with me how far can they go in horizontal way and how far can they com in vertical way. I want to know if there is a pro proporcional relation between X and Y of T1 with X and Y of T2. (X=further position in horizontal; Y:higher position in Vertical) For exemple: Xt1∝Yt1, Xt2∝Yt2 Is it clear now?
as in Hook's law, force is opposite to displacement. f=- kx. Isn't the same phenomenon happening in object falling down (drag force)? Assuming drag force changes linearly with velocity. i.e there should be minus sign. Drag force = -bv, drag force is in upward direction and velocity is downward
>>>is opposite to displacement. f=- kx. Isn't the same phenomenon happening in object falling down (drag force)? Assuming drag force changes linearly with velocity.>>> no it is very very different. with airdrag (in the range that the drag force is prop to the speed f=-k*dx/dt apart from that airdrag is often prop to v^2.
Sir, at 48:08, for the same volume the styrofoam ball has lesser mass than the tennis ball and hence lower density. Since it is regime 2, and c2 is almost equal to density, wont the effect on the styrofoam ball be lesser?
2 balls - same size but different weight. The air drag is the same on both. Thus the lightest ball will be more affected by air drag then the heavy one.
Hello sir, Fres= C1rV + C2(r)2(V)2.... If v is less than critical velocity, how the flow will be in viscous regime? (Because pressure term has square of velocity and thus pressure term dominates for any velocity). Sir, kindly clarify my doubt please.
@@lecturesbywalterlewin.they9259 Sir .. Though you share the pdf link of your book *For the ❤️of physics* But reviving a picture of you with your sign is something which cannot be explained in words... Sir can you please help me to know how to get your book with your picture and signed it by you? Please reply me sir.. I'm eager to get that book..
@@jayantachoudhury4397 Before you decide to buy a book from me *I strongly suggest you get a free copy using the pdf file.* fiisikis.weebly.com/uploads/5/4/9/3/54939617/for_the_love_of_physics.pdf You should also consider to buy my book at amazon in your country. *I have been told that in India you can get my book from Amazon for $7!!!!* If you can afford it, of course, I am still willing to send you a signed copy of my book with *a special note for you + a signed picture of me, which will also have your name on it,* but the price is high as I pay 30% tax here in the US of your payment. Postage for me is also high (e.g., India $41). Your total costs for 1 paperback copy: US $45; Europe, South America, Brazil, Canada, Pakistan, Nepal, Bangladesh & India $65; Middle East & Asia (Japan, Philippines, China & Thailand) $80; Australia $85. You should transfer the money to my PayPal account (see below). *Since we are friends, please mark "family and friends", PayPal will then not charge me a 5% fee* and please let me also know in PayPal your name and complete address and *which name I should use for my note in your book and on my picture.* *For those of you who only want a signed copy of the picture of me, swinging from the pendulum,* with your name on it as well, it will cost you *$12 for 1 copy* (that includes postage). It's the same price for all countries. *You should send me on PayPal your complete address and the name you want me to write on the picture.* Please transfer the $12 to *my PayPal account which is* lewin-physics@physics.comcastbiz.net. *Since we are friends, please mark "family and friends", PayPal will then not charge me a 5% fee.* *Keep in mind that you must first enter PayPal and then my account.*
These equations assume buoyancy is either negligible, or is accounted for in another force. For most solids in air, or for steel balls in water, neglecting it is a reasonable assumption. If the densities are close, or if the object is positively buoyant, you'd have to account for the buoyant force as well. Archimedes principle only applies to fluid statics, when acceleration is zero. Dynamic buoyancy with acceleration involves what I like to call Atwood buoyancy, because it works like an inverted Atwood machine. The buoyant force replaces tension in the string, and the displaced fluid replaces the counterweight. The bottom of the container replaces the pulley. It gives the immersed object an effective weight of (m - rho*V)*g, and an effective inertia of (m + rho*V*g). The displaced fluid adds to its inertia, since it will accelerate in the opposite direction to fill the gaps as it moves. If you apply dynamic buoyancy to the viscous case he's solving at 26:44, you end up modifying the equation as follows: (m + rho*V)*a = (m - rho*V)*g - c1*r*v
Mahardika Firjatullah hey bro! It is because resistive force depends upon viscosity and pressure independently, that is both viscosity and pressure are not functions of one another, we have to deal with them independently, for example, viscosity of air is very less so that k₁v term is very close to zero, and therefore you are only left with k₂v².
But sir during O to P if both the forces the 'drag' and 'gravitational' are acting downwards which force is causing it to get the net acceleration up? If the force is due to our hand than we will have to take it in newton's second law, right? If don't know if you will reply
You have talked about the friction caused by air but here you mentioned the resistive forces are completely different from the friction. What makes them so?
Sir, in the balloon example, we use only regime two to solve for the terminal velocity, however when we form an equation to solve for time it takes to arrive at terminal velocity, we also include regime 1 in the equation (hence making the differential equation more complicated). Why is that? Are we not operating in regime 2 only even before the balloon reaches the terminal velocity?
I gave this lecture in 1999. A program was written by my graduate student who calculated the speed of the balloon as a function of time and how long it takes for the balloon to hit the ground. *I am sure he took all relevant issues into account.*
@@lecturesbywalterlewin.they9259 I have this issue of getting stuck in details until each and everything makes sense to me. Thank you for the reply and I'm blessed to have found you online - not just in terms of physics but I see it on an existential level - in complete awe of your personality, sense of humour and vitality.
Sir, I guess it will take a longer time to move from O to P, than from P to S. The reason I guess is, between O and P both Resistive fore and gravity are decelerating y component of velocity but between P and S, gravity is causing acceleration and the resistive force is decelerating. Since in P to S there is a assist in the direction of motion time of travel should be less. Is this correct?
Professor Lewin, is the force in regime 2 (which looks very similar to k(v^2) equivalent to F=1/2CdA(rho)v^2 since they are both proportional to velocity squared?
When you were showing us the demonstration in which you were dropping different sizes of ball bearings, as the size of ball bearings increased, the time they took to breakthrough decreased.........I THINK THAT'S INTUITIVE because there was more amount of surface to breakthrough but for quarter inch ball bearing, OPPOSITE thing happened, it took less time to breakthrough( ie. IT DID NOT FOLLOW THE TREND).......... Is it because of the GRAVITATIONAL FORCE coming into play?
by "breakthrough" I meant as enters the surface of the syrup...... as the size of ball bearings increased, the time they took to enter the syrup increased.........I THINK THAT'S INTUITIVE because there was more amount of surface to breakthrough but for quarter inch ball bearing, OPPOSITE thing happened, it took less time to breakthrough( ie. IT DID NOT FOLLOW THE TREND).......... Is it because of the GRAVITATIONAL FORCE coming into play?
1/4 inch has the largest weight! grav force (weight) is proportional to R^3 drag is prop to R^2, the upward force due to surface tension is also prop to R^2. Thus, as a general rule, the heavier the balls the faster they will "breakthrough".
Our teacher told us that that there are 3 forces acting on the object. The weight, the drag force and Archimedes' principle. But you stated only two forces.
yes there are 3. But when you fill a balloon with air, the weight of that air-filled balloon (place it on a scale) automatically takes Archimedes into account. Thus it's enough to deal wth weight and air drag.
When I discuss throwing an object from a tower, I mention the mass, m and the force mg. That is accurate as long as the density of the object is >> density of air. Density of air is about 1 kg/m^3, Density of plastic is 1000 kg/m^3. Thus archimedes can be ignored for all objects except for balloons.
When you place something on a scale, aren't you always taking into account the Archimedes force upwards, due to the air surrounding the object? hence having a resulting mass smaller than you would get in absence of air? so when you use the mass you obtain in such a read at a scale, for your experiment of letting the object fall on air, you never need to introduce the Archimedes force upwards, since it is already introduced within the lesser mass? irrespective to the fact that you are dealing with an iron ball or an helium balloon.
I'm saying this because this is not just accurate as long as the density of the object is way bigger than the air's; I'd say it is always accurate no matter what, as long as the mass is measured on a scale surrounded by air, hence already taking into account Archimedes; and of course, the object is going to be used for the experiment in the same fluid which was previously weighed in.
Sir, I have one thing that confuses me. when throw a ball in air, in which critical velocity is very small, which means it will spend most time in regime2, but before even reaching the critical speed, should I first take it as Regime1, then take it as regime2,am I right?
may i ask you sir a personal question when you were young student how much time you specified for sleeping some times I sleep less I feel tired some times I sleep more I feel lazy !!
Very easily explained. I guess this is the reason why SpaceX starship rockets do a belly-flop maneuvre at the last moment so that they keep the rocket at it's terminal velocity.
may be not right but since during OP (upward motion) =acceleration of the particle(Ac)+g acting hence t goes down BUT during fall ie.. from P to S=Ac -g hence t goes up
+Lectures by Walter Lewin. They will make you ♥ Physics. sir, since you throw ball upward so lets say it has only upward velocity component Voy then total acceleration in upward(OP) direction=acc(acceleration of ball in upward direction)+g hence time decreases and during downward (PS)total acceleration =acc -g hence time increases
how does weight will vary if one is free falling in air drag? I think one will feel normal weight once he/she reaches their terminal velocity since Fres=mg at terminal velocity and weight is proportional to velocity until one reaches terminal velocity (i.e. 0
if something is in free fall it is weightless. However, due to air drag there is no such thing as free fall. If you Mass 80 kg) jump out of a plane at 10,000 ft, you will reach a terminal speed of about 125 mph. Your weight will then be 80g Newton (about 176 US pounds).
Hi, I was wondering if there is some sort of database with the values of c_1 and c_2 for a lot of objects, I've been googling and I can't find anything but I'm hoping yuo would have some knowledge of that.
Comments like that are what makes people afraid to ask questions. They're afraid of getting passive aggressive/ elitist responses. This happens all the time. If you're tired of responding to comments then hire someone. If youre tired of running an educational channel where you get questions like these then end it. Your most common response is " use google, use the internet, did you watch the lecture" I mean he said he looked all over the internet for the answer. Did you even read what he said?
I'm not sure if I'm finding the right book. One site has a book with gears on the cover, but when I click on it, I'm sent to a page that shows the book with a light blue cover and appears to possibly be just an answer book.
Good afternoon sir, the answer to the las question, is that the time to go to the point P to S is not 2. times the time to go from O to P. its much less
+Christian L I do not understand your answer. My question is: Is the time to go from O to P shorter, longer or identical to the time to go from P to S and what is the reason for your answer?
I would argue, professor, that the sign of the time difference from O to P and from P to S varies accordingly as v_{0_{x}} is greater, equal or less than v_{term.}. My reasoning is that the time from O to P depends on the vertical distance from O to P and on the average speed of the object from O to P (and similarly from P to S). As the vertical distance from O to P is the same as from P to S, then the time difference will depend on the average speed from O to P _versus_ the average speed from P to S. This in turn will depend on which of v_{0_{x}} or v_{term.} is greater. We have three cases : v_{0_{x}} > v_{term.}, in which case t_{OP} < t_{PS}, v_{0_{x}} = v_{term.}, in which case t_{OP} = t_{PS}, and v_{0_{x}} < v_{term.}, in which case t_{OP} > t_{PS}. Is this valid, or is the average speed not proportional to the initial/final speed ?
On the way from O to P the acceleration downward in the vertical direction is larger than g. On the way from P to S the acceleration downward in the vertical direction is smaller than g because the air drag component flips over at P. Thus it takes less time to go from O to P than from P to S.
Sir. I am a regular viewer of your lectures. I am preparing for SAT subject test in Physics and normal SAT. Sir I want to know after watching your video from where should I practice questions to get prepared for my Physics SAT subject test.
Lectures by Walter Lewin. They will make you ♥ Physics. Sir I am also preparing for subject test in Math. Can you help me from where I should prepare.. and get sufficient support material to get into MIT
Karo syrup is very useful for many things...it can make salsa texture very enjoyable with a little touch of sweetness and it can be appreciated in wonderful sauces for many exotic foods ranging from the Orient to the far Western Regions and it appears to have a completely adequate range for the use of alcoholic beverages and it will attract ants so be careful about that and where would the world be without it and it cleans up quickly with a damp rag or towel and we can all thank Professor Lewin for the obviously wonderful use it has for the teaching of PHYSICS!!!!!!! 🍰🥧🍦🍨🍧🥮🍡🧁🍭🍫🍫🍩🍪😁😁😁
thank you ,sir ,for your attention,.... by the equation i meant the in which ball bearing in the corn syrup ,here through the round blob of it ,so then how would it change , we got no high pressure ,low pressure , so does it entirely depend on viscosity term ? ,so in that surface tension plays a role ?
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the fact that you include experiments makes it so much easier to get an intuitive understanding of a certain topic!
:)
Lectures by Walter Lewin. They will make you ♥ Physics. oh my god THE walter lewin replied!!!!!!!!! i'm a HUGE fan!!
thanks \\/\///
exactly most physics teachers in my country dont have a lab
@@lecturesbywalterlewin.they9259 Yes
*8.01x Lect 12*
*Conclusion*
*“It’s a difficult lectures, but it’s so amazing”*
*Have fun!*
0:09 *drag force / resistive forces*
Equation (viscous term C1>Temperature/ pressure term C2, 4:30 the terminal velocity
*2 regimes*
6:30 *critical velocity*
1.Vvcrit
9:40 *conclusion*
11:00 syrup experiment + interesting diagram
*important - must-watched*
See more here: ruclips.net/video/K389Fw2U5sk/видео.html
??? *Key question*
1.How long does it take for the terminal speed to reached? In the syrup)
28:25 in air
*Assignment 4* just do it
35:00
Experiment with a big ball
+graph (pebble)
??? *Key Question*
2. how the air drag influence the trajectory (assignment)?
3. Last question 48:50
*8.01x Lect 12*
*Thanks to our teacher, Walter Lewin.*
*Physics is so Beautiful.*
Brilliant lecture. Things are getting harder. Needed more than three hours to have everything figured out. or so I believe/hope. The pause button, ah, such a great thing!
:)
Another masterpiece of a lecture! Beautiful approach using just c1 and c2 instead of the more difficult to understand viscosity for the linear dependence on v and friction coefficient for the square dependence on v of the drag.
Love it!
My guess is that it takes a shorter time to go from O to P than the one it takes to go from P to S.
The reason is considering only the y-component , the ball starts with speed V0y up and decelerates with g + something (more than g) until it reaches 0, while on the way back it accelerates with g - something (less than g). Besides the speed might even stop at a terminal value, so if V0y is big enough, it's clear that the average speed from O to P is bigger than the one from P to S.
+Tarik Rahmatallah correct
I would argue that, although the average deceleration between O and P is somewhere above g and the average acceleration between P and S is definitely somewhere below g, the answer is not as clear-cut as it seems, since dVop is also larger than dVps. So it may well be possible for Top to be smaller/larger/equal to Tps depending on the values of C2 and Vy0. Please correct me if i am wrong!
@@mejente1 if there is no air drag than we get a symmetric parabola and the times are identical: so no dependence of Vy0 on the time difference between going up and down.
I believe they will be the same. The vertical component, which is the only part that matters vis á vis time, is entirely controlled by gravity once the object has its initial vertical upward velocity. There are no outside VERTICAL forces of any kind except for air drag, which at any given velocity will be the same going UP as it is at the same point going DOWN.
@@jwills8606 You're not entirely wrong... but it is gravity (accelleration) not velocity that is constant for both up and down directions. As Tarik has said above this accelleration (or decelleration) is acted on asymmetrically, causing the difference in time taken. You are right that the resistive force is proportional to velocity, but the velocity is determined by the accelleration (and time) and so it is in turn affected by the (asymmetrical) resistive force.
You do make me love physics even more.
:)
Professor Lewin, your lectures make physics so intuitive and so beautiful! Thank you so much for providing your lectures for free!
You're very welcome!
Goodness, Prof. Lewin is fantastic. He explains topics so elaborately yet so easy to understand. Me, a eleven year able to understand what he says is nothing but proof that Prof. Lewin is really good at explaining Physics
It's a pretty interesting feeling to watching your aging over the course of lecture series. The education that I've got from your lectures is not only bounded with physics sir.
After hours of suffering with my attempts to calculate air drag, I remembered to check your lectures. This easily solved my problem. As a high school student, I really appreciate that there are quality lectures online to learn from. Thank you
you are very welcome
the radius of the oil drops should be r
Thank you Sir!!! Very well explained, I purely enjoyed this as well as all your other videos which I have been watching with pleasure.
I think the Dave you mention in this video is the person you recently made a video about who is now a full professor! Fantastic!
Sir, u are the finest creation of GOD because you are someone who inspires millions all over the world and creating an impact, as it is said that "SUCCESS" is all about making a difference and you have done it in millions of student's life.
Richard Feynman sir is Richard Feynman but walter lewin is Walter lewin. You have brought very big change in the we think physics as just a subject that is compulsary if we take science not only in me but millions of people all over the world.
We may not get a chance to meet you because by that u may be alive or not but yes till I die u will always be within our hearts.
Thank you,
PIYUSH CHOUDHARY.
LOVE FROM INDIA🇮🇳🇮🇳🇮🇳.
Thank you very much! Great lesson and it is so brilliant to have an experiment for each topic ! This physics exam I have tomorrow in Computer Engineering is really been tough and altogether very boring, until I have discovered your channel prof. Lewin and I started really to understand what I was up to while demonstrating a thing or solving a problem
Sir, I always hated fluid mechanics in my class, but thanks to you, I have gained interest in this subject. I was lucky to have found your channel :-)
:)
a huge huge fan of yours a lots of love from INDIA and a grate love for the physics........
Sir, I think that from O to P, net force
F=ma=mg+F(res) that is a>g
From P to S net force
F=ma=mg-F(res) that is a
How can you calculate C1 and C2, professor? And can you also work out the relationship between C1 and temperature by experiment too?
I have never calculated them. They are tabulated (see google).
7:30 how we get to know that regime 1 is where when velcity v is much much less than critical velocity causes viscous term dominating over the pressure term?
sir pls reply...
Navneet Mishra Look, at the pressure term the velocity is squared. If vvcrit, the v squared will be extremely high, so the pressure term will be high in comparison with the viscous term. I hope I helped
@@dawidskrok3345 Thanks a lot. Initially I thought the conclusion came from the critical velocity ratio between C1/C2r.
@@dawidskrok3345 if you donot put 10^-x than?ie why not to take positive value?
Awesome!! Finally i can SEE THROUGH THR EQUATIONS because this topic was just a big amount of unclear equations for me thank you professor walter lewin 🙂
Great!
You are amazing teacher Dr. Lewin. Thank you for these lectures they helped me a lot. I am a new Physics teacher and I would like to tell me how you prepare the lecture and thank you in advance.
So while watching this video, I learned that parabolic curves turn into straight lines when plotted on logarithmic coordinates, nice!
:)
Sir,
I have some trouble understanding why it takes less time for the object too travel from O to P, than from P to S, Because when the object is traveling from P to S, it is accelerating due too gravity and the pathlength is shorter, wich makes me think the object encounters less air molecules, which reduces the airdrag. So what am i not getting here ?
Hi professor. At 7:30, you drew conclusions about dominant terms in the drag force expression for a sphere (regime 1 & 2). I'm not entirely sure how you drew those conclusions, I had to take ratios of term 1 & 2 to convince myself of it, why wasnt there more elaborate math to prove the approximations? Was it intendent to be obvious?
no it is not obvious. I decided to give the constants and leave it with that. My graduate student (Dave Pooley) used these constants and I show his results.
@@lecturesbywalterlewin.they9259 got it. Thanks for the reply!
Great lecture as always. Thank you Professor.
You are very welcome
when you said that the resistive forces depend on the shape of the object, did you also meant by the surface of the object?
surface roughness is in general not included in the usual tables (only shapes). But architects should include roughness too
Hello sir. My old math teacher showed you in my class I thought you were really awesome. Now I am planning of watching all of your lectures. I do have a question on this one. What is C1 and C2 specifically and how do you calculate them. I’ve watched the lecture over and over again but can’t wrap my head around it.
C1 and C2 are derived from observations.
why regime 1 is when the speed is much less than v-critial ? 7:36
Very well explained .Sir I'm a huge fan of yours. 😊
does the object at 43:57 has a trajectory as y=a*ln(bx)+(c/x) where a,b,c are constants?
Maybe, I don't know, try to derive it.
@@lecturesbywalterlewin.they9259 i have derived it sir and got this result.
I would be extremely grateful for a specific path -> of topics -> to study -> in order to eventually understand this lecture easier, where each subsequent topic builds on the one before it. Thank you.
take my course 8.01
@@lecturesbywalterlewin.they9259 Thank you! I'll do the relevant search and start from the beginning.
Hello professor. I have started watching your lectures with parallel reading from Ohanian's book (2nd edition - 1989). In the pdf file you suggest reading pages 142 to 146 and page 190 from the book, the topics of which seems to be unrelated to the topic of the lecture.
Hello :)
If we have a vertical shot towards the ground and the initial velocity is not 0 and we take into account the air resistance, could the terminal velociry be less than the initial or is it always greater?
could be less
I think it will take less time to go from O to P than to go from P to S...........because from P to S the deacceleration is greater than the acceleration from P to S......
Is it correct?
answer is correct - reasoning a bit strange
>>>>the deacceleration is greater than the acceleration from P to S>>>>
Correct reasoning:
on the way up both gravity and air drag forces are downward,
on the way down the net force down is the grav force minus the airdrag.
Thus the slow down in speed in vertical direction is smaller on the way down than on the way up.
Perhaps this is what you meant.
Sir, if I may ask, what exactly is the phenomenon at 19:27 called, which gives rise to the air bubble at the surface? I ask because I was having difficulty searching for it and finding any pertaining formulae.
physics is not about equations, physics is about concepts. Math is the language of Physics. To answer your question. it's the result of surface tension. Notice that at first the ball bearing had difficulties the break through the surface tension.
at 32:40 is the value of radius is 3*C1*C1/(4*pi*C2*g*density of oil) = .14412 mm?
Eternal Greetings to you sir and please accept my pranam 🙏🙏🙏 at the beginning of this letter. I beg to introduce myself as a grade 11 student from India. The topics mentioned below are not covered in the 3 audios .
1.Relative velocity (1D and 2D)
2.Constraint relations
3.Few parts of Friction (1.Two block system on a rough/Smooth horizontal surface 2.Minimum angle at which a pushing force should be applied 3.Minimum aangle at which a pushing force should be applied)
I would be obliged if you recommend any lectures on you tube so that I can learn the topics mentioned above. Any advice you give me, will be highly valued by me.Moreover,above all I want to thank you for the lectures you upload for the greatest benefit of mine and others.
Your most debtful subscriber and student
Sudeb Saha.
Sir, the assignments refer to some problems in specific pages, would you kindly tell the name of the book whose page numbers are being mentioned in the assignments of lectures 1 - 12?
Thank You.
Physics
Hans C. Ohanian
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
Thank you so much, Sir, for the prompt reply.
I absolutely love Prof Lewin's lectures. But i am curious... how does he do the dotted line trick on the blackboard?
its easy, search the YT and you'll find the answer.
Doubt : when balancing forces why are we not taking into account upthrust force which is equal to [(volume of solid) * (density of liquid) * g]
If C2 is co-related to density (4:29) and as density is a function of temperature, why is C2 not dependent on temperature (3:26)?
Excatly same question. PV=NRT
sir, why when gravity takes over, then you ultimately end up in regime I.14:20
question not clear - watch my lecture again
Professor Lewin great lecture. May I ask if the Terminal speed of a falling man from the sky ist around 120 Mph, how did man a who jumped from the Stratosphere manages to go with 1300 kmh. It was a word record in 2012 I think.
at 130,000 ft altitude the density of our atmosphere is is only 0.3% of what it is at sea level. Thus the air drag is negligibly small in the early part of the fall. By the time he reaches the ground his terminal velocity is about 120 mph.
why we did not use v=gt for ball bearings? we are not in an accelerated motion? why we use x=vt?!
OP path takes smaller time because until the terminal speed is not achieved , the air resistance is according to term C1rv
Very good lecture Sir. 🙏🙏🙏🙏
Thanks and Regards
I think that, by energy considerations,from O to P the mechanical state loses some K energy by air drag, and then even more for P to S. So, E(O,P) > E(P,S); an object with less energy is going to take more time to travel the same trayectory - taking into account only the y component. Thanks prof. Lewin, sorry if my english is bad.
P.S. But i dont completely understand how the energy is conserved as whole... is correct to say that the mass in his trayectory is making the atoms in the medium vibrate? Therefore transfering his kinetic energy?
Energy can not be created or destroyed. IF one form of energy disappears (e.g. KE) it's converted into another form.
prof Lewin in my book they use another equation for drag force: D=1/2 C ρ A v^2 , is it the general form of the drag force for object of any shape?, in the lecture, F=C2r^2v^2 was used for spherical objects only (3:18)
:)
I cant find any of the equations mentioned in this lecture on the internet either. All i see is the equation R=1/2pCAv^2 for drag and terminal velocity is mg=1/2pCAv^2. I cant study any follow up material if i cant find it elsewhere so ive been forced to learn it that way.
Your smiley face doesnt really help anyone. The original poster had a good question, why didnt you want to answer it? I think you really need some constructive criticism what with all the people who suck up to you on a daily basis
From o to p the ball will take the less time than p to s because when the ball moves from o to p it achieves terminal velo there , is it correct
And professor, what is the final question answer? Will the ball in the first trajectory takes longer time than the other half?
So weightless in free fall (jumping without a parachute) Why not calculate the weight upon meet the ground?
Why is time - 100/d^2 relationship graph is the straight line?
Walter, tell me what is the result of the integral(v_t(1 - e^((-k/m)t)) where v_t is terminal velocity and constant. K is the constant of proportionality in the equation of air drag at low speeds f =kv. And m is mass and constant. It should be : v_t(t + (m/k)e^((-k/m)t). Right? In my book, it says that it is : v_t(t -(m/k)(1-e^((-k/m)t). It includes a -(k/m)v_t. Is that right? Why?
watch my lecture in which I cover terminal velocities or use google
Question does a balloon rising through the air use a simular equation?
a balloon rises when its weight is less than the wait of the air that it dispalces. googgle: Archimedes' principle.
Hi Walter, I have a question about the parabola, when we compare two parabolas with and without air drag or the resistence of a liquid, can I say that the highest point in Y and the further point in X are proporcional, just like P point and S point in 47: 28 ? Thank you, professor!
question unclear. If there is air drag the trajectory is not a parabola.
@@lecturesbywalterlewin.they9259 oh, I see now. But just for sure:
We have two trajectories, one with air drag(1) and other one without air drag(2), let me put like this: trajectory 1 as T1 and trajectory 2 as T2, now, imagine with me how far can they go in horizontal way and how far can they com in vertical way. I want to know if there is a pro proporcional relation between X and Y of T1 with X and Y of T2. (X=further position in horizontal; Y:higher position in Vertical)
For exemple: Xt1∝Yt1, Xt2∝Yt2
Is it clear now?
as in Hook's law, force is opposite to displacement. f=- kx.
Isn't the same phenomenon happening in object falling down (drag force)?
Assuming drag force changes linearly with velocity.
i.e there should be minus sign.
Drag force = -bv, drag force is in upward direction and velocity is downward
>>>is opposite to displacement. f=- kx.
Isn't the same phenomenon happening in object falling down (drag force)?
Assuming drag force changes linearly with velocity.>>>
no it is very very different. with airdrag (in the range that the drag force is prop to the speed f=-k*dx/dt
apart from that airdrag is often prop to v^2.
Sir, at 48:08, for the same volume the styrofoam ball has lesser mass than the tennis ball and hence lower density. Since it is regime 2, and c2 is almost equal to density, wont the effect on the styrofoam ball be lesser?
Or is the difference in c2 is negligible?
2 balls - same size but different weight. The air drag is the same on both. Thus the lightest ball will be more affected by air drag then the heavy one.
Sir, is air drag different from resistive force?
If yes, then F(resistive)=C2[r^2][v^2] and C2 depends on the density. Same size but different weights means different densities right?
Oh sorry i understood. C2 is very close to the density of the medium and not the object. Am i right sir?
Hello sir,
Fres= C1rV + C2(r)2(V)2....
If v is less than critical velocity, how the flow will be in viscous regime? (Because pressure term has square of velocity and thus pressure term dominates for any velocity).
Sir, kindly clarify my doubt please.
25:19 Beautiful! Physics works!
Is there a video that derives the equation for wind resistance, using transfer of momentum concepts?
Prof. Walter lewin,,
When you said that the resistive forces depends on the shape of the object. Will you also meant by the surface of the object?
roughness of the surface could have an effect - use google
@@lecturesbywalterlewin.they9259 Tnq sir.. Now I Understand that what you actually meant
@@lecturesbywalterlewin.they9259 Sir
.. Though you share the pdf link of your book *For the ❤️of physics*
But reviving a picture of you with your sign is something which cannot be explained in words... Sir can you please help me to know how to get your book with your picture and signed it by you?
Please reply me sir.. I'm eager to get that book..
@@jayantachoudhury4397 Before you decide to buy a book from me *I strongly suggest you get a free copy using the pdf file.*
fiisikis.weebly.com/uploads/5/4/9/3/54939617/for_the_love_of_physics.pdf
You should also consider to buy my book at amazon in your country. *I have been told that in India you can get my book from Amazon for $7!!!!*
If you can afford it, of course, I am still willing to send you a signed copy of my book with *a special note for you + a signed picture of me, which will also have your name on it,* but the price is high as I pay 30% tax here in the US of your payment. Postage for me is also high (e.g., India $41).
Your total costs for 1 paperback copy: US $45; Europe, South America, Brazil, Canada, Pakistan, Nepal, Bangladesh & India $65; Middle East & Asia (Japan, Philippines, China & Thailand) $80; Australia $85. You should transfer the money to my PayPal account (see below). *Since we are friends, please mark "family and friends", PayPal will then not charge me a 5% fee* and please let me also know in PayPal your name and complete address and *which name I should use for my note in your book and on my picture.*
*For those of you who only want a signed copy of the picture of me, swinging from the pendulum,* with your name on it as well, it will cost you *$12 for 1 copy* (that includes postage). It's the same price for all countries. *You should send me on PayPal your complete address and the name you want me to write on the picture.* Please transfer the $12 to *my PayPal account which is* lewin-physics@physics.comcastbiz.net. *Since we are friends, please mark "family and friends", PayPal will then not charge me a 5% fee.*
*Keep in mind that you must first enter PayPal and then my account.*
@@lecturesbywalterlewin.they9259 Tnq you sir for your kind information... I will surely buy your book with your sign within 2 months..!!!
Please Sir, is upthrust automatically squeezed into the resistive force formula?
These equations assume buoyancy is either negligible, or is accounted for in another force. For most solids in air, or for steel balls in water, neglecting it is a reasonable assumption. If the densities are close, or if the object is positively buoyant, you'd have to account for the buoyant force as well.
Archimedes principle only applies to fluid statics, when acceleration is zero. Dynamic buoyancy with acceleration involves what I like to call Atwood buoyancy, because it works like an inverted Atwood machine. The buoyant force replaces tension in the string, and the displaced fluid replaces the counterweight. The bottom of the container replaces the pulley.
It gives the immersed object an effective weight of (m - rho*V)*g, and an effective inertia of (m + rho*V*g). The displaced fluid adds to its inertia, since it will accelerate in the opposite direction to fill the gaps as it moves. If you apply dynamic buoyancy to the viscous case he's solving at 26:44, you end up modifying the equation as follows:
(m + rho*V)*a = (m - rho*V)*g - c1*r*v
Sir is there any explanation why the one with linear v is viscous term and the other is pressure. Or i need to memorize it?
use google
Mahardika Firjatullah hey bro! It is because resistive force depends upon viscosity and pressure independently, that is both viscosity and pressure are not functions of one another, we have to deal with them independently, for example, viscosity of air is very less so that k₁v term is very close to zero, and therefore you are only left with k₂v².
When v less than v critical how come we are in regime 1.Also what makes us in regime 2 when v more more than v critical.
But sir during O to P if both the forces the 'drag' and 'gravitational' are acting downwards which force is causing it to get the net acceleration up? If the force is due to our hand than we will have to take it in newton's second law, right? If don't know if you will reply
You have talked about the friction caused by air but here you mentioned the resistive forces are completely different from the friction. What makes them so?
Also watch my 8.01 lecture on friction. Maybe that will help you.
When I was in MIT I used to wait for sir's class
which course of mine did you take and in which year?
Sir, in the balloon example, we use only regime two to solve for the terminal velocity, however when we form an equation to solve for time it takes to arrive at terminal velocity, we also include regime 1 in the equation (hence making the differential equation more complicated). Why is that? Are we not operating in regime 2 only even before the balloon reaches the terminal velocity?
I gave this lecture in 1999. A program was written by my graduate student who calculated the speed of the balloon as a function of time and how long it takes for the balloon to hit the ground. *I am sure he took all relevant issues into account.*
@@lecturesbywalterlewin.they9259 I have this issue of getting stuck in details until each and everything makes sense to me. Thank you for the reply and I'm blessed to have found you online - not just in terms of physics but I see it on an existential level - in complete awe of your personality, sense of humour and vitality.
:)
@48:06 he says F=ma but if a tennis ball has grater mass than it should have experience a much greater force but doesn't........pls help
I'm late, but if a tennis ball has more mass, but the same force, it should have less acceleration.
Professor how should I solve the differential equation at 26:52
dv/dt = g - (Cr/m)v. V(t)=gt-e^((Cr/m)t)
Terminal velocity is inversely proportional to radius in one formula and directly proportional to radius in another formula. Which one is correct?
watch my lecture
Both are correct
Sir, I can't realise the equation of |F res|. Please reply.
What is the need for doing 100/d^2 in that syrup experiment
To show that heavy balls take less time to drop is this reason correct someone can tell me
why is it that we use regime 2 when calculating the terminal velocity
and when do we use regime 1 ?
I explain that in my lecture.
Sir, I guess it will take a longer time to move from O to P, than from P to S. The reason I guess is, between O and P both Resistive fore and gravity are decelerating y component of velocity but between P and S, gravity is causing acceleration and the resistive force is decelerating. Since in P to S there is a assist in the direction of motion time of travel should be less.
Is this correct?
correct
Thank You, Professor!
@@lecturesbywalterlewin.they9259 sir, time to travel from o to p should be less than the time to travel from p to s right??
Where can I find C1 and C2 for different temperatures of air?
+Milan Marjanović Use Google
And where i can find formulas for different shapes of object?
+Milan Marjanović search the web.
Professor Lewin, is the force in regime 2 (which looks very similar to k(v^2) equivalent to F=1/2CdA(rho)v^2 since they are both proportional to velocity squared?
watch the lecture
Lectures by Walter Lewin. They will make you ♥ Physics. I did sir.
I am guessing that the two terms(viscous and pressure) relating to drag force are experimental fact.
When you were showing us the demonstration in which you were dropping different sizes of ball bearings, as the size of ball bearings increased, the time they took to breakthrough decreased.........I THINK THAT'S INTUITIVE because there was more amount of surface to breakthrough but for quarter inch ball bearing, OPPOSITE thing happened, it took less time to breakthrough( ie. IT DID NOT FOLLOW THE TREND)..........
Is it because of the GRAVITATIONAL FORCE coming into play?
what do you mean by "breakthrough"? How many minutes into the lecture?
by "breakthrough" I meant as enters the surface of the syrup......
as the size of ball bearings increased, the time they took to enter the syrup increased.........I THINK THAT'S INTUITIVE because there was more amount of surface to breakthrough but for quarter inch ball bearing, OPPOSITE thing happened, it took less time to breakthrough( ie. IT DID NOT FOLLOW THE TREND)..........
Is it because of the GRAVITATIONAL FORCE coming into play?
How many minutes into the lecture?
19:25
1/4 inch has the largest weight!
grav force (weight) is proportional to R^3
drag is prop to R^2, the upward force due to surface tension is also prop to R^2. Thus, as a general rule, the heavier the balls the faster they will "breakthrough".
What did you mean by when the gravity takes over, then you ultimately end up in regime one? 14:05
Our teacher told us that that there are 3 forces acting on the object. The weight, the drag force and Archimedes' principle. But you stated only two forces.
Then why you didn't state the Archimedes force!
yes there are 3. But when you fill a balloon with air, the weight of that air-filled balloon (place it on a scale) automatically takes Archimedes into account. Thus it's enough to deal wth weight and air drag.
When I discuss throwing an object from a tower, I mention the mass, m and the force mg. That is accurate as long as the density of the object is >> density of air. Density of air is about 1 kg/m^3, Density of plastic is 1000 kg/m^3. Thus archimedes can be ignored for all objects except for balloons.
When you place something on a scale, aren't you always taking into account the Archimedes force upwards, due to the air surrounding the object? hence having a resulting mass smaller than you would get in absence of air? so when you use the mass you obtain in such a read at a scale, for your experiment of letting the object fall on air, you never need to introduce the Archimedes force upwards, since it is already introduced within the lesser mass? irrespective to the fact that you are dealing with an iron ball or an helium balloon.
I'm saying this because this is not just accurate as long as the density of the object is way bigger than the air's; I'd say it is always accurate no matter what, as long as the mass is measured on a scale surrounded by air, hence already taking into account Archimedes; and of course, the object is going to be used for the experiment in the same fluid which was previously weighed in.
Sir, I have one thing that confuses me. when throw a ball in air, in which critical velocity is very small, which means it will spend most time in regime2, but before even reaching the critical speed, should I first take it as Regime1, then take it as regime2,am I right?
And thank you sir Lewin, I love your lessons so much!
may i ask you sir a personal question when you were young student how much time you specified for sleeping some times I sleep less I feel tired some times I sleep more I feel lazy !!
read my book "For the Love of Physics"
Very easily explained. I guess this is the reason why SpaceX starship rockets do a belly-flop maneuvre at the last moment so that they keep the rocket at it's terminal velocity.
may be not right but since during OP (upward motion) =acceleration of the particle(Ac)+g acting hence t goes down
BUT during fall ie.. from P to S=Ac -g hence t goes up
+sweety singh I have no idea what you are talking about.
+Lectures by Walter Lewin. They will make you ♥ Physics. sir, since you throw ball upward so lets say it has only upward velocity component Voy then total acceleration in upward(OP) direction=acc(acceleration of ball in upward direction)+g hence time decreases and during downward (PS)total acceleration =acc -g hence time increases
+sweety singh how many minutes into the lecture?
+Lectures by Walter Lewin. They will make you ♥ Physics. sir at 49 you asked the question? so iwant to know is my answer is correct
+sweety singh your answer is incorrect
are we sure terminal velocity is reached by the time the first linemarker is crossed?
ok my question got answered a bit further ... nevermind
how does weight will vary if one is free falling in air drag? I think one will feel normal weight once he/she reaches their terminal velocity since Fres=mg at terminal velocity and weight is proportional to velocity until one reaches terminal velocity (i.e. 0
if something is in free fall it is weightless. However, due to air drag there is no such thing as free fall. If you Mass 80 kg) jump out of a plane at 10,000 ft, you will reach a terminal speed of about 125 mph. Your weight will then be 80g Newton (about 176 US pounds).
thank you so much sir....
Sir, your lectures are so awesome that in an avg I watch 3-5 lectures a day
Hi, I was wondering if there is some sort of database with the values of c_1 and c_2 for a lot of objects, I've been googling and I can't find anything but I'm hoping yuo would have some knowledge of that.
use the web
Comments like that are what makes people afraid to ask questions. They're afraid of getting passive aggressive/ elitist responses. This happens all the time. If you're tired of responding to comments then hire someone. If youre tired of running an educational channel where you get questions like these then end it. Your most common response is " use google, use the internet, did you watch the lecture" I mean he said he looked all over the internet for the answer. Did you even read what he said?
Did i miss abefore lecture i am unable to get this?
sir is time taken to reach 99 % of terminal velocity is equal to m*ln(100)/(c1*r) for the experiment of sphere dropped in the liquid?
I cannot add to the clarity of this lecture
CAN YOU TELL ME HOW YOU GOT THIS RESULT
Thankyou professor
Sir what book should I follow, as complimentary to your 8.01 lecture??or are lectures and assignment enough??
8.01
Physics
Hans C. Ohanian
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
all assignments and my exams are posted below the video thumb nails
Lectures by Walter Lewin. They will make you ♥ Physics. Thanks sir, so kind of you
I'm not sure if I'm finding the right book. One site has a book with gears on the cover, but when I click on it, I'm sent to a page that shows the book with a light blue cover and appears to possibly be just an answer book.
Sir I didn't understand the term "turbulence". Can you please make it clear to me?
google "turbulence"
Good afternoon sir, the answer to the las question, is that the time to go to the point P to S is not 2. times the time to go from O to P. its much less
+Christian L I do not understand your answer. My question is:
Is the time to go from O to P shorter, longer or identical to the time to go from P to S and what is the reason for your answer?
I would argue, professor, that the sign of the time difference from O to P and from P to S varies accordingly as v_{0_{x}} is greater, equal or less than v_{term.}. My reasoning is that the time from O to P depends on the vertical distance from O to P and on the average speed of the object from O to P (and similarly from P to S). As the vertical distance from O to P is the same as from P to S, then the time difference will depend on the average speed from O to P _versus_ the average speed from P to S. This in turn will depend on which of v_{0_{x}} or v_{term.} is greater. We have three cases :
v_{0_{x}} > v_{term.}, in which case t_{OP} < t_{PS},
v_{0_{x}} = v_{term.}, in which case t_{OP} = t_{PS},
and v_{0_{x}} < v_{term.}, in which case t_{OP} > t_{PS}.
Is this valid, or is the average speed not proportional to the initial/final speed ?
On the way from O to P the acceleration downward in the vertical direction is larger than g. On the way from P to S the acceleration downward in the vertical direction is smaller than g because the air drag component flips over at P. Thus it takes less time to go from O to P than from P to S.
Great explanation!
Sir. I am a regular viewer of your lectures. I am preparing for SAT subject test in Physics and normal SAT. Sir I want to know after watching your video from where should I practice questions to get prepared for my Physics SAT subject test.
There are about 140 problem solving videos on this channel. Watch them ALL
Lectures by Walter Lewin. They will make you ♥ Physics. Sir I am also preparing for subject test in Math. Can you help me from where I should prepare.. and get sufficient support material to get into MIT
Watch some OCW math lectures.
Karo syrup is very useful for many things...it can make salsa texture very enjoyable with a little touch of sweetness and it can be appreciated in wonderful sauces for many exotic foods ranging from the Orient to the far Western Regions and it appears to have a completely adequate range for the use of alcoholic beverages and it will attract ants so be careful about that and where would the world be without it and it cleans up quickly with a damp rag or towel and we can all thank Professor Lewin for the obviously wonderful use it has for the teaching of PHYSICS!!!!!!! 🍰🥧🍦🍨🍧🥮🍡🧁🍭🍫🍫🍩🍪😁😁😁
Greetings professor , from India , does surface tension come into the equation ?,does these work in space ?
which equation? surface tension is always present it does not depend on gravity.
thank you ,sir ,for your attention,.... by the equation i meant the in which ball bearing in the corn syrup ,here through the round blob of it ,so then how would it change , we got no high pressure ,low pressure , so does it entirely depend on viscosity term ? ,so in that surface tension plays a role ?
the ball bearing will not go down in the ISS as everything in the ISS is weightless.
so it will be stuck in the blob?
weightless means it will not move in the ISS (regardless of where you release it) as long as you release it with zero speed.
At terminal speed gravity and drag force becomes equal why the object then falls???
Brush up on the concepts of inertia , that's why, the net force is gonna be 0 so the object remain in either in rest or in constant speed motion