Now I have made my first exam. I scored 88/100 which I am very pleased with. Thank you very much professor Lewin. I am so grateful for all your lessons and the inspiration. I have also had practical use for your tip “Any measuring without knowledge of the uncertainty is meaningless”
47:43 The same EXACT question was asked in JEE Advanced 2020(one of the toughest year of the exam) , this just shows the power and the concepts behind this lectures. Thank you sir, for these beautiful lectures.
I'm learning physics on my own and it's vital to have the solutions since I have no one to verify my work with. Thank you sooo much for posting both the exam AND the solutions! It's really quite helpful.
@@siamsama2983 Assignments 8.01 Ass #1 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/997BD3CB-BDE0-4308-BD36-0A75083DB6AD/0/assign1.pdf #1 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/A9F97BBD-590B-4E5F-9B07-A10F1DEDAB7C/0/sol1.pdf #2 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1D390487-9CDE-4CB3-B499-0D1688E2C5D3/0/assign2.pdf #2 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3920665E-30F7-4B27-9E79-7A45F3432EB9/0/sol2.pdf #3 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3A1C14A6-A41B-4EC2-B5F0-400CFFBA3793/0/assign3.pdf #3 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/6501CF54-92E5-4067-9A5B-3A559749C7DD/0/sol3.pdf #4 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DFBCFABA-40BD-4D4E-99E8-7CF420AA86A4/0/assign4.pdf #4 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/BB822774-4C35-4E20-A54A-14B6793CA800/0/sol4.pdf #5 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3ED571A3-0011-4463-B81F-BE6A18744C12/0/assign5.pdf #5 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/C1EB70E4-3A7A-49BE-81CF-1954C39F2A15/0/sol5.pdf #6 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/0FDE03A1-2828-402F-B6E5-4844E963BE18/0/assign6.pdf #6 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/8D3EA3C0-149F-4212-B926-AE0DAE87D642/0/sol6.pdf #7 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/89B4EED5-F732-4F89-9C7A-2D549A573ACF/0/assign7.pdf #7 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/39983B73-56C6-4005-B622-AB9FCB6BCCB0/0/sol7.pdf #8 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/D174EE47-52FD-41AE-8718-DD86C256566E/0/assign8.pdf #8 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DD5CAB9F-A069-4548-95B7-4057ECE66378/0/sol8.pdf #9 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/26F5ED2D-7A54-4678-99E7-E7EDF2A83C8A/0/assign9.pdf #9 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/FBE3E6D0-0B79-4400-A7DA-3D9BF106BDE6/0/sol9.pdf #10 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/95FB4DDE-8D57-402E-9710-B63D23828E7F/0/assign10.pdf #10 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/4BBD0B5D-73A4-407B-BD2E-95B5C74062DF/0/sol10.pdf -----------------------------------------------------------------
Lecture Notes 8.01 #1 femur bones of mammals mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1257BAA3-1C8C-48FF-81FF-1DA453800B70/0/supplement1.pdf #5 Orbital Information on Planets mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/A0961E36-2E51-48C6-9E2B-0E5CB004BEEF/0/supplement3.pdf #9 Information on Exam 1 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/789B2A1A-A4DD-4C96-A8F3-B47F87435FCE/0/supplement4_2.pdf #12 Resistive Forces and Spheres mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/CAF0F24D-5D22-47F9-B62C-DB864D2906E8/0/sup5_1.pdf #12 Karo Corn Syrup mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/107CFE7F-3A51-4C1A-8B57-5FE429470918/0/sup5_2.pdf #12 Drop Steel Ball Bearings in Corn Syrup mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/95F5D12B-3416-4CF4-8ED5-54F5BEE6896E/0/sup5_3.pdf #12 Falling Balloon with and without Air Drag mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/29D26ACF-2C8E-4AF0-88B4-EA4B4715CC56/0/sup5_4.pdf #12 Falling Pebble with and with Air Drag mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/08EED4D6-7B49-4278-AF04-4A4C18B5053E/0/sup5_5.pdf #14 Energy and Power Consumption mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/D4E5B35A-1B97-41C8-B5F4-E686576A8F36/0/supplement6.pdf #17 Rocket Equations mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/77BBA399-5FAB-4A16-AA53-29A30382EEB2/0/supplement8.pdf #18 Information on Exam Review mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/ED4D3A5C-A164-4766-AB36-8D1CE6585F68/0/sup7.pdf #19 Flywheel Technology mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/9470340C-BABD-4EAC-B760-0B6DA14CCBB7/0/sup_1027.pdf #22 Notes on Kepler Orbits mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1B01D195-CD41-41B0-A0B2-D851DDBC2C17/0/sup1103_1.pdf #22 Problem 8.2 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3A3C2B8E-82EC-4D37-8D9C-7A11C6844988/0/sup1103_2.pdf #22 Computer Demonstration - To Catch or not to Catch - Peter and Mary - Sandwich mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/56E47F6B-FE04-4E57-8345-3439F411656F/0/sup1103_3.pdf #22 Data Used by Kepler mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/883A30E0-7B16-4A58-9613-5F9EE5DCE537/0/sup1103_4.pdf #23 Effects of Inclination of Binary Systems mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/34F12704-06CB-420A-A0F7-47EB9506DD88/0/cor_l1105.pdf #24 Torques on Rotating Disks - Gyroscopic Behavior mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/23E0D37E-AD7B-49C2-BA8E-A6A3571B7401/0/sup1108.pdf #25 Conversion from Linear to Rotational Motion mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/B57C6922-557D-493C-9F93-8A05644FA251/0/sup1110.pdf #26 Elasticity of Metals mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/FFFE09F0-F4AB-4F1B-994B-9CC9A3D788AE/0/sup1112.pdf #29 Information on Exam 3 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/4B2E3AB7-E85B-4049-8E82-233129B5F8FB/0/sup1115.pdf #33 Ideal Gas Law mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DA4EE7E1-BC62-48EB-9DA8-6A5023B85D7E/0/sup1203_2.pdf #33 Comments on Phase Transitions mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/42042F15-1E9E-4C2C-9E10-C3914187BC39/0/sup1203_2.pdf ------------------------------------------- EXAMS 8.01 #1 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/398A22AF-F32D-4603-9629-63B8D263CD97/0/exam_1.pdf #1 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/5735976D-D4CC-44E0-B7A1-226D8BFAC1A1/0/quiz1.pdf #2 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/CDF11F4B-37AC-45AB-95AA-A827F7B1C217/0/exam_2.pdf #2 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1EAC2DBA-DA72-445C-A23E-579BCABE5C34/0/quiz2_sol.pdf #3 mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/44006D83-7DA5-4AE0-996D-B43852286067/0/exam3_V2.pdf #3 sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/E07AFD65-842F-4D5C-8F38-93943E2FC0DC/0/quiz993_sol.pdf Final mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/375AA631-F701-4821-B9FA-A0A6FC74B35A/0/final.pdf Final sol mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/F483BF93-6367-4067-AD0B-AF7ED385EE8B/0/finalsol.pdf
@@lecturesbywalterlewin.they9259 Woah idk what you've done they look like the same links but now it works. Thanks sir! Also to properly do the assignments, will I only need the Principles of Physics textbook by Ohanian? Will this textbook also be used/help with your 8.02 and 8.03 lectures as well?
Sir, after watching few of your lectures I'm actually considering going back to university to study physics. Sadly it won't be with you, but I wanna know much more about such wonderful subject! Thank you very much!!
Sir Walter Lewin: in order for one finger to move (slide) under the yard stick the finger force must overcome the static friction,, it is very unlikely to have both fingers pushing towards the center of the stick equally and very unlikely to have a perfect stick surface and very unlikely to have perfectly symmetrical fingers and touch point/surface on the yardstick,,, so eventually one finger crosses the static friction before the other and begins to slide,, then the kinetic friction takes on which is less than the static one so the finger which is moving does not need the same force as before to continue moving while the other finger still has to exceed the static friction,, and the moving finger gets closer to the center the torque applied on it gets bigger and bigger so the weight it feels from the yardstick is larger than when it starts moving and as this weight increases friction increases proportionally till it halts and then the other finger which has now much less friction takes the ride and so an so on,,, Thank You Sir for being born
I think in the last yardstick experiment, as we move our fingers friction opposes our motion... Now since the center of gravity shifts the normal force also changes... Changing friction... This also implies that no matter from where I start my fingers I always end up below the center...
Net Torque on the stick is 0 since its not rotating. So, R1 * N1 = R2 * N2, where N1, N2 are the normal forces from the fingers & R1,R2 are the distances from fingers.. So, the finger that's far away gets lesser friction than the one thats near since friction is proportional to N, the normal force between the surfaces
Sir is my below explanation is correct, Suppose at start both hand fingers at ends then normal reaction exerted by fingers is half of yard stick weight Let start left move left finger close to right finger since force exerted by left finger is more than max static friction on right finger so stick starts to overhang on right finger side At the same time normal reaction exerted by right finger starts to increase as centre of mass of yard stick moves close to right finger as the result max static friction starts to build up more than force exerted by left finger Movement of yard stick stops and same process continues when u move right finger close to left finger
The exam was much easier than I thought. I expected there to be complex questions involving both *uniform* circular motion and kinematic equations mashed together. Like "assume string A is cut at t = 5.0 s, at what position x does m1 land?" or something. Overall,100/100, thank you for providing the exams as well as the class' exercises. The question at the end of the lecture: I believe it happens because the wood might be grained such that it has a higher static friction at one direction than the other. After you get it moving, you can more easily move the ruler because the low kinetic friction. I may be off (or way off), but that is what I thought about it. Thanks. I also love your guess the artist and painting questions (:
@@lecturesbywalterlewin.they9259 Sir is it because of the varying normal rection force, i mean since the rod is in rotational eqilibrium, torque must be balanced and thus normal reaction on both the fingers vary as they come close , and hence, also the friction. Is this the reason why this happens??
48:43 I'll call 2 arbitrary hands 1 & 2..... A. The 1hand which starts to move towards the center......opposes the motion of the 2hand, adds with the opposing friction. ......................so opposes 2hand a little more...... hence 2hand stops... this explains... THE MOTION OF ONLY ONE HAND AT A TIME.... B. Then as the 1hand moves towards center of mass....... experiences more kinetic friction..... 1hand stops ........now, opposing forces on 2hand decrease.... so it starts to move.... THIS HAPPENS SIMULTANEOUSLY... C. This whole process starts bcs of difference between static friction between 2 points....and randomly anyone hand starts the motion..........I think..... Thanks for these great lectures Professor Lewin!
here is my explaination for the yard stick sliding at the end: as you move your first finger the weight shifts on that finger generating more friction on opposite to the direction of finger movement, then the other finger starts moving untill there friction stops it allowing the first finger to move again, the ending result is that yyour fingers meet at the center of mass
About the yardstick problem----- If we consider torque about centre then torque will be by the 2 forces from our fingers which are pushing the yardstick...... As we move one finger closer to centre, the torque due to that force about the centre decreases AND HENCE WE START TO SEE THE YARDSTICK ROTATING IN THE DOWNWARD DIRECTION FROM THE LEFT SIDE......... But then you immediately move the right-hand closer to the centre so the 2 toques again balance each other at some point but as you move it further towards the centre, WE SEE IT BENDING FROM RIGHT SIDE.......... I can't understand why you said that we should refer to the previous lecture?
how many minutes into the lecture? If you refer to the demo where I have a yardstick resting on one finger of my right hand and one of my left hand, and move the fingers slowly together, then it has nothing to do with torques.
Yes, I am talking about that same experiment...... If you move one finger towards the centre then friction will act in the opposite direction of movement but since the finger is already moving so that means kinetic friction is involved here but I think it's all about having no torque on the yardstick otherwise it will fall.....
Make a free-body diagram of the ruler. One finger 40 cm from the center and the other 20 cm from the center. NO movements! Static equilibrium! Then start moving your fingers calculate the maximal frictional force onto the ruler at each finger. That will then tell you the whole story of the demo. good luck!
Sir,what you love most in physics,i mean you love every part of it we know,but is there something which means a lot to you,or has a deep affection with you.something you feel most connected and adore the most in physics....thanks
Sir, I have noticed coffee cup in right hand pocket of shirt. In another lecture I noticed an egg in the same location. What is purpose? Is breakfast your favorite meal of the day? Thank you (yes I think it is humorous lol)
Sir, first of all thank you for giving this great material and making it available to everyone who is interested. I tried following all your lectures, doing assignments, reading available course material from MIT free course downloads, also did the exam, but I have a question. Is it possible to obtain the physics book that was available for the students in the days these lectures were filmed? I tried to get the free PDF for Physics book by Ohanian so I could fully follow the assignments and read material before it's covered in lectures. Of course, I expect it to be exclusive to MIT students and copyrighted but I have to ask. Thank you again !
Prof. Lewin. I was always wondering, how would physics describe a moving object that comes to a halt. It must go trough all the speed steps, including infinitely slow in order to stop.
Sir, does the yardstick move that way because when the left finger moves toward the right, the force it exerts on the stick increases, causing the force exerted by right finger to decrease? Thus the static friction on the right finger decreases, as friction is directly proportional to the normal force, allowing right finger to move and vice versa. Is my reasoning correct?
@@lecturesbywalterlewin.they9259 for me the tougher question comes after, which is "why does the 1st finger stop?" my idea is: as soon as the 2nd finger begins to move, the horizontal force it exerts drops instantaneously (from maximum static to kinetic). while it may sometimes seem that at this moment, the 1st finger stops instantaneously, this is not true. as soon as the 2nd finger starts to move, both fingers are sliding, but the 1st finger has a greater kinetic frictional force (due to the greater normal force). therefore, there is a NET FORCE on the stick in the direction the first finger was/is moving. since both fingers are moving slow, and the mass of the stick is small, the resulting acceleration quickly brings the stick to the same velocity as finger 1, making their relative velocity 0, and so they are at this moment, at rest with respect to each other. so the horizontal force of finger 1, which WAS enough to slide, is now not enough to start motion and therefore the relative velocity remains zero until the process starts over. is this correct? spent the better part of an hour on this.
I think the explanation is - not sure though - when you move your left finger to the right - assuming it starts first - the rod is almost at equilibrium so when the normal force from your finger comes to the right a little bit the force must increase to keep the moment = 0 at any point since the friction depends on the normal force , the friction on the left finger increases causing it to move to the left and so on i have reached these conclusion after trying to do it very slowly , i have noticed it is not the speed that matters but the position , am i right ?
about 17-25 minutes to the lecture, the undefined instantaneous velocity changes and accelerations are infinitely high. But didn't we also agree that acceleration is dv/dt, and v is dx/dt? I know it's mathematically awkard but maybe if I say an object with a speed hits a wall and stops immediately at that moment, cannot we define it physically that it had an acceleration due to an impulse?
Yes, it would be an impulse, or Dirac function. It is what you get when you take the derivative at a jump discontinuity, or a step function. In reality, it is impossible for any object of non-zero mass to do this, and there ultimately will be a finite acceleration, rather than an instantaneous impulse.
Radians multiplied by radius, will eliminate the angle unit. A radian is defined as the angle formed as 1 radius is wrapped around the circle. 1 Radian * 1 radius = 1 radius worth of arc length.
what you think sir,about prof.feynman i am sure you are inspired by him.what you think he did differently that rest of us not even think.Is there something fundamental in understanding the nature ,we are missing that feynman,einstein and of course you noticed.any tips would be much appreciated sir.thanks a lot
Sir., In your lecture on revision of units , vectors, Newton laws and kinematics and friction you moved both of your fingers towards each other under a stick on them. Only one finger moved towards other one in an alternating manner. Sir is it because your fingers were not applying equal force at same time. We know force of friction acts in opposite direction of Motion so at once the finger exerting greater force moved towards other finger. The alterations I think was because the amount of force was not constant once it was greater in your left finger then at another second it was greater in your right finger. I also think torque was also working as the finger farther than the center of Stik moved first when it came closer to the center than other finger then second finger started moving. But the resultant net torque was zero Sir am I right
If the thickness of the femur is proportional to the length at the power 3/2 that means that the constant of proportionality in not adimensional. Does it mean that we can't always trust results obtained by pure dimensional analysis? Any dimensional analysis will never take into account the possibility to have constants of proportionality with their own dimensions. Is it really so or am I missing something? Thank you if you decide to answer. Thank you anyway for publishing such wonderful lectures.
>>>Any dimensional analysis will never take into account the possibility to have constants of proportionality with their own dimensions. >>> *you miss the point. Constants do not enter into my dimensional evaluation of the bones. Watch the lecture again!!!!*
Lectures by Walter Lewin. They will make you ♥ Physics. Ok, thank you. I got the point for what concerns the argument with the bones. Now it comes clear to me that I've put as correlated two questions that are actually different. One doubt remains (not talking about the bones anymore, just about the method of dimensional analysis). When we derive that a variable must have a certain dipendence from others through a dimensional analysis (for example the period of a pendulum has to be proportional to the square root of its length divided by g because that's the only combination which is dimensionally compatible), how can we be sure that we cannot have any other contribute containing dimensional constants (for example how can we exclude a dipendence from the mass maybe because of an additional term of the form "mass of the pendulum"/"universal constant with dimensions of mass divided by time" - well, not saying it could really exist that dependence in the case of the pendulum, but asking about general case)?
If dimensional analysis allows you find dependencies on other dimensions then you need the constants to get numerical values but that was not the case when I checked Galileo's assumption.
I think that in the NASA centrifuge you are struggling with the acceleration if it would be in a straight line because in any moment you are facing towards the center of centrifuge so you are feeling "push" from chair. Isn't it right professor Lewin?
If you sit in a chair and are accelerated in straight line in the direction that you face, you feel the push in your back. The same is true in the NASA centrifuge - push in your back.
Professor Walter Lewin How will I get the exam questions Physics Materials Classical mechanics, electricity and magnets, vibrations and waves that I did at MIT University?
23:15 why first second's function is lineer, I know there was an acceleration but why you didn't said us or was there an another way to find why function is lineer
The velocity is changing during first second from 0 to -6, so there is actually acceleration of -6m/s^2, otherwise the velocity would be constant. Remember that this is V(t) graph :)
Thanks for replying sir, I self evaluated myself(honestly), and I was getting 95,I deducted 5 marks because I think that I didn't answer the first question showing all the calculations, though my answers were correct, I did not write the small calculations because I was lazy. But then while evaluating I thought may be you would cut my marks for that. Did I do right?
I got it that since the normal force will be changing while we move fingers therefore frictional force will be changing which cause fingers to move and stop alternatively
Take the example of projectiles. For small ranges, the direction and magnitude of acceleration due to gravity can be considered to be constant. Let a particle is projected at an angle of a with the horizontal with velocity u. Let us consider a rectangular coordinate system with it's origin at the point of projection. Therefore the velocity can be written as u cos a î + u sin a j . Acceleration = -g j. So we can see that the acceleration is only along the y axis. So the x component of the velocity is constant while the y component first decreases till it becomes 0 at maximum height, and then the y component of the velocity vector keeps on increasing, but in the downward direction. Formulate the equations of motion and finally you will get the equation for parabola of the form y = mx^2+kx+c.
Simplest answer is probably this: if you have a graph of x as a function of time and it's a parabola, that means it's a quadratic equation. By taking the derivative of it with respect to time, you get velocity equation and the derivative of a quadratic function is a linear function. If you take the derivative of that again, you get acceleration. The derivative of a linear function will no longer have any variable t in it. So it's constant.
when u r calculating the velocity in the y-direction at point S, u took acceleration as -gt but the plane or vector is in the direction of gravitational acceleration. So shouldn't it be V = v + gt ?
+Harsha s n Signs are KEY. I suggest you listen again to my lecture and make a serious effort to understand what I did and why Idid it! If you have a sign wrong you end up with nonsense.
+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for the reply. You have considered the same sign at point S as at the initial point i.e, increasing value of Y (upwards ).
Sir regarding the yardstick problem, If your right finger moves first, it must apply a larger force as the net torque on the stick is 0. As a result, the normal force increases and consequently friction increases so it stops and the other finger moves. Is this correct?
I mean sir we know that two electric line never intersect to each other .in this case at the point of intersection of two field line when suppose cross why I can't use resultant value of two electric line for knowing its directions
Electric field lines of force never intersect each other because, at the point of intersection, two tangents can be drawn to the two lines of force. This means two directions of the electric field at the point of intersection, which is not possible.
sir, why did that happen?? Is it that when the left finger is moved static friction starts acting on the right finger? and the same happens with the left finger when the right finger starts moving??
sir i am a jee aspirant for 2024, i understand concepts, but have a struggle in putting it in equations... please suggest me some tips to improve... I have seen a few of your 8.01 videos but equations are a struggle... One more thing, i dont understand when to stop asking questions, sometimes i ask to deep of questions which isn't in the syllabus for which i get scolded... please help me with that sir... Thank you
you have 2 options option 1: eat yogurt every day but *never on Fridays* option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Sir In yardstick experiment, if i neglect gravity and the surface is same as before then Is there any effect on the result that is fingers alternating process..or it is same as before?
Though I'm no Walter Lewin, perhaps I can help you out. If you neglect gravity (and assume that you and the ruler are not accelerating), there will be no force from your hands pushing up on the ruler (there doesn't need to be - there's no force from gravity pulling it down). If there's no contact force between those two surfaces, can there be any friction? Without friction, would you observe the alternating pattern?
Constant acceleration from rest, means that a graph of velocity vs time will be a straight line, with a slope equal to the acceleration, and intersecting the origin. The area of the triangle between this line and the time axis, will correspond to the change in position. Area of a triangle is 1/2*base*height, and for constant acceleration kinematics this becomes ∆x = 1/2*a*t^2. The 1/2 comes from the fact that the triangle only fills half the area of the rectangle it fits within. Adapt it to freefall, ∆x turns into ∆y because it is vertical, and a = -g by definition of free fall. This gives us ∆y = -1/2*g*t^2.
How many marks would I get for just saying Problem 1: a) 2 seconds, 20 meters b) 20 meters c) 20 meters per second I'm guessing like one for each since I'm missing all the formulas which are actually important :D
The one about the stone!! ^^ At time t=0 sec we throw a stone from the ground level straight up with a speed of 20m/sec (ignore airdrag and assume g=10m/sec squared) A) At what time (in sec) will this stone reach its highest point and how high is it then above the ground? (6 marks potential) B) We now throw a second stone straight up 2 sec after the first. How many meters above the ground is the first stone at that moment? (6 marks potential) C) At what speed should we throw this second stone from the ground if it is to hit the first stone 1 second after the second stone is thrown? (10 potential mark) 22 marks potential I assume I'd score 3/22 (if all are even right...) which is not even a passing grade!
Now I have made my first exam. I scored 88/100 which I am very pleased with.
Thank you very much professor Lewin. I am so grateful for all your lessons and the inspiration. I have also had practical use for your tip “Any measuring without knowledge of the uncertainty is meaningless”
great!
47:43 The same EXACT question was asked in JEE Advanced 2020(one of the toughest year of the exam) , this just shows the power and the concepts behind this lectures.
Thank you sir, for these beautiful lectures.
But, what is the answer????
I'm learning physics on my own and it's vital to have the solutions since I have no one to verify my work with. Thank you sooo much for posting both the exam AND the solutions! It's really quite helpful.
How do you access them? The link doesn't work me, none of the links in any of his videos work for me :( Help please
@@siamsama2983 Assignments 8.01
Ass #1
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/997BD3CB-BDE0-4308-BD36-0A75083DB6AD/0/assign1.pdf
#1 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/A9F97BBD-590B-4E5F-9B07-A10F1DEDAB7C/0/sol1.pdf
#2
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1D390487-9CDE-4CB3-B499-0D1688E2C5D3/0/assign2.pdf
#2 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3920665E-30F7-4B27-9E79-7A45F3432EB9/0/sol2.pdf
#3
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3A1C14A6-A41B-4EC2-B5F0-400CFFBA3793/0/assign3.pdf
#3 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/6501CF54-92E5-4067-9A5B-3A559749C7DD/0/sol3.pdf
#4
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DFBCFABA-40BD-4D4E-99E8-7CF420AA86A4/0/assign4.pdf
#4 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/BB822774-4C35-4E20-A54A-14B6793CA800/0/sol4.pdf
#5
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3ED571A3-0011-4463-B81F-BE6A18744C12/0/assign5.pdf
#5 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/C1EB70E4-3A7A-49BE-81CF-1954C39F2A15/0/sol5.pdf
#6
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/0FDE03A1-2828-402F-B6E5-4844E963BE18/0/assign6.pdf
#6 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/8D3EA3C0-149F-4212-B926-AE0DAE87D642/0/sol6.pdf
#7
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/89B4EED5-F732-4F89-9C7A-2D549A573ACF/0/assign7.pdf
#7 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/39983B73-56C6-4005-B622-AB9FCB6BCCB0/0/sol7.pdf
#8
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/D174EE47-52FD-41AE-8718-DD86C256566E/0/assign8.pdf
#8 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DD5CAB9F-A069-4548-95B7-4057ECE66378/0/sol8.pdf
#9
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/26F5ED2D-7A54-4678-99E7-E7EDF2A83C8A/0/assign9.pdf
#9 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/FBE3E6D0-0B79-4400-A7DA-3D9BF106BDE6/0/sol9.pdf
#10
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/95FB4DDE-8D57-402E-9710-B63D23828E7F/0/assign10.pdf
#10 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/4BBD0B5D-73A4-407B-BD2E-95B5C74062DF/0/sol10.pdf
-----------------------------------------------------------------
Lecture Notes 8.01
#1 femur bones of mammals
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1257BAA3-1C8C-48FF-81FF-1DA453800B70/0/supplement1.pdf
#5 Orbital Information on Planets
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/A0961E36-2E51-48C6-9E2B-0E5CB004BEEF/0/supplement3.pdf
#9 Information on Exam 1
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/789B2A1A-A4DD-4C96-A8F3-B47F87435FCE/0/supplement4_2.pdf
#12 Resistive Forces and Spheres
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/CAF0F24D-5D22-47F9-B62C-DB864D2906E8/0/sup5_1.pdf
#12 Karo Corn Syrup
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/107CFE7F-3A51-4C1A-8B57-5FE429470918/0/sup5_2.pdf
#12 Drop Steel Ball Bearings in Corn Syrup
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/95F5D12B-3416-4CF4-8ED5-54F5BEE6896E/0/sup5_3.pdf
#12 Falling Balloon with and without Air Drag
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/29D26ACF-2C8E-4AF0-88B4-EA4B4715CC56/0/sup5_4.pdf
#12 Falling Pebble with and with Air Drag
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/08EED4D6-7B49-4278-AF04-4A4C18B5053E/0/sup5_5.pdf
#14 Energy and Power Consumption
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/D4E5B35A-1B97-41C8-B5F4-E686576A8F36/0/supplement6.pdf
#17 Rocket Equations
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/77BBA399-5FAB-4A16-AA53-29A30382EEB2/0/supplement8.pdf
#18 Information on Exam Review
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/ED4D3A5C-A164-4766-AB36-8D1CE6585F68/0/sup7.pdf
#19 Flywheel Technology
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/9470340C-BABD-4EAC-B760-0B6DA14CCBB7/0/sup_1027.pdf
#22 Notes on Kepler Orbits
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1B01D195-CD41-41B0-A0B2-D851DDBC2C17/0/sup1103_1.pdf
#22 Problem 8.2
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/3A3C2B8E-82EC-4D37-8D9C-7A11C6844988/0/sup1103_2.pdf
#22 Computer Demonstration - To Catch or not to Catch - Peter and Mary - Sandwich
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/56E47F6B-FE04-4E57-8345-3439F411656F/0/sup1103_3.pdf
#22 Data Used by Kepler
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/883A30E0-7B16-4A58-9613-5F9EE5DCE537/0/sup1103_4.pdf
#23 Effects of Inclination of Binary Systems
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/34F12704-06CB-420A-A0F7-47EB9506DD88/0/cor_l1105.pdf
#24 Torques on Rotating Disks - Gyroscopic Behavior
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/23E0D37E-AD7B-49C2-BA8E-A6A3571B7401/0/sup1108.pdf
#25 Conversion from Linear to Rotational Motion
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/B57C6922-557D-493C-9F93-8A05644FA251/0/sup1110.pdf
#26 Elasticity of Metals
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/FFFE09F0-F4AB-4F1B-994B-9CC9A3D788AE/0/sup1112.pdf
#29 Information on Exam 3
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/4B2E3AB7-E85B-4049-8E82-233129B5F8FB/0/sup1115.pdf
#33 Ideal Gas Law
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/DA4EE7E1-BC62-48EB-9DA8-6A5023B85D7E/0/sup1203_2.pdf
#33 Comments on Phase Transitions
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/42042F15-1E9E-4C2C-9E10-C3914187BC39/0/sup1203_2.pdf
-------------------------------------------
EXAMS 8.01
#1
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/398A22AF-F32D-4603-9629-63B8D263CD97/0/exam_1.pdf
#1 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/5735976D-D4CC-44E0-B7A1-226D8BFAC1A1/0/quiz1.pdf
#2
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/CDF11F4B-37AC-45AB-95AA-A827F7B1C217/0/exam_2.pdf
#2 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/1EAC2DBA-DA72-445C-A23E-579BCABE5C34/0/quiz2_sol.pdf
#3
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/44006D83-7DA5-4AE0-996D-B43852286067/0/exam3_V2.pdf
#3 sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/E07AFD65-842F-4D5C-8F38-93943E2FC0DC/0/quiz993_sol.pdf
Final
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/375AA631-F701-4821-B9FA-A0A6FC74B35A/0/final.pdf
Final sol
mit.sustech.edu/NR/rdonlyres/Physics/8-01Physics-IFall1999/F483BF93-6367-4067-AD0B-AF7ED385EE8B/0/finalsol.pdf
@@lecturesbywalterlewin.they9259 Woah idk what you've done they look like the same links but now it works. Thanks sir!
Also to properly do the assignments, will I only need the Principles of Physics textbook by Ohanian? Will this textbook also be used/help with your 8.02 and 8.03 lectures as well?
All links lead to a 404 error page for me.
I freaking adore the open ended questions you make at the end of the lectures!
Tasos Obscure GREAT!
👋🏻 Hyy Me too, hws you doing now ?
It always feels like you're directly addressing me. You are the greatest teacher I can ever have
Wow, thank you!
Sir, after watching few of your lectures I'm actually considering going back to university to study physics. Sadly it won't be with you, but I wanna know much more about such wonderful subject! Thank you very much!!
;)
Which university did you apply to?
Mankind's knowledge held in my hand. What a time to be alive! Yet most of us are unhappy...I don't understand it.
Sir Walter Lewin:
in order for one finger to move (slide) under the yard stick the finger force must overcome the static friction,, it is very unlikely to have both fingers pushing towards the center of the stick equally and very unlikely to have a perfect stick surface and very unlikely to have perfectly symmetrical fingers and touch point/surface on the yardstick,,, so eventually one finger crosses the static friction before the other and begins to slide,, then the kinetic friction takes on which is less than the static one so the finger which is moving does not need the same force as before to continue moving while the other finger still has to exceed the static friction,, and the moving finger gets closer to the center the torque applied on it gets bigger and bigger so the weight it feels from the yardstick is larger than when it starts moving and as this weight increases friction increases proportionally till it halts and then the other finger which has now much less friction takes the ride and so an so on,,,
Thank You Sir for being born
agree. right on the difference between static and kinetic, and the sum of torque being zero. Like my daughter would say: Your rock mr LEWIN!
the concept of torque came from where?? And how it Increases??
I think in the last yardstick experiment,
as we move our fingers friction opposes our motion... Now since the center of gravity shifts the normal force also changes... Changing friction...
This also implies that no matter from where I start my fingers I always end up below the center...
I appreciate you prof WALTER
Net Torque on the stick is 0 since its not rotating. So, R1 * N1 = R2 * N2, where N1, N2 are the normal forces from the fingers & R1,R2 are the distances from fingers.. So, the finger that's far away gets lesser friction than the one thats near since friction is proportional to N, the normal force between the surfaces
Sir is my below explanation is correct,
Suppose at start both hand fingers at ends then normal reaction exerted by fingers is half of yard stick weight
Let start left move left finger close to right finger since force exerted by left finger is more than max static friction on right finger so stick starts to overhang on right finger side
At the same time normal reaction exerted by right finger starts to increase as centre of mass of yard stick moves close to right finger as the result max static friction starts to build up more than force exerted by left finger Movement of yard stick stops and same process continues when u move right finger close to left finger
The exam was much easier than I thought. I expected there to be complex questions involving both *uniform* circular motion and kinematic equations mashed together. Like "assume string A is cut at t = 5.0 s, at what position x does m1 land?" or something.
Overall,100/100, thank you for providing the exams as well as the class' exercises.
The question at the end of the lecture:
I believe it happens because the wood might be grained such that it has a higher static friction at one direction than the other. After you get it moving, you can more easily move the ruler because the low kinetic friction.
I may be off (or way off), but that is what I thought about it.
Thanks. I also love your guess the artist and painting questions (:
explanation incorrect
@@lecturesbywalterlewin.they9259 Sir is it because of the varying normal rection force, i mean since the rod is in rotational eqilibrium, torque must be balanced and thus normal reaction on both the fingers vary as they come close , and hence, also the friction. Is this the reason why this happens??
Yeah! Me too, I thought these questions are little very easy once you practice enough textbook questions
Thank you professor for giving the feel of physics .
What fact you learned that blew your mind in physics........thanks
48:43
I'll call 2 arbitrary hands 1 & 2.....
A. The 1hand which starts to move towards the center......opposes the motion of the 2hand, adds with the opposing friction.
......................so opposes 2hand a little more...... hence 2hand stops...
this explains...
THE MOTION OF ONLY ONE HAND AT A TIME....
B. Then as the 1hand moves towards center of mass....... experiences more kinetic friction..... 1hand stops
........now, opposing forces on 2hand decrease.... so it starts to move....
THIS HAPPENS SIMULTANEOUSLY...
C. This whole process starts bcs of difference between static friction between 2 points....and randomly anyone hand starts the motion..........I think.....
Thanks for these great lectures Professor Lewin!
here is my explaination for the yard stick sliding at the end:
as you move your first finger the weight shifts on that finger generating more friction on opposite to the direction of finger movement, then the other finger starts moving untill there friction stops it allowing the first finger to move again,
the ending result is that yyour fingers meet at the center of mass
About the yardstick problem-----
If we consider torque about centre then torque will be by the 2 forces from our fingers which are pushing the yardstick...... As we move one finger closer to centre, the torque due to that force about the centre decreases AND HENCE WE START TO SEE THE YARDSTICK ROTATING IN THE DOWNWARD DIRECTION FROM THE LEFT SIDE.........
But then you immediately move the right-hand closer to the centre so the 2 toques again balance each other at some point but as you move it further towards the centre, WE SEE IT BENDING FROM RIGHT SIDE..........
I can't understand why you said that we should refer to the previous lecture?
Why can't the other finger move till the first one stops?
how many minutes into the lecture? If you refer to the demo where I have a yardstick resting on one finger of my right hand and one of my left hand, and move the fingers slowly together, then it has nothing to do with torques.
Yes, I am talking about that same experiment...... If you move one finger towards the centre then friction will act in the opposite direction of movement but since the finger is already moving so that means kinetic friction is involved here but I think it's all about having no torque on the yardstick otherwise it will fall.....
Make a free-body diagram of the ruler. One finger 40 cm from the center and the other 20 cm from the center. NO movements! Static equilibrium! Then start moving your fingers calculate the maximal frictional force onto the ruler at each finger. That will then tell you the whole story of the demo. good luck!
Sir,what you love most in physics,i mean you love every part of it we know,but is there something which means a lot to you,or has a deep affection with you.something you feel most connected and adore the most in physics....thanks
Sir, I have noticed coffee cup in right hand pocket of shirt. In another lecture I noticed an egg in the same location. What is purpose? Is breakfast your favorite meal of the day? Thank you (yes I think it is humorous lol)
Sir, first of all thank you for giving this great material and making it available to everyone who is interested. I tried following all your lectures, doing assignments, reading available course material from MIT free course downloads, also did the exam, but I have a question. Is it possible to obtain the physics book that was available for the students in the days these lectures were filmed? I tried to get the free PDF for Physics book by Ohanian so I could fully follow the assignments and read material before it's covered in lectures. Of course, I expect it to be exclusive to MIT students and copyrighted but I have to ask. Thank you again !
8.01
Physics
Hans C. Ohanian
Physics
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
@@lecturesbywalterlewin.they9259 Thank you, you're great !
Prof. Lewin. I was always wondering, how would physics describe a moving object that comes to a halt. It must go trough all the speed steps, including infinitely slow in order to stop.
Lectures by Walter Lewin. They will make you ♥ Physics.
• 1 second ago
watch my lectures; that's what they are for.
Sir, does the yardstick move that way because when the left finger moves toward the right, the force it exerts on the stick increases, causing the force exerted by right finger to decrease? Thus the static friction on the right finger decreases, as friction is directly proportional to the normal force, allowing right finger to move and vice versa. Is my reasoning correct?
correct
@@lecturesbywalterlewin.they9259
for me the tougher question comes after, which is "why does the 1st finger stop?" my idea is: as soon as the 2nd finger begins to move, the horizontal force it exerts drops instantaneously (from maximum static to kinetic). while it may sometimes seem that at this moment, the 1st finger stops instantaneously, this is not true. as soon as the 2nd finger starts to move, both fingers are sliding, but the 1st finger has a greater kinetic frictional force (due to the greater normal force). therefore, there is a NET FORCE on the stick in the direction the first finger was/is moving. since both fingers are moving slow, and the mass of the stick is small, the resulting acceleration quickly brings the stick to the same velocity as finger 1, making their relative velocity 0, and so they are at this moment, at rest with respect to each other. so the horizontal force of finger 1, which WAS enough to slide, is now not enough to start motion and therefore the relative velocity remains zero until the process starts over. is this correct? spent the better part of an hour on this.
Great job on the content. How long was this exam designed for? Also, have you taught linear algebra and if so where could I find some videos?
Very conceptual lecture. Excellent!!!
I get perfect score, I am so glad!!
I think the explanation is - not sure though - when you move your left finger to the right - assuming it starts first - the rod is almost at equilibrium so when the normal force from your finger comes to the right a little bit the force must increase to keep the moment = 0 at any point since the friction depends on the normal force , the friction on the left finger increases causing it to move to the left and so on
i have reached these conclusion after trying to do it very slowly , i have noticed it is not the speed that matters but the position , am i right ?
Mohamed osama Mohammed, yes you are right!
Lectures by Walter Lewin. They will make you ♥ Physics.
physics works ! :D
last question's answer: torque
47:50 OMG!!! 😱 This was the question asked by JEE Advanced exam in year 2020....and prof. Told this question 5 years ago from 2020.....//amazing//
What is the answer??
2020 ... Same year I attempted Advanced...
about 17-25 minutes to the lecture, the undefined instantaneous velocity changes and accelerations are infinitely high. But didn't we also agree that acceleration is dv/dt, and v is dx/dt? I know it's mathematically awkard but maybe if I say an object with a speed hits a wall and stops immediately at that moment, cannot we define it physically that it had an acceleration due to an impulse?
Yes, it would be an impulse, or Dirac function. It is what you get when you take the derivative at a jump discontinuity, or a step function.
In reality, it is impossible for any object of non-zero mass to do this, and there ultimately will be a finite acceleration, rather than an instantaneous impulse.
The assignments are much more fun and interesting to do in my opinion (compared to the exam)
:)
Sir, at 41:20 since velocity=omega*radius, will the unit of tangential velocity at any instant be radian meter per second or meter per second?
linear vel m/s, ang vel rad/s or degrees/s
Radians multiplied by radius, will eliminate the angle unit. A radian is defined as the angle formed as 1 radius is wrapped around the circle. 1 Radian * 1 radius = 1 radius worth of arc length.
what you think sir,about prof.feynman i am sure you are inspired by him.what you think he did differently that rest of us not even think.Is there something fundamental in understanding the nature ,we are missing that feynman,einstein and of course you noticed.any tips would be much appreciated sir.thanks a lot
Feynman was brilliant and also an outstanding teacher. Einstein was *SUPER DUPER BRILLIANT* but was not a great teacher.
Sir., In your lecture on revision of units , vectors, Newton laws and kinematics and friction you moved both of your fingers towards each other under a stick on them. Only one finger moved towards other one in an alternating manner. Sir is it because your fingers were not applying equal force at same time. We know force of friction acts in opposite direction of Motion so at once the finger exerting greater force moved towards other finger. The alterations I think was because the amount of force was not constant once it was greater in your left finger then at another second it was greater in your right finger.
I also think torque was also working as the finger farther than the center of Stik moved first when it came closer to the center than other finger then second finger started moving. But the resultant net torque was zero
Sir am I right
If the thickness of the femur is proportional to the length at the power 3/2 that means that the constant of proportionality in not adimensional.
Does it mean that we can't always trust results obtained by pure dimensional analysis? Any dimensional analysis will never take into account the possibility to have constants of proportionality with their own dimensions.
Is it really so or am I missing something?
Thank you if you decide to answer.
Thank you anyway for publishing such wonderful lectures.
>>>Any dimensional analysis will never take into account the possibility to have constants of proportionality with their own dimensions. >>> *you miss the point. Constants do not enter into my dimensional evaluation of the bones. Watch the lecture again!!!!*
Lectures by Walter Lewin. They will make you ♥ Physics.
Ok, thank you. I got the point for what concerns the argument with the bones. Now it comes clear to me that I've put as correlated two questions that are actually different.
One doubt remains (not talking about the bones anymore, just about the method of dimensional analysis).
When we derive that a variable must have a certain dipendence from others through a dimensional analysis (for example the period of a pendulum has to be proportional to the square root of its length divided by g because that's the only combination which is dimensionally compatible), how can we be sure that we cannot have any other contribute containing dimensional constants (for example how can we exclude a dipendence from the mass maybe because of an additional term of the form "mass of the pendulum"/"universal constant with dimensions of mass divided by time" - well, not saying it could really exist that dependence in the case of the pendulum, but asking about general case)?
If dimensional analysis allows you find dependencies on other dimensions then you need the constants to get numerical values but that was not the case when I checked Galileo's assumption.
Lectures by Walter Lewin. They will make you ♥ Physics.
Thank you
I think that in the NASA centrifuge you are struggling with the acceleration if it would be in a straight line because in any moment you are facing towards the center of centrifuge so you are feeling "push" from chair. Isn't it right professor Lewin?
If you sit in a chair and are accelerated in straight line in the direction that you face, you feel the push in your back. The same is true in the NASA centrifuge - push in your back.
Professor Walter Lewin How will I get the exam questions Physics Materials Classical mechanics, electricity and magnets, vibrations and waves that I did at MIT University?
they are all on my channel - please make the effort to find them
@@lecturesbywalterlewin.they9259
Is it in playlists?
@@lecturesbywalterlewin.they9259
Please post for me
23:15 why first second's function is lineer, I know there was an acceleration but why you didn't said us or was there an another way to find why function is lineer
The velocity is changing during first second from 0 to -6, so there is actually acceleration of -6m/s^2, otherwise the velocity would be constant. Remember that this is V(t) graph :)
Sir the link in the description does not work. Pls look into it
Wonderful! the exam is super easy!
WHY WERE THERE NO FORCE OR FRICTION QUESTIONS IN THE TEST.
there were 4 tests
Thanks for replying sir, I self evaluated myself(honestly), and I was getting 95,I deducted 5 marks because I think that I didn't answer the first question showing all the calculations, though my answers were correct, I did not write the small calculations because I was lazy.
But then while evaluating I thought may be you would cut my marks for that.
Did I do right?
so???
I got it that since the normal force will be changing while we move fingers therefore frictional force will be changing which cause fingers to move and stop alternatively
yes
Now i am seriously getting way more curious about the chalk!..how do you make the dotted line!??
th-cam.com/video/raurl4s0pjU/w-d-xo.html
Hehehehe!..i found it!
Professor, why is the acceleration constant if it's a parabola? This is around 13 minutes
simple math using Newtons laws.
Take the example of projectiles.
For small ranges, the direction and magnitude of acceleration due to gravity can be considered to be constant. Let a particle is projected at an angle of a with the horizontal with velocity u. Let us consider a rectangular coordinate system with it's origin at the point of projection. Therefore the velocity can be written as u cos a î + u sin a j . Acceleration = -g j.
So we can see that the acceleration is only along the y axis. So the x component of the velocity is constant while the y component first decreases till it becomes 0 at maximum height, and then the y component of the velocity vector keeps on increasing, but in the downward direction. Formulate the equations of motion and finally you will get the equation for parabola of the form y = mx^2+kx+c.
Simplest answer is probably this: if you have a graph of x as a function of time and it's a parabola, that means it's a quadratic equation. By taking the derivative of it with respect to time, you get velocity equation and the derivative of a quadratic function is a linear function. If you take the derivative of that again, you get acceleration. The derivative of a linear function will no longer have any variable t in it. So it's constant.
I guess I'm too late to see the posted documents. The links doesn't work anymore.
th-cam.com/video/TQR6VkYzAeY/w-d-xo.html
Please can someone give me the link of ohanian (thr book that has the questions of the exercises ) cus i did not find it
During the whole lecture Sir, you said one half. Is it better to say just half or not?
Sir. How much time have students for do test? 1 hour or 2 hour?
when u r calculating the velocity in the y-direction at point S, u took acceleration as -gt but the plane or vector is in the direction of gravitational acceleration. So shouldn't it be V = v + gt ?
+Harsha s n Signs are KEY. I suggest you listen again to my lecture and make a serious effort to understand what I did and why Idid it! If you have a sign wrong you end up with nonsense.
+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks for the reply. You have considered the same sign at point S as at the initial point i.e, increasing value of Y (upwards ).
Sir regarding the yardstick problem,
If your right finger moves first, it must apply a larger force as the net torque on the stick is 0. As a result, the normal force increases and consequently friction increases so it stops and the other finger moves. Is this correct?
correct
But what i don't understand is why can't the other fingure move till the first one stops?
Respected sir why cannot use resultant value of the electric line at the point of intersection for the two electric line for knowing its directions
question unclear
I mean sir we know that two electric line never intersect to each other .in this case at the point of intersection of two field line when suppose cross why I can't use resultant value of two electric line for knowing its directions
Electric field lines of force never intersect each other because, at the point of intersection, two tangents can be drawn to the two lines of force. This means two directions of the electric field at the point of intersection, which is not possible.
Why can't use resultant directions of both two line at the crossing point
Resultant of two tangent at points of intersection
Reply sir
That last question came in latest jee advanced
Isn't v=0 and a=0 during the second sec?
Great sir.i want your books.
I just notice that the super plane is on the same horizontal direction for 1 second.
sir, why did that happen?? Is it that when the left finger is moved static friction starts acting on the right finger? and the same happens with the left finger when the right finger starts moving??
yes
Hello Sir I Love Your Lectures And Love For Physics ❤️
But these PDFs are not opening 🤕
Please do something about it
yes they do open but you do not know how to do it. I therefore made a playlist: "8.01 Homework, Solutions, Exams & Notes"
sir i am a jee aspirant for 2024, i understand concepts, but have a struggle in putting it in equations... please suggest me some tips to improve... I have seen a few of your 8.01 videos but equations are a struggle... One more thing, i dont understand when to stop asking questions, sometimes i ask to deep of questions which isn't in the syllabus for which i get scolded... please help me with that sir... Thank you
you have 2 options
option 1: eat yogurt every day but *never on Fridays*
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Sir
In yardstick experiment, if i neglect gravity and the surface is same as before then Is there any effect on the result that is fingers alternating process..or it is same as before?
Though I'm no Walter Lewin, perhaps I can help you out.
If you neglect gravity (and assume that you and the ruler are not accelerating), there will be no force from your hands pushing up on the ruler (there doesn't need to be - there's no force from gravity pulling it down). If there's no contact force between those two surfaces, can there be any friction? Without friction, would you observe the alternating pattern?
Hey guys please can any one tell me the time I should take to finish this exam
Just got 85/100 on the exam.
Weird that it had only kinematics, and no Friction/Forces etc.
why?
great!
The unit of the solution in Problem 2c isn't true?
why assignments are not available, when clicking on the link, it says "404 not found"
ocw.aprende.org/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/assignments/
@@lecturesbywalterlewin.they9259 Sir that link is also not opening.
@@soloonhalo3883 8.01 assignments
ocw.aprende.org/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/assignments/
If wind speed is 10 km/h and move forward, which time exactly it will take to reach at 50 km?
Sir teaches the concepts.....he doesnt solve the questions.....u can ask to Google
It depends of wich kind of vector field the velocity of the wind particles follows. It may vary accordingly to the position of the object as well.
Insufficient information
Good 🙏🙏🙏🙏🙏. Thanks 🙏🙏
I found a_stick=k.g(x_2-x_1)/(x_2+x_1)
(x_1 and x_2 their distance to the centre of mass)
21:09 is how physics looks like to Prof.Lewin
Why minus half g t square.
Constant acceleration from rest, means that a graph of velocity vs time will be a straight line, with a slope equal to the acceleration, and intersecting the origin. The area of the triangle between this line and the time axis, will correspond to the change in position. Area of a triangle is 1/2*base*height, and for constant acceleration kinematics this becomes ∆x = 1/2*a*t^2. The 1/2 comes from the fact that the triangle only fills half the area of the rectangle it fits within.
Adapt it to freefall, ∆x turns into ∆y because it is vertical, and a = -g by definition of free fall. This gives us ∆y = -1/2*g*t^2.
thanks sir
How many marks would I get for just saying
Problem 1:
a) 2 seconds, 20 meters
b) 20 meters
c) 20 meters per second
I'm guessing like one for each since I'm missing all the formulas which are actually important :D
which problem?
The one about the stone!! ^^
At time t=0 sec we throw a stone from the ground level straight up with a speed of 20m/sec (ignore airdrag and assume g=10m/sec squared)
A) At what time (in sec) will this stone reach its highest point and how high is it then above the ground? (6 marks potential)
B) We now throw a second stone straight up 2 sec after the first.
How many meters above the ground is the first stone at that moment? (6 marks potential)
C) At what speed should we throw this second stone from the ground if it is to hit the first stone 1 second after the second stone is thrown? (10 potential mark)
22 marks potential
I assume I'd score 3/22 (if all are even right...) which is not even a passing grade!
this is a high school problem
ask google for help
I understand! It was just in your exam paper, so that's why I asked!
solutions to my exams are posted on this channel
Sir , the assignment pages are not opening..
Use my playlist "8.01 Assignment & Solutions"
what is PIVoT in the assignments?
PIVoT no longer exists
some alternative?
sir, where can I find the answers for the exam-1.
pdf file below the video thumb nail
i got it sir, thank you!
I got 90pts sir.
That suits your name "cool"
@@lecturesbywalterlewin.they9259 I got 100 points sir
Thank you!
You're welcome!
Wish the camera man would stop cutting towards the female students. I don't wanna be getting bricked up mid lecture 😩
WHAT IS THE ESCAPE VELOCITY OF A BLACK HOLE ?
c
@@lecturesbywalterlewin.they9259 Thanks for the answer professor
@@lecturesbywalterlewin.they9259 But, professor light does not escape the black hole so how the answer is "c"
100/100
Julius sumner miller is the best physics teacher ever!
26:06
I am extremely happy that I am done with school....
Any PW student
Give attention 👍👍👍
I think I got 100
Hindi me bhous