Dear professor Lewin, I think you've made an enormous contribution to Science by inspiring thousands of young students and showing the beauty of Physics. I'm studying to become a scientist and your amazing lectures are extremely helpful. Thank you!
I taught many courses during a 30 year career teaching engineering. Dynamics was my favorite course. I learned something new every time I taught it. I find your lectures very informative. I really appreciate how you corrected your mistakes in your lecture and I am sure your students do too.
The transition from mundane earthly objects to awe-inspiring neutron stars and supernovas was incredible! Thank you for another great lecture, Professor!
cheers from Brazil, I've recommended your lessons in 8.01 to all my classical mechanics classmates, and we have emulated almost all your experiments so far.
Lectures by Walter Lewin. They will make you ♥ Physics. th-cam.com/channels/iEHVhv0SBMpP75JbzJShqw.html 300+ videos. Many of them with high resolution. This channel has all my lectures and talks.
This part of mechanics is also considered to be the toughest in the entire JEE syllabus, but Professor Lewin made it crystal clear in my head. Thank you sir!
Thankkk you so much for recording this lectures. I live in Turkey!! At my university, I can't really get inspiring classes to love Physics. But the last 20 minutes of this lecture had a concern to inspire us about the Physics and the nature. I study Physics and you, sir, are making me inspired on your every lecture!
for those of you curious about the calculation preformed at ~34:22. you need to calculate the potential energy of the star (which is ofcourse spherical) in the 2 cases (R = 7*10^5 km and R = 10 km); that is the work required to bring all the particles of the star from infinity to their respective positions on the sphere.
Your lectures make me feel like a kid again... sure the calculations get more and more complex as i go on, but both your infectious enthusiasm and seeing some questions from childhood, that i couldnt get out the back of my head, get answered is all the motivation i need. Thank you sir Lewin.
Sir, I would just like to thank you for your excellent teaching. I was extremely confused after my physics professor taught (a much worse version of) this lecture, but watching your video cleared up all questions I have. Your detail and clarity are second to none; certainly the best teacher of physics I have ever seen!
... and after all these years... a few days ago, Jocelyn Bell finally *was* awarded the Nobel prize she's deserved all this time. As I understand it, she intends to donate the entire cash-award part of her prize. I know nothing of her financial circumstances, but even most well-off people would be unlikely to be so magnanimous with "windfall" wealth. You do know her, so please offer her a *whacking* high-five from me, won't you? :-)
Thank you so much for correcting me, Walter: I won't make that mistake again. She *deserves* the recognition and more all the same, but the Nobel snub still rankles. What amazes me is that (as you and others have described her) she's humble, gracious and generous enough to let all that slide *and* offer up her huge cash prize to support others. Yeah... a woman can be a mensch. :-) :-) :-)
"If you have problems with this, you are not alone" A very good sign off for physics students. I remember having trouble in my classes thinking I was completely missing the point. It developes over time and I learned far too late that everyone else were experiencing the same.
I am a 15 yr old from India ( Bharat ) and preparing for jee. I want to thank you. Your lectures are extremely helpful. Professor, My love for physics started just because of you . 🇮🇳
Dear sir I m from india i wanna say that you are phenomenal u make me love PHYSICS which i hated Thanks a lot loads of respect sir hope u live for million years and make futures of students all around tge world
Dear professor, Even if I fail my physics course I just want to show my gratittude to you: I've always hated physics since 7th grade and I can finally find it interesting and enjoyable, as it is! You really have a gift for teaching, every matter is very clear now, thank you so much!
Sir, it was so good of you to mention Jocelyn Bell in your lecture. I recently watched a documentary of her by The New York times, 'Almost Famous'. Even though she did not get any recognition, she is still happy.
Thankyou Sir, for sharing such a beautiful lecture with us. I live in India, and I am a senior high school student. I have watched many lectures of 8.01x and 8.02x several times over since they all are densely packed with knowledge but they all made my concepts crystal clear.
sir,,,i am a little confused at 24:45 as you are pulling your arms nearer to the axis of rotation you are accelerating as your angular velocity keeps on increasing and finally reaches its maximum value due to conservation of angular momentum but to provide an angular acceleration there must be some external force acting on the system which is absent here, then how you are accelerating,,, like if we take an example of a bomb blast into two fragments each fragment gets some velocity due to internal forces but if we look at each fragment separately then this internal force is actually external for them which makes their velocity from 0 to a certain value but to the whole system the net force is 0.i,e we can actually visualize what forces are giving them velocities here..........but in this case of increasing angular velocity what is going on? or IT IS NOT NECESSARY TO HAVE A NET EXTERNAL TORQUE ON THE SYSTEM TO PROVIDE IT AN ANGULAR ACCELERATION??
Given a body that remains rigid that cannot redistribute its mass, a torque is required to change its angular velocity, i.e. for it to have an angular acceleration. For a system that CAN redistribute its mass internally, an external torque is not the only way to give it an angular acceleration. Redistributing the mass, and changing the moment of inertia, will cause its angular velocity to change. There is a torque that acts on a body as it moves radially in a rotating reference frame, and this is a consequence of the Coriolis effect. It is this internal torque that occurs between Professor Lewin and the two barbells, that enables the barbells to apply a torque to speed him up. Try walking along the radial handrail on a spinning merry-go-round, and you will experience this effect. You will feel an apparent force pulling you tangentially forward, when you walk radially inward, and you will feel an apparent force pulling you tangentially backward, when you walk radially outward. To stay at the same position as the handrail, you experience a constraint force as a reaction to these apparent forces that are a consequence of the Coriolis effect. The constraint force (tension or compression in your arms) is what enables the speed of the merry-go-round to change, as you redistriute the mass of the system by moving radially on it.
SOMEONE IN THE COMMENTS SAID that these lectures was delivered in 2002 when I was not in that world but after a long time in 2019 I'm watching it ..and it is also helpful for me but a little bit because 1st reason is that I don't know english too much and 2nd reason is i am studying physics but i am in lower level e.g 11 class and in our course ,there is no too much details ..BUT SIR YOU ARE GOOD TEACHER...APPRECIATION FROM MY BEAUTIFUL COUNTRY PAKISTAN..............
Love you sir. Sir you are a gem for indian students preparing for IIT JEE. This topic specially is very irritating, but you made it very easy. Thanks a lot sir
Hello professor, I sometimes feel all of this so mind boggling! We humans were able to understand stuff which are sooo far away, by just using the laws of physics! It was a really great lecture! Thank you professor! 😊
Where can you start to derive the 10^46W term that you wrote at around 34:00. Is is the integral of the potential energy over the radius of collapse? And if so is mass m rewritten as rho*V, where V=Volume of the star?
I am from India and I literally love your lectures as they make me feel physics . I am gonna rock my test. I am understanding all the concept....to clear jee advanced its highly recommendable sir to watch your videos Lots of love from India
At 2:49 maybe {r(perp.)c} represents the perpendicular distance of point Q from center of mass of M. And suppose the body is in pure translation , then , we do not need to consider rhe rotational Angular momentum.
if an object is in pure translation then there is angular momentum relative to all points except points on the "straight" line of the movement. Depending on the problem you need to solve this angular momentum can be CRUCIAL and can not be ignored.
If you have finished 8.01, 8.02 and 8.03 then go to MIT OCW and watch 8.04 and after that 8.05. 8.05 is lectured by Barton Zwieback. He is a very good lecturer, I have attended several of his lectures at MIT.
@ 26:18 For others confused like me, Even though the points are at different distances from Q --> r1 & r2 ... the torques are equal in magnitude since r1sin(theta1) = r2sin(theta2) ... the component of distance perpendicular to direction of force
In this case, if m is the mass of the star, then what does M and r refer to? Also, may I know does the PE here have any thing to do with the gravitational binding energy?
The masses are being treated as point masses, rather than a uniform cylinder about the axis of rotation. Even if you did treat the masses in his hands as solid uniform cylinders, the parallel axis theorem would render their own size insignificant, such that they might as well be treated as point masses.
Can I say at 26:29 'since there are no external forces acting angular momentum is conserved'? If so, can I replace external torque with external force at 27:35?
yup...i saw ur playlist ...but the problem is the all have japanese title...nd i don't know japanese language...all i know is ...."watashi wa bhanu te imashu"...:)
Spin angular momentum of the Earth is an intrinsic property it's omega*I. Orbital angular momentum of the Earth is ONLY conserved relative to the Sun, NOT relative to any other point. If you choose a point somewhere on the orbit, when the Earth is at that point the orbital angular momentum relative to that point is ZERO.
Hello Prof Lewin, if there is a mouse on the edge of a rotating disk and we know the moment of inertia of the disk about the centre of mass, then we can find the total moment of inertia which includes the moment of inertia of the mouse measured with respect to the CM of disk. Lets say this disk is moving with constant w. Then the mouse moves towards the center of mass of the disk. Then we can use conservation of Angular momentum to get the new angular velocity. But which moment of inertia do we use for when the mouse is in the centre of mass of the disk. Is it just the moment of inertia about the CM of the disk without the mouse on it, or has the mass of the mouse affected the moment of inertia of the disk about the centre of mass? Thank you!
CAN'T STRESS THE IMPORTANCE OF THE FIRST 17 MINS. OF THIS LECTURE ENOUGH! But, of course, the Earth has no INTRINSIC angular momentum (17:02)! The spinning around its own axis is not intrinsic, but a consequence of the way our solar system is created.
24:44 what happens to the rotational kinetic energy? I understand when you pull your arms in it increases and that comes from reduction of internal chemical energy used for the work to bring arms in. But when you pull your arms out how is energy conserved?? You lose rotational kinetic energy, but you don't gain any internal energy. Where does the lost kinetic energy go? Who gains that energy? I understand you could extend the arms almost effortlessly, but still in this case what balances the loss of KE in that closed system? (from the lab frame there is no centrifugal force)
Good question. To understand what happens to the rotational kinetic energy when his arms extend and his body slows down when rotating, consider a linear motion example. Suppose there is a 1 kg bottle of water on top of a 1 meter table above the floor. You pick it up, and carry it to set it gently on the floor. It starts at rest, and ends at rest, so something must've happened to the 9.8 Joules of potential energy in this example, as that energy cannot be in the bottle's kinetic energy. Your muscles generate thermal energy when they act as brakes for objects in motion, absorbing the work done on them, and dissipating the energy as heat. That's what happens to his rotational kinetic energy as well, when he extends his arms while rotating on that platform. The kinetic energy becomes work done on his muscles, and when work is done on a person's muscles, heat is dissipated. Since our bodies aren't built for regenerative braking of objects in motion, we also get tired from doing negative work on objects in motion, just as we would get tired from doing positive work to put objects in motion.
@35:25 the supernova explosion *did not* occur in the year 1054, that's when it was observed! It happened 5,000 years before that date or whatever the distance to the Crab Nebula is in light years.
At 34:00 you calculated change in gravitational potential energy. But as much I know gravitational potential energy is due to gravitational pull between two objects. so with which object this gravitional pull to the star is being considered?
I cannot add to the clarity of my lecture in which I define Grav PE (zero at infinity). --MmG/r (r being the distance of m to the center of M). Watch my lecture again or use google
If its rotational inertia remains unchanged (individual rigid body), it requires a net torque to act upon it, in order to have an angular acceleration. An object can have angular acceleration without torque, while changing its rotational inertia. Unlike linear inertia (i.e. mass), that requires changing the identity of the body to change its mass (like a rocket loosing its propellant mass, or farm equipment collecting a harvest), rotational inertia can change just by redistributing the mass. As you can see with the turntable demonstration in this lecture.
@@lecturesbywalterlewin.they9259 I couldn't understand. Can you explain this clearly please? how can the linear momentum vector can be in the same direction with velocity which is in the direction of r?
@@lecturesbywalterlewin.they9259 yes but this is the definiton of the component of velocity vector which is in the direction of r. but it is not always in the direction of the linear momentum vector. But is that formula only valid for the objects which is doing circular motion? please help me sir
Sir I am in class 11 from India and I am preparing for jee I really appreciate your work in physics and your ability in teaching I am feeling honoured to get the knowledge of physics I am definitely not so rich to pay the expensive fees here in institutes I will grateful and lucky to attend your lectures thank you sir love from India 🙋
professor, is there any change in medium from space to inner part of black hole that cause the speed of light to become 0 there as it is opticaly very dense?
*This channel was created in Febr 2015 by my Dutch friend Daniel Dekkers.* It has become way more popular than "For the Allure of Physics" (created in Dec 2014) which also carries my 94 MIT course lectures + my Farewell Lecture at MIT "For the Love of Physics" of May 16, 2011. That lecture alone has been viewed by more than 6 million people.
Sir, you have always been an inspiration to me and your lectures always motivate me to pursue my dream of getting into research and studying astronomy.....only the fact that you replied makes me feel so special. Believe it or not but this is like a blessing to me. -Regards
36:58 Supernova not recorded in Europe is not mystery if European history between 4-12 century was counterfeited. Star of Bethlehem could be supernova, meaning 1053 years could be stolen from official history later by clergy in Renessance in 15-16 c.
Namaste Sir, (Indian greetings) Please help me!! A thin rod AB of mass M and length L is rotating with angular speed w about vertical axis passing through its end B on a horizontal smooth table. If at some instant the hinge at end B of rod is opened then, can you please explain why the angular momentum of the rod remains conserved about the center of mass of the rod during the whole process? And I have another one sir A cylinder of height h , diameter h/2 and mass M and with a homogeneous mass distribution is placed on a horizontal table. One end of a string running over a pulley is fastened to the top of the cylinder, a body of mass m is hung from the other end and the system is released. Friction is negligible everywhere. Strings and pulleys can be assumed to be light. At what minimum ratio of m/M will the cylinder lift?
Thank you so much for sharing all these marvelous lectures with us ! I am currently in high school and binge watching through all your videos . I have one question from this lecture and it would be great if you could help me out with it! In 33:39 , you estimated the amount of energy released as the radius shrinks but mass remains unchanged... i was wondering how I could calculate the amount of energy released? Can I get any clue or relations that I could get started with? Thanks!
You said that dr/dt has the same direction as p because dr/dt is v. I feel confused because in the example about the earth and the sun, r and v are two perpendicular vectors, and r (relative to C), I see, doesn't change when the earth keeps moving around.
I watched the lecture near 9:52. I corrected in the video the slip of the pen. r_C has to be r_Q. The motion is NOT radial (as I assumed yesterday without watching the video). r here is a vector which rotates. Thus in the plane of rotation it has an x and y component. Assume that the magnitude of the vector r is R and that at t=0 x=R and y=0. Then the x component of r is R*cos(wt) and the y component is R*sin(wt) and dr/dt =v and v is perpendicular to the vector r.
If an object of mass m is rotating about its com with angular momentum A and if it is also translating with velocity V, then the angular momentum relative to any point P is A + m*dXV, d is the position vector from P to the com. X indicates cross product.
@@lecturesbywalterlewin.they9259 sir only if you have written neutron star in the title of the video google will give me the result. I checked all the titles and nowhere stars were mentioned
Hello Walter Lewin! When you were on the turntable you had your angular momentum L conserved. Your rotational KE was L^2/2I, I being your moment of inertia. When you drew in your hands I decreased so your rotational KE must have increased. Where did you get that KE?
hello Sir,as u mentioned a body(scale in this case) when hit at some point other than CM will rotate about CM.So when a body is hinged then it'll rotate about hinged point.My confusion is related to center of percussion(COP).Is it really that at different impact points with respect to (COP) body will try to rotate either clockwise(CW) or anti clock(ACW)?The force on hinge will be either right or left according to supposed CW or ACW rotation?
Dear professor Lewin, I have a question: is the linear momentum also defined to a specified point as the position vector does? As there is no subscript about the linear momentum in the video. Then how to take derivatives of the angular momentum in a different reference frame to get torques, especially not an inertial frame? Will some fictitious forces come out? Thanks a lot!
I have a question about the right-hand rule, if whatever is in the z-axis is "coming out of the blackboard," does that make it negative, and going into the blackboard positive?
Can you please explain why rotations about the center of mass gives the same value for angular momentum irrespective of the point of origin... I've done my homework on Google but it keeps redirecting me towards QM.
At 27:50-28:07 the radius and moment of inertia go down on a star, so obviously the angular velocity will go up. But he misses a very valuable point which lead to the discovery I made. The star actually loses mass. He is assuming as the star shrinks and its moment of inertia goes down, the mass does not change. Yet, all stars have solar wind and flare out trillions of tons of matter. So they are clearly losing mass. So more correctly, it is the radius shrinks, the moment of inertia shrink, and the mass shrinks, meaning the angular velocity will remain constant. This is a major issue that is not addressed for unknown reasons, I guess it amounts to them assuming they have no evidence for stars that have lost a large majority of their mass. I beg to differ. They call them "exoplanets/planets". Saying there is no evidence for stars that have lost a large portion of their mass by calling them by a different name is very peculiar.
>>>which lead to the discovery I made>>> When did you make that discovery? Mass loss of a supernova explosion (not overlooked by me) has been known for more than half a century. When the core of a 20 solar mass star collapses it leaves behind some 19 solar masses streaming out at velocities of some 20,000 km/sec AND it leaves behind an approx 1.4 solar mass neutron star whose rotation period can be 1 second or even less. en.wikipedia.org/wiki/Supernova
I have one question professor. on the spinning experiment, when you pull your arms in and out, if your angular velocity is changing, you will have angular acceleration. and angular acceleration is a result of net torque. and if you say there is no torque, the moment of inertia at that moment in time must be zero. otherwise it wouldn't be compatible with the relation ( net torque=moment of inertia *angular acceleration). so what is the thing i am missing here.
@@lecturesbywalterlewin.they9259 well, i have watched that. but still i am confused about it. can you make it a little more conceptual how there is an angular acceleration with no net torque?
In the absence of external torques, angular momentum is conserved. Thus when a star collapses its angular frequency will increase. The same will happen when I pull my arms inwards when I am on a rotating platform. *No external torque!*
@@lecturesbywalterlewin.they9259 then if there were no net external torque, where does that angular acceleration come from? because you are accelerating at the moment you pull your arms in as there is some change in angular velocity through a certain amount of time.( thank you so much for taking the time to answer my questions though)
Hello Pr. Lewin... I have a question in this lesson that is Fr that you defined as a gravitational force or attract force from sun to earth in showing of conservation of angular momentum, is equal to centeripital force in the circular motion of earth around the sun?? Thanks a lot!!....
ONLY orbital ang mom is conserved relative to the sun; It's not conserved relative to any other point. The grav force from the sun on Earth is the centripetal force on Earth in its orbit around the sun.
Excuse me Pr.Lewin... I have another question about your answer for my last question... and is that as you say we get center...force equal with gravita....force; so It can be true but I suppose that rotating obj stay without any moving around in that position and hasn't got any rotation; so in that case we have gravity...force for this obj from that another object independent of it's rotation case(mood) so I mean that gravit...force can't be equal with centeri....force.. . ( and I don't suppose that rotating obj ,the earth,because it can't be stop and stay in a certain location to the sun for example...) is my thinking true.....??
@@sinakhorsand6206 In the case of circular orbits the gravitational attraction by the sun is exactly the centripetal force on a planet. The Earth orbit is close to circular. In the case of highly ellipsoidal orbits, watch my 8.01 lecture on that topic. The stress in a planet generates its own centripetal force for the spin rotation of a planet, that has NOTHING to do with the sun.
In 1054 everyone who could write was possibly more concentrated on the great Schisma in the Christian Church between the Latin Church and the Greek Orthodox Church. It could be an explanation, there are however mentions of a similar occurance but 2 month earlier. I cannot explain this, but the mention in the Chinese literature is from 3 centuries later.
Sir , I have confused about angular momentum is not beeing intrinsic property while momentum is an intrinsic property. As the angular momentum depends on the point we chose, aren`t the momentum also depends on frame of reference. It seems to me both of them are not intrinsic property.
My notes : 1.Momentum Intrinsic ,angular momentum is not,it depends on point about which it is calculated .2. Angular momentum about C,representing Centre of circle,is constant,.while momentum is changing,in uniform circular motion . 3.V at every point of disk is different but angular velocity is not. 4.Angular momentum is always constant about any point in space for rotation about center of mass,spin angular momentum is thus intrinsic and unique ,e.g earth rotation about axis. You could then talk about THE Angular momentum.
Hi professor, love your lectures, thanks for sharing :) I wanted to ask a question: at minute 16:20 you say that if a body is rotating about the center of mass then its angular momentum is invariant no matter the choice of the pole, but i thought that it was the case only if the poles that you can choose all lie on the axes of rotation; do you know where can i find a demonstration of this property? (i can't manage to find it myself). Thanks again.
sir in your turn table experiment treating your central body and arms as different parts. if we see your central body will have increased angular velocity and angular moment um should then increase how so.is conservation of angular momentum not valid then
In the absence of external torque, angular momentum is conserved. Watch my "ice skater" demo. I*omega is conserved. If I increase I omega goes down and vice versa.
Dear professor Lewin, I think you've made an enormous contribution to Science by inspiring thousands of young students and showing the beauty of Physics. I'm studying to become a scientist and your amazing lectures are extremely helpful. Thank you!
+Alex Zilio Thank you Alex for your kind words. I am delighted that my lectures are "extremely" ,helpful for you.
I completely agree.
@@frankdimeglio8216 in your way
After filling lakhs of fees, real knowledge is found on TH-cam for free.🔥
Respect for sir🙏
I taught many courses during a 30 year career teaching engineering. Dynamics was my favorite course. I learned something new every time I taught it. I find your lectures very informative. I really appreciate how you corrected your mistakes in your lecture and I am sure your students do too.
Yes, you are right.
The transition from mundane earthly objects to awe-inspiring neutron stars and supernovas was incredible! Thank you for another great lecture, Professor!
My pleasure!
cheers from Brazil, I've recommended your lessons in 8.01 to all my classical mechanics classmates, and we have emulated almost all your experiments so far.
Excellent!
Hi
Lectures by Walter Lewin. They will make you ♥ Physics.
th-cam.com/channels/iEHVhv0SBMpP75JbzJShqw.html
300+ videos. Many of them with high resolution. This channel has all my lectures and talks.
Lectures by Walter Lewin. They will make you ♥ Physics. sir, the link is down. please check.
Prof. Lewin, you are a blessing. Thank you for making it open source.
This part of mechanics is also considered to be the toughest in the entire JEE syllabus, but Professor Lewin made it crystal clear in my head. Thank you sir!
How your jee exam went?
Can i prefer his lectures for my jee prep?
"you passed the course" i love his lectures :D
Thankkk you so much for recording this lectures. I live in Turkey!! At my university, I can't really get inspiring classes to love Physics. But the last 20 minutes of this lecture had a concern to inspire us about the Physics and the nature. I study Physics and you, sir, are making me inspired on your every lecture!
You're very welcome!
Which university do you study
for those of you curious about the calculation preformed at ~34:22. you need to calculate the potential energy of the star (which is ofcourse spherical) in the 2 cases (R = 7*10^5 km and R = 10 km); that is the work required to bring all the particles of the star from infinity to their respective positions on the sphere.
Your lectures make me feel like a kid again... sure the calculations get more and more complex as i go on, but both your infectious enthusiasm and seeing some questions from childhood, that i couldnt get out the back of my head, get answered is all the motivation i need. Thank you sir Lewin.
thanks for your kind words
Sir, I would just like to thank you for your excellent teaching. I was extremely confused after my physics professor taught (a much worse version of) this lecture, but watching your video cleared up all questions I have. Your detail and clarity are second to none; certainly the best teacher of physics I have ever seen!
Thank you for your kind words.
... and after all these years... a few days ago, Jocelyn Bell finally *was* awarded the Nobel prize she's deserved all this time. As I understand it, she intends to donate the entire cash-award part of her prize. I know nothing of her financial circumstances, but even most well-off people would be unlikely to be so magnanimous with "windfall" wealth.
You do know her, so please offer her a *whacking* high-five from me, won't you? :-)
she did not get the Nobel Prize
instead she got a $3.5 million prize which is way more than a Nobel Prize. I have congratulated her!
Thank you so much for correcting me, Walter: I won't make that mistake again. She *deserves* the recognition and more all the same, but the Nobel snub still rankles.
What amazes me is that (as you and others have described her) she's humble, gracious and generous enough to let all that slide *and* offer up her huge cash prize to support others.
Yeah... a woman can be a mensch. :-) :-) :-)
19:59 epic voice crack
"If you have problems with this, you are not alone"
A very good sign off for physics students. I remember having trouble in my classes thinking I was completely missing the point. It developes over time and I learned far too late that everyone else were experiencing the same.
I am a 15 yr old from India ( Bharat ) and preparing for jee.
I want to thank you.
Your lectures are extremely helpful.
Professor, My love for physics started just because of you . 🇮🇳
Best of luck
Dear sir
I m from india i wanna say that you are phenomenal u make me love PHYSICS which i hated
Thanks a lot loads of respect sir hope u live for million years and make futures of students all around tge world
You are really dedicated sir.
Sacrifice for science .
For the love of physics
Dear professor,
Even if I fail my physics course I just want to show my gratittude to you: I've always hated physics since 7th grade and I can finally find it interesting and enjoyable, as it is! You really have a gift for teaching, every matter is very clear now, thank you so much!
class 7 e ami physics er namE jantam na 🙄
Sir, it was so good of you to mention Jocelyn Bell in your lecture. I recently watched a documentary of her by The New York times, 'Almost Famous'. Even though she did not get any recognition, she is still happy.
I was about to study fluid mechanics before i’ve been mesmerized by the explanations in this video.
been a mechanical engineer for over 25 years.. these are great stuff.
How do you get the potential energy change at 34:08? I don't get it.
Thankyou Sir, for sharing such a beautiful lecture with us. I live in India, and I am a senior high school student. I have watched many lectures of 8.01x and 8.02x several times over since they all are densely packed with knowledge but they all made my concepts crystal clear.
You are most welcome
sir,,,i am a little confused at 24:45 as you are pulling your arms nearer to the axis of rotation you are accelerating as your angular velocity keeps on increasing and finally reaches its maximum value due to conservation of angular momentum but to provide an angular acceleration there must be some external force acting on the system which is absent here, then how you are accelerating,,,
like if we take an example of a bomb blast into two fragments each fragment gets some velocity due to internal forces but if we look at each fragment separately then this internal force is actually external for them which makes their velocity from 0 to a certain value but to the whole system the net force is 0.i,e we can actually visualize what forces are giving them velocities here..........but in this case of increasing angular velocity what is going on?
or IT IS NOT NECESSARY TO HAVE A NET EXTERNAL TORQUE ON THE SYSTEM TO PROVIDE IT AN ANGULAR ACCELERATION??
Given a body that remains rigid that cannot redistribute its mass, a torque is required to change its angular velocity, i.e. for it to have an angular acceleration.
For a system that CAN redistribute its mass internally, an external torque is not the only way to give it an angular acceleration. Redistributing the mass, and changing the moment of inertia, will cause its angular velocity to change.
There is a torque that acts on a body as it moves radially in a rotating reference frame, and this is a consequence of the Coriolis effect. It is this internal torque that occurs between Professor Lewin and the two barbells, that enables the barbells to apply a torque to speed him up.
Try walking along the radial handrail on a spinning merry-go-round, and you will experience this effect. You will feel an apparent force pulling you tangentially forward, when you walk radially inward, and you will feel an apparent force pulling you tangentially backward, when you walk radially outward. To stay at the same position as the handrail, you experience a constraint force as a reaction to these apparent forces that are a consequence of the Coriolis effect. The constraint force (tension or compression in your arms) is what enables the speed of the merry-go-round to change, as you redistriute the mass of the system by moving radially on it.
SOMEONE IN THE COMMENTS SAID that these lectures was delivered in 2002 when I was not in that world but after a long time in 2019 I'm watching it ..and it is also helpful for me but a little bit because 1st reason is that I don't know english too much and 2nd reason is i am studying physics but i am in lower level e.g 11 class and in our course ,there is no too much details ..BUT SIR YOU ARE GOOD TEACHER...APPRECIATION FROM MY BEAUTIFUL COUNTRY PAKISTAN..............
Love you sir. Sir you are a gem for indian students preparing for IIT JEE. This topic specially is very irritating, but you made it very easy. Thanks a lot sir
Dear Sir, Thanks is not enough... Now I can understand that what a beauty in Physics! Love from Kishanganj Bihar India...
Hello professor,
I sometimes feel all of this so mind boggling! We humans were able to understand stuff which are sooo far away, by just using the laws of physics! It was a really great lecture! Thank you professor! 😊
You are very welcome
👍🏻
Where can you start to derive the 10^46W term that you wrote at around 34:00. Is is the integral of the potential energy over the radius of collapse? And if so is mass m rewritten as rho*V, where V=Volume of the star?
It's NOT 10^46 W it's J. It's the change in gravitational potential energy.
Sorry! Slip of the mind. But in any case where can you begin in order to find this number?
Take a spherical mass M of radius R2. Then change the radius of the star to R1 (R1
Thank you sir!
Best physics teacher ever!!!!!
Love you from Bangladesh!🇧🇩💝💝😍😍😍😍😍😍
Prof. Lewin's lecture is always the best !
I am from India and I literally love your lectures as they make me feel physics . I am gonna rock my test. I am understanding all the concept....to clear jee advanced its highly recommendable sir to watch your videos
Lots of love from India
Best of luck!
Can you give me suggestion to which video i should watch in this channel😢😢😢
thankyou sir- i loved the supernova explosion picture!! THANKYOU SOS SO SO MUCH- YOU ARE AN INSPIRATION TO ME.
At 2:49 maybe {r(perp.)c} represents the perpendicular distance of point Q from center of mass of M. And suppose the body is in pure translation , then , we do not need to consider rhe rotational Angular momentum.
if an object is in pure translation then there is angular momentum relative to all points except points on the "straight" line of the movement. Depending on the problem you need to solve this angular momentum can be CRUCIAL and can not be ignored.
Dear professor Lewin sir, your are a great concept clearer.
sir, when will 8.04 and 8.05 be uploaded...waiting..
I really started loving physics
If you have finished 8.01, 8.02 and 8.03 then go to MIT OCW and watch 8.04 and after that 8.05. 8.05 is lectured by Barton Zwieback. He is a very good lecturer, I have attended several of his lectures at MIT.
@ 26:18 For others confused like me,
Even though the points are at different distances from Q --> r1 & r2 ... the torques are equal in magnitude since r1sin(theta1) = r2sin(theta2) ... the component of distance perpendicular to direction of force
Thank you
Hello Prof Lewin, may I ask how did you calculate the loss of gravitational potential energy at 34:09?
I used the definition of PE which is zero at infinity and negative at radius r>R. PE = -mMG/r.
In this case, if m is the mass of the star, then what does M and r refer to?
Also, may I know does the PE here have any thing to do with the gravitational binding energy?
Watch th-cam.com/video/9gUdDM6LZGo/w-d-xo.html
Binding energy BE and PE are related. use google.
Watch th-cam.com/video/9gUdDM6LZGo/w-d-xo.html
I have a question: how come that the derivative of r_c(which is indeed r_q) with respect to time is equal to v ? (9:50)
derivating a position vector (r) with respect to time gives a velocity :)
At 22:21 why is it (m)(r^2) and not (1/2)(m)(r^2) for each mass on the hands ?
The masses are being treated as point masses, rather than a uniform cylinder about the axis of rotation. Even if you did treat the masses in his hands as solid uniform cylinders, the parallel axis theorem would render their own size insignificant, such that they might as well be treated as point masses.
Can I say at 26:29 'since there are no external forces acting angular momentum is conserved'? If so, can I replace external torque with external force at 27:35?
no
Lectures by Walter Lewin. They will make you ♥ Physics.
Isn't internal torque 0, because internal force 0 due to Newton's 3rd law?
I am sorry but I cannot add to the clarity of this lecture.
torque=dL/dt
zero torque => no change in L
Lectures by Walter Lewin. They will make you ♥ Physics.
How is newtons third law related to internal torque of the system?
ask google
now thats i call teaching it his best....best teacher i have seen in my life.
Bhanu Sharma Thank you Bhanu
really from dumb to pro .... credit goes to sir walter lewin ... n to ... thnxx for uploading whole course ...good work
Bhanu Sharma click on "playlist" you view wayyyyyyy more than my course lectures alone!
yup...i saw ur playlist ...but the problem is the all have japanese title...nd i don't know japanese language...all i know is ...."watashi wa bhanu te imashu"...:)
all have Japanese titles ????????????
For me all have English titles. Why are your Japanese?
This is absurd.
sir, does the spin angular momentum remains the same for any reference point but orbital angular momentum does not
Spin angular momentum of the Earth is an intrinsic property it's omega*I. Orbital angular momentum of the Earth is ONLY conserved relative to the Sun, NOT relative to any other point. If you choose a point somewhere on the orbit, when the Earth is at that point the orbital angular momentum relative to that point is ZERO.
Hello Prof Lewin, if there is a mouse on the edge of a rotating disk and we know the moment of inertia of the disk about the centre of mass, then we can find the total moment of inertia which includes the moment of inertia of the mouse measured with respect to the CM of disk. Lets say this disk is moving with constant w. Then the mouse moves towards the center of mass of the disk. Then we can use conservation of Angular momentum to get the new angular velocity. But which moment of inertia do we use for when the mouse is in the centre of mass of the disk. Is it just the moment of inertia about the CM of the disk without the mouse on it, or has the mass of the mouse affected the moment of inertia of the disk about the centre of mass? Thank you!
the answer is in my 8.01 lectures. WATCH them or use google
CAN'T STRESS THE IMPORTANCE OF THE FIRST 17 MINS. OF THIS LECTURE ENOUGH! But, of course, the Earth has no INTRINSIC angular momentum (17:02)! The spinning around its own axis is not intrinsic, but a consequence of the way our solar system is created.
24:44 what happens to the rotational kinetic energy? I understand when you pull your arms in it increases and that comes from reduction of internal chemical energy used for the work to bring arms in. But when you pull your arms out how is energy conserved?? You lose rotational kinetic energy, but you don't gain any internal energy. Where does the lost kinetic energy go? Who gains that energy?
I understand you could extend the arms almost effortlessly, but still in this case what balances the loss of KE in that closed system?
(from the lab frame there is no centrifugal force)
Good question. To understand what happens to the rotational kinetic energy when his arms extend and his body slows down when rotating, consider a linear motion example. Suppose there is a 1 kg bottle of water on top of a 1 meter table above the floor. You pick it up, and carry it to set it gently on the floor. It starts at rest, and ends at rest, so something must've happened to the 9.8 Joules of potential energy in this example, as that energy cannot be in the bottle's kinetic energy. Your muscles generate thermal energy when they act as brakes for objects in motion, absorbing the work done on them, and dissipating the energy as heat.
That's what happens to his rotational kinetic energy as well, when he extends his arms while rotating on that platform. The kinetic energy becomes work done on his muscles, and when work is done on a person's muscles, heat is dissipated. Since our bodies aren't built for regenerative braking of objects in motion, we also get tired from doing negative work on objects in motion, just as we would get tired from doing positive work to put objects in motion.
İlk defa bir fizik dersinde eğlendiğimi düşündüm..
Thanks for sharing with us.
@35:25 the supernova explosion *did not* occur in the year 1054, that's when it was observed! It happened 5,000 years before that date or whatever the distance to the Crab Nebula is in light years.
At 34:00 you calculated change in gravitational potential energy. But as much I know gravitational potential energy is due to gravitational pull between two objects. so with which object this gravitional pull to the star is being considered?
I cannot add to the clarity of my lecture in which I define Grav PE (zero at infinity). --MmG/r (r being the distance of m to the center of M). Watch my lecture again or use google
sir I have a question. sir whenever there is angular accelration on particle there is always an torque acting on it. is am I right
yes, that is correct
torque=I*alpha. alpha is dw/dt
If its rotational inertia remains unchanged (individual rigid body), it requires a net torque to act upon it, in order to have an angular acceleration. An object can have angular acceleration without torque, while changing its rotational inertia.
Unlike linear inertia (i.e. mass), that requires changing the identity of the body to change its mass (like a rocket loosing its propellant mass, or farm equipment collecting a harvest), rotational inertia can change just by redistributing the mass. As you can see with the turntable demonstration in this lecture.
Hi good sir. What is the appx age of these students? Which class is it. Is it 11th standard? Or first year in college?
1st year college
17 years appx age I guess?
@@sadhgurusfunniestandwittie3620 Around 19 or 20
And yes congratulations for 1 million subscribers 👌👌
thankeew soo much
love from india
9:57 Sir, I couldn't understand that how can dr/dt is in the same direction as P? isn't it in the same direction with position vector r?
what I have is correct
@@lecturesbywalterlewin.they9259 I couldn't understand. Can you explain this clearly please? how can the linear momentum vector can be in the same direction with velocity which is in the direction of r?
9:57 dr/dt=v it's the definition of v
@@lecturesbywalterlewin.they9259 yes but this is the definiton of the component of velocity vector which is in the direction of r. but it is not always in the direction of the linear momentum vector. But is that formula only valid for the objects which is doing circular motion? please help me sir
@@lecturesbywalterlewin.they9259 ok, Sir. I understood my mistake, It was so elementary😭. Thank you😀
It's been an honour to study from you sir , thus I'm obliged to srudy from you for my jee exams
Sir I am in class 11 from India and I am preparing for jee I really appreciate your work in physics and your ability in teaching I am feeling honoured to get the knowledge of physics I am definitely not so rich to pay the expensive fees here in institutes I will grateful and lucky to attend your lectures thank you sir love from India 🙋
How much are you scoring in Physics in JEE Advanced Mock Tests per paper of around 66 marks , and which coaching institute?
This lectures are pretty cool. I was a math student (working now) and i find myself watching this lectures. Cheers from portugal.
:)
Sacrifice for the sake of science. Doesn't matter being a finance student, still I love Physics. Respect from INDIA 🇮🇳
professor, is there any change in medium from space to inner part of black hole that cause the speed of light to become 0 there as it is opticaly very dense?
physics.stackexchange.com/questions/26515/what-is-exactly-the-density-of-a-black-hole-and-how-can-it-be-calculated
How can we determine the direction of angular momentum with just the vectors of velocity and direction? 6:43
do the cross product using the right hand rule
I really thank the person who has made this channel......feel so privileged to watch Mr.Walter Lewin's lectures at a click of a button.
*This channel was created in Febr 2015 by my Dutch friend Daniel Dekkers.*
It has become way more popular than "For the Allure of Physics" (created in Dec 2014) which also carries my 94 MIT course lectures + my Farewell Lecture at MIT "For the Love of Physics" of May 16, 2011. That lecture alone has been viewed by more than 6 million people.
Sir, you have always been an inspiration to me and your lectures always motivate me to pursue my dream of getting into research and studying astronomy.....only the fact that you replied makes me feel so special. Believe it or not but this is like a blessing to me.
-Regards
Sir how can i get that assignment which are in description..
thank you very much for your videos!
you make a giant difference in the world.
thank you Mark!
36:58 Supernova not recorded in Europe is not mystery if European history between 4-12 century was counterfeited. Star of Bethlehem could be supernova, meaning 1053 years could be stolen from official history later by clergy in Renessance in 15-16 c.
well , professor does black hole is spinning very fast as it''s moment of inertia is negligible , i think?
physics.stackexchange.com/questions/310881/do-black-holes-have-a-moment-of-inertia
www.quora.com/Does-inertia-exist-inside-black-holes
Excellent lecture Sir. Thanks and Regards 🙏🙏🙏🙏🙏🙏🙏🙏🙏
Most welcome
still watching these lectures! i love it, I am 32 and everyday learning something new.
Wonderful!
Namaste Sir, (Indian greetings) Please help me!!
A thin rod AB of mass M and length L is rotating with angular speed w about vertical axis passing through its end B on a horizontal smooth table. If at some instant the hinge at end B of rod is opened then, can you please explain why the angular momentum of the rod remains conserved about the center of mass of the rod during the whole process?
And I have another one sir
A cylinder of height h , diameter h/2 and mass M and with a homogeneous mass distribution is placed on a horizontal table. One end of a string running over a pulley is fastened to the top of the cylinder, a body of mass m is hung from the other end and the system is released. Friction is negligible everywhere. Strings and pulleys can be assumed to be light. At what minimum ratio of m/M will the cylinder lift?
Thank you so much for sharing all these marvelous lectures with us !
I am currently in high school and binge watching through all your videos .
I have one question from this lecture and it would be great if you could help me out with it!
In 33:39 , you estimated the amount of energy released as the radius shrinks but mass remains unchanged...
i was wondering how I could calculate the amount of energy released? Can I get any clue or relations that I could get started with?
Thanks!
total energy (KE + PE) is conserved. When the star implodes the PE decreases (the smaller R the less PE) thus KE goes up.
@@lecturesbywalterlewin.they9259 so to work out the loss in PE, work out the increase in rotational KE?
Can you please explain why derivative of r is v. (You mentioned it in 9:52)?
You said that dr/dt has the same direction as p because dr/dt is v. I feel confused because in the example about the earth and the sun, r and v are two perpendicular vectors, and r (relative to C), I see, doesn't change when the earth keeps moving around.
I watched the lecture near 9:52. I corrected in the video the slip of the pen. r_C has to be r_Q. The motion is NOT radial (as I assumed yesterday without watching the video). r here is a vector which rotates. Thus in the plane of rotation it has an x and y component. Assume that the magnitude of the vector r is R and that at t=0 x=R and y=0. Then the x component of r is R*cos(wt) and the y component is R*sin(wt) and dr/dt =v and v is perpendicular to the vector r.
how do we find the angular momentum of an object(rotating about it's centre of mass) about a point other than it's com
If an object of mass m is rotating about its com with angular momentum A and if it is also translating with velocity V, then the angular momentum relative to any point P is A + m*dXV, d is the position vector from P to the com. X indicates cross product.
Professor Lewin, which book was being used in this course?
8.01
Physics
Hans C. Ohanian
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
Wow! Thanks so much for replying :)
sir can you please in which lecture of 8.02 do you deal with magnetic fields around a neutron star
use google
@@lecturesbywalterlewin.they9259 sir only if you have written neutron star in the title of the video google will give me the result. I checked all the titles and nowhere stars were mentioned
please learn how to use google
en.wikipedia.org/wiki/Neutron_star
section 2.3
Thanks sir, this is extremely useful for me, these videos are treasure for me
Glad to hear that
Hello Walter Lewin! When you were on the turntable you had your angular momentum L conserved. Your rotational KE was L^2/2I, I being your moment of inertia. When you drew in your hands I decreased so your rotational KE must have increased. Where did you get that KE?
yes it increased. As I pulled my arms in, I had to do work.
You are great sir....physics is real magic.
awesome lecture sir . god bless you 💖💖💖
hello Sir,as u mentioned a body(scale in this case) when hit at some point other than CM will rotate about CM.So when a body is hinged then it'll rotate about hinged point.My confusion is related to center of percussion(COP).Is it really that at different impact points with respect to (COP) body will try to rotate either clockwise(CW) or anti clock(ACW)?The force on hinge will be either right or left according to supposed CW or ACW rotation?
I cover all this in my 8.01 lectures
Dear professor Lewin, I have a question: is the linear momentum also defined to a specified point as the position vector does? As there is no subscript about the linear momentum in the video. Then how to take derivatives of the angular momentum in a different reference frame to get torques, especially not an inertial frame? Will some fictitious forces come out? Thanks a lot!
watch my 8.01 lecture where I cover linear momentum
These lectures surely do have a classical feel about them
I have a question about the right-hand rule, if whatever is in the z-axis is "coming out of the blackboard," does that make it negative, and going into the blackboard positive?
x X y = z that's a righthanded coordinate system. PERIOD it does not matter how you draw it as long as x X y = z.
@@lecturesbywalterlewin.they9259 okay, thank you
Can you please explain why rotations about the center of mass gives the same value for angular momentum irrespective of the point of origin... I've done my homework on Google but it keeps redirecting me towards QM.
use google, simple math
At 27:50-28:07 the radius and moment of inertia go down on a star, so obviously the angular velocity will go up.
But he misses a very valuable point which lead to the discovery I made. The star actually loses mass. He is assuming as the star shrinks and its moment of inertia goes down, the mass does not change. Yet, all stars have solar wind and flare out trillions of tons of matter. So they are clearly losing mass.
So more correctly, it is the radius shrinks, the moment of inertia shrink, and the mass shrinks, meaning the angular velocity will remain constant.
This is a major issue that is not addressed for unknown reasons, I guess it amounts to them assuming they have no evidence for stars that have lost a large majority of their mass. I beg to differ. They call them "exoplanets/planets". Saying there is no evidence for stars that have lost a large portion of their mass by calling them by a different name is very peculiar.
>>>which lead to the discovery I made>>>
When did you make that discovery?
Mass loss of a supernova explosion (not overlooked by me) has been known for more than half a century. When the core of a 20 solar mass star collapses it leaves behind some 19 solar masses streaming out at velocities of some 20,000 km/sec AND it leaves behind an approx 1.4 solar mass neutron star whose rotation period can be 1 second or even less. en.wikipedia.org/wiki/Supernova
You know this is going to be hard when there are two corrections in the first 9 minutes
Professor, What is the 26 100 you keep referring to, is it a frame of reference?
lecture hall
So, it is easy to specify the frame of reference and compare it with other frames,isn't?
31:29 8.05x?
Walter, your words about Jocelyn Bell were very nice.
24:54 I am surprised He is perfectly spinning without feeling dizzy
I have one question professor. on the spinning experiment, when you pull your arms in and out, if your angular velocity is changing, you will have angular acceleration. and angular acceleration is a result of net torque. and if you say there is no torque, the moment of inertia at that moment in time must be zero. otherwise it wouldn't be compatible with the relation ( net torque=moment of inertia *angular acceleration). so what is the thing i am missing here.
watch the 8.01 lecture where I do this. Conservation of angular momentum.
@@lecturesbywalterlewin.they9259 well, i have watched that. but still i am confused about it. can you make it a little more conceptual how there is an angular acceleration with no net torque?
In the absence of external torques, angular momentum is conserved. Thus when a star collapses its angular frequency will increase. The same will happen when I pull my arms inwards when I am on a rotating platform. *No external torque!*
@@lecturesbywalterlewin.they9259 then if there were no net external torque, where does that angular acceleration come from? because you are accelerating at the moment you pull your arms in as there is some change in angular velocity through a certain amount of time.( thank you so much for taking the time to answer my questions though)
sir u took moment of inertia of dish has mr^2 but the moment of inertia is 1\2mr^2 ????
The moment of inertia of a hoop is Mr^2, and the Cylinder is 1\2mr^2.
Hello Pr. Lewin...
I have a question in this lesson that is Fr that you defined as a gravitational force or attract force from sun to earth in showing of conservation of angular momentum, is equal to centeripital force in the circular motion of earth around the sun??
Thanks a lot!!....
ONLY orbital ang mom is conserved relative to the sun; It's not conserved relative to any other point. The grav force from the sun on Earth is the centripetal force on Earth in its orbit around the sun.
Thanks a lot!!!!!
Pr.Lewin
Excuse me Pr.Lewin...
I have another question about your answer for my last question...
and is that as you say we get center...force equal with gravita....force;
so It can be true but I suppose that rotating obj stay without any moving around in that position and hasn't got any rotation; so in that case we have gravity...force for this obj from that another object independent of it's rotation case(mood)
so I mean that gravit...force can't be equal with centeri....force.. .
( and I don't suppose that rotating obj ,the earth,because it can't be stop and stay in a certain location to the sun for example...)
is my thinking true.....??
@@sinakhorsand6206 In the case of circular orbits the gravitational attraction by the sun is exactly the centripetal force on a planet. The Earth orbit is close to circular. In the case of highly ellipsoidal orbits, watch my 8.01 lecture on that topic. The stress in a planet generates its own centripetal force for the spin rotation of a planet, that has NOTHING to do with the sun.
In 1054 everyone who could write was possibly more concentrated on the great Schisma in the Christian Church between the Latin Church and the Greek Orthodox Church. It could be an explanation, there are however mentions of a similar occurance but 2 month earlier. I cannot explain this, but the mention in the Chinese literature is from 3 centuries later.
Sir , I have confused about angular momentum is not beeing intrinsic property while momentum is an intrinsic property. As the angular momentum depends on the point we chose, aren`t the momentum also depends on frame of reference. It seems to me both of them are not intrinsic property.
Absolute legend, Walter Lewin.
My notes : 1.Momentum Intrinsic ,angular momentum is not,it depends on point about which it is calculated .2. Angular momentum about C,representing Centre of circle,is constant,.while momentum is changing,in uniform circular motion . 3.V at every point of disk is different but angular velocity is not. 4.Angular momentum is always constant about any point in space for rotation about center of mass,spin angular momentum is thus intrinsic and unique ,e.g earth rotation about axis. You could then talk about THE Angular momentum.
Professor Lewin, I wonder why non-conservative internal force of a rigid system do no work?
Hi professor, love your lectures, thanks for sharing :)
I wanted to ask a question: at minute 16:20 you say that if a body is rotating about the center of mass then its angular momentum is invariant no matter the choice of the pole, but i thought that it was the case only if the poles that you can choose all lie on the axes of rotation; do you know where can i find a demonstration of this property? (i can't manage to find it myself). Thanks again.
+SilverAppleMan It's invariant. You may choose any point anywhere.
sir in your turn table experiment treating your central body and arms as different parts. if we see your central body will have increased angular velocity and angular moment um should then increase how so.is conservation of angular momentum not valid then
In the absence of external torque, angular momentum is conserved. Watch my "ice skater" demo. I*omega is conserved. If I increase I omega goes down and vice versa.
th-cam.com/video/BPizQDIcQ70/w-d-xo.html