8.01x - Lect 21 - Torques, Oscillating Bodies, Physical Pendulums

[@namanhtu4938]

"Physics is not difficult, physics is there to make very difficult things easy"

[@SAHZ-xe3pz]

Dramatic but the truth

[@thienthanhtranoan6723]

*8.01x lecture 21* ---------- 0:47 *review equation number 1+2+3* ⁃ angular momentum ⁃ Torque ⁃ The change in angular momentum 2:59 *Review equation 4+5+6 (relative to inertia I)* Everyone must focus on some examples which Prof. Walter Lewin mentioned. WL: *Physics is not equation, Physics is concept* 4:20 *Example 1 (the earth around the sun)* ->equation 1,2,3 7:45 *Example 2 (the ruler/rob)* -> equation 4,5 12:50 *Example 3 (centre of mass - spin angular momentum)* ->equation 6 14:18 harder application (1) -> *ruler on frictionless table* (always use your intuition and Physical thinking, maybe it’s not true, but it help you understand more) 25:53 harder application (2): -> *simple harmonic oscillation* WL *Physics is not difficult, Physics is there to make very difficult things easy* And you must check your result after every exercise, it will bring to you a lot of intuition and knowledge about the system. 36:57 Harder application (3): *Hula Hoop~pendulum* 44:35 *Challenging* (which make life very interesting) ~sleepless night: Restore force??? ----------- *Assignment in the description box:* Assignments Lecture 21, 22, 23 and 24: freepdfhosting.com/2e96daf94f.pdf Solutions Lecture 21, 22, 23 and 24: freepdfhosting.com/86109d309b.pdf ----------- *See more here* zyzx.haust.edu.cn/moocresource/data/20080421/U/01220/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/Video-Segment-Index-for-L-21.htm ----------- Thanks a lot, my best teacher. I have cried when i saw this video. Because i could recognize the beauty of Physics , whenever I see your video lectures.

[@lecturesbywalterlewin.they9259]

🌺

[@che5764]

dear mr lewin, i ve become regularly watching your lectures series, just to say thanks a lot and do remember there are many physics lovers around the globe doing the same. best regards from malaysia

[@lecturesbywalterlewin.they9259]

Thanks, Jo, for your kind words.

[@mehmetalivat]

Thank you Professor Lewin for lecture series. 46:40 Last question was about elastic material. Since the friction is too much, the force is reflected in the reverse direction. A spinning super ball changes direction of rotation when hit ground.

[@LazZanZaz]

That plastic piece actually has a name "RattleBack", the trick lies in the design of that piece and the direction that you rotate it in... if you rotate it counterclockwise it will be a good boy but if yu rotate it in a clockwise direction it will behave "abnormally" ...it's all about the Rolling and pitching motions. Legends has it that some of those students never slept again... I like your tricks at the end of each lecture Sir :)

[@pierocahuanovera7899]

Thanks for you aclaration :')

[@BleakStarshine]

Wow! these are really useful. Thank you, professor Lewin!! My university has recently closed all face-face lectures, the physics lectures have been discontinued and replaced with a series of short clips, which in my opinion are difficult to learn from. These lectures are helping me to keep up to date, and learning efficiently. Thank you, professor and good luck getting through these times.

[@sachinraut5729]

At 19:18 the dependency of 'd', the velocity of CG changes with 'd' cause as d is greater the most of the impulse is used in rotational motion and if d is smaller used in translational motion like if I hit a pen at free end and at the centroid. "Did'nt get the CG velocity is independent of 'd'" I watches ur lectures, the reasoning you are providing behind every thing is very great which i can relate. "You become one of my greatest teacher which i'll never forget" 🙏🙏

[@amitagrawalchd]

sir referring to that plastic rotation problem in the last of the lecture 46:40 I think it behaved so because it was not rotated about its centre of mass (which was somewhere inside the hump)so it experienced unequal forces and to conserve the angular momentum it had to rotate in the opposite direction. thank you ir for your selfless service.

[@nalloastro6353]

You are awesome! I always watch your lectures. They are clear and entertaining.

[@lecturesbywalterlewin.they9259]

:)

[@mehulverma9496]

16:15 If anyone still have _confusion_ you can see that *Impulse=Change in momentum* so I=MV (V= -Velocity of COM-) where I is vector so the direction of V vector should same as I vector :)

[@lecturesbywalterlewin.they9259]

torque = dL/dt

[@masterx5828]

Professor Lewin, I believe you may have confused some people in the comments when you said at 22:15 that angular momentum is conserved for any point P on the line you drew. This is ONLY true if P is not on the rod. I've seen many people in the comments who take point P on the rod, calculate the angular momentum, set it to zero, and get nonsensical results. If P IS on the rod, then the torque certainly isn't zero. Why? Because P is accelerating in C's reference frame. THEREFORE, the whole rod is accelerating in P's reference frame. If the rod is accelerating, then there MUST be a torque. This doesn't apply if P isn't on the rod because then P wouldn't be accelerating and torque would indeed be zero. You can try to calculate the torque about point P on the rod. You'll find that it is M*alpha*d^2 - Id/(delta t) where alpha is the time derivative of omega. If you multiply by delta t and set that equal to the angular momentum for p (omega*(Md^2 + 1/12Ml^2), you get the correct value for omega, namely omega = -12Id/Ml^2 Just a minor detail in an otherwise perfect lecture :)

[@lananhtran1124]

Thank you so much! I was so confused.

[@jaiviswesh6030]

in the asymmetry case of 10:25, the rotation with a larger radius has more force than a smaller radius. but centripetal force decrease with increase in radius

[@lozy497]

Centripetal force increase with the radius mate

[@shyamvijay8985]

Im a keen golfer and i stumbled upon your videos in order to understand the swing which i had read somewhere as being a double pendulum swinging around the left shoulder joint and the wrist joint.. Im quite excited about some of the ideas and im going to try it out on the range!

[@VickysTuition]

The plastic at 47:30 is called a rattleback or celtic stone if you want to google about it

[@Kurzy123]

I only wanted to say thanks for these lectures, they had helped me a lot through my Physics 1 course. Also, I'm finding the effort you put in responding to most questions on your videos, and really trying to help random youtubers quite an honorable deed. To sum it all up, thanks for you :)

[@lecturesbywalterlewin.they9259]

:)

[@yourguide5561]

I AM watching regularly and improving my my knowledge

[@nandeeshvirdi7975]

According to my guess: In last challenge, at first the spin provided was nice and smooth. So the contact point of plastic from the surface is nice and single, whereas after that the 2nd and 3rd time rotation provided to plastic was turbulent and contact points are different and in rotation too. Therefore to counter that rotation, It spins backward. Kindly correct me If I am wrong or lacking something in explanation.

[@baklavatatli7301]

At time 28:30 there is a discussion of an oscillating rod. Center of mass theorem says that you can think of the total external force acting on an object as if it is acting on the center of mass of that object. You don't just consider gravity acting on CM. The total external force on the oscillating rod is gravity plus the force exerted by point P. I assume this force at point P is along the line of the rod since the rod is free to rotate around point P. Let's call this force FpointP. So we can now show two forces acting at the center of mass. Mg, and Fpoint P. Mg is vertical. Fpoint P is along the line of the rod. Its contribution to the torque =0 because T=rXF=b.F.sinTheta = 0 because F and r are in the same direction and sinTheta =0. Dr. Lewin only shows Mg acting at the center of mass and disregards the external force on point P saying the torque produced by the force at point P is = zero because T=rXF = 0 xF =0 since r=0. This is correct and it is also how it is explained in many books, but it seems the other interpretation showing sinTheta=0 to cancel the torque due to FpointP more closely follows the center of mass theorem.

[@lecturesbywalterlewin.they9259]

the force on mass m at the center of Earth is zero. The gravitational PE of mass m at the center of Earth is half of what it is at the Earth's surface.

[@baklavatatli7301]

​@@lecturesbywalterlewin.they9259 Professor Lewin, thanks for all your work. Very good intro Physics lectures. p.s. Did you mean to reply to another comment? My comment was about torques on the rod.

[@shivaniits]

Hello Sir, At 47:14 when the plastic reverses it's direction after coming at halt I am assuming that since the plastic only reverses it's direction when initially rotated in clockwise direction so when the plastic is rotated clockwise it comes to halt and then reverses it's direction and starts rotating anti clockwise ..is it because that plastic in constructed in a way that it's equilibrium position is anticlockwise or it's due to some other reason ?

[@lecturesbywalterlewin.they9259]

There is no such thing as an "anti-clockwise" equilibrium position. Google "rattle back".

[@shivaniits]

Lectures by Walter Lewin. They will make you ♥ Physics. Oh alright , I will sir ..thank you :)

[@bhanu8391]

@@lecturesbywalterlewin.they9259 is that because of 3rd law of motion?? When you pushes plaatic your PE makes plastic to rotate and then when come to halt, cm may be on the other side made it rotate in opposite direction??

[@dmasmathematics7599]

Mashallah sir you are awsome . if you were our teacher we will be the futher scientists ...

[@lecturesbywalterlewin.they9259]

:)

[@muhhamadgulzar460]

Please tell me professor how we are getting diff amount of kinetic energy in the rod problem (15:00) if we strike at center we get only linear kinetic energy ,but if we strike at little higher or lower to center then we get additional rotational KE from the very same Impulse

[@lecturesbywalterlewin.they9259]

>>>if we strike at center we get only linear kinetic energy ,but if we strike at little higher or lower to center then we get additional rotational KE from the very same Impulse>>> *correct* and I cover that in detail in this lecture. Watch it again!

[@AL-op3ue]

remember impulse doesnt equal energy. work does. and the work is indeed greater. you didnt just push the rod by the amount the center of mass has moved. you also pushed an additional distance s, as in r times theta. do you understand?

[@ale8088]

you should think that is the same of leverage effect: there's no violation of conservation of energy. A force or an impulse produce different effect with respect to the geometry (e.g.: the lever arm) of the mass involved. Less force is applied, more space you have to run across and VICEVERSA. Due to additional distance from the CM, with a fixed amount of force, you get the extra rotational work. The real mistery, maybe, is how space and time are in somehow like diluited energy.

[@nachogonzalez8346]

why the object do that at the final? 46:10

[@annefiftythree]

I never studied physics and now almost aged 70, it is difficult to learn: my mind needs more time to process information. My best guess: he spins the object CW but the object is probably "rigged up" to spin CCW. Therefore given the CW torque, it will move CW but then stop and return to CCW as it was conceived to do. Our solar system spins CCW therefore the torque exerted by the sun will ultimately overcome any CW induced spin. In the solar system, it would be the friction of the solar wind, bombardement of CCW spinning photons, resp. the CCW vortex. Just a guess!

[@akhilanr1233]

I am just 13 and I really want to become a physicist, So I wanted to start to learn from MITOCW last year and now I know more math and physics than a 12th-grade student. THANKYOU MIT!

[@emanuelelorenzano5206]

Dear Professor, I'd like to thank you for your enthusiasm and for all the insights on the wonderful world of Physics. Unfortunately, I've found out about your lectures and book too late, they would've been of great help during my academic studies. I'm a high school Physics teacher, trying to spread out my love for this discipline. I always talk to my students about you and your works - even show them some of your real-life experiments! Let me also take advantage of this comment to ask you about the solutions to the "non intuitive" situations shown in class, such as the apparent violation of angular momentum conservation for the spinning plastic object: could you tell me about any source (document, video, website, etc.) discussing them? I'd like to analyze them with my students. Thank you. Best regards, Emanuele Lorenzano

[@tchan4214]

Sir, I was wondering if it's appropriate to simplify the Mg just on the center of mass while we are calculating the torque, since the mass is distributed all along the rod or ruler. At 29:09.Thanks!

[@lecturesbywalterlewin.they9259]

try it

[@tchan4214]

@@lecturesbywalterlewin.they9259 Got it! Thank you!

[@juanmanuelespanabolacuenta4048]

I wish I knew more English to fully understand everything :) Great class :D

[@mohammedanaskhan5772]

@40:08 centre of mass of disk is at C. howsoever it we compare the value of omega with pendulum, it occurs that centre of mass is situated at 2R. where am i making mistake ?

[@lecturesbywalterlewin.they9259]

My derivation is correct. Watch it again, and listen what I say about an ideal pendulum with length 2R. Yes 2R.

[@mohammedanaskhan5772]

GEOMETRY :)

[@lecturesbywalterlewin.they9259]

The center of mass of the "hoop" is at the middle of the circle. It would be naieve to expect that the period of the hoop would be like an ideal pendulum with length R. The geometry is so very very different!

[@mohammedanaskhan5772]

thank you sir.

[@joelthoorengstrom]

Hello Professor and thank you for your magnificent lectures! I can't get my head around what is happening with the energy when you hit the ruler in your example @17:00? My intuition tells me that some energy from the impulse will go to KE of the centre of mass and some KE will go to the rotation. And yet you say that v of center of mass is independent of the distance d, (while rotational KE obviously isn't). With the conservation of mechanical energy in mind I must be missing something. Can you help me? Thank you again for your wonderful work and inspiration and for making it so available for people throughout the world! I have watched all the previous 20 lectures in 8.01 and enjoyed them thoroughly.

[@lecturesbywalterlewin.they9259]

Impulse is NOT energy. They are as different as apples and coconuts. If an object of mass m has zero speed and you apply an impulse J then its speed increases by v (J=mv). If v=1, then its KE has increased my 0.5*m. Now we apply the same impulse to that object when it has a speed of 10, then its speed will again increase by 1 thus it becomes 11. What now is the increase in KE? It's 0.5*m(11^2 - 10^2) which is 0.5*m*21. Thus the same impulse now increased its KE 21 times more!

[@joelthoorengstrom]

Thank you for your reply! Been thinking about this the whole day but I think I can go to sleep now :). Let me just check if I understand this correctly. So the equation of Impulse is I=F*time, and the equation of energy is E=F*distance (work). If an impulse is given closer to the centre of mass, there will be a higher resistance on that force of the impulse since the "lever" relative to centre of mass is shorter, meaning that the force will do less work, since it will travel a shorter distance, thus resulting in lower energy transfer. Because please say it is so that it IS you hitting the ruler that gives it KE, I mean it must come from somewhere right? Also, hypothetically, would this mean that if you build a wind-turbine with ridiculous radius and, for the sake of the argument has the same mass as the ruler, and then you give it the same tap you gave the ruler at the far end of the wings you would produce energy to supply a whole nation? Or am I, as we say in Sweden, "out on a bike-ride"? ;) Thank you for your time, Professor!

[@joelthoorengstrom]

ps. because if you have the same work input in the system, it would no longer matter at which distance d you hit the ruler, the KE would be the same, right?

[@lecturesbywalterlewin.they9259]

The best way to think is: no matter where you hit the center of mass of the ruler will ALWAYS move with a speed pf J/m. J here is momentum, m the mass of the rod. If you hit away from the center of mass NOT ONLY will the com move with speed v=J/m with KE 0.5*m*v^2 but the ruler will also start to rotate, There is an ADDITIONAL rotational KE. END OF story.

[@joelthoorengstrom]

Thank you sir. I will continue to follow your lectures!

[@En_theo]

What was the answer for the last question at the end of the video please ?

[@juanmanuelespanabolacuenta4048]

have you found it?

[@annefiftythree]

Watch carefully the spins; the refers to outer space: the first spin is CCW like our planet. The next ones are CW which is against the ambient field... causing friction... causing the halt and then return CCW like the ambient field... Venus is currently slowing down its spin. Some time in the future, it will stop and spin CCW like the ambient field.

[@superduperstaar]

17:44, dL/dt = torque, so dL( change in angular momentum of rod about com) would be equal to (torque)(dt) = (Force)(d)(dt) = (Impulse)(d), hence angular velocity = (impulse)(d)/(moment of inertia about com) is it correct?

[@superduperstaar]

one more question, As rod rotates about it's com after impulse it's angular momentum must be same about any point (at p also) but torque about point' p' is zero so angular momentum about p should be zero but it's not!? how?

[@lecturesbywalterlewin.they9259]

I watched at 17:44 Angular momentum at all points on the line through the impulse is ZERO. That's true before the hit and after the hit. You can solve this problem by using the impact point as your reference point. Ang mom is conserved and is ZERO. No need to use the com. Students prefer the com as they have a better feel for it.

[@lecturesbywalterlewin.they9259]

I cannot add to my lecture. If you include the impulse before the hit then angular momentum about any point you choose before hit is the same as after the hit. But, of course, the amount of ang momentum depends of the point you choose. If you choose any point on the line of the impulse then the total any mom is ZERO before and after. But if you choose a point somewhere else, it has a different value but it IS conserved. If you ONLY view the situation after the hit. You may choose ANY point (don't move it) ang mom after the hit is then conserved. But its value depend on the point you choose.

[@franciscogabriel8108]

On 11:27 why isn't angular momentum conserved about point C? If F is the only force acting on the rod then, relative to C, torque = r x F = 0 (since r is parallel to F) am I missing something? Is there some other force I'm not considering?

[@lecturesbywalterlewin.they9259]

I watched 11:00 - 12:30 I cannot add anything to the clarity of this lecture.

[@franciscogabriel8108]

@@lecturesbywalterlewin.they9259 Thank you for such a quick response, and many thanks for the work and devotion that must have gone into these lectures, you can really feel your love for the subject and we can't help but be infected by it too. But can you please tell me, when the rod rotates about point P, what is the torque relative to point C, just the result then I can, probably, work out the reasoning behind it, because on 11:27 you say angular momentum ins't conserved relative to point C meaning there is a torque relative to that point but I don't see how, since r_c//F. Shouldn't all points on the line of action of the force, F, also have torque=0?

[@xnonqme3716]

​@@franciscogabriel8108 there are other forces that act at the pivot (point p). Namely the normal force & friction (if it isn't negligible). When you take the torque around point P, you can disregard those forces as their position vector is 0.

[@adarshchaturvedi3498]

i dont want to go for dinner when i watch your lectures Sir ...

[@lecturesbywalterlewin.they9259]

:)

[@sidhardhansunilkumar8763]

can someone please tell me why the angular momentum at point C is not conserved like in point P? it is shown at 11:30 (I mean the positions vector is in the same direction as that of F so the Torque must be 0 right?)

[@solounomas0]

best teacher ever

[@williamleung8588]

Professor, I am confused at the result at 19:00 that "the velocity of center of mass is independent of where I hit the ruler", does it mean the translational velocity, and therefore, the translational kinetic energy of the ruler is the same no matter where I hit it?

[@lecturesbywalterlewin.they9259]

yes the translational KE of the cm is independent of where you hit. The rotational KE does depend on where you hit.

[@williamleung8588]

Lectures by Walter Lewin. They will make you ♥ Physics. But where is the rotational KE come from if all the impulse is “converted” to translational KE.

[@lecturesbywalterlewin.they9259]

impulse is NOT energy.

[@williamleung8588]

Lectures by Walter Lewin. They will make you ♥ Physics. I mean the energy generated by the impulse. The system energy(rotational KE + translational KE) will increase if I hit it further away from the center of mass, by the same amount of input. Is that possible?

[@lecturesbywalterlewin.they9259]

YES that is possible. *Impulse is NOT energy.* The change of KE depends on where I hit.

[@ismailsevimli6424]

Hi from Turkey, These guys are so lucky, really. And also I am since I learned this channel :))

[@lecturesbywalterlewin.they9259]

Thanks! 😃

[@AuCeVi]

How i wish you were my physics teacher Professor Lewin. I've understood so much from you. Thanks.

[@aman2426]

I didn't understand at 22:55 when you said take the origin at P. How do I take the Angular Momentum about P? For particles it is r x mv but in case of rigid body we have only studied I.Ω about CG where body rotates cg only.

[@lecturesbywalterlewin.they9259]

angul mom of a rotating object can be determined (with some math) relative to any point P.

[@yourguide5561]

Physics is mind blowing subject which blows our mind and will change our world

[@abhishekpatil3105]

Pr. Lewin you're really amazing teacher and gift to the physics world. I'm preparing for UPSC Exam (India) where i opt. for Physics subject as one optional. You're lectures made it easier to prepare. Thank you so much 😃

[@lecturesbywalterlewin.they9259]

:)

[@manuelsorianogaitero2434]

Dear Mr Lewin, in the case of bar spinning around P point, you say that angular momentum is constant cause no torque is produced by centripetal force but I think that you must to observe the force Mg acting vertically descending in centre of mass that obviously creates a torque in P. Thanks, I love physics too.

[@lecturesbywalterlewin.they9259]

if a bar is rotating about a fixed point P then the angular momentum of the rotating bar relative to point P is ZERO as there is No torque relative to point P.

[@backyard282]

At 29:50 you said that the minus sign is necessary because we have a restoring force, just like with a spring. However you took the absolute value of the torque so you shouldn't be dealing with negative signs at all, no?

[@lecturesbywalterlewin.they9259]

what I did and what I said was correct

[@kathandesai5896]

30:07 mgbsin(theta) is also restoring torque so why we don't put minus sign there and on both side's minus cancle out each other ?

[@saidioussama2837]

Good morning sir, I also want to thank you for being an inspiring physicist for all mankind and for raising new questions and also for giving us new insights by looking differently at nature ! I think you are one of the greatest teachers ever!

[@lecturesbywalterlewin.they9259]

thanks for your kind words

[@ashwina5472]

sir @11:56 if we take c as a reference point will the torque be zero since force and position vector lies on the same plane

[@lecturesbywalterlewin.they9259]

you cannot take point C as you do not know the force at P.

[@simrandeepbahal7594]

Lectures by Walter Lewin. They will make you ♥ Physics. Dear Prof Lewin Firstly Thanks for your fantastic lectures. At 23:07 you told that the same problem can be solved by taking point P as a reference. How? I am unable to solve it. About point P there is no Torque. So angular momentum before and after will be zero. How to proceed further??

[@youquestion9249]

Prof. Lewin, At around 22:40 you say that if we take a point P which the impulse I goes through, the change in angular momentum is zero therefore L = 0 before and after. But isn't the object rotating about its center of mass? And doesn't that mean that its angular momentum should be the same everywhere, and equal to its spin angular momentum? So why is the momentum I different about 2 different points in this case?

[@lecturesbywalterlewin.they9259]

The angular momentum relative to point P of the rotating ruler after the hit is ZERO. You have overlooked that the angular momentum of the center of mass has then the same magnitude as the angular momentum of rotation but with opposite sign.

[@arunjhalak755]

at 21:29 in the 'rod' problem..i have a thing to add..i need you tell me if I am right or wrong.. when you hit the rod with a certain impulse you give it a certain energy, when you hit at the centre of mass, some of that energy gets eaten up in giving the rod a certain translational velocity and the rest is getting up used to heat the rod, beacuse masses are balanced so the rod can't rotate, however if you hit it away from centre of mass, the energy which was lost in the form of heat in 'hitting at centre of mass' case, is now responsible for rotational kinetic energy of the rod.. FINAL CONCLUSION:- the rod is most hot if you hit it at the centre of mass,and least hot if you hit it at one of the edges.. does this seems right to you?

[@lecturesbywalterlewin.they9259]

Impulse is NOT energy!!!!! energy is Joules; impulse is NOT. Nothing gets eaten up when I hit the rod at its center. Change in impulse can lead to different changes in KE. TAke an object of mass 1 kg and zero speed. Apply an impulse of 1 units SI. calculate the increase in KE. THen take the same object when it has a speed of 10 m/sec now apply an impulse of 1 unit SI. Calculate again the increase in KE! It's 21 times higher now!!! Of course if I apply the impulse of 1 unit SI I have to do 21 times more work.

[@arunjhalak755]

Hmmm.. I got it...if there are two objects of same masses,one is moving and the other is not...and if I were to increase their velocities by one unit each...then I have to apply a greater force in case of moving mass cause it's always moving and therefore time of contact between me and the mass will be lesser and therefore I better apply greater force so that I give impulse equal to that I have given to standing mass...and hence more force equals , more work i.e I gotta do more work in case of moving mass

[@youngbloodbear9662]

The understanding hurts. Mr. Lewin you're making me see so much more in the world around me.

[@lecturesbywalterlewin.they9259]

great

[@kathandesai5896]

10:20 who is responsible for the centripetal force?

[@mohammedanaskhan5772]

@11:50 how are we going to calculate the torque along point C ? should i take component of force and cross product it with the perpendicular distance of individual axis ?

[@lecturesbywalterlewin.they9259]

You can choose C but you cannot move C. It must be a fixed point. If you choose it where I placed it in my drawing on the blackboard, there is no torque relatve to C as the force that P exerts on the rod goes through C. But the rod rotates, and thus the force that P exerts will no longer go through C. Thus there will be a changing torque relative to a fixed point C. Clearly angular momentum relative to any point that you choose other than P will not be conserved.

[@lecturesbywalterlewin.they9259]

If you choose a point anywhere in the plane of the rotation (call that point C) other than at P, then during one rotation about P the torque relative to C will twice be zero. If you choose C not in the plane of rotation then the torque relative to C will never be zero but, of course, it will also change in time as the ruler rotates.

[@arunbhardwaj2744]

@Madhukar B R BECAUSE FORCE APPLIED BY HINGE IS ALWAYS IN THE PLANE OF ROTATION AND IF YOU CHOOSE A POINT NOT IN THAT PLANE THEN,,THE DISTANCE FROM THAT POINT AND THE FORCE APPLIED BY HINGE WILL ALWAYS HAVE SOME ANGLE BETWEEN THEN THUS THEY ALWAYS HAVE A TORQUE....

[@abhishekcherath2323]

Sir I can't find the solution for the problem at 23:29 taking the origin on the line of the impulse. Also, I don't understand how it can be zero if the spin angular momentum is nonzero. Since the object is rotating around CoM shouldn't measuring L always give the same value regardless of where I do it from? EDIT: OK, I think Angular momentum about P = Angular momentum about CoM + Angular momentum of CoM wrt. to P. This is giving me the correct answer in this case, but this is clearly not the same value as the spin angular momentum...

[@lecturesbywalterlewin.they9259]

I watched again. It's al straightforward Newtonian mechanics. I cannot add to the clarity of that lecture. I suggest you brush up on the meaning of F=ma, on momentum, impulse, the conservation of momentum, on angular momentum and on the conservation of angular momentum. You may want to consult google or quora.

[@abhishekcherath2323]

Thank you for replying sir, but I'm just a big idiot. I thought that SPIN angular momentum being conserved also implied ORBITAL angular momentum being conserved. Just confused because I zoned out for about 30 seconds right after the spin angular section in the last lecture. Which is precisely where you clarified that. Got it now. Anyhow, moral of the story being listen better and take more complete notes. Thank you very much for your lectures!

[@abhishekcherath2323]

Oh but sir, is the equation I got above correct? That is to say, is the angular momentum of the earth about the sun equal to the sum of the spin angular momentum of the earth and the orbital angular momentum of the earth about the sun? edit, better put Lp = Lcom + Rcom x MVcom where Rcom is vector from p to CoM.

[@lecturesbywalterlewin.they9259]

>>> I thought that SPIN angular momentum being conserved also implied ORBITAL angular momentum being conserved.>> orbital angular momentum of planets is conserved but ONLY about the Sun, about no other point.

[@lecturesbywalterlewin.they9259]

question unclear. The spin ang mom of the Earth is conserved since there is no external torque on the Earth about the Earth spin axis. The orbital ang mom of the Earth is conserved ONLY about the sun as the torque on Earth is then always zero. .

[@toygarozel6261]

Thank you from Turkey

[@TheDinobeg]

Hi , 11:27 , how come the momentum is not conserved in point C when angle between the force and r is 180?

[@TheDinobeg]

PS 11:46 nice

[@lecturesbywalterlewin.they9259]

momentum of center of mass is changing all the time.

[@jabedumar9656]

Your lecture makes physics very easy and logical 😊😊 Thank you sir ❤️ .

[@BroadeningHorizonsos]

at 7:25 you said that angular momentum about any other point is not conserved.... my question is that ......Is angular momentum in this case, a harmonic function of time or something else?

[@lecturesbywalterlewin.they9259]

I leave it up to you to answer this.

[@herzikkimolog]

11:13 in lecture, you said that torque relative to "P" is zero; but isn't there the weight of ruler itself? If "C" is center of mass, there should be a force downwards my intuition says, am I wrong?

[@utkarshtripathi2755]

That's balanced by the normal reaction I suppose

[@prakharbhalla9461]

at 11:26 you told that on taking point c angular momentum is not conserved but torque about that point c would be zero since position vector and force are in same direction hence about c it must be conserved. please reply. thank you

[@lecturesbywalterlewin.they9259]

what I said is correct

[@manassrivastava7448]

sir in last part of your video you made an experiment with a plastic piece you give some momentum to it it rotated in One Direction then it reversed its direction of motion I found similar plastic speech in my home I observed it carefully and I found get the hump or the curve part was not on centre of mass so when you rotated it the angular momentum was not conserved and in one specific direction because of unequal division of torque it reversed it's direction. it was only in case of anticlockwise not in clockwise direction Sir am I correct

[@lecturesbywalterlewin.they9259]

yes very cool, not so easy to explain

[@manassrivastava7448]

Lectures by Walter Lewin. They will make you ♥ Physics. Thanks Sir . You are​ very cool that you shared such a different situation and enforced me to think about it.

[@nna23101989]

You are genius, manas srivastava. Even after watching the explanation from professor Walter Lewin here: kzbin.info/www/bejne/np6rh2B-e5JlaNU, I still don't understand the mechanism of this toy (the rattleback). Of course, I understand that its behaviour comes from the asymmetry of mass distribution and appropriate design. However, I cannot explain it to myself using my knowledge on torque, angular momentum that I learned from the lectures of prof. Walter Lewin. That is I still don't understand anything, I just know the name of this toy.

[@omozaas]

sir can u please show me the solution by taking point P as reference point. see timing 23:10

[@lecturesbywalterlewin.they9259]

ang mom relative to Pis 0 = sum of ang mom of the com and the rotation (rel to P). Please work it out for yourself

[@hponesanaung8221]

The work done on the homework question would changes depending on where the impulse is applied? Because if translation KE stay the same no matter where it is hit, but the angular momentum changes depending on where it is hit, the work done has to be different .

[@gerhardimmanuel]

SIR THANK YOU SO MUCH FOR YOUR LECTURE!!! Because of you now i can fully understand the concept of the material 😅😅

[@canned_heat1444]

Hello! At 11:23 you said it wasn't conserved for point c. Why is that, being that it is parallel to the force?

[@canned_heat1444]

by "it" I mean angular momentum

[@lecturesbywalterlewin.they9259]

Put yourself at position C on the blackboard. It's a fixed point. You must stay at C. As the rod rotate about P, the torque about C is ONLY zero at the moment that C and P are aligned with the rod (which is the case in my drawing). Pick ay random point in time and there is a torque relative to C (and this torque even varies in time). Angular momentum is not conserved relative to C.

[@canned_heat1444]

Thank you!

[@sohithvulipe7577]

Professor, How is the angular momentum into the board while the torque is in its opposite direction. Is it the right hand rule. At approximately 2:00 professor you said that the direction of angular momentum is through the board and that of torque is out of the board.

[@lecturesbywalterlewin.they9259]

watch again at 2:00. I define ang mom and I define torque. *Thus angular momentum is into the board and the torque is out of the board*

[@sohithvulipe7577]

Lectures by Walter Lewin. They will make you ♥ Physics. Professor, Because of cross product the direction is perpendicular but why are both not in the same direction. Why is ang mom in negative z-direction and torque in positive z-direction when the board is considered as the x and y directions. Why aren't both in the same direction.

[@lecturesbywalterlewin.they9259]

>>>>Why aren't both in the same direction.>>> *rXF is in the opposite direction as rXp.* Watch my lecture and listen to what I said. I cannot add to this. *When I stretch both my arms, why are they in opposite directions?*

[@sohithvulipe7577]

Lectures by Walter Lewin. They will make you ♥ Physics. Professor, I got it now but a small doubt. Is it fixed that the ang mom is in the negative z or you just said it for explanation purpose.

[@lecturesbywalterlewin.they9259]

It's NOT important what you call x and what you call y and what you call z . *What counts here is the rXp and rXF are in opposite directions.*

[@jothieswaranp5660]

Dear Professor, It is claimed that torque about the point p is zero at 11:23. Though may the centripetal force about that point becomes zero, What about the weight that passes through the center of mass? It would surely produce torque about the point p?

[@lecturesbywalterlewin.they9259]

>>>It is claimed that torque about the point p is zero at 11:23.>>> *yes that is correct, I cannot add to the clarity of this lecture*. I suggest you watch it again. >>>>What about the weight that passes through the center of mass? It would surely produce torque about the point p?>>>> *incorrect* bcoz the mg force (weight) downwards is canceled precisely by the normal force upwards from the table

[@jothieswaranp5660]

ok sir got it!

[@heramb575]

Professor why did the piece of plastic behave in that weird (seemingly unnatural) manner at the end?

[@prakharbhalla9461]

in the problem of ruler on friction less horizontal surface at 22:17 you told that on choosing point p as origin the angular momentum after the impulse is zero since torque is zero about P. But since ruler is rotating about C.M hence its angular momentum should be spin angular momentum which will be (I about center of mass * omega). hence change in angular momentum is spin angular momentum. And this change should be equal to zero since torque is zero which gives omega as zero which is not correct. please help thanks in advance. and i have seen your solution notes but in that point P is not chosen as origin .

[@lecturesbywalterlewin.they9259]

>>> you told that on choosing point p as origin the angular momentum after the impulse is zero since torque is zero about P. >>> *that is correct* The angular momentum after the hit of the system is again ZERO. You overlooked the angular momentum of the translating center fo mass is in the opposite direction of the am of the rotation. The two cancel.

[@prakharbhalla9461]

thank you very much sir. you are a genius. in india it is very difficult to get a chance to ask question from great prof. so you are doing a great help to student like us. thank you very much.

[@Dudeguymansir]

Is it correct to say that omega at P is zero? That's where I'm getting confused. I think my intuition is leading me astray, but I feel like since the bar is still spinning, omega at P shouldn't be zero. If I followed point P with a "zoomed-in camera", I'd still see the bar making rotations (although my camerawork would be pretty jostlely as I try to keep P in the center of the "screen"). After some amount of time, it will complete a revolution and return to its original orientation, which would imply a non-zero omega about point P, no? What am I missing? Thank you, Dr. Lewin, in advance!

[@lecturesbywalterlewin.they9259]

@@Dudeguymansir how many minutes into the lecture?

[@Dudeguymansir]

@@lecturesbywalterlewin.they9259 I've rewatched it in its entirety twice, and some parts I've watched as many as four times.

[@canmumcu9804]

43:50 that excitement tho :)))

[@lupolupo2317]

Hi prof. First of all sorry for my english, it's not my native language. Thank you very much for your lessons. I have a question about the problem in 22:51 by taking point P as reference point. I understand the first part, the torque realtive to p is 0 and therefore the initial and final angular momentum relative to p are the same, both 0. I come to the conclusion that the final angular momentum relative to p is equal to the angular momentum relative to the CM plus the vectorial product (position of the CM relative to P x linear momentum of the CM). *L(p)* = *L(CM)* + *r(Cm-p)* x *p(Cm)* I understand: 1) From the point of view of p the CM *apparently* rotate relative to p with an angular momentum equal to *r(Cm-p)* x *p(Cm)* 2 ) *BUT* you dont have a particule, you have a rod and you need to add the angular momentum of the rod relative to the CM. *L(CM)* I obtain the same solution that if i choose the CM point of view. I feel that my interpretation it is not correct because you say in the lecture 20 that the angular momentum if an object rotate around an axis that cross the CM is *THE SAME* for all points of view kzbin.info/www/bejne/qX_Ekn9nbtSjrtk#t=16m02s. Could you help me with my interpretation? Thank you very much

[@lecturesbywalterlewin.they9259]

ang mom rel to point P is zero and it is conserved. Thus after the impulse there is a POSITIVE rotational ang mom of the rod about its center of mass but there is a NEGATIVE ang mom about P due to the translation of the center of mass of the rod. The sum of the 2 are ZERO. That should help you enough. good luck

[@betty_like_butarrr6247]

Sir, is spin angular momentum also valid for 3-D objects ??

[@mohammedanaskhan5772]

@30:02 Sir i cant digest the idea of torque being negative .can you please explain.I will be obliged.

[@lecturesbywalterlewin.they9259]

Torque is a vector. Velocity is a vector, Forde is a vector. Thus they all can be positive, zero and negative.

[@mohammedanaskhan5772]

why mgrsin(theta)= - I x alpha

[@lecturesbywalterlewin.they9259]

torque=I*alpha this is the rotational version of F=ma

[@mohammedanaskhan5772]

sir why is the negetive sign comming

[@lecturesbywalterlewin.they9259]

watch my lecture! It's all there if you are familiar with vectors.

[@BroadeningHorizonsos]

how can same impulse transfer different energies??

[@lecturesbywalterlewin.they9259]

Change in impulse is a change in Mv, change in KE is a change in 0.5Mv^2. They are not the same.

[@yashdesai8372]

Sir at 8:48 you wrote Lp=Ip*w, but sir 'w' is about center of mass 'c', so we should require 'wp', so that we can write Lp=Ip*wp. From equation (5) in the same video

[@lecturesbywalterlewin.they9259]

>>> Lp=Ip*w>>> *that is correct.* >>> 'w' is about center of mass 'c'>>> *that is incorrect*

[@sweetysingh7113]

hello sir at 22:51 you said tp =0 bot this would be only zero if we choose position vector direction same as that of force(I) so my question is : that to solve problem we specifically choose the above case or tp=0 even for any other position vector (except from center)thank you for wonderful vedios love you sir

[@lecturesbywalterlewin.they9259]

+sweety singh If I hit the rod at point P, then the angular momentum relative to point P is zero and thus angular momentum about P is conserved. If I draw a line through P which indicates the direction of my hit, then angular momentum relative to any point on that line is also zero. In my case that line is perpendicular to the rod.

[@tanaykumar3493]

Please help!! Sir, when you were explaining the problem 7.9, how should we proceed the problem by taking point P as the origin. I mean what will be the angular momentum about the point P after impulse because when I conserved angluar momentum about point P then the equatuion which I got was Lp= Ip*w 0=(Ml^2/12 + Md^2)*w this would give w=0 which is not possible. Lp is angular momentum about P and w is angular velocity.

[@lecturesbywalterlewin.they9259]

freepdfhosting.com/86109d309b.pdf

[@tanaykumar3493]

But sir in the solution the reference point taken is C and not P and I did not understood the approach using P as reference.

[@Galenus0]

I'm stuck at the exact same question, how would you solve the problem for point P? (22:49) Can't find the solution to this in the Assigment.

[@arunbhardwaj2744]

@@tanaykumar3493 ANGULAR MOMENTUM ABOUT A POINT,,,IS THE SUM OF ANGULAR MOMENTUM OF C.O.M ABOUT THE POINT+ANGULAR MOMENTUM OF ROD ABOUT C.O.M

[@RakibulHasan-zg6zh]

​@@arunbhardwaj2744can you please show in details??

[@arafsartistry5421]

Hello sir, When a disk rotates on its CM and also moves forward (Like a tire on the road, A disk rolling over a slope) How can we calculate It's Moment of inertia? (Does this case follow Parallel Axis Theorem? I think it does not.)

[@lecturesbywalterlewin.they9259]

you can caculate it's moment of enertia about the rotating axis. The moving axis is then your frame of reference. The moment of enertia relative to a point on the road or above the roaad or below the road will be different.

[@pawan120989]

sir I want to ask if natural frequency is same on earth and moon.

[@lecturesbywalterlewin.they9259]

If you tell me what is the natural frequency is of the Earth, I will answer your question.

[@pawan120989]

+Lectures by Walter Lewin. They will make you ♥ Physics. sir like in time period of simple pendulum T=2*pi* sqrt(L/g) we have 'g' term which is different for earth and moon ...on moon g/6 ..doesn't it make any difference in time period....as you said angular frequency is not going to change. Thanks

[@lecturesbywalterlewin.they9259]

I now understand your question. Yes g' on the surface of the moon is about 9.8/6 m/s^2. Thus the period of a simple pendulum on the moon is 2pi*sqrt(L/g') about 2.5 times longer than on Earth.

[@pawan120989]

+Lectures by Walter Lewin. They will make you ♥ Physics. Thanks sir for such a valuable information. Now this the best place for all my queries thanks again.

[@lecturesbywalterlewin.they9259]

> YES

[@رهامسرور-م9ه]

Will the answer to question at 20:00 equal 12Id/Ml^2 ?

[@naixin-lg6tc]

Yes

[@masterx5828]

Correction: it's -12Id/Ml^2. The object is rotating clockwise so omega is negative. This is accounted for when we take the cross product

[@sudiptosarkergarbo3674]

Sir, can we calculate the time when the Leaning Tower of Pisa will lean to the extreme point that its center of mass will be out of its balance...?

[@lecturesbywalterlewin.they9259]

try it

[@yash29210]

What is PIVOT, you mentioned it DURING THE VIDEO....(18:00)??

[@lecturesbywalterlewin.they9259]

PIVoT was a site with my lectures in 2000 and thereafter. That site does not exist anymore as it has been replaced by better sites like this one.

[@yash29210]

Just wanted to tell you that I was born at 9:15 on the day when you gave this lecture...............and by the way, this is me, YASH (just changed the name of my account).......... . . . . . . Didn't knew that when I was born, such wonderful things were happening inside 26-100!!!!

[@lecturesbywalterlewin.they9259]

:)

[@atharvsakhala9469]

Professor, thank you for the amazing lecture. For the HW problems and readings from the book in the attached pdf, where can we find the book?

[@pheonixnova4383]

JEE

[@surendrakverma555]

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

[@lecturesbywalterlewin.they9259]

Most welcome

[@BentHestad]

Sir, your lectures are awsome! Thanks! Where can i donate to this fantastic channel? Grateful regards from Trondheim, Norway!

[@lecturesbywalterlewin.they9259]

yes you can donate - please do! PayPal: [e-mail]

[@BentHestad]

Thanks, a little donation on its way across the ocean. Thanks again and please keep up great work for the benefit of students all over the globe. God bless.

[@lecturesbywalterlewin.they9259]

Thank you kindly, Bent - I received it!!!

[@正正气常存]

does the spin angular momentum only apply for symmetric objects that the forces all cancel out?

[@lecturesbywalterlewin.they9259]

watch my 8.01 lectures to know how angular momentum is defined.

[@WilfredoAbudeye]

Excellent Videos! We really appreciate it!

[@mohammedsulaimanusman525]

love from india!🖤❤

[@gouravnishad6829]

consider there was a marble colling perfectly elastically with the ruler (head on)..if the marble hit the end of the ruler..if the ruler hit the centre of mass.there would be just translational kinetic energy in the rod then after collision..but if the marble hit the end of the rod..there would be some extra energy in the form of rotational kinetic energy..where does that energy come from?

[@lecturesbywalterlewin.they9259]

use google

[@trilokimaurya1479]

11:50 why ever do that????

[@fabriziotabasso1694]

Ok that little piece of plastic really confuses me. After watching it at least 100 times and after two sleepless nights I still don't have an answer but I do not intend to give up! Ok...If I've learnt something from these lectures, when an object rotates about its center of mass, angular momentum should be conserved, and if an object changes its angular momentum there must be a torque acting on it. So, my best guess is: In the first trial it was rotating about an axis through its center of mass, so angular momentum is conserved, and it came to a hold because of friction forces. Anyway I think the mass is not uniformly distributed so the direction of the force applied to make it rotate DOES matter. In the clockwise rotation the object was forced to rotate about a different axis and that's why it appeared to be shaking a little, maybe that motion could provide the torque necessary to invert the direction of rotation. Am I on the right track?

[@fabriziotabasso1694]

shreyas Sudhaman Thank you!

[@lecturesbywalterlewin.they9259]

+Fabrizio Tabasso Google "rattleback"

[@tehyonglip9203]

+Lectures by Walter Lewin. They will make you ♥ Physics. why did it come to a stop when push it slower, and reverse direction when you spin it harder?

[@prakharbhalla9461]

sir, can't we solve for SHM of ruler by assuming that whole mass is concentrated at center of mass and then aplying F=-ma?

[@lecturesbywalterlewin.they9259]

try it - if you get my answer, then the answer is YES if you don't get my answer then the answer is NO I already know what you will find.

[@prakharbhalla9461]

i have solved and every time i get the time period of pendulum of length equal to the distance between pin and center of mass which is not matching your ans. am i doing something wrong in solving or this is not the way to solve.

[@lecturesbywalterlewin.they9259]

>>> can't we solve for SHM of ruler by assuming that whole mass is concentrated at center of mass and then aplying F=-ma?>>> *try it* - if you get my answer, then the answer is YES if you don't get my answer then the answer is NO I already know what you will find.

[@prakharbhalla9461]

sir i got the time period 2*pi*sqrt(b/g) which is not same as your ans. what i am doing wrong?

[@lecturesbywalterlewin.they9259]

I sent you the correct solution. That should be enough for you to figure out what you did wrong. If you cannot do that, then watch more of my lectures or ask quora.

[@ashwina5472]

sir is the angular velocity for that problem is I*d/Inertia

[@lecturesbywalterlewin.they9259]

which problem. how many minutes into the lecture?

[@ashwina5472]

@20:10

[@lecturesbywalterlewin.they9259]

Id/Inertia ???? is that d?

[@ashwina5472]

position vector from c to p is d ,so i think that is correct.did i make any mistake sir

[@ashwina5472]

mvd=w*inertia ,mv=impulse so w=impulse×d/inertia

[@arunavasarkar3508]

hello professor i have a question.. can a rigid body like the rod in your lecture oscillate if the hinge was at its centre of mass?

[@lecturesbywalterlewin.they9259]

no

[@michaelz5633]

professor please tell my why the plastic spins back I have insomnia now

[@jmart474]

Thanks for your videos professor they are really amazing. I have a doubt related to the ruler movement when you apply an impulse to it. I understand that the ruler's CM has a certain velocity after the hit independently of where the hit has been applied. This gives a certain kinetic energy to the ruler. If you hit the ruler at its CM the ruler won't rotate, but if you hit the ruler at any other point the ruler will have a combined rotation-translation movement. The translation KE will always be the same, but you will also have a rotation KE. I cannot understand how you can have two different energies with the same applied force.

[@lecturesbywalterlewin.they9259]

momentum is not energy. If you change the momentum of an object, the change in KE can differ. m1=m2 v1=1 v2 = 10. I add momentum 3 SI units to both. No calculate the change in KE in these 2 cases!!!

[@jmart474]

So this means that depending on where you hit you will have different energy losses, and the less "efficient" would be the hit on the CM?

[@lecturesbywalterlewin.they9259]

@@jmart474 the words "energy losses" is incorect. any change in momentum can give rise to a different cange in KE - remember momentum is NOT energy.

[@jmart474]

I think that I have finally demonstrated what I was trying to explain. I considered a system consisting of two point masses m joined by a massless rod of length 2R at rest that is hit perpendicularly by a point mass m with speed v at a certain distance d from the center of mass. The hitting mass remains at rest after the collision. Analyzing this problem with conservation of linear and angular momenta I find that the decrease of total kinetic energy is (mv^2)((d/R)^2-1)/4 which confirms that the maximum decrease corresponds to d=0, this is, the impact at the center of mass. The minimum corresponds to the collision with one of the rod ends (d=R). In this last case the collision will be elastic.

[@beratgonultas5493]

In video, you have asked how the object can be supposed to move after the impulse I applied. And you have also said that because center of mass can't do any absurd movements -in case of if it rotated about any other point-, the object must rotate about its center of mass. It is reasonable even there is no concrete proof. But i don't understand why the object has to rotate. There is no friction, and so i feel like there won't be any reason to exist for a point which behaves like a pivot and about which the object will rotate. Why the object must rotate? For object, moving without rotating seems to me more logical since there is no friction. Can you help me intuit and understand why the object must rotate?

[@lecturesbywalterlewin.they9259]

how many minutes into the lecture?

[@beratgonultas5493]

Lectures by Walter Lewin. They will make you ♥ Physics. around 16th minute

[@lecturesbywalterlewin.they9259]

It HAS to rotate because of the conservation of angular momentum. The entire motion rotation and translation are the resut of Newton's Laws.

[@beratgonultas5493]

Thank you Sir. I think I am more comfortable with it now.

[@efeguleroglu]

Is the phonomenon at the last about that the plastic contains a kind of fluid in it?

[@lecturesbywalterlewin.they9259]

question unclear

[@efeguleroglu]

@@lecturesbywalterlewin.they9259 I mean I don't know how to explain the phenomenon but is it about fluid mechanics? It looks from the camera that it has fluid in it.

[@mohammedbinalimaqqavi6599]

Sir, your lectures are very inspiring. Can i know which is the reference book the students use? Thanking you in advance...

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian Physics Volume 1 2nd edition W.W. Norton & Company ISBN 0-393-95748-9 8.02 Physics for Scientists & Engineers by Douglas C. Giancoli. Prentice Hall Third Edition ISBN 0-13-021517-18 8.03 Vibrations and Waves by Anthony French CRC Press ISBN 9780748744473 8.03 Electromagnetic Vibrations, Waves and Radiation by Bekefi and Barrett. The MIT Press ISBN 0-262-52047-8

[@magsdechert4647]

Dear Mr. Lewin, you say that the velocity of the ruler's center of mass (COM) is only dependent of the Force you apply, not the position. But the torque you apply, does depend on the point, naturally. Now, is it not so that in case of Force to the COM you raise the (kinetic) energetic level by a certain amount, but if you choose another point you have the same kinetic energy as in the COM case, but also a a rotational energy? Since you assume no friction and assuming I haven't got a twist in my logic, where would that energy go or come from? Best regards and i wish you good health, Max

[@lecturesbywalterlewin.they9259]

I cannot add to the clarity of this lecture. You are confusing "force" with "impulse, J". Reardless of where I apply the impulse the velocity of the COM after the impulse is always the same. J/m Thus the translational KE is always the same. But the rotational KE does depend on where I apply J. *Keep in mind that J is NOT energy.*

[@QCYTB]

Is there a way we can access PIVoT right now?

[@lecturesbywalterlewin.they9259]

My lectures on PIVoT were moved in 2003-2005 to OCW and in 2014 to the KZbin Channel "For the Allure of Physics" and in 2015 also to my KZbin Channel.

[@shivamkrishnan9009]

Sir , can you please tell me the book that you use for 801 or where will i find it ? Thank you

[@lecturesbywalterlewin.they9259]

8.01 Physics Hans C. Ohanian Physics Volume 1 2nd edition W.W. Norton & Company ISBN 0-393-95748-9

[@abesaale9135]

Mr. Lewin, you didnt explain the blue object which you showed which reverses its direction after coming at rest. Could you please throw some light on that? I'd be very thankful to you.

[@annefiftythree]

That was very easy to spot. You should pay better attention to what he did. Look again!