8.01x - Lect 21 - Torques, Oscillating Bodies, Physical Pendulums
HTML-код
- Опубликовано: 1 окт 2024
- Torques - Oscillating Bodies - Hoops
Assignments Lecture 21, 22, 23 and 24: freepdfhosting....
Solutions Lecture 21, 22, 23 and 24: freepdfhosting....
"Physics is not difficult, physics is there to make very difficult things easy"
Dramatic but the truth
*8.01x lecture 21*
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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???
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*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
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*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
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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.
🌺
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
Thanks, Jo, for your kind words.
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 :)
Thanks for you aclaration :')
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.
i dont want to go for dinner when i watch your lectures Sir ...
:)
43:50 that excitement tho :)))
You are awesome! I always watch your lectures. They are clear and entertaining.
:)
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.
I wish I knew more English to fully understand everything :) Great class :D
What was the answer for the last question at the end of the video please ?
have you found it?
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.
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 :)
Thank you so much! I was so confused.
why the object do that at the final? 46:10
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!
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.
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
Mashallah sir you are awsome . if you were our teacher we will be the futher scientists ...
:)
Sir, your lectures are awsome! Thanks! Where can i donate to this fantastic channel? Grateful regards from Trondheim, Norway!
yes you can donate - please do!
PayPal: lewin-physics@physics.comcastbiz.net
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.
Thank you kindly, Bent - I received it!!!
Professor why did the piece of plastic behave in that weird (seemingly unnatural) manner at the end?
Sir, is spin angular momentum also valid for 3-D objects ??
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.
I AM watching regularly and improving my my knowledge
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 .
Thank you from Turkey
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?)
best teacher ever
Subjects are more enjoyable than just giving exams of subjects
professor please tell my why the plastic spins back I have insomnia now
how can same impulse transfer different energies??
Change in impulse is a change in Mv, change in KE is a change in 0.5Mv^2. They are not the same.
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.)
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.
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.
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!
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!
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?
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.
Thank you sir. I will continue to follow your lectures!
Interessante
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?
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
yes very cool, not so easy to explain
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.
You are genius, manas srivastava.
Even after watching the explanation from professor Walter Lewin here: ruclips.net/video/hmHV-HF-22s/видео.html, 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.
Dear sir, please explain the last phenomenon of reverse rotation. Please sir. I can’t sleep sir.
love from india!🖤❤
oh man, is this the 1st biweekly problem ;)
Sir, your lectures are very inspiring.
Can i know which is the reference book the students use?
Thanking you in advance...
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
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...?
try it
@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 ?
My derivation is correct. Watch it again, and listen what I say about an ideal pendulum with length 2R. Yes 2R.
GEOMETRY :)
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!
thank you sir.
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
>>>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!
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?
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.
How can sin(a) be equal to alfa in case alfa is in radian. GOT İT NOW :)
try 2 degrees which is 0.03491 radians, sin of 2 degrees is 0.034899
@@lecturesbywalterlewin.they9259 Yes sir, you are right. By the way I really love you, even though I don't know you. I won't forget you sir :)
Dear sir , Will you plz tell me what is the reason behind the rotation of plastic in opposite direction after coming to a hault? I didn't get it 😋😋😅
Sir, as per guess there may be different mass distribution on the plastic
May be there is more heavier masses on diagonally opposite area .
Am i correct sir ?
@@udits9767 You did not pay attention: look at the rotation the first time, then the second and third time. It is not the same
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?
shreyas Sudhaman Thank you!
+Fabrizio Tabasso Google "rattleback"
+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?
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 ?
There is no such thing as an "anti-clockwise" equilibrium position.
Google "rattle back".
Lectures by Walter Lewin. They will make you ♥ Physics.
Oh alright , I will sir ..thank you :)
@@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??
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!
try it
@@lecturesbywalterlewin.they9259 Got it! Thank you!
SIR THANK YOU SO MUCH FOR YOUR LECTURE!!! Because of you now i can fully understand the concept of the material 😅😅
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!
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?
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?
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.
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.
Angular momentum became easy for me
sir I want to ask if natural frequency is same on earth and moon.
If you tell me what is the natural frequency is of the Earth, I will answer your question.
+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
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.
+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.
>
YES
Levers lecture would be good before this. Wheel properties are special case of Lever where for one edge r=0. Also for other rotating objects such as gears, screwdriver as torque example... we see Lever principle of mass, lenght, speed, torque... trade off.
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.
freepdfhosting.com/86109d309b.pdf
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.
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.
@@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
@@arunbhardwaj2744can you please show in details??
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!
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?
no
2:26
2:53
Sir you are GREAT
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 :)
:)
sir can u please show me the solution by taking point P as reference point.
see timing 23:10
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
does the spin angular momentum only apply for symmetric objects that the forces all cancel out?
watch my 8.01 lectures to know how angular momentum is defined.
Sir , can you please tell me the book that you use for 801 or where will i find it ?
Thank you
8.01
Physics
Hans C. Ohanian
Physics
Volume 1
2nd edition
W.W. Norton & Company
ISBN 0-393-95748-9
sirr u are just awesome can u help me by giving your precious time for few days
*Take all my 3 MIT courses, do all the homework and take all my exams. Also solve all 140+ problems I cover in my Help Sessions. All are on this channel. Your understanding of and performance in physics will go sky high!*
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
you cannot take point C as you do not know the force at P.
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??
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?
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.
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.
>>> 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.
sir i got the time period 2*pi*sqrt(b/g) which is not same as your ans. what i am doing wrong?
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.
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 ?
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?
yes the translational KE of the cm is independent of where you hit. The rotational KE does depend on where you hit.
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.
impulse is NOT energy.
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?
YES that is possible. *Impulse is NOT energy.* The change of KE depends on where I hit.
Is there a way we can access PIVoT right now?
My lectures on PIVoT were moved in 2003-2005 to OCW and in 2014 to the RUclips Channel "For the Allure of Physics" and in 2015 also to my RUclips Channel.
10:20 who is responsible for the centripetal force?
love you sir
Hello sir.what is the secret of that piece of plastic ?And actually if I see at rotatinal mechanics I will get fear and after listening to your class I think I can enjoy that. Thank u sir for your excellent lecture.
Physics is there to make very difficult things tooooo easy not only easy by prof Walter lewin sir
The best
The plastic at 47:30 is called a rattleback or celtic stone if you want to google about it
12Id/Ml²
Physics is mind blowing subject which blows our mind and will change our world
Listening to your lecture alone amounts to 1 semester of my shitty prof's teaching, thank you sir.
:)
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.
That was very easy to spot. You should pay better attention to what he did. Look again!
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
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.*
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.
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.
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.
angul mom of a rotating object can be determined (with some math) relative to any point P.
Hey Professor, the question you asked at the end of the lecture, that why the piece of plastic is moving in opposite direction after coming to halt. I think it is due to center of mass, the center of mass is not at the center of the body. Am i right ??!!!!
Hey Professor, am i right ??
@@pratikshinde2121 take any object with COM not at the center, and it will not reverse direction of rotation.
I need explaination of theory of relativity
take a course in SR
How i wish you were my physics teacher Professor Lewin. I've understood so much from you. Thanks.
Hello Sir. Namaste from India. I know my comment is way too late but this is the first time I am watching your lectures. I really enjoyed it.
For the spinning plastic material, I am guessing that it might have something to do with impact. When you were giving it a slight push, the impact time was high. But when you gave it a stronger push, the impact time was less which causes the impact due to friction high giving it a push in the opposite direction.
If you see this, please do respond. Thank you very much sir.
ask google
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?
what I did and what I said was correct
Can someone explain the ending
Mr. Walter lewin I have some I am from India your explanation is perfect to understand and learnt somany. Please can you send your mail for asking doubts or any website that you can answer
my email is secret
Sir, I wonder how would I calculate angular momentum relative to point P without looking at the formula. Can you please help me out?
angular momentum is I*omega yoou have to know this to calculate it
Sir, when i see you making predictions with uncertainties,,i just love it....i remember your word "MEASUREMENTS WITHOUT THE KNOWLEDGE OF UNCERTAINTY IS USELESS". But i dnt really had become accustomed to this technique...so i feel worried about it :( ..Would you suggest me how to learn to apply it?( I managed a book but 300 pages...way way boring.)
By the way, you may had given a vacation to your cute badges that day.
I am stuck in the same situation too
sir is the angular velocity for that problem is I*d/Inertia
which problem. how many minutes into the lecture?
@20:10
Id/Inertia ???? is that d?
position vector from c to p is d ,so i think that is correct.did i make any mistake sir
mvd=w*inertia ,mv=impulse so w=impulse×d/inertia
Sir, What is the reason for the plastic to stop and turn the other way round? Is it conservation of angular momentum?
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?
I leave it up to you to answer this.
Sir i just cant get the idea of spin angular momentum being constant......I clearly understand that angular momentum will change with reference point, howsoever cant get the idea of spin angular momentum. Kindly explain !!
Please watch my 8.01 lecture where I prove that angular momentum is conserved in the absence of an external torque. If you take a conical pendulum which is suspended from point P and of which the bob goes a round in a circle in the horizontal plane with center M, then angular momentum is conserved about M but not about P.
Your lecture makes physics very easy and logical 😊😊 Thank you sir ❤️ .
What is the answer of your last question ? Thanks a lot for these excellent lectures.
That class is fully silent.
Sir, please explain can a particle in vertical circular motion be an SHM
the projection of a uniform circular motion on a plane perpendicular to the plane of the circle is a SHM
@@lecturesbywalterlewin.they9259 Sir Thank you for the reply,
my question was about non-uniform circular motion in a vertical circle completing full revolutions ( gravitational force of earth) and the equilibrium position is the lowest point, so will that be an oscillatory motion?
And another question is, can planetary motion in an elliptical orbit be considered oscillatory?
Hi from Turkey, These guys are so lucky, really. And also I am since I learned this channel :))
Thanks! 😃
Hello sir can you tell me what is the secret of that piee of plastic?
Proffesor lewin. If you have an initial angle in the pendulum for example 45 degrees. The period will be the same?
no - watch my 8.01 lectures
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?
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.