my estimate is that I have 4 +/- 4 more years to live. Not clear whether I will be able to continue to be productive in teaching physics in the years to come.
Professor, I think the answer's g/3. Btw, I feel so good after finding your channel! Greetings Professor!! Method - 1)Drew a fbd 2)Balanced the moments 3)Apply D'Alembert's Principle 4)Solve the equations 5)Done Thank you
Real great teachers like you are the reason why students like me enjoy physics and choose to study it at a deeper level. If any of us students manages to achieve something great in science then please know that it couldn't have happened without you sir . Thank you for your continued contributions towards science, education and us students.
Sir, would you suggest me books about Physics? I have just read your book. By the way i have got a question: Is gravitational energy endless? For example whenever i drop something from a height, it falls. And it's been happening for years. Won't gravitational energy come to an end? Can we make an engine with two magnets (not electromagnets)?
>>>Won't gravitational energy come to an end?>>> gravity is not energy, it's due to the existence of mass. >>>Can we make an engine with two magnets>>>> yes, but you need a lot more. You also need a power supply to drive the engine. YOU NEED an ENERGY SOURCE!
I think I did it right, but it entirely possible there is a silly math error along the way... If we picture the yo-yo as a bunch of three thin discs then we can simplify the problem, as the width of the discs has no effect. Each full size disc has mass M so the area density constant will be rho= M/(R^2)pi Using I=Integral of r^2dm we can define dm as a thin ring of the disc which when unrolled would have about an area of dr(2*pi*r) Thus the integral is now rho Integral (from 0 to R) (r^2(2*pi*r))dr or as the pi cancels with the pi in rho, M/R^2times Integral (from 0 to R ) (r^3) dr this will yield the expected I of a thin disc = I = (MR^2)/2. There are two large discs, so the I there is 4MR^2 + (R^2M)/8 or (33R^2M)/8 =I total about the center of mass parallel to the axis of rotation. The yo-yo is undergoing smooth rolling so Vcom=Rw thus Acom=RAlpha . For angular acceleration, Torque = I(alpha). Torque here is R cross 9/4Mg (9/4M is total mass of the yo-yo) Thus, 9/4MgR=((33R^2M)/8) alpha or alpha = 18g/33R which multiplying this by R gives us Acom, so Acom should thus be 18/33 or 5.345454... M/S^2 I am going to check over the area density constant I defined, I think that could be the root of any error. I think I at least approached this correctly.
yo my dude, I love this explanation, don't get me wrong, but as an average person who doesn't know physics to the nth degree like you might, I have no fucking clue what your saying. I wish I did homie, because this shit is cool in my mind, but after two paragraphs, I decided it wasn't worth the mental energy if I don't understand the terminology and context in the first place. I'm a silly billy who has a smooth little marble brain and I can't shove too much info in at once. just maybe think about simplifying for simpletons like me so we can understand you my friend
Sir i really enjoy your lecture you are such an amazing teacher i just have a question about the spin of the electron . Do electrons really spin like classical rotation ? Thank you so much
I can send you a signed copy of my book for $25 + postage. Postage depends on where you live. It is $10 to the US, $28 to Europe, $48 to India. When you transfer the money to my PayPal account let me know your full name and address. I will write a special note for you in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). Keep in mind that PayPal charges 6% for each currency transition. PayPal: lewin-physics@physics.comcastbiz.net
I can send you a signed copy of my book for $25 + postage. Postage is $10 in the US, $30 to Europe and $48 to India. I will write a special note for you in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). For an additional $25 I will send you a signed and dated 24x18 inch sheet of paper with my original handwritten calculations as I post on my Bi-Weekly Physic Problems. They are the original - *thus one of a kind - there are NO copies.* You can see 3 such sheets in the video in which I solved the yo-yo problem of 2 weeks ago: ruclips.net/video/uAOSKkdVxck/видео.html When you transfer the money to my PayPal account let me know your full name and your address (make sure it is correct and complete). Keep in mind that PayPal charges 6% for each transition. PayPal: lewin-physics@physics.comcastbiz.net
that the kinetic energy of the center of mass, which is converted into spin as the string unwinds, results from being thrown, rather than from falling through a gravitational potential.
Walter sir, i am an 17 yr old boy ,i often get confused when to integrate or differente in a particular problem which requires calculus.. please help me...
There are practically two forces and one torque applied on this system: the force of gravity and the force of the tension (T) on the string. The torque (Tau) on the system is due to the rotation of the system around the string. The total mass of the system is 5m (two cylinders of mass 2m, and one of 1m). The inertia (I) of the system is 1/2 mR**2 + 2(1/2(2m)(2R)**2)=17/2 mR**2. The two equations that govern the system are Torque and Summation of Forces (Tau and T, respectively). Tau=I * Alpha. Alpha is the tangential acceleration (a) at the edge of the disks with 2R where Alpha=a/2R. Tau is also equal to the tension (T) times the distance from the edge of smaller cylinder to point of rotation (R); so, T*R=17/2 mR**2 * a/2R. Summation of the forces is the force of gravity (positive) and the Tension (T) negative and substituting, 5mg - T= 5m*a. Solving for T and substituting in the torque equation above, (5mg-5ma)R = 17/2 mR**2 * a/2R. Now solving for "a" gives the solution to the problem, the acceleration. a = 20g/37, which is pretty close to half the value of gravity.
I bet you I made a mistake in the tangential acceleration Alpha. I was not sure whether to use Alpha= a/2R or Alpha= a/R. Should have used the last one. Anxiously awaiting for your answer.
Hmmmm.... is this where Billy Gaede got his Rope Model idea from? Cause I am telling you right now Prof. Lewin! IT'S NOT MAGNETS ALL THE WAY DOWN! IT'S YOYOS, CAROSELS AND ROPES!
release the yoyo from rest then 2 2 1 y at where y is the distance fallen and t is the time of fall, t can be measured with a stop watch or a photogate timer. Videotaping the falling yoyo is also useful in obtaining position versus time data. The tension in the string of the falling yoyo can be measured with a spring force meter
Using torque due to gravity about the point of contact with the string, which I assumed stays at velocity 0 relative to the ground(no slipping on the string), I got that the acceleration is (6/11)g where g is the acceleration due to gravity. I'm not too sure about this since I haven't done this sort of question in a while. So it does not depend on the mass or the radius!
"which I assumed stays at velocity 0 relative to the ground" that right there tells you that the answer is wrong because that is not true. The yoyo is falling so velocity relative to the ground is not 0.
Sir can you please help me with this question A bullet of mass 50 g moving horizontally with a velocity 100 m/s hits and gets embedded in a wooden block of mass 450 g placed on a vertical wall of height 19.6 m, if the bullet gets embedded in the wall does the block fall on the ground?
Use the conservation of linear momentum to solve the problem. The initial velocity of the block is zero, and that of the bullet is 100m/s, once the bullet is embedded in the block the block and the bullet becomes one system, you can calculate the final velocity.
First Attempt: a = 10g/(2+3(PI)r**3) my first stab at it which is likely very wrong. I probably have to go back and rethink about how I summed linear forces and angular momentum. Second Attempt: Found a math error. My second attempt is a = 10g/ (10 + 33R**3) Third Attempt: Set potential energy = kinetic energy Solving for acceleration assuming the YO YO drops for exactly 1 second over a distance, h, gives: a = 5g/11 Fourth Attempt: Math error in 3rd attempt corrected: a = 9g/21 = .43g so if the correct answer is .5 g, would my answer be close enough?
consider a neutral cube placed in a non uniform external field...the flux through the cube would be non zero...but the charge enclosed in the cube would still be zero...how,would gauss law hold then in this case?
Walter, I had a very bad week, I'm totally tiered of this stupid life. For better feeling I bought a Yo Yo, because I'm a very good YoYo player. Problem is, because of my Yo-Yo play I forgot the cleanings here. Now I'm still working. Walter, I'm not a useful character for our Society. My interests aren't normal.
@@lecturesbywalterlewin.they9259 could be, you are right. My Yo-Yo was made in Asia I think. I bought it in a fair trade shop. The Yo-Yo was cut of and I repaired it in the shop. The owner was so surprised, that he gave me the Yo-Yo for less money. It is a very beautiful thing, wooden with a snake on it.
Prof. I have a competitive exam IES 40 days are left but the main problem is memory.i just forgot the theory part so please please please help me how to prepare? ??
I hope I got this one without stupid math errors. The middle cylinder has the mass 1/9 m and the outer cylinders each have the mass 4/9 m. The moment of inertia on a cylinder is I=mr^2/2. So the moment of inertia of the yo-yo is 1/9 m x R^2/2 + 2 x 4/9 m x (2R)^2/2 = 33/18 mR^2. The torque is mgR. So the angular acceleration is α = mgR / (33/18mR^2) = 18g/33R. The linear acceleration is a = αR = 18/33 g ~ 0.545g
a=alpha.R, where alpha=torque/I. The torque of the centre of the mass is (9/4)Mg, I is (17/8)MR^2. That means that alpha is (18/17)(g/R^2) and a=(18/17)(g/R). That's the solution if I'm not wrong with the moment of inertia I.
*My problem has no relation with the video* My question is, Why the biker doesn't fall down while on a spiral like structure? My teacher told it is because, mg is responsible for falling him down and mg is exhausted by acting as centripetal force. I want to know answer in your words, sir. And also, Is he was correct?
Hey Professor Walter Lewin. Can you do a video on a 'step by step how to make and publish a scientific refereed journal"? i think that would be very interesting and useful for me and many other people are doing science projects who are on this channel. Thank you.
I think this can be solved by applying the law of conservation of energy. Potential energy = Kinetic Energy + Rotational Energy...I need to figure out how to calculate the rotational energy.
First Mg-T=Ma Then i found a torque relation. This gave a relationship between linear and angular acc. Then combined with moment of inertia. So i found 18g/83
Sir can u please explain a similar phenomenon of " a disc with a a cylindrical rod passing through the centre and both the ends of the rod tied to stringsbof equal length from a rigid support the system looks somewhat like this l_,_l the comma is the disc the underscores depict the rod through its centre and the l s are the two strings that connect the system to a rigid support....Now on winding it all way upto max height I release it then it comes down and then it moves up but to height less than hmax say hmax' and then again it comes down due to gravity and again goes up but to a height less than hmax' say hmax" please can u explain this entire thing in one of ur lectures using proper mathematics to describe why the system finally comes to rest
Walter, when you name this video "yo-yo" youtube will automatically suggest a similar video with the same author, in this case the solution you have done before. Just name the problem by numbers if you dont want this connection next time. :)
sir I think the answer is a = (3g)/17 sir. I used the fact that the yo-yo rolls along the thread and hence got three equations involving a, α and T. solving I got the required answer sir.
Akash! don't be too much active on social media while preparing for IIT-JEE. Download and watch lectures of Professor Lewin and focus on what your teachers say.
Sir I am from India... I'm in love with maths but I observed d fact tat... Actually Physics gives true meaning to the mathematical theories like limits and derivatives...
I think there's a contribution to the acceleration of 9.81ms/kg as there is tension in the string which contributes to its acceleration. although theirs allot of friction within the coils of string apposing that torque. The friction slows it down but the tension in the string accelerates it. Because the yo yo is spinning then it will hold a horizontal equilibrium. It will still accelerates due to the unequal forces in the vertical components.
m x g - T = m x a T x R = I x alpha a = alpha x R = (m x g) / (I / R^2 + m / R) I = 2 x I_2R + I_R = 9 / 2 x m x R^2 a = 2 x g x R / (9 x R + 2) relating angular acceleration to tangential acceleration of a solid body and using free body diagram
sorry I think I made a mistake when clearing the tangential acceleration, it should be a = alpha x R = (m x g) / (I / R^2 + m ) and since the yoyo has moment of inertia 9 / 2 x m x R^2 the answer would be a = g / (1 + 9/2)
I guess that the answer would be 8/17g since the yoyo is consisted of two solid disk and one connection between them, If we consider the rotational inertia of the connection we will then proceed to the answer of 8/17g
Lectures by Walter Lewin. They will make you ♥ Physics. Sir,could you offer just a bit hint of this question~ Am I on the right track of solving it or should I modified the method? Much Appreciation ~
Sorry sir , I am just voracious to know whether method of combining rotational inertia is a step of the solution. Could you please reply. Sorry if bother , Sir . Thanks in advance!
Sir ,if I use energy conservation, then is this expression correct :- (Mg-T)x = ½Mv² + ½Iw² (T is tension in string) where I is the Moment of Inertia perpendicular to the plane of yo - yo passing through its COM, w is its instantaneous angular velocity?
Ok. This is my last try. I forgot to consider central cylinder mass and moment of inertia. I am sorry if I am wrong again. I got a=g/5. Thank you for your patience.
Lectures by Walter Lewin. They will make you ♥ Physics. Hehehe I didnt thought you would reply to my comment I am 13 years old and i love your lectures :)
At MIT I had 3 cameras (run by 3 different people), 2 sound experts, and 1 light expert during my lectures in a lecture hall with 9 HUGE blackboards that I could move up and down; the lecture hall could seat 600 students. The videos that I have been posting since January this year I do at home in my hall. I have no black boards. The space I have on the wall is less than 1 of my 9 boards at MIT. There is room for ZERO students in that hall. I have only 1 light in the ceiling. I have a $150 Canon camera that can take simple videos for no longer than 30 minutes (limited by battery power). I have ZERO help; I do the taping on my own. I place the camera on a ladder and tape it down with scotch tape to make sure it will not fall on the floor. I hope you will appreciate this. Preparation and selection of each problem takes me roughly 16 hours per problem. To prepare and tape the solutions takes an additional 10 hours. I often have to tape a video twice or even 3 times before I am happy with it (slips of the tongue are common). That alone can take more than 2 hours.
Lectures by Walter Lewin. They will make you ♥ Physics. Felt really sad after reading this sir. But I love your lectures. I wish my physics teacher could explain me like this :*(
yes, you cheated and looked it up on the web what the acceleration is of a solid disk rolling off a string. that is (2/3)g. But this is not a sold disc. *STOP cheating. You got it WRONG*
I have a pretty decent set of headphones, and a decent amp, and I am not hearing much noise. Also considering Professor Lewin is taking his own time to make these videos for you I do not see how you can complain if you really want to watch these.
At MIT I had 3 cameras (run by 3 different people), 2 sound experts, and 1 light expert during my lectures in a lecture hall with 9 HUGE blackboards that I could move up and down; the lecture hall could seat 600 students. The videos that I have been posting since January this year I do at home in my hall. I have no black boards. The space I have on the wall is less than 1 of my 9 boards at MIT. There is room for ZERO students in that hall. I have only 1 light in the ceiling. I have a $150 Canon camera that can take simple videos for no longer than 30 minutes (limited by battery power). I have ZERO help; I do the taping on my own. I place the camera on a ladder and tape it down with scotch tape to make sure it will not fall on the floor. I hope you will appreciate this. Preparation and selection of each problem takes me roughly 16 hours per problem. To prepare and tape the solutions takes an additional 10 hours. I often have to tape a video twice or even 3 times before I am happy with it (slips of the tongue are common). That alone can take more than 2 hours.
Sir I am very blessed to view your online lectures thank you for your efforts ...
I like the enthusiasm you have.
:)
I'm glad to see you still making videos like this... Hope your health is good and you continue making videos like this for a long time!
my estimate is that I have 4 +/- 4 more years to live. Not clear whether I will be able to continue to be productive in teaching physics in the years to come.
I love the problem and it's physics as a kid.
correct answer ! I made it invisible for others
Lectures by Walter Lewin. They will make you ♥ Physics. ☺
there were 2 correct answers so far
Pritam Sarkar ☺🙋🙋🙋🙋🙋🙋🙋
is ur answer 0.5922N????????????????????? @Pritam ji
Very interesting. Thanks and Regards 🙏🙏🙏🙏🙏
Professor, I think the answer's g/3. Btw, I feel so good after finding your channel! Greetings Professor!!
Method -
1)Drew a fbd
2)Balanced the moments
3)Apply D'Alembert's Principle
4)Solve the equations
5)Done
Thank you
I like that whistle when he says: you are ready? And the Yo-Yo goes down.
Where can i find the 8.02 textbooks?
(In one website i found says that access is forbidden )
8.02
Physics for Scientists & Engineers by Douglas C. Giancoli.
Prentice Hall
Third Edition
ISBN 0-13-021517-18
Amazon sells them
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you! ;)
Real great teachers like you are the reason why students like me enjoy physics and choose to study it at a deeper level. If any of us students manages to achieve something great in science then please know that it couldn't have happened without you sir . Thank you for your continued contributions towards science, education and
us students.
😊💝💝💝 love physics
:)
I seen ua explanation in Newton biography film....😘😘😘
:)
Thank u so much sir..u explain very well.
:)
Hes over 80 but is more Alive then me 😊
:)
Hi sir I love the problems of Newtonian mechanics. Super sir
Sir, would you suggest me books about Physics? I have just read your book. By the way i have got a question: Is gravitational energy endless? For example whenever i drop something from a height, it falls. And it's been happening for years. Won't gravitational energy come to an end? Can we make an engine with two magnets (not electromagnets)?
>>>Won't gravitational energy come to an end?>>>
gravity is not energy, it's due to the existence of mass.
>>>Can we make an engine with two magnets>>>>
yes, but you need a lot more. You also need a power supply to drive the engine. YOU NEED an ENERGY SOURCE!
Lectures by Walter Lewin. They will make you ♥ Physics. Thank you, sir :)
I think I did it right, but it entirely possible there is a silly math error along the way...
If we picture the yo-yo as a bunch of three thin discs then we can simplify the problem, as the width of the discs has no effect. Each full size disc has mass M so the area density constant will be rho= M/(R^2)pi Using I=Integral of r^2dm we can define dm as a thin ring of the disc which when unrolled would have about an area of dr(2*pi*r) Thus the integral is now rho Integral (from 0 to R) (r^2(2*pi*r))dr or as the pi cancels with the pi in rho, M/R^2times Integral (from 0 to R ) (r^3) dr this will yield the expected I of a thin disc = I = (MR^2)/2.
There are two large discs, so the I there is 4MR^2 + (R^2M)/8 or (33R^2M)/8 =I total about the center of mass parallel to the axis of rotation.
The yo-yo is undergoing smooth rolling so Vcom=Rw thus Acom=RAlpha .
For angular acceleration, Torque = I(alpha). Torque here is R cross 9/4Mg (9/4M is total mass of the yo-yo)
Thus,
9/4MgR=((33R^2M)/8) alpha or alpha = 18g/33R which multiplying this by R gives us Acom, so Acom should thus be 18/33 or 5.345454... M/S^2
I am going to check over the area density constant I defined, I think that could be the root of any error. I think I at least approached this correctly.
yo my dude, I love this explanation, don't get me wrong, but as an average person who doesn't know physics to the nth degree like you might, I have no fucking clue what your saying. I wish I did homie, because this shit is cool in my mind, but after two paragraphs, I decided it wasn't worth the mental energy if I don't understand the terminology and context in the first place. I'm a silly billy who has a smooth little marble brain and I can't shove too much info in at once. just maybe think about simplifying for simpletons like me so we can understand you my friend
I might actually give this a try!
Sir i really enjoy your lecture you are such an amazing teacher i just have a question about the spin of the electron . Do electrons really spin like classical rotation ?
Thank you so much
use google, NOTHING about electrons is related to classical physics
Lectures by Walter Lewin. They will make you ♥ Physics. God bless you sir
Where can i buy your book and how much does it cost?
I can send you a signed copy of my book for $25 + postage. Postage depends on where you live. It is $10 to the US, $28 to Europe, $48 to India. When you transfer the money to my PayPal account let me know your full name and address. I will write a special note for you in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). Keep in mind that PayPal charges 6% for each currency transition.
PayPal: lewin-physics@physics.comcastbiz.net
Lectures by Walter Lewin. They will make you ♥ Physics. Sir, this link is not working
I can send you a signed copy of my book for $25 + postage. Postage is $10 in the US, $30 to Europe and $48 to India. I will write a special note for you in the book. I will also include a signed colored picture of me swinging on the pendulum (with your name written on it). For an additional $25 I will send you a signed and dated 24x18 inch sheet of paper with my original handwritten calculations as I post on my Bi-Weekly Physic Problems. They are the original - *thus one of a kind - there are NO copies.* You can see 3 such sheets in the video in which I solved the yo-yo problem of 2 weeks ago: ruclips.net/video/uAOSKkdVxck/видео.html
When you transfer the money to my PayPal account let me know your full name and your address (make sure it is correct and complete). Keep in mind that PayPal charges 6% for each transition.
PayPal: lewin-physics@physics.comcastbiz.net
Lectures by Walter Lewin. They will make you ♥ Physics. I think I made the transaction correctly
I have not received it.
Isn't it amazing how many different answers are there in the comments?
Nice hair cut,Professor
:)
that the kinetic energy of the center of mass, which is converted into spin as the string unwinds, results from being thrown, rather than from falling through a gravitational potential.
this is incorrect - in my problem it's ONLY gravitational PE that is released.
Sir why avg KE of two gas different samples are equal at same temprature
teacher.pas.rochester.edu/phy121/lecturenotes/Chapter18/Chapter18.html
Lectures by Walter Lewin. They will make you ♥ Physics. Thank u sir ❤️ for physics
Walter sir, i am an 17 yr old boy ,i often get confused when to integrate or differente in a particular problem which requires calculus.. please help me...
I am very pleased you replied sir..i live in india, our physics teacher used to show us your videos,and we used to love it..
take a course in calculus
There are practically two forces and one torque applied on this system: the force of gravity and the force of the tension (T) on the string. The torque (Tau) on the system is due to the rotation of the system around the string. The total mass of the system is 5m (two cylinders of mass 2m, and one of 1m). The inertia (I) of the system is 1/2 mR**2 + 2(1/2(2m)(2R)**2)=17/2 mR**2. The two equations that govern the system are Torque and Summation of Forces (Tau and T, respectively). Tau=I * Alpha. Alpha is the tangential acceleration (a) at the edge of the disks with 2R where Alpha=a/2R. Tau is also equal to the tension (T) times the distance from the edge of smaller cylinder to point of rotation (R); so, T*R=17/2 mR**2 * a/2R. Summation of the forces is the force of gravity (positive) and the Tension (T) negative and substituting, 5mg - T= 5m*a. Solving for T and substituting in the torque equation above, (5mg-5ma)R = 17/2 mR**2 * a/2R. Now solving for "a" gives the solution to the problem, the acceleration. a = 20g/37, which is pretty close to half the value of gravity.
incorrect
I bet you I made a mistake in the tangential acceleration Alpha. I was not sure whether to use Alpha= a/2R or Alpha= a/R. Should have used the last one. Anxiously awaiting for your answer.
Hmmmm.... is this where Billy Gaede got his Rope Model idea from?
Cause I am telling you right now Prof. Lewin!
IT'S NOT MAGNETS ALL THE WAY DOWN!
IT'S YOYOS, CAROSELS AND ROPES!
release the yoyo from rest then 2
2
1
y at where y is the distance fallen and t is the time of fall, t
can be measured with a stop watch or a photogate timer. Videotaping the falling yoyo is also
useful in obtaining position versus time data. The tension in the string of the falling yoyo can be
measured with a spring force meter
Using torque due to gravity about the point of contact with the string, which I assumed stays at velocity 0 relative to the ground(no slipping on the string), I got that the acceleration is (6/11)g where g is the acceleration due to gravity. I'm not too sure about this since I haven't done this sort of question in a while. So it does not depend on the mass or the radius!
"which I assumed stays at velocity 0 relative to the ground" that right there tells you that the answer is wrong because that is not true. The yoyo is falling so velocity relative to the ground is not 0.
I need just one clue- how to calculate the moment of inertia of the yo yo?
watch my solution video
Lectures by Walter Lewin. They will make you ♥ Physics. Ok
Sir can you please help me with this question
A bullet of mass 50 g moving horizontally with a velocity 100 m/s hits and gets embedded in a wooden block of mass 450 g placed on a vertical wall of height 19.6 m, if the bullet gets embedded in the wall does the block fall on the ground?
I do not solve problems for viewers. I teach Physics.
Use the conservation of linear momentum to solve the problem.
The initial velocity of the block is zero, and that of the bullet is 100m/s, once the bullet is embedded in the block the block and the bullet becomes one system, you can calculate the final velocity.
sello
sir i hav a doubt...........a very confusing term............
what is spontaneity........?????
use dictionary
meaning of spontaniety in thermodynamics......????????
I have no clue
that was a good problem
:)
First Attempt:
a = 10g/(2+3(PI)r**3)
my first stab at it which is likely very wrong.
I probably have to go back and rethink about
how I summed linear forces and angular momentum.
Second Attempt:
Found a math error. My second attempt is
a = 10g/ (10 + 33R**3)
Third Attempt:
Set potential energy = kinetic energy
Solving for acceleration assuming the YO YO drops for exactly 1 second over a distance, h, gives:
a = 5g/11
Fourth Attempt:
Math error in 3rd attempt corrected:
a = 9g/21 = .43g
so if the correct answer is .5 g, would my answer be close enough?
a=(6/11)g
The terms you add to each other do not have the same units. This is usually a good indicator that you made a mistake somewhere.
consider a neutral cube placed in a non uniform external field...the flux through the cube would be non zero...but the charge enclosed in the cube would still be zero...how,would gauss law hold then in this case?
Gauss' Law ALWAYS holds. It's one of the 4 Maxwell's equations. All 4 eqs ALWAYS hold. However, you have to know how to use these eqs.
okk professor..:)
Walter, I had a very bad week, I'm totally tiered of this stupid life. For better feeling I bought a Yo Yo, because I'm a very good YoYo player. Problem is, because of my Yo-Yo play I forgot the cleanings here. Now I'm still working. Walter, I'm not a useful character for our Society. My interests aren't normal.
>>>My interests aren't normal.>>> *most normal people are boring*
@@lecturesbywalterlewin.they9259 could be, you are right. My Yo-Yo was made in Asia I think. I bought it in a fair trade shop. The Yo-Yo was cut of and I repaired it in the shop. The owner was so surprised, that he gave me the Yo-Yo for less money. It is a very beautiful thing, wooden with a snake on it.
@@lecturesbywalterlewin.they9259 Now I love it more. And you are the best I ever met. I know you have some understanding.
Prof. I have a competitive exam IES 40 days are left but the main problem is memory.i just forgot the theory part so please please please help me how to prepare? ??
watch all my MIT course lectures, do all the homework and take all my exams.
I hope I got this one without stupid math errors. The middle cylinder has the mass 1/9 m and the outer cylinders each have the mass 4/9 m. The moment of inertia on a cylinder is I=mr^2/2. So the moment of inertia of the yo-yo is 1/9 m x R^2/2 + 2 x 4/9 m x (2R)^2/2 = 33/18 mR^2. The torque is mgR. So the angular acceleration is α = mgR / (33/18mR^2) = 18g/33R. The linear acceleration is a = αR = 18/33 g ~ 0.545g
you're wrong.
Your torque and moment of inertia are about different axes.
Well, at least it wasn't a stupid math error but a proper physics error :-)
Yep, I see it now. My rope needs to be pulled upwards for it to work like that.
It's not stupid math errors its a physics error.
a=alpha.R, where alpha=torque/I. The torque of the centre of the mass is (9/4)Mg, I is (17/8)MR^2. That means that alpha is (18/17)(g/R^2) and a=(18/17)(g/R). That's the solution if I'm not wrong with the moment of inertia I.
a=(18/17)(g/R).
g/R is NOT an acceleration.
Sir, I am a young Physics Student. May I ask you your age😅
stone age
sir your hlw hlww language soo cool
*My problem has no relation with the video*
My question is, Why the biker doesn't fall down while on a spiral like structure?
My teacher told it is because, mg is responsible for falling him down and mg is exhausted by acting as centripetal force.
I want to know answer in your words, sir.
And also, Is he was correct?
use google
Hey Professor Walter Lewin. Can you do a video on a 'step by step how to make and publish a scientific refereed journal"? i think that would be very interesting and useful for me and many other people are doing science projects who are on this channel. Thank you.
use google
I think this can be solved by applying the law of conservation of energy. Potential energy = Kinetic Energy + Rotational Energy...I need to figure out how to calculate the rotational energy.
do u know matt anderson??? prof
no
Lectures by Walter Lewin. They will make you ♥ Physics.
:-)
is a=g/3 .i first found the position of centre of mass since they are not uniform .
incorrect
I am sorry l make mistake the answer is 3 . 3 3 m/s2 or g/3
a = g/k = g/(1+(R^2/2r^2)). R is the radius of the outer discs, r is the radius of the inner disc.
I got it as (13/22)g
First Mg-T=Ma
Then i found a torque relation.
This gave a relationship between linear and angular acc.
Then combined with moment of inertia.
So i found 18g/83
incorrect
once again I asking a doubt to you professor what is centrifugal barrier
I would never use the words "centrifugal barrier". Centrifugal force is a fictitious force. use google
a = 4 m/s 2 thank you very much
Sir can u please explain a similar phenomenon of " a disc with a a cylindrical rod passing through the centre and both the ends of the rod tied to stringsbof equal length from a rigid support the system looks somewhat like this l_,_l the comma is the disc the underscores depict the rod through its centre and the l s are the two strings that connect the system to a rigid support....Now on winding it all way upto max height I release it then it comes down and then it moves up but to height less than hmax say hmax' and then again it comes down due to gravity and again goes up but to a height less than hmax' say hmax" please can u explain this entire thing in one of ur lectures using proper mathematics to describe why the system finally comes to rest
so as i am not so good in problem solving i get g/2 as answer.... waiting for your solution
incorrect. This answers shows that you have NO CLUE how yo deal with this problem. Wait till Sat when I post the solution
Walter, when you name this video "yo-yo" youtube will automatically suggest a similar video with the same author, in this case the solution you have done before. Just name the problem by numbers if you dont want this connection next time. :)
Yes I see the suggested solution videos. what a pain. I'll be more careful with titles from now on. Perhaps ONLY the problem number.
sir I think the answer is a = (3g)/17 sir. I used the fact that the yo-yo rolls along the thread and hence got three equations involving a, α and T. solving I got the required answer sir.
Is there is same surface mass density on disks??
question unclear
Is the answer 3/14g???
incorrect - please stop this, it's NOT a fishing expedition. Wait till I post the solution next Saturday.
Akash! don't be too much active on social media while preparing for IIT-JEE. Download and watch lectures of Professor Lewin and focus on what your teachers say.
Sir can you please make video on radius of gyration and centre of gravity
watch my 8.01 lectures. It's all there already.
Lectures by Walter Lewin. They will make you ♥ Physics. Tq sir
I got it and i m pretty sure about it.
a=(9g)/26
Please reply sir?????? Please!!!!!!!!
incorrect
Akash Verma he said not to ask if it is right or wrong
a=(6/11)g
I end up with (16*g*PI)/16.5
Too bad, that's over 10 m/s^2. So must be wrong. But no more time to figure it out. Sorry. Next time maybe.
incorrect
Sir I am from India... I'm in love with maths but I observed d fact tat... Actually Physics gives true meaning to the mathematical theories like limits and derivatives...
Math is the language of Physics
How to send the solution to you?
send them HERE
I have only one Idea, The Tension in string reduces the acceleration
that is NOT your idea - I mentioned that in my video
I think there's a contribution to the acceleration of 9.81ms/kg as there is tension in the string which contributes to its acceleration. although theirs allot of friction within the coils of string apposing that torque. The friction slows it down but the tension in the string accelerates it. Because the yo yo is spinning then it will hold a horizontal equilibrium. It will still accelerates due to the unequal forces in the vertical components.
I think you have no clue
32/45g
Is it g/2 =a?
incorrect - how on earth did you come up with g/2?
I am in 11th I saw the motion of yoyo posted by you so I compared both the equations and got this value
I knew that the answer is wrong waiting for the answer on Saturday
:)
m x g - T = m x a
T x R = I x alpha
a = alpha x R = (m x g) / (I / R^2 + m / R)
I = 2 x I_2R + I_R = 9 / 2 x m x R^2
a = 2 x g x R / (9 x R + 2)
relating angular acceleration to tangential acceleration of a solid body
and using free body diagram
(9 x R + 2)
what are the units of this 2? Is it kg, or meters, or Newtons?
sorry I think I made a mistake when clearing the tangential acceleration, it should be a = alpha x R = (m x g) / (I / R^2 + m )
and since the yoyo has moment of inertia 9 / 2 x m x R^2
the answer would be a = g / (1 + 9/2)
incorrect
Hello.are you real Walker
yes I Walk every day 2 miles.
lol
Lectures by Walter Lewin. They will make you ♥ Physics. Only one letter difference in Walker ,Walter and water .
Cause you dont want any solutions. Im not gonna send in one this time. :(
(5/18)g
incorrect
And if a bullet shot from a gun gets into a wooden block with initial velocity 100 m/s how do we calculate the total length it has gone in the block
watch my 8.01 lectures - you will then be able to do your problem.
(18/51)g
4.978 m/s
incorrect
a=3g/5
a=g/4
hope its correct :)
yes it's correct
Lectures by Walter Lewin. They will make you ♥ Physics. Didn't expect your reply. Anyways, thank you sir, your videos/problems are really helpful :)
I guess that the answer would be 8/17g since the yoyo is consisted of two solid disk and one connection between them,
If we consider the rotational inertia of the connection we will then proceed to the answer of 8/17g
incorrect
Lectures by Walter Lewin. They will make you ♥ Physics.
Sir,could you offer just a bit hint of this question~ Am I on the right track of solving it or should I modified the method?
Much Appreciation ~
relax - I will post the solution tomorrow morning.
Sorry sir , I am just voracious to know whether method of combining rotational inertia is a step of the solution.
Could you please reply. Sorry if bother , Sir . Thanks in advance!
you will have to calculate the moment of inertia of the yo-yo. tomorrow you will see the solution.
18Rg/(18R+33)
Just from the dimensions this is clearly incorrect
a=12g/13
(2/3)g
incorrect
your 2nd try is correct
Thank you for replying sir! :)
I think i got it.
If Total mass=9M/4
I=33/4MR^2
Making equations of Force and torque and solving them, I got
accn=3/14 g. ...
Am i right???? 😯😯😯😯
incorrect
a=12/23 g.
incorrect
Sir ,if I use energy conservation, then is this expression correct :-
(Mg-T)x = ½Mv² + ½Iw²
(T is tension in string)
where I is the Moment of Inertia perpendicular to the plane of yo - yo passing through its COM, w is its instantaneous angular velocity?
I'll post the solution on Saturday Nov 25
Please tell sir, because I am getting negative acceleration, solving it ,
a = -20/33 g
incorrect - stop it, this is not a fishing expedition. Wait till Saturday
a=g/3.
incorrect
I guess I got the question wrong. My second attempt gave a=g/2. Am I right?
how on earth did you come up with g/2? please stop it. this is not a fishing expedition. wait till Saturday when I will post the solutions.
Ok. This is my last try. I forgot to consider central cylinder mass and moment of inertia. I am sorry if I am wrong again. I got a=g/5. Thank you for your patience.
wrong again. It's clear that you do no know how to deal with the physics Wait till Saturday and study my solution. You may then learn some physics.
a=(6/11)g
Is time travel possible in 2017
ask google
Google is saying different point of views but it is not telling about yours
Sir, can you describe angular momentum about com of system which consist of a rod and a particle before and after collision
the com is moving in my yoyo problem, ang mom at time t is I*w_t
according to my opinion g/2. where g is acceleration due to gravity.
very incorrect
Sir I prefer you to use a better mic
the mic is part of my camera. I have no plans to buy a new camera.
Lectures by Walter Lewin. They will make you ♥ Physics. Hehehe I didnt thought you would reply to my comment
I am 13 years old and i love your lectures
:)
At MIT I had 3 cameras (run by 3 different people), 2 sound experts, and 1 light expert during my lectures in a lecture hall with 9 HUGE blackboards that I could move up and down; the lecture hall could seat 600 students. The videos that I have been posting since January this year I do at home in my hall. I have no black boards. The space I have on the wall is less than 1 of my 9 boards at MIT. There is room for ZERO students in that hall. I have only 1 light in the ceiling. I have a $150 Canon camera that can take simple videos for no longer than 30 minutes (limited by battery power). I have ZERO help; I do the taping on my own. I place the camera on a ladder and tape it down with scotch tape to make sure it will not fall on the floor. I hope you will appreciate this. Preparation and selection of each problem takes me roughly 16 hours per problem. To prepare and tape the solutions takes an additional 10 hours. I often have to tape a video twice or even 3 times before I am happy with it (slips of the tongue are common). That alone can take more than 2 hours.
Lectures by Walter Lewin. They will make you ♥ Physics. Can't Thank you enough!.. All your efforts in the background are incredible!
Lectures by Walter Lewin. They will make you ♥ Physics. Felt really sad after reading this sir. But I love your lectures. I wish my physics teacher could explain me like this
:*(
2/3(g)
incorrect
Lectures by Walter Lewin. They will make you ♥ Physics. Sir.......But ,I'm damn sure about that
yes, you cheated and looked it up on the web what the acceleration is of a solid disk rolling off a string. that is (2/3)g. But this is not a sold disc. *STOP cheating. You got it WRONG*
Sir.....I've got my mistake ....I've understood that my answer is wrong but my absence of mind said that,not web
Every high school and college physics book discuss this problem *with a disc* and find a=(2/3)g.
Please get rid of background noise.
I suggest you get better equipment to listen to my videos.
I have a pretty decent set of headphones, and a decent amp, and I am not hearing much noise. Also considering Professor Lewin is taking his own time to make these videos for you I do not see how you can complain if you really want to watch these.
At MIT I had 3 cameras (run by 3 different people), 2 sound experts, and 1 light expert during my lectures in a lecture hall with 9 HUGE blackboards that I could move up and down; the lecture hall could seat 600 students. The videos that I have been posting since January this year I do at home in my hall. I have no black boards. The space I have on the wall is less than 1 of my 9 boards at MIT. There is room for ZERO students in that hall. I have only 1 light in the ceiling. I have a $150 Canon camera that can take simple videos for no longer than 30 minutes (limited by battery power). I have ZERO help; I do the taping on my own. I place the camera on a ladder and tape it down with scotch tape to make sure it will not fall on the floor. I hope you will appreciate this. Preparation and selection of each problem takes me roughly 16 hours per problem. To prepare and tape the solutions takes an additional 10 hours. I often have to tape a video twice or even 3 times before I am happy with it (slips of the tongue are common). That alone can take more than 2 hours.