Dear Dr. Walter Lewin, I hope this letter finds you in great health and high spirits. My name is Rajarajan, and I am a Class 11 student from India. I am writing to express my heartfelt gratitude for the incredible lectures you have shared with the world. Your passion for physics and your unique teaching style have had a profound impact on me. Your ability to simplify complex concepts and bring them to life with demonstrations has made me fall in love with the subject. Watching your lectures has been an inspiring journey that has deepened my understanding and appreciation of physics. Your dedication to teaching and your belief that "physics works, and it's beautiful" resonate deeply with me. You have shown me that learning is not just about passing exams but about exploring the wonders of the universe with curiosity and joy. Thank you for igniting a spark of curiosity and motivating students like me across the globe to pursue knowledge with enthusiasm and perseverance. You have truly been a beacon of inspiration in my academic journey. With warm regards and immense gratitude, Rajarajan Class 11, India
Hello Sir, I am in my Final Year, I have secured an IT JOB, I used to watch your lectures which was really amazing, I would like to thank You for all the free education your provided and that too in an understanding manner... wherever I am a part of it credit goes to you...Thanks sir !
Here's my first try , Let T be the tension in the string , 'a' be the acceleration of mass m in downward direction and 'b' be the angular acceleration of the disc . mg - T = ma -(1) TR = MR^2/2 × b -(2) a = Rb -(3) Using (3) in (2) , we get T = Ma/2 mg = ma + Ma/2 a = mg/m + 0.5M
Dear Professor, merry Christmas and Happy New Year. I am an old (73 y.o.) medical doctor with love for physics and i follow your video from some years. A great hug from Italy - Sardinia island 🤗
Hey Prof, Happy holidays, and love your content. Here's my go at it: Apply 𝜏 = I α to find: mgR = I α Assuming the mass M is uniformly distributed on the pulley, we find I = MR²/2 Plugging this in, we find: α = 2mgR /MR² The R's cancel upstairs and downstairs α = 2mg/MR This gives the angular acceleration. One last step remains, and it is to find the downward acceleration. Luckily, it is related to α by a = αR Hence, a = 2mg/M Notice that the heavier the pulley, the less acceleration there will be, and the heavier the object, the more it will accelerate. I hope this satisfactorily solves the problem
Sir if we let the change in potential energy equal to the rotational kinetic energy of the disk + the k.e. of the mass m we get a = mg/(m+ M/2) Also, happy new year!
Welp, judging by the warning of "not being able to solve it", we can throw the 9.8 m/s out the window. The massĺess string factory didn't crack the zero friction yet.
m*g/(0,5M+m) ..... I took m equivalent to a point mass on the rim. I think that's correct. Then moment of intertia for disk and point mass are added. torque is m*g*R. angular acceleration is torque over moment of intertia in rad/s². Multiply by R to give m/s². Writing that down, the R² cancels out, resulting in above solution.
Radius cancels out in the derivation so only leaves g, M and m to worry about. If we assume both have a mass of 1 unit, then the downward acceleration of the mass m is approximately 6.53 m/s/s.
Walter Lewin Sir...i am in 11th std ,one of your big fan sir, your classes are really amazing, I am also a JEE Aspirant, i wish i need your all blessings with me as much your gorgeous classes❤️❤️✨
Solution: a = 2*m*g/(2*m + M) Supporting calculations: Assess forces acting on both bodies. Hanging mass: Weight (m*g): downward Tension (T): upward Disk: Weight (M*g): downward Tension (T): downward Bearing support (B): upward, and not of interest to find. B will be as large as necessary, to hold the disk's axle in place. Assume hanging mass accelerates downward. Construct N's 2nd law on the hanging mass: m*g - T = m*a Construct rotational form of N's 2nd law on disk: T*R = I*α Geometric constraint between a & α: a = α*R Solve 2nd equation for T, & substitute: T = I/R * α m*g - I/R * α = m*a Solve 3rd equation for α, & substitute: α = a/R m*g - I/R^2 * a = m*a Solve for a: a = m*g/(m + I/R^2) For a uniform solid disk: I = 1/2*M*R^2 Thus, after simplifications, we have our solution: a = 2*m*g/(2*m + M)
τ = Iα = TR, a = Rα, I = ½ mR² where τ = torque, I = moment of inertia of the disc, α = angular acceleration, T = tension in the string, a = linear acceleration. => (mg - ma)R = ½ MR²a/R mg -ma = ½ Ma ½ Ma + ma = mg => a = mg/(½ M + m) = 2mg/(M + 2m) Answer: a = 2mg/(M + 2m)
Hello sir, I am a student that just started 11th grade a few months ago and I finished learning kinematics and I thought of a question that came up in my mind from previous questions that I had it was that was, if a ball with certain mass and velocity is thrown on a wall and then the wall absorbs some energy then the ball returns so the ball would be in horizontal motion if I am correct and it would be alright to suppose that I can calculate the values like usual. Thank you sir for making these various videos public, my teacher also referenced and mentioned your video in class while teaching us :)
I cover this in my 8.01 lectures. bounce a tennis ball (mass m, speed v) off a wall with HUGE mass M. The momentum transfer to the wall is 2mv, but the Kinetic energy of the wall after the bounce is zero.
@@lecturesbywalterlewin.they9259 sir, so the speed the ball now rebounds with how would it be calculated? and well the rebound speed wouldn't be the initial speed but i'd like to think we will consider that initial speed then with the angle given we'd be able to calculate range and height?
my answer is acceleration = g/(1+(M/2m)) Sir, in your country, do students go to college right after high school? Because I googled and saw that high school is equivalent to grade 9 and 10 in my country (India) and then we also have grade 11 and 12 (called senior high school?) before we go to college. This problem would've been of 11th grade. Its confusing a bit, we aren't familiar with terms like high school and all (at least i'm not). Also merry christmas and a happy new year to you and your family! the way you teach and your questions, it cheers me up. thanks professor
The United States education system is made up of several levels of education, including: Early childhood: Nursery to pre-K, for children ages 2-5 Elementary school: Grades 1-5, for children ages 6-10 Middle school or junior high school: Grades 6-8, for children ages 11-14 High school: Grades 9-12, for children ages 15-18
@@lecturesbywalterlewin.they9259 Oh we have equal number of grades. I didn't know that. In india we have Secondary 9 and 10, and Senior Secondary 11 and 12. So this isn't a high school problem? Also professor, I've changed my answer. I forgot that the masses were different. (Sorry!)
Here's my corrected solution. Edit: I had a misplaced parenthesis typo in the final equation, which I have fixed. Let F be the string tension. Moment of inertia of the disc is MR^2/2 Angular acceleration dw/dt = FR/I For the mass m, dy/dt = -R dw/dt = F/m - g (from the length of string spooling out per second) (note I have switched the direction of y in the previous post to be positive upward) Then dy/dt = (2FR^2)/(MR^2) = 2F/M We have (F/m) - g = 2F/M from which we derive F = g((mM)/(M-2m)) Dividing by m, the acceration component due to F is then g(M/(M-2m)) and the total acceleration is: a = dy/dt = g ((M/(M-2m)) - 1) Sanity checks: As M approaches infinity, acceleration approaches zero. As M approaches zero, acceleration approaches -g
PLse simplify the math. I cannot figure out what this is a = dy/dt = g ((M/(M-2m)) - 1) what is the meaning of the -1? *I think what you have is probably wrong*
@@lecturesbywalterlewin.they9259 Thank you for taking the time to reply! Yes, I was wrong because I lost a minus sign in the middle of my calculations, so the factor (M-2m) should be (M+2m). (M-2m) meets a sanity check for M = 0 or M = infinite, but not for finite M, as it implies acceleration faster than g. With the minus sign corrected and with my assumption of up = positive, the force of gravity = -mg (this is the minus sign you asked about). The string tension force = (mgM)/(M+2m) Divide both forces by m to get the corresponding components of acceleration of the mass m, and add those components for the final answer of net acceleration. It's hard to write clearly on a single text line because of the parenthesis nesting. Here's a version with paired brackets and braces instead of all parentheses: Als note: dy/dt (velocity) was a typo and should have been the second derivative (acceleration). Eqn 1: a = g [{M/(M+2m)} - 1] If you define downward as positive, it becomes: Eqn 2: a = g [1-{M/(M+2m)}] Sanity checks of eqn 2: M = 0 ---> a = g(1-0) = g M => infinite ---> a => g(1-1) => 0 M = m ---> a = g(1 - m/3m) = (2/3)g M=2m ---> a = g(1 - 2m/4m) = (1/2)g
@@lecturesbywalterlewin.they9259 I saw Ulf Haller's solution, and with all the corrections I made, my answer, which still contained separate terms for gravity and string tension, can be reduced to the same answer as his. I kind of like the separate terms though, as it will be a reminder of how to get to the result if I read it in the future. The reduced answer hides the play of opposing forces.
I get a=g*m/(1/2M+m). Sanity check: If m=0, a=0. If M=0, a=g. Relations used: J=1/2MR^2, (J+mR^2)theta(double dot)=mgR. Note: The mass m "contains" inertia too! That's why I added mR^2 on the left hand side.
@@lecturesbywalterlewin.they9259do you refer to my first line? a=d^2y/dt^2=g*m/(1/2M+m) which can be expressed with the ratio M/m as a=g*2/(2+M/m) or using the inverse: a= g* 2m/M /( 2m/M + 1) ? Last but not least: Merry Christmas!
Here's my first try: Moment of inertia of the disc: I = (MR^2)/2 Tension on the string: F = mg, where g is the local gravitational acceleration. Angular acceleration of the disc: dw/dt = FR/I = 2FR/(MR^2) Acceleration of m: dy/dt = R(dw/dt) =2gm/M (downward, of course) EDIT: There's a mistake! Do you see it? I didn't until I did a quick sanity check in my head as I was going to bed. I'll post a full correction later today.
@ actually, it is about my physics teacher, we wanted you to make a video about him and his students,if you would like it and have time, also speaking of which he is always watching your videos ever since he’s so into physics.
Hello sir i am jee aspirants i have a question in my mind what if gravitaional force become zero means , all matter move up or stay on surface of earth .?? 🌎
gravitational force can be very very low but never zero. In the absence of any other force (only very very small gravity) you will then be in free fall (you have then no weight). Also remembr that in our gakaxy there is no up. The grav force on Earth cannot change.
How do I submit a video solution? I almost made a big mistake, but I believe I corrected it A= mxG / m+.5M Where little m is mass of hanging weight, and big M is mass of disk
First we will solve translation motion mg - T = ma (block is not in equilibrium so T will not be equal to mg) , Now , we move to disk part doing rational motion τ = TR = Iα we get the equation , TR = MR²/2 * α here , R will get cancel on both side , Now , we have 2 equations, mg - T = ma -------- (1) T = MR/2 * α -----------(2) solving these 2 equation , we get acceleration = mg/m + 0.5M
Sir, a phenomenon in a sports called Cricket keeps perplexing me. herein a fast bolwer throws a ball ( a small spherical object) at high velocity. When they do so, sometimes the ball makes a sideways movement (called swing) even when force has been applied in forward direction. To explain this phenomenon, cricket experts use the concept of Aerodynamic. They say, when one hemispherical surface is relatively rough, and the other smooth. The air friction is more on the rough side of ball compared to the smoother side, this imbalance caused by friction leads to the movement of the ball. I find this idea very unreliable. Because when i personally did experiments my cricket-ball, even the brand new ball with perfectly symmetrical sides swing more. Aerodynamics fails to explain this. Another, phenomenon is reverse-swing , here the ball moves in towards the smoother side where the friction force is low. This violates the aerodynamic principles. Sir, cricket experts are not educated in physics. So they'll never be able to explain some non-intuitive phenomenon. So my personal explanation behind this phenomenon is : gyroscopic movement. Gyroscope and movement it generates are very non intuitive. Only a good physist can work their head around such phenomenon. Hence, i decided to communicate with you regarding this. I believe cricket bowlers are utilising the phenomenon of gyroscope with fully understanding it. If only they understood it well, they can more accurately determine the movements and trajectory of the ball. So pls make a video on this sir. But pls analyse the phenomenon using concept of Gyroscope. Because explanation using Aerodynamic is very inconsistent .
I cover in 8.01 bouncing a ball speed v, mass m off a wall (HUGE mass M). The mometum transfer to the wall is 2mv but the KE of the wall after the bounce is zerlo.
If my parents gave birth to me and their parents gave birth to my parents and if this cycle goes on then who gave birth to first person? There must be a starting point?
this is a misconception - 2 billion years ago there were single cells in our oceans and they evolved into millions of double cell creatures etc etc - millions of creatures got extinct for many differenrt reasons (that's what Darwin is all about). Homo Sapiens evolved in Africa between 200,000 and 300,000 years ago. This occurred during a period of significant climate change. Chances are that we too will become extinct
Sir i am studying in 10th grade i have doubt. Sir do electrons gain energy as they move across the battery because battery is doing work against coloumb force of othe electrons and this work get stored as potential energy and is utilised by the electron in the resistors. Sir am i right
I cover this in one of my 8.02 lectures. Watch it, that's what it is for. Yes, when electrons move through a battery, they gain energy, as the battery converts chemical energy stored within it into electrical energy, effectively "pushing" the electrons through the circuit and increasing their potential energy.
Sir in plane definition said that it is a two dimensional flat surface extended infinitely and It has no thickness and curves in it and it contains infinite number of points. Now my question is any material upper most part which we can see is said to be surface. But in this plane definition plane itself a flat surface. now i not able to understand this concept very clearly please explain this concept and why we use plane in any definition of shapes . Kindly make a video in your channel in english language . Thank you.
@@lecturesbywalterlewin.they9259 thanks a lot sirr, your physics lectures motivated me to become an astrophysicist i always wanted to read this book too!
Maybe. But how is this an easy problem ? Not only the disk but also m is to be taken into account. I think you'll get some naive answers. Not sure I can find the time. I'll travel to egypt 3e xmas day for a week. My wife, she likes to travel. That ok. I've seen much of the world.
Dear Dr. Walter Lewin,
I hope this letter finds you in great health and high spirits. My name is Rajarajan, and I am a Class 11 student from India. I am writing to express my heartfelt gratitude for the incredible lectures you have shared with the world.
Your passion for physics and your unique teaching style have had a profound impact on me. Your ability to simplify complex concepts and bring them to life with demonstrations has made me fall in love with the subject. Watching your lectures has been an inspiring journey that has deepened my understanding and appreciation of physics.
Your dedication to teaching and your belief that "physics works, and it's beautiful" resonate deeply with me. You have shown me that learning is not just about passing exams but about exploring the wonders of the universe with curiosity and joy.
Thank you for igniting a spark of curiosity and motivating students like me across the globe to pursue knowledge with enthusiasm and perseverance. You have truly been a beacon of inspiration in my academic journey.
With warm regards and immense gratitude,
Rajarajan
Class 11, India
Hello Sir, I am in my Final Year, I have secured an IT JOB, I used to watch your lectures which was really amazing, I would like to thank You for all the free education your provided and that too in an understanding manner... wherever I am a part of it credit goes to you...Thanks sir !
You are very welcome.
@@lecturesbywalterlewin.they9259 thanks again sir ❤
I’m a retired engineer love your lectures. Brings me back to when I was a carefree college student. Merry Christmas happy Hanukkah happy new years!
same to u
Cmon…it can’t be that everyone forgot about him..
What do you mean?
Fr man the views have gone down soo much 😢
Thanks for everything Mr. Lewin :) merry christmas!!!!
Merry Christmas, Professor Lewin. Thank you for everything that you do.
you are most welcome
may God bless you with health, wealth and prosperity
Happy Holidays Mr.Walter sir
lots of respect from my side sir!
Lots of respect for real!
Thank you so much sir 👍♥️ you are brilliant 💯
Merry Christmas Walter Lewin sir.
Love from India ❤❤❤
This is one of the motivation ❤😊
Here's my first try ,
Let T be the tension in the string , 'a' be the acceleration of mass m in downward direction and 'b' be the angular acceleration of the disc .
mg - T = ma -(1)
TR = MR^2/2 × b -(2)
a = Rb -(3)
Using (3) in (2) , we get
T = Ma/2
mg = ma + Ma/2
a = mg/m + 0.5M
Hey Walter, just a quick message to wish you a Merry Christmas and a Happy New Year! Hope you have a good one!
HAPPY CHRISTMAS and HAPPY HANUKKAH SIR .AND THANK YOU FOR ALL YOUR WORK
Happy holidays!
Sir although you are retired at the age, but not still in the brain and voice ❤
I am not retired in spirit!
And that's what counts.@@lecturesbywalterlewin.they9259
@@lecturesbywalterlewin.they9259 You are an inspiration my good sir!!!!
Dear Professor, merry Christmas and Happy New Year. I am an old (73 y.o.) medical doctor with love for physics and i follow your video from some years. A great hug from Italy - Sardinia island 🤗
nice to hear that
@ ❤️
Sir we wish you merry Christmas !!
Im your biggest fan sir
You're the best physics teacher 🎉
Merry Christmas…Walter😊
Thank you for the holiday wishes.
Hey Prof, Happy holidays, and love your content. Here's my go at it:
Apply 𝜏 = I α to find:
mgR = I α
Assuming the mass M is uniformly distributed on the pulley, we find I = MR²/2
Plugging this in, we find:
α = 2mgR /MR²
The R's cancel upstairs and downstairs
α = 2mg/MR
This gives the angular acceleration. One last step remains, and it is to find the downward acceleration.
Luckily, it is related to α by a = αR
Hence, a = 2mg/M
Notice that the heavier the pulley, the less acceleration there will be, and the heavier the object, the more it will accelerate.
I hope this satisfactorily solves the problem
a= mg/(m+M/2), thanks for the problem prof
Sir if we let the change in potential energy equal to the rotational kinetic energy of the disk + the k.e. of the mass m we get a = mg/(m+ M/2)
Also, happy new year!
a = 2mg/(M+2m)
Sir You are my favorite physics teacher ❤ always forever
Welp, judging by the warning of "not being able to solve it", we can throw the 9.8 m/s out the window.
The massĺess string factory didn't crack the zero friction yet.
Big fan sir ❤
I love physics
Marry Christmas sir.
I wish this will be the best Christmas for you and all of you family members . 🎄 🎁
same to u
Merry christmas professor🌲🎂💫🎉🎉
Hello sir i love your videos❤…
That's great!
Merry Christmas Sir.
May God bless you
m*g/(0,5M+m) ..... I took m equivalent to a point mass on the rim. I think that's correct. Then moment of intertia for disk and point mass are added. torque is m*g*R. angular acceleration is torque over moment of intertia in rad/s². Multiply by R to give m/s². Writing that down, the R² cancels out, resulting in above solution.
Radius cancels out in the derivation so only leaves g, M and m to worry about.
If we assume both have a mass of 1 unit, then the downward acceleration of the mass m is approximately 6.53 m/s/s.
Merry Christmas sir🎉🎉🎉🎉
The downward acceleration of the mass m is (2mg)/(2m+M)
Walter Lewin Sir...i am in 11th std ,one of your big fan sir, your classes are really amazing, I am also a JEE Aspirant, i wish i need your all blessings with me as much your gorgeous classes❤️❤️✨
Solution:
a = 2*m*g/(2*m + M)
Supporting calculations:
Assess forces acting on both bodies.
Hanging mass:
Weight (m*g): downward
Tension (T): upward
Disk:
Weight (M*g): downward
Tension (T): downward
Bearing support (B): upward, and not of interest to find. B will be as large as necessary, to hold the disk's axle in place.
Assume hanging mass accelerates downward. Construct N's 2nd law on the hanging mass:
m*g - T = m*a
Construct rotational form of N's 2nd law on disk:
T*R = I*α
Geometric constraint between a & α:
a = α*R
Solve 2nd equation for T, & substitute:
T = I/R * α
m*g - I/R * α = m*a
Solve 3rd equation for α, & substitute:
α = a/R
m*g - I/R^2 * a = m*a
Solve for a:
a = m*g/(m + I/R^2)
For a uniform solid disk:
I = 1/2*M*R^2
Thus, after simplifications, we have our solution:
a = 2*m*g/(2*m + M)
τ = Iα = TR, a = Rα, I = ½ mR²
where
τ = torque, I = moment of inertia
of the disc, α = angular acceleration,
T = tension in the string,
a = linear acceleration.
=> (mg - ma)R = ½ MR²a/R
mg -ma = ½ Ma
½ Ma + ma = mg
=> a = mg/(½ M + m) = 2mg/(M + 2m)
Answer:
a = 2mg/(M + 2m)
Hello sir, I am a student that just started 11th grade a few months ago and I finished learning kinematics and I thought of a question that came up in my mind from previous questions that I had it was that was, if a ball with certain mass and velocity is thrown on a wall and then the wall absorbs some energy then the ball returns so the ball would be in horizontal motion if I am correct and it would be alright to suppose that I can calculate the values like usual. Thank you sir for making these various videos public, my teacher also referenced and mentioned your video in class while teaching us :)
I cover this in my 8.01 lectures. bounce a tennis ball (mass m, speed v) off a wall with HUGE mass M. The momentum transfer to the wall is 2mv, but the Kinetic energy of the wall after the bounce is zero.
@@lecturesbywalterlewin.they9259 thank you very much sir :)
@@lecturesbywalterlewin.they9259 thank you sir i'll cover the 8.01 lectures as i have my breaks rn :) thank you sir
@@lecturesbywalterlewin.they9259 sir, so the speed the ball now rebounds with how would it be calculated? and well the rebound speed wouldn't be the initial speed but i'd like to think we will consider that initial speed then with the angle given we'd be able to calculate range and height?
Merry Christmas Sir.
The downward acceleration of the mass is mg/ m+ 0.5m
my answer is acceleration = g/(1+(M/2m))
Sir, in your country, do students go to college right after high school? Because I googled and saw that high school is equivalent to grade 9 and 10 in my country (India) and then we also have grade 11 and 12 (called senior high school?) before we go to college. This problem would've been of 11th grade. Its confusing a bit, we aren't familiar with terms like high school and all (at least i'm not). Also merry christmas and a happy new year to you and your family! the way you teach and your questions, it cheers me up. thanks professor
The United States education system is made up of several levels of education, including:
Early childhood: Nursery to pre-K, for children ages 2-5
Elementary school: Grades 1-5, for children ages 6-10
Middle school or junior high school: Grades 6-8, for children ages 11-14
High school: Grades 9-12, for children ages 15-18
@@lecturesbywalterlewin.they9259
In India it's quite similar but 9th class start at age of 14 (I'm in)
@@lecturesbywalterlewin.they9259 Oh we have equal number of grades. I didn't know that. In india we have Secondary 9 and 10, and Senior Secondary 11 and 12. So this isn't a high school problem?
Also professor, I've changed my answer. I forgot that the masses were different. (Sorry!)
Im first sir from DHUPGURI HIGH SCHOOL COEDUCATIONAL ,THANK YOU from BHARAT ❤❤
Merry Christmas sir ,i have a question is there any topic which u find very difficult in physics till now
Here's my corrected solution.
Edit: I had a misplaced parenthesis typo in the final equation, which I have fixed.
Let F be the string tension.
Moment of inertia of the disc is MR^2/2
Angular acceleration dw/dt = FR/I
For the mass m, dy/dt = -R dw/dt = F/m - g (from the length of string spooling out per second)
(note I have switched the direction of y in the previous post to be positive upward)
Then dy/dt = (2FR^2)/(MR^2) = 2F/M
We have (F/m) - g = 2F/M from which we derive F = g((mM)/(M-2m))
Dividing by m, the acceration component due to F is then g(M/(M-2m)) and the total acceleration is:
a = dy/dt = g ((M/(M-2m)) - 1)
Sanity checks:
As M approaches infinity, acceleration approaches zero.
As M approaches zero, acceleration approaches -g
PLse simplify the math. I cannot figure out what this is a = dy/dt = g ((M/(M-2m)) - 1) what is the meaning of the -1? *I think what you have is probably wrong*
@@lecturesbywalterlewin.they9259
Thank you for taking the time to reply!
Yes, I was wrong because I lost a minus sign in the middle of my calculations, so the factor (M-2m) should be (M+2m).
(M-2m) meets a sanity check for M = 0 or M = infinite, but not for finite M, as it implies acceleration faster than g.
With the minus sign corrected and with my assumption of up = positive, the force of gravity = -mg (this is the minus sign you asked about). The string tension force = (mgM)/(M+2m)
Divide both forces by m to get the corresponding components of acceleration of the mass m, and add those components for the final answer of net acceleration.
It's hard to write clearly on a single text line because of the parenthesis nesting. Here's a version with paired brackets and braces instead of all parentheses:
Als note: dy/dt (velocity) was a typo and should have been the second derivative (acceleration).
Eqn 1: a = g [{M/(M+2m)} - 1]
If you define downward as positive, it becomes:
Eqn 2: a = g [1-{M/(M+2m)}]
Sanity checks of eqn 2:
M = 0 ---> a = g(1-0) = g
M => infinite ---> a => g(1-1) => 0
M = m ---> a = g(1 - m/3m) = (2/3)g
M=2m ---> a = g(1 - 2m/4m) = (1/2)g
@@lecturesbywalterlewin.they9259 I saw Ulf Haller's solution, and with all the corrections I made, my answer, which still contained separate terms for gravity and string tension, can be reduced to the same answer as his. I kind of like the separate terms though, as it will be a reminder of how to get to the result if I read it in the future. The reduced answer hides the play of opposing forces.
I get a=g*m/(1/2M+m). Sanity check:
If m=0, a=0. If M=0, a=g.
Relations used: J=1/2MR^2, (J+mR^2)theta(double dot)=mgR. Note: The mass m "contains" inertia too! That's why I added mR^2 on the left hand side.
a = dy/dt = g ((M)/(M-2m)-1) pls write this differently
@@lecturesbywalterlewin.they9259do you refer to my first line? a=d^2y/dt^2=g*m/(1/2M+m) which can be expressed with the ratio M/m as a=g*2/(2+M/m) or using the inverse: a= g* 2m/M /( 2m/M + 1) ?
Last but not least: Merry Christmas!
SIR It was Amazing.
what was amazing?
Sir everything from approach to solution you are an inspiration for millions of students across world like us Indians.
....what
Here's my first try:
Moment of inertia of the disc:
I = (MR^2)/2
Tension on the string:
F = mg, where g is the local gravitational acceleration.
Angular acceleration of the disc:
dw/dt = FR/I = 2FR/(MR^2)
Acceleration of m:
dy/dt = R(dw/dt) =2gm/M (downward, of course)
EDIT: There's a mistake! Do you see it? I didn't until I did a quick sanity check in my head as I was going to bed. I'll post a full correction later today.
Professor can i ask you a favor?
you can always ask but I may not do you that favor.
@ actually, it is about my physics teacher, we wanted you to make a video about him and his students,if you would like it and have time, also speaking of which he is always watching your videos ever since he’s so into physics.
Hello sir i am jee aspirants i have a question in my mind what if gravitaional force become zero means , all matter move up or stay on surface of earth .?? 🌎
gravitational force can be very very low but never zero. In the absence of any other force (only very very small gravity) you will then be in free fall (you have then no weight). Also remembr that in our gakaxy there is no up. The grav force on Earth cannot change.
@lecturesbywalterlewin.they9259 thanks sir and happy Christmas day 💐
Very nice dear professor
thanks
How do I submit a video solution?
I almost made a big mistake, but I believe I corrected it
A= mxG / m+.5M
Where little m is mass of hanging weight, and big M is mass of disk
Hello sir please make a solution on the topic namely quantam gravity
nice joke - I suggest you ask Quora
😂😂@@lecturesbywalterlewin.they9259
sorry sir
First we will solve translation motion
mg - T = ma (block is not in equilibrium so T will not be equal to mg) ,
Now , we move to disk part doing rational motion
τ = TR = Iα
we get the equation , TR = MR²/2 * α
here , R will get cancel on both side ,
Now , we have 2 equations,
mg - T = ma -------- (1)
T = MR/2 * α -----------(2)
solving these 2 equation , we get acceleration = mg/m + 0.5M
Sir, a phenomenon in a sports called Cricket keeps perplexing me.
herein a fast bolwer throws a ball ( a small spherical object) at high velocity. When they do so, sometimes the ball makes a sideways movement (called swing) even when force has been applied in forward direction.
To explain this phenomenon, cricket experts use the concept of Aerodynamic.
They say, when one hemispherical surface is relatively rough, and the other smooth. The air friction is more on the rough side of ball compared to the smoother side, this imbalance caused by friction leads to the movement of the ball.
I find this idea very unreliable.
Because when i personally did experiments my cricket-ball, even the brand new ball with perfectly symmetrical sides swing more.
Aerodynamics fails to explain this.
Another, phenomenon is reverse-swing , here the ball moves in towards the smoother side where the friction force is low.
This violates the aerodynamic principles.
Sir, cricket experts are not educated in physics. So they'll never be able to explain some non-intuitive phenomenon.
So my personal explanation behind this phenomenon is : gyroscopic movement.
Gyroscope and movement it generates are very non intuitive.
Only a good physist can work their head around such phenomenon.
Hence, i decided to communicate with you regarding this.
I believe cricket bowlers are utilising the phenomenon of gyroscope with fully understanding it.
If only they understood it well, they can more accurately determine the movements and trajectory of the ball.
So pls make a video on this sir. But pls analyse the phenomenon using concept of Gyroscope. Because explanation using Aerodynamic is very inconsistent .
I cover in 8.01 bouncing a ball speed v, mass m off a wall (HUGE mass M). The mometum transfer to the wall is 2mv but the KE of the wall after the bounce is zerlo.
I had difficulty understanding the exercize.
watch the solutions
Sir I am a JEE Aspirant.. Any advice?
If my parents gave birth to me and their parents gave birth to my parents and if this cycle goes on then who gave birth to first person? There must be a starting point?
this is a misconception - 2 billion years ago there were single cells in our oceans and they evolved into millions of double cell creatures etc etc - millions of creatures got extinct for many differenrt reasons (that's what Darwin is all about). Homo Sapiens evolved in Africa between 200,000 and 300,000 years ago. This occurred during a period of significant climate change. Chances are that we too will become extinct
a = (mg) / (0.5M + m)
Merry Isaacmas!
1st comment sir ❤
First comment❤😊😊
ain't it g professor??
Sir i am studying in 10th grade i have doubt. Sir do electrons gain energy as they move across the battery because battery is doing work against coloumb force of othe electrons and this work get stored as potential energy and is utilised by the electron in the resistors. Sir am i right
I cover this in one of my 8.02 lectures. Watch it, that's what it is for. Yes, when electrons move through a battery, they gain energy, as the battery converts chemical energy stored within it into electrical energy, effectively "pushing" the electrons through the circuit and increasing their potential energy.
a=2mg(M+2m).
Happy holidays!
incorrect
@@lecturesbywalterlewin.they9259 Oh my goodness! I forgot "/"!: a=2mg/(M+2m). Thank you Prof Lewin!
Sir in plane definition said that it is a two dimensional flat surface extended infinitely and It has no thickness and curves in it and it contains infinite number of points.
Now my question is any material upper most part which we can see is said to be surface. But in this plane definition plane itself a flat surface. now i not able to understand this concept very clearly please explain this concept and why we use plane in any definition of shapes . Kindly make a video in your channel in english language .
Thank you.
I cover this problem in my 8.01 lectures watch them, that's what they are for. All subtitles are in English and I also speak English.
Thank you sir @@lecturesbywalterlewin.they9259
My answeri is Professor: 2.5 m/s^2 Anyway, I wish you and your family a Merry Christmas and a Happy New Year. Greetings from Norway.
Downward acceleration = mg/(M-m)
If M equals m, what happens...?
a=(2mg)/(M+2m)
When m=M then
a=(2/3)g
You are too cute guru ji
🤓🤓
Ans is 2mg/M
a=2mg/M
a = g / (1 + (M / 2m)).
Merry Christmas!
Sir can you give me a Hai😊❤ today's my birthday
Happy birthday
@lecturesbywalterlewin.they9259 thank you sir ♥️🥹
Merry Christmas sir i am from India sir can you give me love of physics book free on this Christmas please sir
fiisikis.weebly.com/uploads/5/4/9/3/54939617/for_the_love_of_physics.pdf
@@lecturesbywalterlewin.they9259 thanks a lot sirr, your physics lectures motivated me to become an astrophysicist i always wanted to read this book too!
@lecturesbywalterlewin.they9259 the link Is not working
Merry Christmas
Happy holidays!
I am getting 2g/3 m/s as answer
my bet:
a = (2mg)/(M+2m)
2g/3
2/3g
g/(1 + M/2m)
a=mg/m+1/2M
g/(1+M/(2m))?
Hello sir Priyanshu this side
From Uttar Pradesh India ❤
U've implemented so many positive energies in all of us
Thank you for everything sirr
my pleasure
Maybe. But how is this an easy problem ? Not only the disk but also m is to be taken into account. I think you'll get some naive answers. Not sure I can find the time. I'll travel to egypt 3e xmas day for a week. My wife, she likes to travel. That ok. I've seen much of the world.
a = mg / (m + 1/2M)
Acc to me
mg-t=ma
a=R@
tR=(0.5MR^2)@
t=0.5Ma
Thus
a= g÷[1+0.5(M÷m)]
Thus when M tends to zero we get a=g
Merry Christmas…Walter😊