Solution Problem 201 - Kepler Orbits

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Lectures by Walter Lewin. They will make you ♥ Physics.

Lectures by Walter Lewin. They will make you ♥ Physics.

Күн бұрын

Пікірлер: 26
@Drarmalviy
@Drarmalviy 2 ай бұрын
Sir your the reason why i love physics ❤
@African.Football
@African.Football 5 ай бұрын
Love you from Kenya professor,you made me love physics
@Ceferov_82
@Ceferov_82 5 ай бұрын
Yes me too
@duckyoutube6318
@duckyoutube6318 5 ай бұрын
You are not alone in that sentiment!
@World_Complex_The
@World_Complex_The 5 ай бұрын
I love your videos sir they are actually priceless❤
@gaming_with_bhadola
@gaming_with_bhadola 5 ай бұрын
Love you sir , love from hills of uttarakhand ❤️❤️❤️
@garfielddexter6224
@garfielddexter6224 5 ай бұрын
Thank you for very interesting video🙂 very impressive 😃👍
@Ceferov_82
@Ceferov_82 5 ай бұрын
Hello Mr. Genius. Why can't you eat yogurt on Friday? He said it happens every other day. Please answer the question when you see it. Thanks in advance
@DharaniDharan-rj6ps
@DharaniDharan-rj6ps 5 ай бұрын
Sir, i love your teaching and approach to physics❤❤... Currently I am preparing for NEET medical entrance exam 2025 so I have a complete year in my hand. but still, here in my language there is no proper physics lectures in KZbin and also I can't afford paid courses😢......can I watch your lectures for physics and solve your example problems and test for my preparation??.... I have been in this dilemma for watching your lectures for my physics preparation.... 😕😕
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, SolutionsY & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@DharaniDharan-rj6ps
@DharaniDharan-rj6ps 5 ай бұрын
@@lecturesbywalterlewin.they9259 Thank you very much sir, for responding to my request💖💖..i will be grateful for your help✨😊
@AyanHaider_5mdcat
@AyanHaider_5mdcat 5 ай бұрын
Thank u sir g I watched your videos interesting
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
Most welcome
@Nikhil-gh7qr
@Nikhil-gh7qr 5 ай бұрын
Sir please answer my query. The question goes as follows : A block of mass m is connected to another block of mass M by a massless spring of springs constant K initially the blocks are at rest and the spring is unstreatched when a constant force F starts acting on the block of mass M to pull, Find the max extension of the spring. If I try to solve it using centre of mass frame then the elongation comes out to be 2mF/(m+M) but when I solve it using inertial frame using equations, F-Kx = Ma ma = kx where x is max elongation then it is mF/(m+M) where the hell is 2 going?
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
I do not solve problems - I teach Physics. This is an easy high school problem
@carultch
@carultch 5 ай бұрын
The non-intuitive part about this question, is the blocks do NOT have the same acceleration when the spring is stretched at the maximum. The center of mass will accelerate at F/(M + m), but the blocks will oscillate indefinitely (for the ideal case). In the steady state solution in reality, they settle to a common acceleration. But at a point before this, block m will lag behind. For the steady state case, the spring extension settles to x_ss = F*m/(k*(M + m)). For the transient behavior, it oscillates between 0 and twice this value. Assign capital A as the acceleration of M, and little a for m's acceleration. In the lab frame: F - k*x = M*A k*x = m*a The setup in the center of mass frame, has to account for a force term of -M*a_cm and -m*a_cm in each equation on the left side, due to an accelerating system. The accelerations on the RHS would be replaced with (A - a_cm) and (a - a_cm) respectively. We need a kinematic constraint to relate A to a. To do this, relate A & a to the 2nd derivative of x. The acceleration of the spring length is the difference of the two accelerations: x" = A - a Multiply the 2nd equation by M/m, so we can generate (A - a), and replace it with x" when combining equations. This gives us the differential equation: M*x" + k*(1 + M/m)*x = F The solution with all initial conditions of zero, for x(t) is: x(t) = F*m/(k*(M + m) * [1 - cos(ω*t)] where ω = k*(M + m)/(M*m) Its maximum value occurs when cos(ω*t) = -1. Which is therefore: x_max = 2*F*m/(k*(M + m))
@pranav.mishra
@pranav.mishra 5 ай бұрын
Sir please make a video on electric dipole As its not covered in your playlist It will be very helpful
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
it is covered in my 8.02 lectures
@pranav.mishra
@pranav.mishra 5 ай бұрын
@@lecturesbywalterlewin.they9259 oh thank you sir I think I missed that part
@qinga8
@qinga8 5 ай бұрын
Great and beautiful, I will do it again
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
Have fun
@mrxzod20
@mrxzod20 5 ай бұрын
Walter sir can u suggest the best book for physics fir jee examination India
@lecturesbywalterlewin.they9259
@lecturesbywalterlewin.they9259 5 ай бұрын
ask your teachers - they know you
@studytosuccess6501
@studytosuccess6501 5 ай бұрын
Sir , on full moon day, we can see full moon whole night but when it's crescent moon day we don't see it for longer period why?
@carultch
@carultch 5 ай бұрын
During a full moon, the moon is on the opposite side of the sun, and is visible to every part of the Earth that has nighttime. You can see a full moon at sunrise or sunset due to atmospheric refraction, but not during the middle of the day. During a crescent moon, the moon is on the same side of the Earth as the sun, at an acute angle from the line-of-sight to the sun. The vast majority of the land area that can see the moon at any given time, is experiencing daylight. During a waxing crescent, you can see it for a few hours after sunset. During a waning crescent, it rises a few hours before the sun.
@saaaalimmm
@saaaalimmm 5 ай бұрын
walter sir like my comment
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