I'm a little bit unsure. My intuition tells me that for the vertical major axis that if you smush the ellipse a lot then the top basically turns into a super tight corner. So I would think that there would need to be a lot of force needed to change the direction of velocity. Also in the same example of the super smushed ellipse, the sides will approach vertical walls so would the minimum velocity for not falling off at the top even get up the sides to get to the top(since the derived expression seems to say the min velocity just keeps decreasing. )
@PhysicsMath-my3cjКүн бұрын
Hey thanks for this wonderful video ❤
@DrBenYelverton11 сағат бұрын
Glad you enjoyed it!
@SkyDarmos-uh9nxКүн бұрын
Only someone who doesn't believe that balances measure mass instead of weight could ever believe in the validity of this experiment. The two sides of a balance incline as one. You can't treat the different sides as independent. More force is needed to incline one side because there is all the mass on the other side.
@SkyDarmos-uh9nxКүн бұрын
Except that it actually always rotates because of the rotation of the earth and so one has to calculate out any potential effects. You didn't show any of the data. It actually disproves the equivalence principle.
@SkyDarmos-uh9nxКүн бұрын
A really misguided experiment. It does not at all prove what it pretends to prove. It is pretending that the two masses are moving independent from each other.
@wheatfieldcolor6 күн бұрын
Hi ben, I am a physical chemistry teacher. I tried to derived a function for the electric field lines for days and finally got it independently by a different method from yours. Then I find your video. I am curious if your method is you own discovery or you used a reference? Another questions is that I haven't figured out what values should A take. To draw a few lines in a figure, they should be reasonablely spaced. Do you have any idea?
@CFOptimusPrime7 күн бұрын
Thanks for this, your explanation was very clear! I have two questions: 1. Does the graph imply that the balloon's pressure actually goes down as the balloon's radius increases past that tipping point? 2. If you were to tie off the balloon after filling it to a given pressure/radius, and then increase the external pressure (say, in a pressure chamber), the balloon should shrink, right? What equation governs the decrease in radius? How does it interact with the ideal gas law?
@Naman_shukla4108 күн бұрын
I have a doubt that u²=(y²+σ²)/y² then u =+ or - √(y²+σ²)/y so why you rejected -√(y²+σ²)/y ? and only taken +√(y²+σ²)/y in place of u in arccot(2zu/a) ?????
@duetothefacthat8 күн бұрын
Sir at 03:13 it shouldn't be (u.t).t - (u.n).n due to the n towards the opposite ( not exactly but almost ) way ?
@Naman_shukla4108 күн бұрын
Please don't stop posting these videos i know these are gaining less views but believe you are actually clearing doubts of many people related to real physics.🎉
@DrBenYelverton7 күн бұрын
Thanks for your support! I will get back to making videos soon.
@tomaszchrobak-pv5yh8 күн бұрын
What a great explanation. This should go viral one day
@DrBenYelverton7 күн бұрын
Thanks for your support, let's hope so!
@curtischee253210 күн бұрын
Best proof of this I have ever seen.
@DrBenYelverton10 күн бұрын
Excellent, thanks for saying so!
@hypatiakovalevskayasklodow919510 күн бұрын
Neat! I really liked it , it is such a nice observation and much faster than integration by parts
@DrBenYelverton10 күн бұрын
I'm glad you enjoyed it!
@hypatiakovalevskayasklodow919510 күн бұрын
@@DrBenYelverton Thank you for the video!! I hope I get to use it on an exam tomorrow 🤞🏻
@DrBenYelverton10 күн бұрын
Good luck, hope it goes well!
@Staycalm6914312 күн бұрын
Great explanation
@DrBenYelverton10 күн бұрын
Thank you!
@josphatmuriuki18913 күн бұрын
excellent, i finally understand
@DrBenYelverton10 күн бұрын
I'm glad this helped!
@simratkaur413214 күн бұрын
THANKU SIR 🙏
@parthahalder395815 күн бұрын
If you observe a electron, it becomes a particle, so if you observe a phonon, will it become a particle?
@mujahidali698816 күн бұрын
Thanks Ben, clear physics and clear mathematics,
@DrBenYelverton16 күн бұрын
Glad to hear it, thanks!
@surry9919 күн бұрын
Very nice! It is interesting you did not use the term "no slip condition" at the top surface but tried instead to emphasize its source which is fluid friction (viscosity).
@DrBenYelverton19 күн бұрын
Thanks for watching. Yes, very much an intuition-focused video!
@Aeist_OG19 күн бұрын
Haven't learnt integration that well yet, however can't we integrate electric field due to line wires(finite) and then as we integrate wouldn't it give us E.F at general point?... If i knew integration that clearly i would definitely try every possibility ..but as i dont know it that well .. can you tell me if what im thinking is possible or not..
@DrBenYelverton19 күн бұрын
That works in principle but I'm not sure if the resulting integral can be done analytically - it's hard enough finding the on-axis field!
@thewhyquestion558920 күн бұрын
Plz provide the proof
@ANNOYMOUS90822 күн бұрын
I believe this problem is in lev landaus book it's funny all the other classical mechanics texts like morin and goldstein are 600 pages long but lev gets right into he's book is 200 or less
@geraldspina492326 күн бұрын
Thank you so much this was unbelievably helpful.
@shors584126 күн бұрын
Unbelievably well-made! Thank you.
@suryavardhansinghshekhawat86526 күн бұрын
It would be nice if you solved an example question
@DrBenYelverton25 күн бұрын
See the following video for an example: kzfaq.info/get/bejne/qsp7idGjqNbInXk.html
@djionzir27 күн бұрын
Hello, for the value of x, why do we utilise both values of lambda? Sorry if its a bad question.
@OrmondOtvos29 күн бұрын
Thank you for this exposition. My name is Ormond Otvos (Ëotvös) and my roots are in the small town Visc formerly in Hungary, but now called Vishkovo in Ukraine. There is some question about our relations, but my father was a brilliant electronics inventor who contributed greatly to several classified projects in WWII -- sonar, mostly, including the meter-diameter barium titanate nuclear submarine sonar lens.
@DrBenYelverton28 күн бұрын
Interesting, thanks for sharing!
@the_excocistАй бұрын
Thanks sir, it helped in my BSC degree mathematical problems. Really helped a lot
@DrBenYelverton28 күн бұрын
Glad to hear it!
@dwinsemiusАй бұрын
I still remember a problem in my second semester physics class that had a coin of radius r atop a coin of radius R and the request was to determine the angle at which the upper coin would leave the lower coin. So you would need to consider the moment of inertia and the angular velocity of the upper coin in the process. The angular momentum would be "soaking up" some of the energy gain in the rolling fall. Pretty sure I got a complete zero on that question but I did get an A in the course. I think the prof was just trying to humble some of us who thought we knew everything. My next course was Lagrangian mechanics, but I never solved the earlier question to my satisfaction. I see that you have solved at least two similar problems but not that exact one.
@DrBenYelverton28 күн бұрын
I've solved basically the same problem in an older video, you'd just need to use a suitable value of α, which would be 1/2 for a typical coin: kzfaq.info/get/bejne/p8CWaJSh2puVnqM.html
@dwinsemius28 күн бұрын
@@DrBenYelverton Yep. That's the exact question at least as I understood it 2024-1968= 56 years ago this month! . Thanks very much. I had asked it in a couple of venues since then without any answer, and the only specific non-answer I'd gotten from fellow students of physics on the Interweb was "oh, sure, that question probably appeared in Sears and Zemansky." Well I went out and got a copy of S&Z and no, it's not in that hallowed 1st year undergrad physics text.
@DrBenYelverton17 күн бұрын
I'm glad you finally got the answer!
@anatolykhinaАй бұрын
Thanks for the nice video (still watching...). At around 19:00, you neglect multiple terms but some may not be negligible AFAIU. For example, the terms containing g may in fact be larger than the ones containing A omega^2 whenever cos(omega t) ~= 0. Similarly, the second derivative of theta_0 may be temporarily larger than that of delta for specific times. It seems that there is some averaging across time that is implicitly applied...
@DrBenYelvertonАй бұрын
Thanks for watching. For ω²Acos(ωt) to be comparable with g, cos(ωt) needs to be infinitesimally small. For most of its cycle, cos(ωt) is not infinitesimal, and while it does reach zero at specific times, the proportion of time it spends being infinitesimal is negligible.
@amirhosseinhosseinmardi4848Ай бұрын
Nice
@kierkegaard54Ай бұрын
Could I aplied this prescription to solve the shape of a cabe hanging from 2 poles (i.e. the catenary problem)?
@dylendye7410Ай бұрын
Thanks for the explanation, this came for my test today couldn't solve it
@DrBenYelverton28 күн бұрын
It's one of those problems that looks easy once you've seen how to do it, but it's definitely not obvious at first!
@Johnny-tw5prАй бұрын
What would happen if the sphere was accelerating?
@DrBenYelvertonАй бұрын
You could investigate this by including a fictitious force pointing in the opposite direction to the sphere's acceleration!
@AmeerHamza-bc4fhАй бұрын
So theoretically there should be a vertical reaction R on the surface element because of net downward inward forces of attraction. How that vertical R is not there and only the surface tension which is a horizontal force comes into the surface? I'm confused about this situation and my mind is not clear. Thanks for your reply.
@erwinmanalastas5827Ай бұрын
This is just only for 45 degree angle of inclination ...
@DrBenYelvertonАй бұрын
The angle θ is arbitrary here!
@chriszenker6890Ай бұрын
Shouldn't the potential energy term be -mgRcos(theta)?
@DrBenYelverton28 күн бұрын
There are two ways to handle constraints in Lagrangian mechanics. Either you make the constraint implicit in the Lagrangian (i.e. use R instead of r, as you suggested), or use an unconstrained Lagrangian and apply the constraint later via a Lagrange multiplier. In this problem we specifically want to find the point where the Lagrange multiplier becomes zero, so the first method won't work as it doesn't give any information about the constraining force. It will still give the correct equation of motion up until the point where the particle actually loses contact, but assumes that the particle always sticks to the sphere even beyond this point.
@abbasibrahim9435Ай бұрын
Its too good. Now, i am gonna share this with my classmates. There was an exam on this topic an nobody understood the lecture. So, everyone just memorized the proof and wrote in the exam
@DrBenYelvertonАй бұрын
Thanks for your support, I'm glad this helped with your understanding!
@varun0904Ай бұрын
3:39 Aren't Newton's laws valid only in inertial frames? Wouldn't it then be wrong to say "The object isn't accelerating even though there is force T on it. Therefore there must be another force balancing T" and instead say "Newton's laws don't hold in non inertial frames. Hence the object as seen in this non inertial frame doesn't accelerate even though it has a net non zero force acting on it"? In other words, can we tell whether we're in an inertial frame or not by just observing an object?
@samuels1123Ай бұрын
have a code problem where I want to translate a target distance and height into a firing angle
@BoxxsАй бұрын
Using this for American football
@HoangAnh-tr6bnАй бұрын
Hello Teacher, I have a question about this analysis. You choose a reference system with Oy pointing up so the gravity g in the system will have a negative value?. Hmnn 17:36 . Therefore omega squared will have a negative value. .. I hope to receive feedback from the teacher. ........Sorry for using Google translate. I hope you understand what I am saying. Thanks a lot
@DrBenYelvertonАй бұрын
By definition, g is the magnitude of the gravitational field and is therefore positive, regardless of which coordinate system you use.
@HoangAnh-tr6bnАй бұрын
I hope to receive feedback from the teacher...... I have a test with similar content but choose the Oy axis to point down so there are some changes in the sign in the expression. Detail : 3:55 : y₂ = l. cos𝜃 4:25 _ y dot = - l. sin 𝜃 7:00 _ v = m₂gl.cos𝜃 7:35. L = T- V = .... - m₂gl . cos𝜃 9:20 : 𝜎L/ 𝜎𝜃 = .... + m₂gl.sin𝜃 11:15 : ..... -m₂gl. sin𝜃 13:00 : ..... ẍ. cos𝜃 + l.𝜃 double dot - g.sin𝜃=0 15:00 . ...... ẍ + l.𝜃 double dot - g𝜃 = 0 17:40 . Ẅ = -{g/l .[(m₁ + m₂)/ m₁ ]} i know it's a mistake but i don't know how to find the solution because omega squared is always positive. Hope to receive your feedback. Thank you so much
@DrBenYelvertonАй бұрын
Your expression for V needs a minus sign in front, the same as in the video. When 𝜃 increases, the pendulum bob is moving up, gaining GPE and therefore V should increase. However, cos𝜃 is a decreasing function, so the minus sign is necessary regardless of your choice of coordinate system!
@justmax8145Ай бұрын
Great video
@HoangAnh-tr6bnАй бұрын
Cảm ơn thầy rất nhiều ạ . I am from Vietnam ❤
@DrBenYelvertonАй бұрын
Thanks for watching!
@lunam7249Ай бұрын
your vid doent show the good thumbnail graph is very disappointing
@ayaan_maanАй бұрын
Will you be making a series on properties of solids like elasticity, stress and strain? I really can't get my head around these concepts or the use of tensors in them (I tried reading Landau-Lifschitz but that got me nowhere).
@DrBenYelvertonАй бұрын
Will try to cover this some day, it's on my to-do list. I have never read the Landau & Lifshitz books but my impression is that they're probably not the best resource for gaining intuition!
@ayaan_maanАй бұрын
Never thought about this before. Really goes beyond the standard sqrt(5gl) result in string pendulums 😅
@Archers.creed.Ай бұрын
Hi Dr. Ben, Where do you recommend I should look for 2D projectile motion with quadratic drag? I'm trying to take this into account for a very delicate ballistics project.
@playenz3103Ай бұрын
What are the additional steps if you are given v0≠0 and some z0 ?
@DrBenYelvertonАй бұрын
The equations of motion would be the same, you'd just end up with different constants of integration when you apply your initial conditions.