Math is the universal language, and pursued by a lot of Asians in higher education. Well, that is to say a far higher proportion of Asians will go on to a mathematics-related career than physics or even chemistry.
@@tinafeyalienActually the real olympiad question is not shown in this video but if you go look at it, the question is 2 pages long with 5 sub-questions that slowly lead you to the final solution. As a person smart enough to solve that question, I can confirm that I am not smart enough to design that question the way it is presented in the original paper. It's really a great way to design the question so that students can solve it without being confused as to what to assume and what not to assume.
This trig identity is considered common knowledge even for a lot of regular high school students in the USA, which doesn't have the hardest math education. It's hard for students representing their countries at the IPhO to not remember this.@@AttilaAsztalos
@@skimmelsvamp9531 sure is - for a physics olympiad contestant. Not so much for the average mathematically-inclined UA-cam viewer who might reasonable be able to follow/tackle simple maths and reasoning but last saw "trivial trig identities" years and years ago.
For the ones who are curious about the loss of kinetic energy: it happens when the axis of rotation (the edge in contact with the ground) changes. Basically at the instant the axis change there is conservation of angular moment (i.e. the force from the ground plane has no torque about the axis) . So, you can calculate angular moment right before impact with the equation that relates the angular moment of the axis about the center of mass with an arbitrary one and the angular moment right after impact simply by angular moment about the new axis of rotation. Due to the conservation of angular moment you know both of those must have the same result, but the first one is a function of angular velocity before impact and the second one a function of angular velocity after impact. Those velocities are different and by equating the angular moments you can find a constant that relates them. Say we call that constant b. Now we see how this affect kinetic energy: at any given moment it is proportional to the square of the angular velocity of the moment. So from a moment right before impact to a moment right after it is multiplied by our constant b but squared. That b squared must be 0.42 and represents how much of the energy is conserved. 1 - 0.42 = 0.58, that is the 58% that is lost. To verify it all we would need is the moment of inertia of the pencil about its center of mass and about the edges Hope my explanation was understandable, I'm not very good at English :p
Couple of questions. Why wouldn't the normal force of the surface (ground) produce a torque around either axis point when the face hits the ground. And wouldn't the moment of inertia about both of these axes be the same by symmetry (they're the same distance from the centre of mass). So If the angular momentum was conserved as we go from one axis to the other, wouldn't angular velocity also be conserved.
@markkennedy9767 Hey, I've went and read the solution: ipho.olimpicos.net/pdf/IPhO_1998_S1.pdf so I may be able to help. The angular momentum is conserved *around the edge of an impact* specifically - the solution assumes that an impact force is applied only along the impact edge, which is a reasonable assumption (you can inset the faces of the pencil by a tiny amount, and you wouldn't expect the behaviour to change). Moreover we know the directions of the center of mass right before and right after the impact. We find that the direction before contributes an extra 1/2 M R^2 w_before angular momentum, but the direction after contributes M R^2 w_after, since the angle between velocity and direction to center changed from 30deg to 90deg. Knowing that moment of inertia of a hexagonal prism is 5/12 M R^2 (which you can calculate by integration), we then equate 11/12 M R^2 w_before with 17/12 M R^2 w_after, obtaining that angular velocity changes by a factor of 11/17. (Note that the first angular velocity is about the "old" edge and the new one is about the "new" edge!) Then, since the situation is symmetric, we find that kinetic energy can be expressed as rotational kinetic energy about the edge (1/2 17/12 M R^2 w^2, but that doesn't matter), therefore kinetic energy change is a square of angular velocity change.
@@Kuwustilam It may happen due to the bounce after roll, if the floor is incline enough to create enough horizontal force to roll on the air, but not for long.
After looking at the vid and before looking at the comments, I thought that you had at least 100k subscribers. You will grow a lot in the future, your content is so good
Damn, UA-cam JUST had it in mind to recommend this channel to me, and your animations and awesome explanations got me hooked. Sad there are only 5 videos, but I can tell it requires a lot of work, and you may not be getting the traction you need and deserve. Gonna do my part and spread these "limited-edition" videos to more of my nerdy friends for sure. These are seriously amazing!
@@overthinkingdude3462 Can you please tel how to prepae? My goal is not till IPHO but till INPHO(conducted in india for selection in camp from where team is decided). It is basically more maths than physics in it.
This channel needs more attention. You will probably not believe what happened. I solved thos question on the 1998 paper, was not sure what part (d) of the question (about limit of KE meant) so I asked my friend to explain the question. Then, a day after it was all done, he sends me this video. Quality content
I took the click bait designed to capture exceptional and brilliant mind’s attention. At the 9:23 mark of this fascinating video I awoke from a 90 second nap and thought. “This is rather interesting but I must have coffee to continue”. I have a rule (well one of very many) if I start watching a video and it has any merit at all I am obligated to hit the like button if by chance I die before coming back to it (I am a man who takes rules and duty very seriously and dying is on the table i’m 78). Just above the like button was the channel’s name “Overthinking Dude”. I laughed thinking well he does admit to this weakness as I admit to mine, For some very odd reason I believe what I think or write is worth the time to read (but most likely not) but you have gotten this far and I salute you. Your coffee is getting cold.
I didn't watch the full video but found the problem quite intriguing after the point where he says it's not 30°. So I searched for 1998 physics olympiad and found the question paper. And I must say it's designed so nicely that the 5 sub-questions subtly nudge you towards the final solution. I haven't solved many physics problems for 5-7 years now, but I could solve it in a couple of hours because of all the sub-questions guiding me in the right direction. It felt very satisfying to solve it myself and I must say I wouldn't have been able to do it if the questions weren't framed so beautifully.
I saw a similar question in SS Krotov, but in that question the wedge was at an inclination α with horizontal, and the hexagonal pencil made an angle φ from the plane of the incline. It asked the value of φ for which the pencil just stays in equilibrium. That question was interesting too
Note that there will usually be some sliding if coefficient of friction isn't literally infinite. Also it can spontaneously transition to a runaway-accelerating hopping gate above a 24-27 degree angle, and maintain such a gate down to 14 degrees.
This means that the 14-27 degree region is bistable, or rather, has one stable state (rolling normally at a speed under 166 mm/s) and one unstable state (accelerating to an infinite speed).
@@overthinkingdude3462 In the 1998 Physics Olympiad paper . I was glancing through it and saw this question. Though it was out of my capacity to solve it.
@@rishinaik4326 Don't be scared. Try to imagine the problem in every detail you can. Try applying some common sense. And after that apply some physics laws. That's generally how I solve all the problems. If you really think you are not at the level you can solve this problem, you can start with something easier such as Irodov.
@@rishinaik4326 Back in high school I solved all Irodov mechanics and thermodynamics problems and would have solved electromagnetism if I had the time. Some people suggest Krotov instead of Irodov, but personally, I didn't even know it existed. Basically, after solving Irodov you should be able to progress to past IPhO problems. Start with the older ones since they are easier. And you can probably pick up some theory along the way. :)
I enjoyed the presentation. Good to hear the physics reasoning and see the mathematical analysis. Combined, they provide a reasonable explanation of what we experience. Unfortunately, the problem and solution links don't work for me.
He says its somewhere else but I can't find it, its effectively the "bounce" as the pencil hits the table on the flat side which comes down to the materials of the pencil
4:26 Due to conservation of energy, with perfectly elastic collisions, the pencil should never stop rolling down no matter the angle if the pushing force is enough to tip it over once.
@@untruegamer8079 I do understand friction prevents sliding but in a ideal scenario it should not consume energy from a rolling non-sliding pen. Just like how ground to rubber friction helps the a wheel roll and does not slow it down (again, in an ideal scenario).
@@untruegamer8079 Friction does not count as an energy loss unless the pencil is sliding. Which it isn't. Energy equals force times distance, without any distance moved you don't have energy loss.
You are exactly right, but the guy doing the video just pulls 58% out of his ass with no justification. The pencil should theoretically keep rolling even on a flat surface.
It's a really good question...me being in high school have learnt the concept of toppling in chapters like centre of mass and rotational dynamics..which were arguably helpful here
I am a neet aspirant even tho I can understand by your simple explanation.... And I did completed the rotational motion chapter so I was connecting the chapter with the video.. Thank you❤
More reasons for why this first level of consideration is incomplete: * the materials in the game are plastic and elastic, so a pencil tipping over might see a noticeable return of energy. * the pencil not only has forward driving momentum, but it also has a rotational momentum (and this can be very strong). so in effect it can be the case that it will only shortly tip with its edges to the surface until it "floats" in air for the next edged to find contact. in other words, it does not necessarily fall down fast enough to just have at all any sort of flat contact with the surface. also air trapped might help it to get lifted quickly to such a "hovered" position. and guess what? a hovered motion will not have any sort (or at least much) of lifting up energy included - that would be able to stop a pencil from rotation. so the overall run length might be much increased. rationale: there might be much more in play that impacts on a real world setup. dont trust things to behave in a way the still simple formula above would predict.
i tried to solve before watching, and ended up thinking only about how to get the answer to this energy loss without calculating the rotational inertia of the hexagonal shape around its center and its edge. To somehow get the energy loss on impact. I disn't get there in a few minutes and watched to find out. And then he just gives a number 😢
Amazing vid. I wonder tho... how the idea of rotational kinetic energy would affect this entire calculation. I reckon KE you give to the centre of mass is not just doing work to overcome the peak of potential energy and energy loss during a hit. But also to give the pencil it's rotational kinetic energy.!
Thank you for the video.. i was just going through some ipho papers and i saw the same problem but i didn't tried because i was developing my problems solving skills still developing .... But this video was great it was not very tough just didn't understood how 58% energy got lost after the impact🤔🤔.... Thank you anyways and please upload more 🙏
Home-boy here pulled a gangsta 58 out of the back pocket and managed to smuggle in a calculator at an Olympiad. 10:30 . 17:13 . Show your work or get out. Also, solution "assumes" stuff at the end like Friction, drag, density. Give solution considering those variables or can I just "easily use a calculator" / LLM and send it?
Before continuing the video I wanna write my initial guess of “anything greater than 30 degrees”. A hexagonal pencil has interior angles of 60 degrees on each angle. The pencil would perfectly balance on an edge at the center point of the angle, which is 30 degrees. At greater than 30 degrees, the pencil will no longer be stable, and fall towards the steep side. This will result in the pencil being unstable balancing on the next edge, causing it to fall to the side again. This would make the pencil role.
There's something no one took into account. What material is he pencil made from ? And the table ? Depending on that, the pencil will roll before or after it's tipping point. So this exercice is void. Thanks to physics, you can save on brain power 👌🏻
I want to see how you calculated that the pencil lost that procent of kinetic energy while he's falling down from his highest point, make the full-of-math video, please!!! I really like your video, thank you very much!
The calculation was not included in the video, but it is shown in the links. If you follow the link for "The solution" you will get a 404 error, but if you google the filename 1998_Iceland_p1Sol.pdf , you can see the full calculation. The video just skips it and pretends to have shown the full solution to the problem. Inattentive listeners believe it. 58% is just an approximation, the exact value is 168/289, which can give you a clue about to the complexity of the problem.
Very awesome video. You really took a complicated physics problem and animated it into real life. I would say it is sometimes slow. Mainly when it comes to repetitive simulations like 13:30 and 6:30. If you say "let's track the center of mass" it should not be a painful wait for the center of mass to be tracked. Animate then explain or animate while explaining. This channel will be golden.
Not sure I understand you right, but don't we need to have both inequalities theta > 6.58° K_i > K_min (K_i = initial kinetic energy) for the pencil to at least start rolling and keep rolling?
There's a bit of an issue here. We are assuming that thehexagonal pencil is "walking" when rolling. That is, there is constant contact with the ground, and briefly there are two separate points of contact once every circumradius of the pencil. However, this will only be the case for a pencil that can make that sharp of an arc. That is, one where the vertical launch speed of the pencil can be cancelled within the time it takes to roll. Otherwise it will transition to either a very efficient running gate, or a less efficient bchaotic hopping/bouncing gate. The vertical launch speed of a pencil in a walking gate must be enough to reach 0 velocity at an apex of 0.134 circumradii under gravity within half a circumradius of horizontal travel. If we make the slightly wrong but helpful approximation that way the transition point the cycloidal arc is actually parabolic, for a 3.5mm radius, we get a time to apex of slightly under 9.8 milliseconds. This caps the step frequency of our walking gate at about 1 step per 19.6 ms, or a speed of 3.5 mm per 19.6 ms, or 178.9 mm/s. However, remembering that our steps are in fact cycloids and not parabolas, we end up with higher vertical velocity on every launch than necessary. This means that we will get too high an apex once our vertical speed ever exceeds approximately 96 mm/s. Which will actually happen at a speed of 166.3 mm/s. What we see is that we will transition to a running or hopping gate at somewhere around this speed. Basically once we reach about ~950 rpm it will either jump around like crazy or roll on the corners while being fully airborne at least some of the time. Which will actually occur is another question.
It turns out that the efficient running gate is unstable and will always transition to the chaotic hopping gate. This will nonetheless lead to runaway acceleration above about a 14 degree angle.
Roughly 58% is because the angular momentum is dependent on the pivot. The pivot changes when the next edge hits the table, so we "lose" a bunch of angular momentum. I think.
in a perfectly rigid environment with 0 friction, to my undesrtanding, even on a 0 slope surface the hexagon will continue rolling, as its not losing any energy to heat, and just bounces back as high as was required to initialize the spin. i think delving into more detail as for the friction model would make sense. just saying theres 58% reduction on impacte assums the environment is not perfect, in which i would like to see where it is stated in the question you are answering.
Friction does not make the pencil lose energy unless it is sliding. Energy = Force * Distance. No distance sliding means no energy is lost due to friction.
static friction still needs to be explicitly defined by the question, i am asking for the definition of friction conditions, where static friction would be one of those.@@-ZH
Except the most handwavy assumption - the 58% lost in each bump - was totally glossed over. I don't even know how I'd approach figuring that out, I would probably give the answer in terms of some constants like the coefficient of restitution kind of analolgue and maybe make an estimate based on a few "what if" values.
Amazing video! I'd love to understand how the recursion equation is solved by hand as I'm guessing graphing simulators wouldn't have been allowed at the Olympiad? Thanks!
The pencil colides with the pad in each step. If the colision is elastic, no energy is lost. That means, 58% is some serious guesswork. Also, the mass distribution doesn't matter, contrary to the claim at the end of the video, as long as mass is not movable and the center of gravity stays the same.
Quality is amazing, just a thing- don't relly on simulation observations.. try using the things a student can use whilst in an exam, use simulation to depict already proven things by calculating. For example - You said- for smaller angles like 5° you can see it stops.. well not without a simulation I can't see just like that.
firction is a variable force so its not really a way to get the critical angle. the more precise one is, normal force will shift to the tipping vertex, so friction and normal will be about that vertex, and their torque will be zero about the vertex, so only torque acting would be gravity.
At the 3:00 when the center of gravity is directly above point O, depending on the static friction, wouldn't the pencil start to slide? I know this is not the question being asked or the point of the video, I just thought it was an interesting argument that could be made for having a lower desk angle to maintain rolling if the rolling was initiated by external force.
Shouldn't this simply be MGH to MVV/2? Potential to Kinetic Energy? Unless you're making allowance for friction, thermal losses, air resistance etc, then it's no different from a pendulum in space - What it gains in kinetic energy from falling is sufficient to raise it's CoG to the same height again. That being the case, the angle is anything > 0. The shape should be irrelevant, whether a square, hexagonal, an oval or any other shape. If the collision with the surface does not lose energy, then any angle will result in acceleration, not stability. If potential energy is lost and the only place for it to go is kinetic energy, I think you may have overthought this one and added formulaic components that are not relevant to the calculation. Without losses, a round pencil and a hexagonal pencil of the same mass and rotational distribution on average should start rolling at the same time, and reach the same speed at the same time, though of course you should get a pendulaic result where the shaped pencil is sometimes winning and sometimes losing... Energy cannot be created or destroyed. Exchanging potential energy for kinetic energy means that the speed is the same in both the case of a round pencil and a hex pencil, and both will roll forever if there is any vector of thrust assuming no losses.
I am just shocked how less of physics and chemistry channels are on UA-cam as compared to maths keep going!!! Uniqueness is the key to success
Math is the universal language, and pursued by a lot of Asians in higher education. Well, that is to say a far higher proportion of Asians will go on to a mathematics-related career than physics or even chemistry.
fi you havent seen a platypus doesnt mean it doesnt exists,
there are more no of physics than math
I think maybe because of the problems of those are long and the solution as well.
Search insp on UA-cam its a brilliant channel and u will learn a plenty of new concepts
physics videos are easy to find, chemistry is basically impossible, sadly
The best or worst fact you'll learn today is that no matter how good you are at an olympiad, remember someone had to design those problems first.
I coulda designed that problem, but solving it not so much
@@tinafeyalien 🤣👍
@@tinafeyalienpfp checks out
well, people like Jaan kalda get paid for this stuff only
@@tinafeyalienActually the real olympiad question is not shown in this video but if you go look at it, the question is 2 pages long with 5 sub-questions that slowly lead you to the final solution. As a person smart enough to solve that question, I can confirm that I am not smart enough to design that question the way it is presented in the original paper. It's really a great way to design the question so that students can solve it without being confused as to what to assume and what not to assume.
The quality is amazing for such a small creator, love it
Thank you. It did really take quite some effort. Cheers!
What do you think about my content?🤔
Yeah such a small creator innit mate
@@overthinkingdude3462please come back mate :(
holy handwave, batman. that 58% is the most important part of this calculation but it's "left as an exercise to the reader" haha
...also, that final equation is only simple to solve if one happens to still remember that cos (a - b) = cos a * cos b + sin a * sin b...
This trig identity is considered common knowledge even for a lot of regular high school students in the USA, which doesn't have the hardest math education. It's hard for students representing their countries at the IPhO to not remember this.@@AttilaAsztalos
@@AttilaAsztalos That’s literally one of the most trivial trig identities…
@@skimmelsvamp9531 sure is - for a physics olympiad contestant. Not so much for the average mathematically-inclined UA-cam viewer who might reasonable be able to follow/tackle simple maths and reasoning but last saw "trivial trig identities" years and years ago.
@@AttilaAsztalos bruh I am not physics olympiad contestant but learned it in high school
For the ones who are curious about the loss of kinetic energy: it happens when the axis of rotation (the edge in contact with the ground) changes.
Basically at the instant the axis change there is conservation of angular moment (i.e. the force from the ground plane has no torque about the axis) . So, you can calculate angular moment right before impact with the equation that relates the angular moment of the axis about the center of mass with an arbitrary one and the angular moment right after impact simply by angular moment about the new axis of rotation.
Due to the conservation of angular moment you know both of those must have the same result, but the first one is a function of angular velocity before impact and the second one a function of angular velocity after impact. Those velocities are different and by equating the angular moments you can find a constant that relates them. Say we call that constant b.
Now we see how this affect kinetic energy: at any given moment it is proportional to the square of the angular velocity of the moment. So from a moment right before impact to a moment right after it is multiplied by our constant b but squared. That b squared must be 0.42 and represents how much of the energy is conserved.
1 - 0.42 = 0.58, that is the 58% that is lost.
To verify it all we would need is the moment of inertia of the pencil about its center of mass and about the edges
Hope my explanation was understandable, I'm not very good at English :p
It’s already unbeatable, my brain couldn’t even comprehend it
Couple of questions. Why wouldn't the normal force of the surface (ground) produce a torque around either axis point when the face hits the ground.
And wouldn't the moment of inertia about both of these axes be the same by symmetry (they're the same distance from the centre of mass). So If the angular momentum was conserved as we go from one axis to the other, wouldn't angular velocity also be conserved.
@@markkennedy9767 Maybe due to the normal force passing through the centre of mass ie axis of rotation so the torque is zero( as theta is 0)
@markkennedy9767 Hey, I've went and read the solution: ipho.olimpicos.net/pdf/IPhO_1998_S1.pdf so I may be able to help. The angular momentum is conserved *around the edge of an impact* specifically - the solution assumes that an impact force is applied only along the impact edge, which is a reasonable assumption (you can inset the faces of the pencil by a tiny amount, and you wouldn't expect the behaviour to change). Moreover we know the directions of the center of mass right before and right after the impact. We find that the direction before contributes an extra 1/2 M R^2 w_before angular momentum, but the direction after contributes M R^2 w_after, since the angle between velocity and direction to center changed from 30deg to 90deg. Knowing that moment of inertia of a hexagonal prism is 5/12 M R^2 (which you can calculate by integration), we then equate 11/12 M R^2 w_before with 17/12 M R^2 w_after, obtaining that angular velocity changes by a factor of 11/17. (Note that the first angular velocity is about the "old" edge and the new one is about the "new" edge!) Then, since the situation is symmetric, we find that kinetic energy can be expressed as rotational kinetic energy about the edge (1/2 17/12 M R^2 w^2, but that doesn't matter), therefore kinetic energy change is a square of angular velocity change.
Instructions unclear, pencil rolled on the air
Had the same issue, apparently you had to "follow the laws of physics" or it won't work
@@Kuwustilampff, what a bummer. I wanted to skip doing that...
blame the fluid viscosity and gravity not working
@@Kuwustilam It may happen due to the bounce after roll, if the floor is incline enough to create enough horizontal force to roll on the air, but not for long.
@@bansibhavesh@themannishow
After looking at the vid and before looking at the comments, I thought that you had at least 100k subscribers. You will grow a lot in the future, your content is so good
Wow, thank you! Maybe someday :)
Still 6.2k 😢
7.4k
Criminal how the algorithm works in highlighting crap content and not highlighting actual content like this to viewers.
@@yatogami1783 8.4k
Damn, UA-cam JUST had it in mind to recommend this channel to me, and your animations and awesome explanations got me hooked. Sad there are only 5 videos, but I can tell it requires a lot of work, and you may not be getting the traction you need and deserve. Gonna do my part and spread these "limited-edition" videos to more of my nerdy friends for sure. These are seriously amazing!
Great video and explanation !!! Many more success to you .
Thanks a lot!
Guruji 🤩🥴
Sir aap yaha 🥳
Pranam Sir 🙏🙏
Are sor orz 🙏🙏
Great animations. I believe it will help me a lot in learning physics for Olympiads.
Glad to hear that!
Lagrange wrote the whole mechanics without a single image. Physics speaks math, not photos.
@@florincoter1988shut up
Proffesional video analysis!!! Its a wonder that you have not yet reached 100 k subs! You definetly deserve more
just watched it now pretty good tbh
Great video.. dude. You have no idea how awesome a job you have done. How does pencil lose 58% of KE... Can you explain that part?
This is so cooool, I seriously encourage you to do more tough IPHO problems, you are really good at it, well encouraged!
Thank you! To tell you a secret, I am a IPhO medalist XD
@@overthinkingdude3462that's more than fantastic! Felt so nice to hear that! Keep it up man, I love the IPHO animations.
@@overthinkingdude3462 phenomenol, I would love to interact with the people who have reached the milestone which is my inspiration.
@@overthinkingdude3462 Can you please tel how to prepae?
My goal is not till IPHO but till INPHO(conducted in india for selection in camp from where team is decided).
It is basically more maths than physics in it.
The most difficult and crucial part, the 58% KE loss, is not explained and the solution link is a 404. Fail.
Oh wow I didn’t notice, thanks for pointing that out for us
how did we get the 58% energy loss estimate? wonderful video! Unfortunately, the link to question/solution is not working :(.
This channel needs more attention. You will probably not believe what happened. I solved thos question on the 1998 paper, was not sure what part (d) of the question (about limit of KE meant) so I asked my friend to explain the question. Then, a day after it was all done, he sends me this video. Quality content
The most underrated Physics Channel on UA-cam. You deserve more brother. De
Glad you think so! I only have 4 videos now so it makes sense. Each video takes a lot of effort :)
@@overthinkingdude3462 I'm personally gonna publicise it.
@@godfxth4104 Thank you, I appreciate it!
I took the click bait designed to capture exceptional and brilliant mind’s attention. At the 9:23 mark of this fascinating video I awoke from a 90 second nap and thought. “This is rather interesting but I must have coffee to continue”. I have a rule (well one of very many) if I start watching a video and it has any merit at all I am obligated to hit the like button if by chance I die before coming back to it (I am a man who takes rules and duty very seriously and dying is on the table i’m 78). Just above the like button was the channel’s name “Overthinking Dude”. I laughed thinking well he does admit to this weakness as I admit to mine, For some very odd reason I believe what I think or write is worth the time to read (but most likely not) but you have gotten this far and I salute you. Your coffee is getting cold.
Dude please continue these videos. These are sooo good
That's the plan! Just had a pneumonia and my video production got delayed. The next video is almost complete.
I didn't watch the full video but found the problem quite intriguing after the point where he says it's not 30°. So I searched for 1998 physics olympiad and found the question paper. And I must say it's designed so nicely that the 5 sub-questions subtly nudge you towards the final solution. I haven't solved many physics problems for 5-7 years now, but I could solve it in a couple of hours because of all the sub-questions guiding me in the right direction. It felt very satisfying to solve it myself and I must say I wouldn't have been able to do it if the questions weren't framed so beautifully.
dude don't Stop making videos about ipho question. You're amazing... I love your video thanks dude ❤!!!
Thank you, good video! Did you ever make the video to explain how to get the 58% figure at 8:20?
I saw a similar question in SS Krotov, but in that question the wedge was at an inclination α with horizontal, and the hexagonal pencil made an angle φ from the plane of the incline. It asked the value of φ for which the pencil just stays in equilibrium. That question was interesting too
Yesssss!!!!
That was the problem which brought me here!
Note that there will usually be some sliding if coefficient of friction isn't literally infinite. Also it can spontaneously transition to a runaway-accelerating hopping gate above a 24-27 degree angle, and maintain such a gate down to 14 degrees.
This means that the 14-27 degree region is bistable, or rather, has one stable state (rolling normally at a speed under 166 mm/s) and one unstable state (accelerating to an infinite speed).
Overclarified by the overthinking dude. You re an amazing explainer
Great Quality :) May u reach to millions !!
Thank you! I hope so
Man, the video is just MIND BLOWING. Every steps are so deeply explained.
If it's so well explained please provide the timestamp for the explanation on the 58% mentioned at 10:06
I actually saw this question a week or two and now I have the solution .
Thank You , please keep this coming
That's quite a coincidence, where did you see it?
@@overthinkingdude3462 In the 1998 Physics Olympiad paper . I was glancing through it and saw this question. Though it was out of my capacity to solve it.
@@rishinaik4326 Don't be scared. Try to imagine the problem in every detail you can. Try applying some common sense. And after that apply some physics laws. That's generally how I solve all the problems. If you really think you are not at the level you can solve this problem, you can start with something easier such as Irodov.
@@overthinkingdude3462 Thank You I do solve Irodov seldomly . Can you suggest a progression I can follow to be able to solve IPho level problems.
@@rishinaik4326 Back in high school I solved all Irodov mechanics and thermodynamics problems and would have solved electromagnetism if I had the time. Some people suggest Krotov instead of Irodov, but personally, I didn't even know it existed. Basically, after solving Irodov you should be able to progress to past IPhO problems. Start with the older ones since they are easier. And you can probably pick up some theory along the way. :)
beauty of physics cannot be ignored
Sir please make a video on the mathematical derivations concerned with the amount of energies lost. One of the best videos I've watched so far!
Wait... where did 58% come from?
I had the same question. He hand waved over the most important part.
180/pi, probably
This is so awesome, your explanation is truly easy to grasp. Would love to see how you found out how much energy was lost each time.
Brother love from India
I am very amazed to see this that's the reason why I love physics ❤ Thanks for making such a video
I enjoyed the presentation. Good to hear the physics reasoning and see the mathematical analysis. Combined, they provide a reasonable explanation of what we experience.
Unfortunately, the problem and solution links don't work for me.
Very good quality... InsAllah,You can go further more.Doing such a good job.Love from Bangladesh.
well, i found 3b1b's brother
Thank you! I use the animation engine that he has developed. It's true that I'm quite inspired by his work!
@@overthinkingdude3462will you return to making videos?
Man everyone's getting recommended after he stopped uploading for 3 years.Its so random
Why the energy lost in collision is 58%??
He says its somewhere else but I can't find it, its effectively the "bounce" as the pencil hits the table on the flat side which comes down to the materials of the pencil
4:26
Due to conservation of energy, with perfectly elastic collisions, the pencil should never stop rolling down no matter the angle if the pushing force is enough to tip it over once.
not if there's friction, which is needed for it to roll in the first place.
@@untruegamer8079 I do understand friction prevents sliding but in a ideal scenario it should not consume energy from a rolling non-sliding pen. Just like how ground to rubber friction helps the a wheel roll and does not slow it down (again, in an ideal scenario).
@@untruegamer8079 Friction does not count as an energy loss unless the pencil is sliding. Which it isn't. Energy equals force times distance, without any distance moved you don't have energy loss.
You are exactly right, but the guy doing the video just pulls 58% out of his ass with no justification. The pencil should theoretically keep rolling even on a flat surface.
Well, wood's not elastic.
I remember trying to solve this problem :) This brings back memories! Nice
10:30 I don't understand where the 58% comes from. Does anyone know? Still, nice video.
It's a really good question...me being in high school have learnt the concept of toppling in chapters like centre of mass and rotational dynamics..which were arguably helpful here
I am a neet aspirant even tho I can understand by your simple explanation....
And I did completed the rotational motion chapter so I was connecting the chapter with the video..
Thank you❤
More reasons for why this first level of consideration is incomplete:
* the materials in the game are plastic and elastic, so a pencil tipping over might see a noticeable return of energy.
* the pencil not only has forward driving momentum, but it also has a rotational momentum (and this can be very strong). so in effect it can be the case that it will only shortly tip with its edges to the surface until it "floats" in air for the next edged to find contact. in other words, it does not necessarily fall down fast enough to just have at all any sort of flat contact with the surface. also air trapped might help it to get lifted quickly to such a "hovered" position. and guess what? a hovered motion will not have any sort (or at least much) of lifting up energy included - that would be able to stop a pencil from rotation. so the overall run length might be much increased.
rationale: there might be much more in play that impacts on a real world setup. dont trust things to behave in a way the still simple formula above would predict.
I want to know how to find the exact amount of energy (In this case 58%) that is lost when the pencil collides with the surface while rolling down.
i tried to solve before watching, and ended up thinking only about how to get the answer to this energy loss without calculating the rotational inertia of the hexagonal shape around its center and its edge. To somehow get the energy loss on impact. I disn't get there in a few minutes and watched to find out. And then he just gives a number 😢
It's totally OK to include mathematical analysis in the videos too.
Please upload more videos! Subscribed to your channel from 3 devices already,mate! Consistency is key!!!
Amazing vid. I wonder tho... how the idea of rotational kinetic energy would affect this entire calculation. I reckon KE you give to the centre of mass is not just doing work to overcome the peak of potential energy and energy loss during a hit. But also to give the pencil it's rotational kinetic energy.!
Such a high-quality video and explanation!
Love this video! Question though, where did you get that 58% loss?
Thank you for the video.. i was just going through some ipho papers and i saw the same problem but i didn't tried because i was developing my problems solving skills still developing .... But this video was great it was not very tough just didn't understood how 58% energy got lost after the impact🤔🤔.... Thank you anyways and please upload more 🙏
Why is he using benzene in place of pencil ? Chemistry is everywhere. 💀😞
Bruh
Home-boy here pulled a gangsta 58 out of the back pocket and managed to smuggle in a calculator at an Olympiad. 10:30 . 17:13 .
Show your work or get out.
Also, solution "assumes" stuff at the end like Friction, drag, density. Give solution considering those variables or can I just "easily use a calculator" / LLM and send it?
We need more of this and less of the dumb stuff we get suggest from UA-cam. Great analysis!
Before continuing the video I wanna write my initial guess of “anything greater than 30 degrees”.
A hexagonal pencil has interior angles of 60 degrees on each angle. The pencil would perfectly balance on an edge at the center point of the angle, which is 30 degrees. At greater than 30 degrees, the pencil will no longer be stable, and fall towards the steep side. This will result in the pencil being unstable balancing on the next edge, causing it to fall to the side again. This would make the pencil role.
Overthinking dude? Perfect name for me. Subscribed.
Reminds me of one of my undergrad mechanical engineering courses about 60 years ago in kinetics and kinematics.
There's something no one took into account.
What material is he pencil made from ? And the table ?
Depending on that, the pencil will roll before or after it's tipping point.
So this exercice is void.
Thanks to physics, you can save on brain power 👌🏻
I want to see how you calculated that the pencil lost that procent of kinetic energy while he's falling down from his highest point, make the full-of-math video, please!!!
I really like your video, thank you very much!
Off course make the math for that, that’s why we’re here, also to be indulged by Mathematics equations, please 🙏
where did the 58% loss calculation come from?
The calculation was not included in the video, but it is shown in the links. If you follow the link for "The solution" you will get a 404 error, but if you google the filename 1998_Iceland_p1Sol.pdf , you can see the full calculation. The video just skips it and pretends to have shown the full solution to the problem. Inattentive listeners believe it.
58% is just an approximation, the exact value is 168/289, which can give you a clue about to the complexity of the problem.
Seriously your curiosity is just like anything
The first thing I thought when I saw this thumbnail is, the pencil is not perfectly hexagonal.
Very awesome video. You really took a complicated physics problem and animated it into real life. I would say it is sometimes slow. Mainly when it comes to repetitive simulations like 13:30 and 6:30. If you say "let's track the center of mass" it should not be a painful wait for the center of mass to be tracked. Animate then explain or animate while explaining. This channel will be golden.
Thanks for the feedback!
The animation quality is very good.
How did u get 58%??
keep going dude! sincerely, an interested physics university student :D
Not sure I understand you right, but don't we need to have both inequalities
theta > 6.58°
K_i > K_min (K_i = initial kinetic energy)
for the pencil to at least start rolling and keep rolling?
There's a bit of an issue here.
We are assuming that thehexagonal pencil is "walking" when rolling. That is, there is constant contact with the ground, and briefly there are two separate points of contact once every circumradius of the pencil.
However, this will only be the case for a pencil that can make that sharp of an arc. That is, one where the vertical launch speed of the pencil can be cancelled within the time it takes to roll. Otherwise it will transition to either a very efficient running gate, or a less efficient bchaotic hopping/bouncing gate.
The vertical launch speed of a pencil in a walking gate must be enough to reach 0 velocity at an apex of 0.134 circumradii under gravity within half a circumradius of horizontal travel. If we make the slightly wrong but helpful approximation that way the transition point the cycloidal arc is actually parabolic, for a 3.5mm radius, we get a time to apex of slightly under 9.8 milliseconds.
This caps the step frequency of our walking gate at about 1 step per 19.6 ms, or a speed of 3.5 mm per 19.6 ms, or 178.9 mm/s.
However, remembering that our steps are in fact cycloids and not parabolas, we end up with higher vertical velocity on every launch than necessary. This means that we will get too high an apex once our vertical speed ever exceeds approximately 96 mm/s. Which will actually happen at a speed of 166.3 mm/s.
What we see is that we will transition to a running or hopping gate at somewhere around this speed. Basically once we reach about ~950 rpm it will either jump around like crazy or roll on the corners while being fully airborne at least some of the time. Which will actually occur is another question.
It turns out that the efficient running gate is unstable and will always transition to the chaotic hopping gate. This will nonetheless lead to runaway acceleration above about a 14 degree angle.
Roughly 58% is because the angular momentum is dependent on the pivot. The pivot changes when the next edge hits the table, so we "lose" a bunch of angular momentum. I think.
in a perfectly rigid environment with 0 friction, to my undesrtanding, even on a 0 slope surface the hexagon will continue rolling, as its not losing any energy to heat, and just bounces back as high as was required to initialize the spin. i think delving into more detail as for the friction model would make sense. just saying theres 58% reduction on impacte assums the environment is not perfect, in which i would like to see where it is stated in the question you are answering.
17:36
Well it says there is static friction, so it’s not a 0 friction environment.
Friction does not cause the pencil to lose energy unless it is sliding.
Friction does not make the pencil lose energy unless it is sliding. Energy = Force * Distance. No distance sliding means no energy is lost due to friction.
static friction still needs to be explicitly defined by the question, i am asking for the definition of friction conditions, where static friction would be one of those.@@-ZH
really interesting approach to solve the problem, it's a great video. ♥️
Except the most handwavy assumption - the 58% lost in each bump - was totally glossed over. I don't even know how I'd approach figuring that out, I would probably give the answer in terms of some constants like the coefficient of restitution kind of analolgue and maybe make an estimate based on a few "what if" values.
Amazing video! I'd love to understand how the recursion equation is solved by hand as I'm guessing graphing simulators wouldn't have been allowed at the Olympiad? Thanks!
3 years late, but we need more videos from this channel.
2 minutes ago
The pencil colides with the pad in each step. If the colision is elastic, no energy is lost. That means, 58% is some serious guesswork. Also, the mass distribution doesn't matter, contrary to the claim at the end of the video, as long as mass is not movable and the center of gravity stays the same.
Alright I got 4 more years! So let's get started and Man You're Awesome Keep Up The Good Work !
Actually I don't like physics but u r truly good in explaining great work ull become awesome one day like 3b1b
This gotta be the best compliment I have received for my work yet! :)
Nice problem. Still trying to figure out where the 58% value comes from.
Love it ❤thanx dear
Do u use Manim for animation ?
Quality is amazing, just a thing- don't relly on simulation observations.. try using the things a student can use whilst in an exam, use simulation to depict already proven things by calculating.
For example - You said- for smaller angles like 5° you can see it stops.. well not without a simulation I can't see just like that.
you forget friction, otherwise the pencil slides
I guess it's presumed that friction is sufficient for the body to not slip
17:36
That's why their not made round, also they save some wood shavings which are reused
Physics problems man.... all the chickens are spherical and in a vacuum.
Thank god I found this channel
其实没有那么复杂,
几何学定律:三角形的内角和是180度,N边形的内角和是360度,
铅笔:(360/6)/2=30度,
如果是立方体(360/4)/2=45度,
如果是等腰三角形,(180/3)/1=60度,
长条形的东西是(180/2)/1=90度,
To find the minimum angle, we must equate the torque by friction with the torque by gravity
firction is a variable force so its not really a way to get the critical angle. the more precise one is, normal force will shift to the tipping vertex, so friction and normal will be about that vertex, and their torque will be zero about the vertex, so only torque acting would be gravity.
Every time when there time to aware our mind for this aptitude,we just say HARD,, how we making fool....
Dude come back the algorithm is finally there!!!
Amazing video my dude. I want more from you.
In other words, to stop the pencil from rolling from a slight impulse. We made it a triangle.
ok but where do we get the 58% energy loss on each step? WHERE DOES THAT NUMBER COME FROM????? how do we calculate that number?
Nice content brother 👌👏👍 continue and you will definitely be successful some day.
I subscribed
To invest in the future
great video! what software/library did you use for the animations?
Probably Manim, the maths animation engine developed by 3B1B
At the 3:00 when the center of gravity is directly above point O, depending on the static friction, wouldn't the pencil start to slide? I know this is not the question being asked or the point of the video, I just thought it was an interesting argument that could be made for having a lower desk angle to maintain rolling if the rolling was initiated by external force.
A great content from a great channel
Shouldn't this simply be MGH to MVV/2? Potential to Kinetic Energy? Unless you're making allowance for friction, thermal losses, air resistance etc, then it's no different from a pendulum in space - What it gains in kinetic energy from falling is sufficient to raise it's CoG to the same height again. That being the case, the angle is anything > 0. The shape should be irrelevant, whether a square, hexagonal, an oval or any other shape. If the collision with the surface does not lose energy, then any angle will result in acceleration, not stability. If potential energy is lost and the only place for it to go is kinetic energy, I think you may have overthought this one and added formulaic components that are not relevant to the calculation. Without losses, a round pencil and a hexagonal pencil of the same mass and rotational distribution on average should start rolling at the same time, and reach the same speed at the same time, though of course you should get a pendulaic result where the shaped pencil is sometimes winning and sometimes losing...
Energy cannot be created or destroyed. Exchanging potential energy for kinetic energy means that the speed is the same in both the case of a round pencil and a hex pencil, and both will roll forever if there is any vector of thrust assuming no losses.
yay! found the 3B1B for physics
8:20 yess we want
Why is pure translation not happening??
High roughess is assumed, I think.
Nice video bro!