Will a falling pencil hit the table? We do the maths!

[@daveayerstdavies]

Would be nice to know why downward acceleration due to gravity is not a factor in that calculation.

[@orlandonerz2999]

Yes, in space it would never hit the table

[@Jason-ke4jf]

@@orlandonerz2999 I don't think so. It hits the table because it rotates, not because it falls. It all happens so fast that the fall of the pencil due to the force of gravity should be negligible.

[@bloodspatteredguitar]

They did the working on the assumption that the tip touches the table. Working out how hard you need to hit it to get that result would, I assume, require g.

[@Macieks300]

Yeah, I don't get it too. For example if you take your formula v=(1/3)*a*ω and ω=v/a from the definiton of ω (where radius of rotation r=a) then you get 1=1/3 which is a contradiction.

[@Mmmm1ch43l]

@@Macieks300 v is the velocity of the center of mass, not the tip, which is what pops up in the definition of omega

[@adamsbja]

"The bluetack doesn't affect the physics." *Pencil fails to move because of bluetack*

[@TheMajorJF]

I thought the same thing. For the most part, the bluetack can be approximated to be a constant reduction to the initial impulse. So, the i that is used is defined as i ~ i(initial) - b(luetack). That is how you can get a no movement impulse.

[@Danilego]

They could also just use a pencil that's flat at the bottom

[@cigmorfil4101]

The blu-tak affects the friction of the initial contact, and how that friction changes once the impulse has been given - unlike the "normal" pencil-table friction, blu-tak has a flowing ability: notice with too little impulse how the pencil ended up leaning slightly.

[@ciarfah]

@@Danilego I assumed they had until he mentioned blue tac

[@marceelino]

What is bluetack?

[@albertemc2stein290]

12:25 is the most perfect hand drawn Sinecurve I've ever seen.

[@bibektg]

Have you been to any math class above 10th grade lol , it's not THAT impressive

[@albertemc2stein290]

@@bibektg Of course... but in university the quality of hand drawn curves drops rapidly

[@Gabu_]

One of the classic party tricks of an engineer.

[@ViratKohli-jj3wj]

@@bibektg Ok Boomer

[@FLScrabbler]

Something maths-liking Zen monks pratice, no doubt..! 😉

[@pyrobryan]

Matt: Does it change the physics? Hugh: No. Pencil: Doesn't move

[@Karttrak2]

Whoever photoshopped and animated the scanned notes did a great job! It was much easier to follow along line by line rather than looking at a completely filled out paper full of formulas and drawings. And I didn't even notice that other lines/shapes would overlap what was already written until I saw the animation, rewound, and scrutinized the "past" version of the paper. Beautiful!

[@whoeveriam0iam14222]

"why are all the pencils broken inside???"

[@janhetjoch]

Why am I broken inside?

[@wierdalien1]

Why is the rum always gone.

[@rich1051414]

I could only afford cheap pencils as a kid, and nearly all of them were broke about 4 times inside.

[@alansmithee419]

@@wierdalien1 hide the rum!

[@Real2Reaper]

@@janhetjoch (left ear) aren't we all?

[@xgozulx]

4:00 just some ducks drinking in the background

[@NickStallman]

It's quite a flock of drinking birds!

[@badlydrawnturtle8484]

@@NickStallman One drinking bird is a novelty. Four is a statement. I'm not sure what it's saying, but it's saying something.

[@kennethflorek8532]

The engineer is harvesting free energy using dunking-duck novelties. He's on the threshold of revolutionizing the energy industry, and overthrowing the pesky law of conservation of energy.

[@taysem321]

@@badlydrawnturtle8484 i know what it's saying. "I zarking love drinking birds!"

[@dogdrovenorth]

It's all go in the kitchen.

[@alessi4249]

What a brilliant way of showing the working out on paper very clean! Must have took some editing skills, Brady would be proud lol

[@standupmaths]

Glad you like it! Was only because we lost the first GoPro file.

[@olipolygon]

@@standupmaths Honestly, I kinda prefer the way you did it here! It makes the notes that much easier to look at

[@GetMeADrinkNow]

@@standupmaths yes, that looked great! How was it done? Video editing or some other way?

[@d1amxnd]

@@GetMeADrinkNow he probably scanned the sheets and then edited it the way it looks.

[@ditzfough]

Miss the brown paper and sharpie

[@Tumbolisu]

sin θ = 1/3 θ ⇒ θ = 0 :) Cool how the initial state is technically a situation in which one end of the pencil is "hitting" the edge of the table.

[@sergey1519]

There is also a negative solution

[@aleksandersuur9475]

@@sergey1519 well if you were to hit the pencil from the other side the tip could also hit the edge of the table so yes, there is also a negative solution.

[@squidcaps4308]

I am deeeply disappointed for the pen length being 'a' and not 'b'.. "The pen is 2B long" would've been a perfect pun..

[@jrbleau]

He was wondering if was worth it just for a pun. "2B or not 2B, that is the question", and he decided it was not meant to be.

[@bjarnivalur6330]

I wonder how meany people would get it

[@jw41538]

@@jrbleau nicely done

[@Vgamer311]

Bjarni Valur idk, but put me on the list of people who don’t.

[@bjarnivalur6330]

@@Vgamer311 I don't remember it completely but B and H mark the "hardness" / "softness" of the pencil lead

[@LowPolyPixel]

Matt Parker, answering the questions we never thought to ask.

[@zym6687]

3:22 "there's a center of mass somewhere in the very middle of the pencil" Clearly not as the topside is sharpened and given a sharpening angle of 75 degrees on a typical 18 cm long, 6mm thick pencil would result in the top half being ~13% lighter than the bottom half. As such the center of mass on such a pencil no longer coincides with the 9cm midpoint and is actually only ~7.8cm from the surface of the table.

[@goatmeal5241]

Good point, and that might actually influence the result, since the tip-side is longer (i.e. farther from the center of rotation).

[@DoctorRedstone72]

they also dont include the 3rd dimension, they will always hit it at a little bit of an angle, causing the axis of rotation to not be parallel to the edge of the table. im sure there are other things, but thats not really the point.

[@RodelIturalde]

This is physics which means an abundance of approximations. Such as, centre of gravity is in the middle of pencil. That the pencil is rod. That the radie of the rod doesn't matter, only length. Even though we know that for a very wide rid, we should treat it as a cylinder instead. We can use pi=3 aswell if needed (though not here).

[@f.n.2649]

This is actually needed for the tip to hit the table. If the center of Mass was in the middle and would only move horizontally, the tip will never hit the table.

[@GregorShapiro]

@@DoctorRedstone72 And they do not hit it exactly at the bottom but at least half a pencil diameter above the bottom.

[@porkeyminch8044]

Finally another episode of "Matt and Hugh play with a thing and do some working out" I'm still waiting for a followup on the 3 sided coin video, though.

[@jeffreyblack666]

A fundamental issue with the three sided coin, which I believe they didn't cover, was if you want a 3 sided coin, or a 3 sided die. The 2 are fundamentally different as you roll a die put flip a coin.

[@cigmorfil4101]

@@jeffreyblack666 I like the way you can normally ignore the possibility of the coin landing on its edge, but size and density can affect this probability as Ian Stewart found in his Royal Institution Christmas lectures in 1997 (The Magical Maze) where a volunteer flipped a large pound coin and it ended up on its side. But then again, having seen Matt's video about probability where (I think/suspect) he did many takes to ensure the result he wanted with regards probability, I'm beginning to wonder how many takes Ian did, but his look of surprise does look genuine.

[@HonkeyKongLive]

Matt's sheer elation at 1:39 is my favorite reaction to anything ever right now.

[@karljohnsson1531]

Two grown up men happy like a couple of children :)

[@thejiminator8816]

I love these videos with Hugh. Hugh actually reminds me of Arthur Weasley (Ron's Dad) of HP.

[@helloarigato]

I cannot get enough of you two together! More Matt and Hugh please!

[@chchaitanya]

Sure, I can try. I have exactly 72 pencils lying around from a project I am yet to do :)

[@lombre9149]

chchaitanya ahh good video i remember

[@jasonharding9490]

I love how the table has three legs, it is clearly made for that job (or similar).

[@wompastompa3692]

"You wanna see a magic trick? I'm gonna make this pencil disappear."

[@robinisomaa]

Matt "Close Enough" Parker

[@1stGruhn]

It never ceases to astonish me that you can scribble on a piece of paper some numbers and come up with a fairly accurate forecast of what happens in reality.

[@konstantinkh]

Back when I used to teach physics labs, I would always do a setup with a spring-loaded cannon. Ask a student to place a cup on the other side of the room from a spring-loaded cannon, measure the distance, have the class work out the angle from range formula given initial velocity, aim the cannon based on the result, and have a ball bearing travel across the class room, landing perfectly in the cup on the first try. They all know the formula. They've used it in the class a bunch of times. But for some reason, they still look at it like it's black magic. It doesn't matter how much you learn the subject. The idea that numbers can explain reality seems to be very hard to accept on emotional level to a lot of people.

[@koenth2359]

A very nice experiment! However it may leave the viewer with some questions: ▪Is the correct formula used for moment of inertia for a rod rotating about the centre of mass (which is I=ma²/3), or should we use that for rotation about an end point (which is I=4ma²/3)? [Note that a is the *half* length of the pencil] ▪What is the role of gravity? ▪What rotation speed do we need so that the tip does not touch the table? Let's do the physics a bit more rigorously, avoiding the formula issue altogether. We assume - the pencil to be an ideal rod (stiff, uniform mass distribution, no thickness) - it is hit at its very bottom point - no air resistance - gravity does play a role After it is hit, the motion of the pencil will move in such a way that 1. the center of mass (com) will move uniformly in the x-direction 2. the com is accelerated downward due to gravity 3. the pencil will uniformly rotate in the xy-plane When talking about angular momentum, we have to be very clear and careful about our reference point. In the video the com is just taken as a reference point, the motivation is not made very clear. Below I we will consider angular momentum with respect to the origin, which will be at the point of contact (which is at the table edge, where the pencil is kicked at its bottom). The positive x-axis is horizontal (away from the table), the positive y axis vertical, upward. Positive rotation will be anticlockwise. Angular momentum ( *L* ) of any particle at a location *r* and and a momentum *p* is defined as the vector product *r* × *p* . For an object we have to sum this quantity over its constituent part(icle)s. In our reference frame: The pencil at rest has *p* = *o* for all of its parts, so it must have *L* = *o* . The blow or impulse (= transfer of momentum) has *r* = *o* , so it must also have *L* = *o* . So, because of conservation of angular momentum, we know that the combined motion of the pencil must also have *L* = *o* . Immediately after impact: The com will be moving horizontally to the right. This gives rise to angular momentum of L=r×p=-amv (clockwise, so negative sign!). The rotation of the pencil itself will be anticlockwise with respect to its centre of mass, which constitutes an angular momentum of L=ωI. So we must have amv=ωI. We can also integrate total angular momentum for the rod. Using v=p/m-(y-a)ω, dm=ρdy, ρ=m/2a, and dp=vdm we write 0 = L = ∫ r×dp = ∫ r×vdm = ρ ∫ r×(p/m-(y-a)ω)dy Directly upon impact motion is all horizontal and r=y is vertical, so that the vector products line up. 2a = ρ ∫ y(p/m-(y-a)ω)dy = ρ (½(2a)²(v+aω) - (2a)³ω/3) = 0 which simplifies to v = aω/3. 0 Now aω is the rotation velocity of the tip about the com. This way we find the motion of the tip of the pencil: x(t) = vt - a sin(ωt) y(t) = a + a cos(ωt) - ½gt² Setting v = aω/3 and θ=ωt, indeed the x-coordinate of the tip of the pencil is 0 when θ/3 = sin(θ) or θ =2.2789. No matter how hard we hit the pencil, as long as the tip has y>=0 (otherwise it would have hit the table beforehand). Yes, we arrive at the same result. Back to Matts question @2:18: 'If you hit it so it just, just nips the end, what angle is the pencil on?', but later he says 'So we dont care how hard we hit it.' We should care, because it is fun. It just touches the table edge when y=0, too. What is the condition required for this? Set y(t) = a (1+ cos(θ)) - ½gt² = 0. We get t = √( 2a (1+ cos(θ))/g), and ω=θ/t As a realistic estimate: for a 0.15 m long pencil, t=73ms, and it needs to make at least 4.97 rps.

[@mrphlip]

The next step, now that we have a value for theta would be to calculate y = -a cos theta = 1/2 g t^2 and substitute in all the previous equations and end up with an formula for the impulse which would result in the pencil exactly hitting the corner of the table. And then go one step further and change it to an inequality: -a cos theta > 1/2 g t^2, and end up with what range of impulses will result in a table hit, and what range will have it miss entirely. Left as an exercise to the reader, naturally.

[@bdr420i]

Loved the flock of drinking birds 😂

[@aidanoleary6514]

The close enough graphic at the end is the best part

[@serkif1]

I just finished taking calculus-based physics last semester. It's interesting seeing a video on some topics we covered in class!

[@BandanaDrummer95]

For people complaining about gravity not being factored in, it's in the assumption that the pencil just barely touches the table. Working from that, you can find that it then brings in the length and mass of the pencil in order to find out what impulse is required to make that situation happen. Having done that, the impulse required is: I=m*sqrt(ag(1+cos(θ))/2)

[@toastedcheese50]

Interesting that the experiment would work equally well on the moon or Jupiter. In space the pencil would just spin, but the tip would align with the edge of the table at the same angle. As to why 1/3ma^2 This is the inertia around the centre of mass since the formula is 1/12m m(2a)^2=1/3 ma^2

[@alfiechenery4146]

Think it’s worth noting why it only works in radians. Because it seems like an arbitrary restriction. Angular velocity, omega, can be thought of as the speed at which an angle is covered. This is (2 x pi)/T where T is the time period, the time taken to complete a full rotation of 2 x pi radians. By using this formula they have implicitly restricted the equation to only work in radians. It is convention that an angular velocity can only be denoted by omega, if it is in radians per second. However if we used a different angular velocity which we defined as 360/time period, then our later step involving theta would only work in degrees and not radians. So it isn’t that the formulas only work in radians, it’s that you must be consistent with degrees or radians throughout the method, and the method they used, implicitly took angular velocity in terms of radians not degrees.

[@mikefochtman7164]

True, but then wouldn't the formula for finding the moment of inertia also have to be modified? A particular impulse against an object of a given moment of inertia must cause a particular angular velocity. If you measure angular velocity in degrees/time, then the moment of inertia must be in some compatible units. Instead of 1/3ma^2, it might have to be something like 120ma^2/pi

[@alfiechenery4146]

Mike Fochtman possibly, not gonna lie in my a level course in further maths and in physics we don’t really do inertia so I low key don’t fully understand what inertia is, I’m just accepting the formula as true. But we do circular motion quite heavily and I know it works in degrees but is conventionally in radians. My point still stands that degrees would work if you were consistently in degrees throughout, but you might be right that inertia is also dependent on keeping angle measure consistant

[@rpyrat]

What amazes me is how simple of a reasoning you used to solve this. I was thinking parametric curves would be the way to go but this is much more efficient. Well done!

[@reznovvazileski3193]

Now do keep in mind velocity is a vector here so while it may not be a matter of impulse once you assume the 1 case where it hits the table perfectly, it's still very much a matter of angle. If you shoot the pencil under an angle alpha from the centerline of the table with the same impulse, the pencil will still go the same velocity obviously because we know for sure the mass is a constant (unless you break the pencil ;) ). However, the useful parallel component to this centerline that's effectively moving away from the table will be shorter. This means you'll have to introduce a higher impulse to fire at a higher velocity so the pencil can still barely hit the table. That will increase Omega, so I'd expect that changes the angle Theta at which the pencil ends up hitting the table perfectly. When we consider an experiment in the 0

[@munjee2]

I would've clicked on this video 40 minutes ago had I know it was an episode of Matt and hugh play with a thing then do some working out , but alas that was absent from the title yet again

[@CJonestheSteam72]

I would have clicked 4 hours ago except work got in the way again

[@michaelslee4336]

Add blue tac “Does it change the physics?” “No” Well strictly yes, ok I’m being pedantic.

[@TheZoltan-42]

Very _stickly_ yes :)

[@NovaOnce]

Being pedantic, the physics remain the same in our lifetime, regardless what you do to this pencil. Physics won't change :)

[@petertaylor4980]

Hugh Hunt's an engineer, not a physicist. For practical purposes...

[@hikmatullahpakhtoon3694]

this one video has more information then one whole chapter of mechanics. thank you for this video.

[@donerickson1954]

Would also depend on how high it is struck and how "stuck" to the table it is? Both of those would affect the angular velocity. Could you hit the pencil at the perfect height for there to be no angular movement?

[@maxihihi]

Don Erickson if you hit it in the center of mass there would be no angular momentum. They did their calculations for hitting the Pencil at the end, otherwise the would need to rework the formula for applying force to the pencil

[@adityak1231]

I think the moment of inertia is 4/3ma^2 because the total length of the pencil is 2a at 5:17

[@Bigger-Than-Jesus]

i personally think the sticky stuff makes the whole test inconclusive because when you strike the pencil you arent just overcoming gravity. You are overcoming the sticky stuff first. However weak or strong it is. Also you are having way too much fun. This reminds me of the 'magic' trick where you yank a table cloth out from under a set of plates and glasses fast enough and they dont come flying off the table

[@thebigmacd]

The sticky stuff just reduces the impulse effectively applied, which can be rectified by increasing the total impulse applied. The resulting angle will be identical.

[@alternativeduck9920]

I definitely prefer the animated scans. It conveys all the necessary information without an odd camera angle and unnecessary background details.

[@otakuribo]

Somewhat counterintuitive result with a surprisingly intuitive explanation!

[@JFra472449]

Hugh Hunt: Excuse me while I do calculus by site. God I have forgot so much since I got out of school.

[@gabor6259]

It's always good to see 2 grown men playing with their pencils.

[@SuperVstech]

Gábor Králik no... not always...

[@MrL314]

That is the most beautiful hand-drawn sine graph I have ever seen

[@MiniBetrayal]

I’m curious to know if there’s a second valid solution where the pencil is hit just very slightly, so the bottom is just nudged off the edge of the table. The pencil would drop further and have an angle less than 90° when the tip clips the edge

[@pontifier]

Yes, wondering why they missed it when drawing their solution. It's clear from their drawing that the line 1/3 theta intersects sin theta at zero. There is also a negative solution.

[@SuperVstech]

MiniBetrayal nope... because the blue tack prevents that scenario... the pencil just sticks to the table.

[@apathyisFTW]

Good catch! This second hit would come out in the wash from the quadratic nature of gravitational acceleration if they actually tried to answer the question in the title of this video. On a side note, if you tried to do this in a 0g environment, the pencil would not hit the table.

[@nitfumble]

Oh man, I do love a video where I can watch at Matt's face in 4K in slow motion!

[@cschmi9624]

I find it awesome that there is a group of people on earth that find it fun to watch people do math. Its a good sign for the future of the earth.

[@isoul22]

Got humble pi for my b-day loving it!

[@marcoronzani7197]

The formula can be improved: if the pencil hit the table not right on the edge, but slightly before it, we can call k the distance from the edge of the table to where the pencil hits, and we can do your exact same calculation, but with x+k being used instead of x, since now x is smaller and we need k to use sin!

[@iagocasabiellgonzalez7807]

4:00 that flock of drinking birdies in the background lmao

[@JavierAlbinarrate]

1:48 Beautiful tower clock in the background! With gravity escapement! wow... a real jewel.

[@landsgevaer]

For all the people who actually want to know the answer to the original title question "will a falling pencil hit the table" that was left dangling in the video, or - more precisely - how strong gravity has to be in order for the pencil to hit the table: the critical (minimum) value equals g = 1.2118523026900971579476924133.. * v^2 / a Where v equals the imparted center-of-mass velocity of the pencil (v=i/m) and a equals the half-length of the pencil.

[@Richster64]

Excellent - gotta love how things cancel out and that implies certain physical predictions. Maths is great!

[@1prshark]

I was actually in the mood for a standupmaths video today and I was rewarded!

[@standupmaths]

Wish and you shall receive.

[@LHJC10]

his hand drawn sine wave is amazing

[@rewolff2]

Gravity comes in as y = 1/2 g t^2 . Then there is another equation involving sin or cos that tells us that the tip of the pencil is at the table height. From there you can calculate the time when this happens. And then you can calculate the X position where the pencil hits the table. When that x position is less than zero the tip hits the table, if not it flies off. On the boundary you get what you guys calculated. Now the impulse and g matter, only when you assume it precisely hits the corner of the table, then g doesn't matter and the impulse has to be precisely the value that works to get that. Just for fun lets check a few boundary cases. Hit it with impulse epsilon (so small you can ignore it), you just knock the pencil of the table and it simply starts falling vertically. The solution here has a theta on the order of epsilon and in reality the pencil will hit with its side to the corner of the table. In the mathematical model the tip is at the level of the table at y = 1/2 g t^2 = 2a. That should be a solution for very small impulse.

[@flakmagnet9357]

Fun follow up question: What is the minimum velocity that you would need to hit the pencil so that it does not hit the table?

[@kaushikiyer9445]

Its the same as the velocity needed to just hit the table. A 0.00001 increase in that velocity will make it not hit the table. Hope im making sense

[@danielmalo1753]

I love how Matt asked Hugh at the end to just hit as many pencils off the table as he could

[@davicorosello1588]

The pencil (his varicenter) describes a parabolic "falling" outside de table that depend of the initial impulse => horizontal force applyed to de pencil. That initial force is obstructed by the scotch tape at the base, because some of the force applyed get lose while breaking that friction and initiate movement. If you graph that parabolich equiation, dependant of the initial velocity/ force applyed, you can measure the minimun distance between the parabola and the edge of the table. Then, you can measure the time that toke thenpencil to get in that position, and calculating de amount of spinning in that period of time, check if the pencil hit the table or not.

[@freshrockpapa-e7799]

You mean barycenter? Not only you spelled it wrong, it isn't even what you wanted to say. What you mean is center of mass, barycenter is the center of mass of multiple discrete masses. There would be friction at the base of the pencil regardless, you just need more force, which doesn't effect anything that happens afterwards.

[@RobertKreegier]

I have no idea how they came up with what the did, but the whole thing is dependent upon the acceleration of gravity for one, and the ratio of the distance between the half length of the pencil and the distance between the midpoint and the point of impact. In a perfect system without gravity where the pencil has no thickness and the impulse happens on the exact end of the pencil, the other end of the pencil will never hit the table. When you introduce thickness into the pencil, you change the tangent that the impulse exerts upon the pencil, angling it into the table. Just the same, when you use a striker with thickness, you change where the impulse happens upon the length of the pencil, which means - because there is pencil above and below the impulse point - the mass below the striker acts as a counter-balance to the mass above the impulse point, which endows the center of mass with forward velocity. Add the acceleration of gravity and you have a three variable system (with one constant) that you need to film multiple times to match your hypothesis.

[@LaurentNittler]

If the bottom of the pencil is moving with the velocity v and the center of gravity is moving with v, the the pencil is not spinning at all. The bottom has to move faster ! Therefore I can not agree on the assumption x=v*t made at 9:56 , because this v must be smaller than the v at the bottom of the pencil. The ratio of the v of the center of the pencil and the one at the bottom depends on the location where the pencil is hit. (If you hit it at a distance "a" from the center then the center of gravity of the pencil should barely move in horizontal direction)

[@LaurentNittler]

Although the equation indicating between the angular velocity omega and the velocity of the tip of the pencil v (at 7:15) might be correct but the interpretation at 7:30 making a relation between angular velocity and linear veloity of the center is incorrect.

[@anthonycannet1305]

By calculating for equal you get the angle for when it hits by a point, if you use an inequality like

[@oskaramos5624]

1:44 so excited about flying pencils☺️

[@maximilianmuller5699]

Well, you're technically correct in what you did, but I was still a bit disappointed, since you didn't really answer the question that the title implies, which is "how hard you would have to hit it?". Instead you answered the question with "depends" and then asked "what would the angle be if it just barely hit?". I would also like to add that while I do understand how linear momentum and angular momentum are linked, you failed to mention that the given ratio heavily depends on WHERE you hit the pencil or how the mass is DISTRIBUTED in the pencil, which might lead to wrong conclusions for some viewers.

[@Kycilak]

It really is a Parker square of an answer.

[@kumoyuki]

and also depends on the modulus of elasticity (which can probably be simplified into some kind of bending moment), &cet. The impulse contains a fixed amount of energy, that is translated into kinetic energy of the pencil in motion which has 3 components: angular momentum, linear momentum, and a mechanical vibration (which is how the linear momentum arises from the off-center impulse). while its an interesting problem the answer is surprisingly weak...

[@TheBlueArcher]

I didn't have much prior knowledge, I understood the video, but I to am disappointed he didn't actually answer the question. I'm fine with clickbait to a certain extent, but if you have a question as a video title, and not "can we figure out" I expect a concrete answer =(

[@Anankin12]

@@TheBlueArcher well you can't have a precise answer anyway because you have to take into account how heavy the pencil is, how long it is, how the mass is distributed, where you hit it and how flexible both the pencil and what you hit with are.

[@TheBlueArcher]

@@Anankin12 yeah but I was expecting a general formula to answer it. Not a general formula that... Doesnt answer it.

[@Vares65]

May I also say: God I love nerds so much! You guys are so fun. Thanks for making these videos and sharing your brilliance with all of us. =)

[@NovaOnce]

I prefer video like this one, when you can clearly see the solving process and not only the results, as in a lot of popular blogs about math.

[@julienroux8413]

The value of theta that you compute is the angle of the pen when the tip is above the edge of the table!!! That is because in the equations, you only considered the horizontal and rotational component of motion. In this case, they are not influenced by gravity. Note that theta=0 is also a solution of the equation sin(theta)=theta/3 and indeed when theta=0 the tip of the pen is above the edge of the table ;) I think you can confirm this theory by looking at the slow motion footage when the pen is hit too hard and flies away without touching the table: the pen should make an angle of around 130.5° when the tip is above the edge! Touching the table or not only depends on how much the pen could fall before the tip was above the edge and thus depends on the gravity, mass of the pen and impulsion. Pushing even further, I think you could hit at any angle depending on where you hit the pen. There is however one thing that bothers me about the video: at the end, you only show one successful case. Coming from someone who flipped a coin 10 000 times, I am a little bit disappointed... Is there a reason for this? In the end, thank you for your videos, they are great and very instructive.

[@DeJay7]

If someone wanted to calculate the impulse needed for such an event, gravity would have an effect but IT would depend on factors like the mass and length of the pencil and the time frame. Because a pencil this size or even smaller really would experience significant change in vertical velocity, whereas a large rod (several meters long and many kilos) would experience a _relatively_ less significant change. Because gravity will only accelerate the rod so much in a specific time frame, whereas the rod can be too long and heavy and already have a very substantial speed that gravity will make little difference. And don't tell me "mass doesn't affect gravity's acceleration", it's the fact that as it gets heavier, we can afford to increase the impulse we cause and therefore increase the velocity gained, directly minimising gravity's role. Relatively speaking. I just wanted to feel like I did something, because it is a problem I could solve myself (as a physics student), it's not hard, but I didn't do anything, I just sat in my chair and watched the video, so this is my fairly useless contribution.

[@life42theuniverse]

1:30 vs 1:27 The differences between follow-through and no follow-through on a golf swing? The striking pencil makes a greater follow-through when it's normal to the table is in front of the struck pencil.

[@stephenbeck7222]

Timely. My physics class at school is testing momentum this week. Linear and angular versions. We look up moments of inertia in tables too.

[@terrysalava6981]

Now do it for a pencil that does not have an even mass distribution and show the derivation of the moment of inertia.

[@arrrg3846]

What sort of geometrical shape, or distribution of mass, would allow: 1) the bottom of the shape to hit the table, (almost 360 degree rotation) or (different shape) 2) whacking the top causes the object to strike the table?

[@JotoCraft]

Just my guess on the topic of how hard do you need to hit regarding gravity. It should be, that we need to calculate how much the center of mass needs to drop (y). Using this we can calculate how long this will take (t = sqrt(2y/g) ). Then we need to hit the pencil so hard, that it travels x in the calculated time (v = x/t). Now this can be put into the impulse formula (i = mv). So overall: i = m x / sqrt (2y / g) + (some thing to overcome the sticky tape)

[@ILsupereroe67]

I'm pretty sure the blue track makes a huge difference

[@nymalous3428]

Finally, a reason to use radians in the real world! (I'm talking about myself here, I don't ever need radians in my normal life.)

[@discoveryofphysics9303]

Awesome the derivation is just awesome sir

[@askepihl7610]

If you but the point of rotation at the center of the pencil the moment of inertia is 1/12*m*L^2. If it's at the end of the pencil the moment of inertia is1/3*m*L^2. The drawn hinge suggest that the rotation is about the center hence 1/12 and not 1/3?

[@KieranMetcalfe]

Does gravity not play a part? Or is it just a case of knowing at what anglr the tip of the pencil passes the edge of the table regardless of height above it? It was mentioned that speed does matter to whether it hits or not - even if it doesn't hit, is it still at the same angle as it passes the edge? Meaning that hitting the table is really a question of whether the speed was low enough to allow gravity to pull the circle of the spinning pencil low enough to intersect the table? From that, I assume its possible to calculate the speed required, and for a pencil of known mass, the force you need to apply?

[@Gustbear]

Animated scans! More of that! 👍🏼

[@IamCrusaderRUS]

15:37 sure, close enough, i guess...

[@The_J485]

An impulse can be over any length of time, with any force. Not just a big force over a small time.

[@ronaldramos8873]

The math and physics of the scenario honestly didn't matter to me--Matt's visible happiness and joy throughout made this video worth watching.

[@juriaanv]

For people wondering why gravity does not play a role, this is because you are assuming from experiment that the point of the pencil hits the edge of the table after some time. After this you can deduce the angle from the relation between the forward velocity and the angular velocity as shown in the video.

[@blindleader42]

But gravity plays a critical role. Consider the case where g = 0. No matter how much or little impulse is applied, the pencil can never hit the table.

[@juriaanv]

@@blindleader42 Gravity plays a role because it allows the pencil to drop to a position where the point can hit the edge of the table. However, regardless of what kind of downward force is applied or how strong it is, the angle at which it hits is the same.

[@legofan2284]

Confusing to follow the explanation, because "momentum" in German is "Impuls" and "impulse" is "Moment"

[@sehabel]

I actually had the exact same problem 😂

[@peter_castle]

I can see here that momentum in german is de.wikipedia.org/wiki/Impuls, howeveer, could you provide me, for my curiosity, with a reference or link to impulse being "Moment"?

[@LucaBl]

Isn‘t torque the equivalent of Moment in German though?

[@genau14zeichen]

@@LucaBl exactly. Talking about the pencil's moment of inertia, we would say "Massenträgheitsmoment", torque is "Drehmoment" and the second moment of inertia is "Flächenträgheitsmoment" So the answer is: the more you know about physics, the more ambiguous the word "Moment" becomes

[@zyaicob]

Danke, Deutsch.

[@hillaryclinton2415]

Started with the extremes..so hard it does not hit the table, then so soft it doesn't leave the table

[@gigaprofisi]

I like that the video length is the Golden Ratio

[@hikmatullahpakhtoon3694]

here they are only considering the horizontal component of the motion, since their applied force is along horizontal direction, so it is not necessary to include the vertical gravitional force and the aceleration due to it that is gravitation aceleration.horizontal and vertical compnent of motion under gravitional field can be delt separately without any flaw in the mathematical results.

[@puncherinokripperino2500]

The gravity must be considered, because without gravity it obviously could only touch the table at zero or flat (180°) angle as there is no vertical force. But your answer doesn't have g in it.

[@standupmaths]

Nice point! But we don’t need to explicitly include g because we assume it has fallen under gravity such that the top hits the edge of the table. You can see the centre of mass has dropped.

[@solakv]

At 2:27 Professional Mathematicians with a Professional Pencil Sharpener.

[@axelschafer2934]

The moment of inertia of a thin rod about its center of mass is 1/12 ma^2 -- not 1/3 ma^2. The corresponding angle then should be 166 deg.

[@SquidofCubes]

The moment of inertia is the variance of the mass from its center

[@KuraIthys]

A yes. I remember this stuff. Didn't get anything to do with angular velocity, momentum and so on until University level stuff though. I actually once ran into a university version of the Physics textbook I used in High School. It was pretty much completely identical but the High School level one was missing about 1/3 of the pages and many entire chapters...

[@koolguy728]

Would have liked a bit more detail about how you solved sin(x) = x/3 analytically. did you use the series expansion of sin(x)?

[@TheOskay00]

It can't be solved analytically, only by numerical methods like iteration or Newton Raphson.

[@bibektg]

Not native English speaker , but if I understand correctly "analytically " the guy did show what he did and thats pretty solid

[@TheOskay00]

@@bibektg The equation sin(x) = ax (for any non-zero constant 'a') has no non-trivial* analytic solution, meaning you can't write an exact solution like sqrt(2) or 3.23. But it can be approximated numerically. Most equations cannot be solved exactly but there is a rich body of numerical methods for solving equations approximately to some desired accuracy. These typically involve iteration, where each iteration gets you closer and closer to the answer but never reaches an exact answer (though it may appear to once your calculator/computer runs out of RAM). Hugh presented a graphical method, which is useful for determining whether or not a solution exists and for getting a rough interval over which a solution might lie, but more detailed numerical analysis (such as iteration, Newton Raphson etc...) should be used to obtain an answer. As Matt points out in the video, it's easy to eyeball an estimate from a graph if you already know the answer (because you've previously done numerical analysis or looked it up...) Edit: non-trivial*, since there is of course the trivial solution x=0

[@QuentinHudspeth]

Neat. Would like to see a follow up showing how the able changes as a function of gravitational acceleration.

[@ninjanayr]

I love how this guy is planting words so that we can think about something right before he says it just so we can feel smart too.

[@stuartmcconnachie]

5:38 It’s a moment of inertia moment.... boom boom! 15:47 Yep, he was a huge help!

[@fishwinker]

It seems that the answer (~135) shows that it all happens so fast that the centre of the pencil can't drop enough dur to gravity to make a difference to the calculation in this instance. If you assume that the centre of the pencil will not fall a measurable distance, it looks like it becomes a simple question of right angle triangles with the two sides equal to half the length of the pencil.

[@NotHPotter]

Really appreciated that moment of inertia moment.

[@juliangorski5760]

1:53 love this shot

[@xgozulx]

the blue tack actually would make it spin less and have more forward momentum, if you did't have the blue tack you could hit the pencil just at the bottom and make it spin without going forward. of course then it would fall but you know

[@IAssemble]

Yes, the friction of the blue tac will increase the forward momentum transferred to the pencil but no, the pencil would still have some forward momentum as well as angular momentum even if there were no friction (i.e. without blue tac). That is what the calculations showed in the video.

[@stevenmellemans7215]

I don’t think so. The friction force is opposite to the impuls force so the resulting net impuls force will be lower but it should have no effect on the relation between the rotation and linear velocity as calculated. You just have to hit harder to compensate for the friction.

[@IAssemble]

@@stevenmellemans7215 of course you are right! If the pencil being used to provide the impulse is negligible in diameter so it hits the point of the blue tac that is in contact with the table and the blu tac immediately loses contact with the table after the impulse is applied ;-)

[@ahmaran4894]

Really interesting topic. Keep it up Matt.

[@afifakimih8823]

Thats beautiful!!