@@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.
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.
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.
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.
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.
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!
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.
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.
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.
@@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.
@@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.
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.
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.
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).
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.
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.
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.
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.
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)
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.
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.
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
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.
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
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
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.
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...
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 =(
@@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.
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
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
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
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.
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.
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!
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
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.
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?
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
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 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.
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.
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.
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!
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.
I don't get this derivation. 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.
In your second equation you are using the velocity of the tip of the pencil. Whereas in the first equation, and our working out, v is the velocity of the centre of mass.
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.
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.
This bothered me too, but the 1/3 is correct. The rotational inertia of rods are typically written in terms of the total length "L" of the rod, but they are using half the length "a". Substituting a=1/2 L we get that 1/12 as expected.
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 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
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.
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.
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?
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?
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...
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.
they worked under the assumption that the tip of the pencil hits the table exactly at the edge. Thus the speed at which you have to hit it to get this results depends on the length and mass of the pencil and gravity, but the angle at which the pencil is if it hits the table right at the edge is independent of these factors
This is all just first-order accurate, due to several simplifications like omitting friction and how the distance from the point of impact to the center of mass will be shorter than the distance from the tip to the center of mass (because 1. you can't hit the pencil perfectly at its end and 2. the tip has less mass compared to an equally long part at the other end, shifting the center of mass slightly away from the tip). And the location of an object that just started freefalling is, up to first order in time, constant (y(t)=1/2 g t^2 = 0 + a quadratic term). This is not a mistake, it's efficiency: if all formulas are first-order accurate in time and your experiment takes place over a short span of time, your calculations will be a pretty good approximation of the real results from the experiment. Anything beyond that wouldn't improve the quality of the RUclips video enough to be worth the tedium.
Yeah, this video doesn't answer the question in the title, whether the pencil hits or not. For that question, you need gravity. They started with the assumption that it barely hits and went from there. The impulse required for hitting vs missing will depend on gravity.
@@franzluggin398 I disagree. I want the tedium, please! Otherwise I'm left with frustration, just knowing this isn't entirely correct but not knowing what must be done to correct it.
Would be nice to know why downward acceleration due to gravity is not a factor in that calculation.
Yes, in space it would never hit the table
@@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.
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.
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.
@@Macieks300 v is the velocity of the center of mass, not the tip, which is what pops up in the definition of omega
"The bluetack doesn't affect the physics."
*Pencil fails to move because of bluetack*
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.
They could also just use a pencil that's flat at the bottom
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.
@@Danilego I assumed they had until he mentioned blue tac
What is bluetack?
12:25 is the most perfect hand drawn Sinecurve I've ever seen.
Have you been to any math class above 10th grade lol , it's not THAT impressive
@@bibektg Of course... but in university the quality of hand drawn curves drops rapidly
One of the classic party tricks of an engineer.
@@bibektg Ok Boomer
Something maths-liking Zen monks pratice, no doubt..! 😉
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!
Matt: Does it change the physics?
Hugh: No.
Pencil: Doesn't move
"why are all the pencils broken inside???"
Why am I broken inside?
Why is the rum always gone.
I could only afford cheap pencils as a kid, and nearly all of them were broke about 4 times inside.
@@wierdalien1 hide the rum!
@@janhetjoch (left ear) aren't we all?
What a brilliant way of showing the working out on paper very clean! Must have took some editing skills, Brady would be proud lol
Glad you like it! Was only because we lost the first GoPro file.
@@standupmaths Honestly, I kinda prefer the way you did it here! It makes the notes that much easier to look at
@@standupmaths yes, that looked great! How was it done? Video editing or some other way?
@@GetMeADrinkNow he probably scanned the sheets and then edited it the way it looks.
Miss the brown paper and sharpie
I am deeeply disappointed for the pen length being 'a' and not 'b'.. "The pen is 2B long" would've been a perfect pun..
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.
I wonder how meany people would get it
@@jrbleau nicely done
Bjarni Valur idk, but put me on the list of people who don’t.
@@Vgamer311 I don't remember it completely but B and H mark the "hardness" / "softness" of the pencil lead
Matt's sheer elation at 1:39 is my favorite reaction to anything ever right now.
Two grown up men happy like a couple of children :)
4:00 just some ducks drinking in the background
It's quite a flock of drinking birds!
@@NickStallman
One drinking bird is a novelty. Four is a statement. I'm not sure what it's saying, but it's saying something.
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.
@@badlydrawnturtle8484 i know what it's saying. "I zarking love drinking birds!"
It's all go in the kitchen.
Matt Parker, answering the questions we never thought to ask.
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.
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.
@@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.
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.
There is also a negative solution
@@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.
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.
Good point, and that might actually influence the result, since the tip-side is longer (i.e. farther from the center of rotation).
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.
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).
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.
@@DoctorRedstone72 And they do not hit it exactly at the bottom but at least half a pencil diameter above the bottom.
Sure, I can try. I have exactly 72 pencils lying around from a project I am yet to do :)
chchaitanya ahh good video i remember
I cannot get enough of you two together! More Matt and Hugh please!
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.
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.
Matt "Close Enough" Parker
I love these videos with Hugh.
Hugh actually reminds me of Arthur Weasley (Ron's Dad) of HP.
Loved the flock of drinking birds 😂
"You wanna see a magic trick? I'm gonna make this pencil disappear."
It's always good to see 2 grown men playing with their pencils.
Gábor Králik no... not always...
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.
I love how the table has three legs, it is clearly made for that job (or similar).
The close enough graphic at the end is the best part
Add blue tac
“Does it change the physics?” “No”
Well strictly yes, ok I’m being pedantic.
Very _stickly_ yes :)
Being pedantic, the physics remain the same in our lifetime, regardless what you do to this pencil. Physics won't change :)
Hugh Hunt's an engineer, not a physicist. For practical purposes...
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)
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.
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.
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
I would have clicked 4 hours ago except work got in the way again
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.
1:48 Beautiful tower clock in the background! With gravity escapement! wow... a real jewel.
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.
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
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
I definitely prefer the animated scans. It conveys all the necessary information without an odd camera angle and unnecessary background details.
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.
It really is a Parker square of an answer.
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...
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 =(
@@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.
@@Anankin12 yeah but I was expecting a general formula to answer it. Not a general formula that... Doesnt answer it.
this one video has more information then one whole chapter of mechanics.
thank you for this video.
Hugh Hunt: Excuse me while I do calculus by site.
God I have forgot so much since I got out of school.
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
I love how Matt asked Hugh at the end to just hit as many pencils off the table as he could
Oh man, I do love a video where I can watch at Matt's face in 4K in slow motion!
his hand drawn sine wave is amazing
That is the most beautiful hand-drawn sine graph I have ever seen
I just finished taking calculus-based physics last semester. It's interesting seeing a video on some topics we covered in class!
It would be a more accurate experiment if the pencil was unsharpened...
1:44 so excited about flying pencils☺️
4:00 that flock of drinking birdies in the background lmao
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
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?
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
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
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.
MiniBetrayal nope... because the blue tack prevents that scenario... the pencil just sticks to the table.
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.
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!
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
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.
Got humble pi for my b-day loving it!
I think the moment of inertia is 4/3ma^2 because the total length of the pencil is 2a at 5:17
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?
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
Somewhat counterintuitive result with a surprisingly intuitive explanation!
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.
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.
@@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.
I was actually in the mood for a standupmaths video today and I was rewarded!
Wish and you shall receive.
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.)
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.
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.
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!
Excellent - gotta love how things cancel out and that implies certain physical predictions. Maths is great!
Horrific pencil torture. A fallen pencil is a dead one.
At 2:27 Professional Mathematicians with a Professional Pencil Sharpener.
15:37 sure, close enough, i guess...
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.
I liked the animated scan better than the GoPro
It has to fall, but it does not depend on the gravity! The same angle would take it on the moon! This is awesome!! Congrats
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.
anyone else get entranced by the turkey/bird drinking bowl in the background
Frogman Arts The drinking bird is another famous video.
@@gordonrichardson2972 thought i saw it before
Timely. My physics class at school is testing momentum this week. Linear and angular versions. We look up moments of inertia in tables too.
By calculating for equal you get the angle for when it hits by a point, if you use an inequality like
Animated scans! More of that! 👍🏼
Matt has the best face for slowmo
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.
I don't get this derivation. 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.
In your second equation you are using the velocity of the tip of the pencil. Whereas in the first equation, and our working out, v is the velocity of the centre of mass.
I like the constantly drinking ducks in the background. I left mine for months and they got horribly mouldy...!
5:38 It’s a moment of inertia moment.... boom boom!
15:47 Yep, he was a huge help!
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.
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.
I think there is a mistake here which is that the moment of inertia of the pencil about its center of mass is not 1/3mr^2 but actually 1/12mr^2
This bothered me too, but the 1/3 is correct. The rotational inertia of rods are typically written in terms of the total length "L" of the rod, but they are using half the length "a". Substituting a=1/2 L we get that 1/12 as expected.
Confusing to follow the explanation, because "momentum" in German is "Impuls" and "impulse" is "Moment"
I actually had the exact same problem 😂
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"?
Isn‘t torque the equivalent of Moment in German though?
@@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
Danke, Deutsch.
Give me a place to stand and I will move the earth.
I started watching for the pencils, I kept watching for the dimples.
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.
1:46 if only I could be that happy...
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.
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?
Nothing has even happened at 1:38 and matt still looks like a kid at Christmas.
More people should be like matt.
Now do it for a pencil that does not have an even mass distribution and show the derivation of the moment of inertia.
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?
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...
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.
I like that the video length is the Golden Ratio
The moment of inertia is the variance of the mass from its center
you forgot the gravity matt, tut tut
they worked under the assumption that the tip of the pencil hits the table exactly at the edge. Thus the speed at which you have to hit it to get this results depends on the length and mass of the pencil and gravity, but the angle at which the pencil is if it hits the table right at the edge is independent of these factors
@@Mmmm1ch43l The pencil did not always hit the corner right, but only these results were used due to the assumptions made in the calculations :)
This is all just first-order accurate, due to several simplifications like omitting friction and how the distance from the point of impact to the center of mass will be shorter than the distance from the tip to the center of mass (because 1. you can't hit the pencil perfectly at its end and 2. the tip has less mass compared to an equally long part at the other end, shifting the center of mass slightly away from the tip). And the location of an object that just started freefalling is, up to first order in time, constant (y(t)=1/2 g t^2 = 0 + a quadratic term).
This is not a mistake, it's efficiency: if all formulas are first-order accurate in time and your experiment takes place over a short span of time, your calculations will be a pretty good approximation of the real results from the experiment. Anything beyond that wouldn't improve the quality of the RUclips video enough to be worth the tedium.
Yeah, this video doesn't answer the question in the title, whether the pencil hits or not. For that question, you need gravity. They started with the assumption that it barely hits and went from there. The impulse required for hitting vs missing will depend on gravity.
@@franzluggin398 I disagree. I want the tedium, please! Otherwise I'm left with frustration, just knowing this isn't entirely correct but not knowing what must be done to correct it.
The birds in the background...LOL!!!
Now everyone has an excuse for flicking pencils across the classroom.
I'm so confused what this has to do with a three-sided coin...
I'm pretty sure the blue track makes a huge difference
No pencils got hurt in this video...
Started with the extremes..so hard it does not hit the table, then so soft it doesn't leave the table