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The solution to 4 rotation-related riddles, including the mystery cylinder, bike pedal pulling puzzle, track problem, and train part going backwards. Thank you to everyone who responded, liked, shared, or made a video response.

Please fill out this short survey for research: ve42.co/Rresearch

Special thanks to:

Mathematician George Hart: georgehart.com/

For allowing me to use excerpts from his pedal pulling puzzle solution: ve42.co/ppp

Petr Lebedev for combing through thousands of comments and providing the stats I gave in this video.

Video responses I used in this video (or watched):

everWonder? svfrom.info/history/video/ZbibZnvZydGhaa4

A Random Nerdy Channel svfrom.info/history/video/eZyLdnG42L-AqNQ

The Physics DoJo svfrom.info/history/video/gNGsZISnrbGuccw

Oblivious Jim svfrom.info/history/video/YZWQjoGxq7WiaaI

Armchair Explorers svfrom.info/history/video/g5SyjJewutatpdQ

MrEngineeringGuy svfrom.info/history/video/dLV_bGq23p6Ykao

Professor Cubers svfrom.info/history/video/obKdaafWrpKjnq4

Scoop Science svfrom.info/history/video/pd2Fg26r19KfoKo

A few notes on the puzzle:

1. A half-full container of honey does pretty well in reproducing the behaviour of the mystery cylinder. I wonder if the motion is a little smoother or more periodic with the ping-pong balls because they move as organized objects - also the delays between motion seemed to be longer with them than without ping pong balls.

2. For the average speed track problem, every time I said velocity I meant speed. Sorry to the pedants out there who are perhaps looking for some trick answer due to displacement being zero when you run around a track.

3. Although a lot of people identified it was something about a train's wheels that move backwards, fewer identified that specifically it was the part of the flange below the rail. Some simply said the bottom half of the wheel.

4. The bicycle question is perhaps the most complex of these riddles. If you tried it with a bike you likely found that it went backwards. But what happens if you sit on the bike and only push backwards on the bottom pedal. The answer might surprise you so give it a shot!

This video just made me think of What if you ride a bike at a particular speed and run over a travelator (the kind you see in the airports) that has the same speed. will the bike appear stationary with its wheels negating the travelator??

You said "a part of the train that is going backwards with respect to the ground, that is NOT backwards with respect to the train". Maybe it's just my bad englisch that causes me to search for something obviously impossible. But what about some outer parts of the motor, which is spinning so fast, that these parts move backwards faster than the train is moving forward. They are also moving backwards with respect to the ground. That brings me to the Bike question. OK I missed the fact, that you can pull it backwards this way with the pedal going forward while pulling it backwards. That was confusin but is logical. but I said it depends ;-)

I regret filling my bike tires with honey.

Thank god I thought you were gonna leave me hanging

What do riddles have to do with rolling bottles?

Attached it to my bike so I can hook random objects while cycling. That will give me free accelerating.

Is there a ratio where the bike doesn't move?

I got all the answer right

I go around thinking I'm smart, but sadly I am too lazy to challenge myself to become smart. Therefore, I'm average. How depressing.

This video (the average-velocity-riddle) just sparked a discussion about special relativity in my friend group...

im sorry man but your mountain bike should NEVER go backwards when pedaling backwards. It should only ever be one thing, pedaling backwards activates coaster which makes the pedals disconnect from the wheel and freely spin without turning the wheel. Rolling backwards on the other hand, should indeed rotate you pedals arms, but vis versa should never happen on an anywhere near decent bike as its extremely dangerous.

If im running at 2 kms at the first round I'll have to run 6 kms at the second round so that will give me a 4 kms average which is 2 times the first one

3:56 Rather than running twice the distance, the equation will hold true if same distance is covered in half the time; i.e. 2V = d/t/2

Pfff, at least got the wheels right. Thanks for this!

Ok, i was two times right and the third question maybe i went for coffee, so that i did not recognice this. The fourth question answer i do not credited. In real drive-situation, this phenomenon (backwards) did not take place. If the Bike has no freewheel (no idling), then there is no other direction, as forwards - independent of gear ratio or lever length. And if you look accurate on your Video-sequence, then you would see, that this happends: the force of backwards-pull pulls the boika backwards, but only so long, as far as the gear and the chain-drive traction begins, and then the pedal moves forwards, while the bika moves backwards. What means this? That there is an unrealistik Force closure (in german: Kraftschluß/Kraftübertragung), wich is feasible only in this special constellation.

But wait! Didn’t Richard Feynman just tell us that the flanges on the train wheel do not keep it on the track. And that’s not what those flanges are for?

svfrom.info/wiki/PL2D30B1DEFFDA0310.html

This and the other video were oddly some of the most exciting videos I've seen in a long time :D

Normally flanges are not needed for stability of rail wheels---this is achieved by making them a bit conical.

he ask about speed and then explain about velocity , thats not fair

Riddle 2 seems a bit of a problem. for example distance for a lap = d and time taken for 1st lap = t if v1 is the speed for 1st lap v1=d/t now if v2 is the speed of 2nd lap then (v1+v2)/2=vag=2v1 substituting v1 = d/t (v2 + d/t)/2=2d/t :. v2=3d/t now since we cant travel 3d as we need to complete our task of v average = 2v1 in just a single lap we can say we need to complete 2nd lap in 1/3 of the time of what it took to do the first. :. v2 = 3d/t = d/(t/3) but now if we take total time i.e t + t/3 = 4t/3 and total distance = 2d we get vag=total distance /total time = 2d/(4t/3) = 3d/2t = 3v1/2 ???????????????????????????????????????????????????????????????

Your race track math is off. You’re not changing the distance, fool. The track length remains constant. The time it takes to complete the revolution is precisely the thing that changes. If you complete the distance in one third the original time, the initial velocity is irrelevant. The result will be an average of twice the original velocity. Why on earth would you set time as a constant?

Your confusion is perplexing to me because you even WROTE OUT the fact that distance was the constant, but then went on to go 2d instead of .5 t1

Here’s the proof you’re looking for: Vave = (v1 + v2) /2 Two is the number of times around the track. The distance of the track doesn’t matter. The time it takes to get there on the first run doesn’t matter. The distance is the constant, not the time. 2(V1) = (V1 + V2)/2 4(V1) = V1 + V2 3(V1) = V2 Saying it can’t be done because 2v1 = 2d over t. You’re not running the same track. You’re running a track that is twice the length of the track of the first run.

You’re basically saying that you can’t do this problem because in order to get twice the velocity, you need to run the same exact track twice on the second go, and do it in the same time. No wonder you can’t solve it.

3:55 wrong because we don't what the one way of the speed of light is and you said in a recent video (I thought) that the one way speed of light may be unlimited or instant

@Anonymous Secret ye that's also true

Anything with mass can't travel that fast. And presumably we are the target runners, so no speed of light :(.

give me only one answer of this question below of experiment, Based on real experiment if done before but not by calculations. If we take two clocks set time together at same place.and one clock is placed 10metre away from point M and another clock 1km away from point M . A light bulb is glowed at point M now would the reading of the clocks when light reaches them be different or same if path for light was through vacuum. Would they show different or same reading as when light reaches them the watch goes off. And saves that time when light reaches them through vaccum.!.!.!

That running riddle is not impossible if the velocity need not be constant for each lap. Oh, 3 years ago video, anyone already explained this?

You pull the pedal of a bike backwards then the bike moves forward... that's the bottom line. Unless it's a bike with a gazillion gears and you set up the gear ratios so it actually moves backwards... f*ck off...

Them toes look tasty

I finally caught it 3 years later

Ok I know its 3 years late but I got the ping pong ball thing right with the honey

1=1/1,. ≈3=1/.33,. ≈3+1≈4,. ≈4/2≈2 aka double the average speed no? 🤷🏽♂️

ur fooling people

I tried to find The avg speed using V1 and V2 ..and the came up V1+V2=V2 and V1=0 ..and I was like WTF!!

Is there anyone with a more expansive explanation to number 3? I seem to weigh my average differently or something - I just can't quite wrap my head around why we don't try a different method instead of saying its impossible

The Train scenario reminded me of the time a Virginia Tech Mechanical Engineering professor told our class that car tires had to be incredibly tough because each part of the tire on a car going 50 mph (for example) cycles from 0 to 50 mph and back to zero, because the part of the tire in contact with the ground is going zero (because the ground is 'going' zero) and the top of the tire is going 50. This struck me as the stupidest thing I'd heard an educated person say, but I couldn't articulate a rebuttal. Can you provide a good explanation? P.S. The first time I saw an advertisement for a tire & rubber company seeking a Tribologist, I laughed my ass off.

I NEED A BETTER EXPLANATION FOR THE SECOND PROBLEM. I THINK THAT IT IS POSIBLE TO DOUBLE THE TOTAL AVARAGE !!!!!!! PLEASE MAKE A VIDEO EXPLAINING!!!!!!!

You helped me confirm that I am not as smart as I thought I was😂 I really love your channel!!

Jesus loves you brochacho

riddle number 3 got me good, thats such a cool riddle

I have Dutch pedals so the last riddle would be false for me

What is your you youtube income

doesn't time stop if you travel at the speed of light?

What about in the running riddle 1?

Where is riddle #3? 1. cylinder 2. race lap 4. bike ???

All those train videos when I was 3 really paid off.

My bike has a function where the pedals move backwards but the wheels stay still (I tried this myself with my bike)

I was thinking instatanous velocity

Well... Mathematically you could run Vavr=2V1... If you run at speed infinity in both laps Then it mathematically it is possible, physically though, no it is not possible

If you use a fixed gear bike that won't work even with the same ratio

I still don't understand the track one (number 3?), not even a little bit Usually I understand things =p

You have to run twice the distance in the time it takes you to run a single lap, but you already ran the first lap so the second lap is instant

To the 3rd riddle: It is possible if v1=v2

Actually, what am I know about physics..??? 😅 I'm a kid in physics. You're amazing!!!

You're assuming both laps cover equal distance. let's say you're running on a track with n lanes. Each lane out from the center will have a greater total length than the last, making a solution at least possible as the distance of the second lap could be greater than the first.

I ran the second lap. Jus that no one saw me.

The bike one depends on gearing and how long you pull cuz you could end up just pulling it back forever

Im unsatisfied with these answers

Oooooohhhhhhh...... Revolutionary....... Ooooooohhhhhhhhh.......

I consider the multiple components of the cylinder to be a bit of a cheat. I was always intuitively suspicious about the circuit running problem but lack the maths to pin down the problem. I already knew about the train wheels My guess was the bike wouldn't move. I guess that all amounts to a fail... but I learned more than I would have had I known them all, so it's also a win, yay!

Ohhh man I got all 4 of them right... Although I'm an engineering student....😅😅🙄🙄

Where is the vortex spinning riddle :(

4:08 "This may seem like a trick question, but actually, it's a trick question"

It’s not a trick question, it’s an inaccurate response to that question. He simply isn’t correct. Sorry

lol yeah

The track riddle can be done! You gave us an object to run around, not a distance. Assuming the track is a standard 400m, if you spend your second lap running in a 5.2m wide x 2m long zig-zag, you could effectively run 1200m on your second lap, adding making it take just as long to run thrice as fast, equating to a 2v average.

huh, I ...don't actually know if I can say I got 1 wrong or not since it works on the same principle: I was thinking an oblong on an axis but that actually reverses a little bit each "step" down the hill. thaaaat one was admitably not instantly obvious until I stopped and looked at that 2, I just went right into trying to find a ratio on autopilot. I mean, yeah ,wheels, I certainly couldn't think of anything else that would make sense but that was REALLY obvious so I kept thinking it couldn't count as a "riddle" so there might be something else. wooo! I am the mostest correct!

With the bike thing I mean if you pull hard enough no matter what its going backwards because the force would be so great it would overpower the force pushing it forward lol

its matematickly imposible ??? Well im whud point to the fact that you can run infinite speed lap..... since your speed is infinite your time is practicly 0 ....... Well eat that .....

Answer to everything: It depends... (auto correct)

why not 3x? You double the time but quadruple distance. 4d/2t=2d/t. There is no limitation on more time because the question is about average velocity of both trials and not about increasing the average velocity of the original trial.

Why would you quadruple the distance when you only want to run an extra lap

4:02 , the avg velocity can be equated to 2V1 if you run the same distance d, in the time period of t/2.

i was way too confident on my 3(V1)

Apologies if this is a stupid question, but what is meant by "forward in respect to the ground"?

It’s just said to make sure there’s no loopholes

You can either be stationary with things moving around you or you moving with other things being stationary (forward in respect to the ground)

5:14 Watch closely! It's crazy!

I feel stupid now

I dont get the second riddle, cant you Just complete the second lap in half the time?

That would make the total time for both laps 1.5 times the time for the first lap thus lowering the total average speed

For a guy with a Ph.D. in Physics, it's astonishing you're confused about the very definition of "average velocity." By definition, Vavg = (V1 + V2)/2 unless, of course, every physics book I've ever owned has been lying to me. If Vavg = 2V1 = (V1 + V2)/2 4V1 = V1 + V2, solve for V2 V2 = 4V1- V1 = 3V1 ...it ain't rocket science. I think what you're getting balled up on is the difference between arithmetic and geometric mean...?

This one took a while for me to get, but you're incorrect. The factor you're neglecting is the time spend at each speed. For easy numbers, imagine you ran at 1m/s for 100 seconds, then ran at 2m/s for 1 second, for that speed and time, you would have ran 102 meters in 101 seconds. Now, by your definition, the average speed is (1+2)/2=1.5m/s. But that's obviously wrong. If you had been running at an average of 1.5m/s for 101 seconds, you'd have travelled 151.5meters, not 102. The correct equation for these numbers is ((1m/s*100s)+(2m/s*1s))/(100s+1s)= about 1.001m/s. As a sanity check, just do the normal velocity calculation. You know the distance you ran(102m), and the time you ran it in(101s). So, distance over time, 102/101= about 1.001m/s. I think Derek did a really bad job of explaining this, but hopefully this has helped.

Anyone notice how he didn't use the original video for the bike riddle? My guess is that he was unaware that the gear ratio affected the answer until the 5% who said it depends pointed it out.

Your bikes are bad they should allow for back pedaling what kind of bike allows you not to paddle backwards and doing nothing because it rachets

hold up, you have havaianas??? that's cool

For the velocity question could you not just run the same distance in half the time?

Then the average speed would be 1.3333 (V1)

Ok my guesses was not bad: i guessed there is a ball in cylinder; 3v XD; train wheels (I 3D model a lot so yeah XD); move forwards

Wow. O-O

Who is here after watching 4 revolutionary riddles?

What if you walk sooooooooo slow on that first lap? Like a 30 minute lap

What happens when the pedal is almost halfway between bottom and top dead centre? It will no longer be able to turn, therefore not driving the chain. The string will not be able to lift the pedal to top dead centre and push it forwards. The freewheel would stop the cycle's momentum pulling the pedals with it. I'm willing to be proven wrong, if you're willing and able to try making the pedal go 720 degrees by pulling the string backwards only.

Your bicycle question/answer for some reason reminded me of an old WWII pilot's bet. He bet his fellow pilots that he could move his B-17 BACKWARDS with just the engines (which did NOT have a reverse prop setting back then.) He always won his bet. How did he do it?

Dp you think it's worth it to learn physics and math after 30 with no background in it?

run the second lap in 0 seconds

I was just thinking you run 1 mph the first lap then 3 mph the second wouldn’t that average to 2 mph which is double 1 mph

I don't get it: If I ran the first lap of say 10m in 4 secs (so v1 = 2.5m/s). The equation to run the second lap so average speed is 2*v1 (5m/s) = 5 = (10 + 10) / (4 +x) 20 + 5x = 20 x = 0 So I have to run the second lap in 0 seconds????? HOW???

Isn't 2V1=2d/t1 the same as 2V1=d/t1/2 Meaning to travel twice as fast you dont have to travel twice as far in the same time but rather the same distance in half the time? If so, then that would mean running at 3V1 will in fact be the correct answer.

Derek did a really bad job explaining this one, and I think that line in particular just makes it more confusing, so ignore the "You can't run twice the distance" explanation for now. Here's my suggestion: Go into excel and use this formula Vave=((V1*t1)+(V2*t2))/(t1+t2) No matter how slow you make V1, or how fast you make V2, the average only approach 2V1, and since there's a universal speed limit, even if you could make the second lap infinitely quickly... you can't. The important thing is that the average isn't just (v1+v2)/2. You can see this yourself easily. Imagine you ran at 10m/s for 100seconds, then 20m/s for 1second. Using that calculation the average speed for your total journey was 15m/s. But that makes no sense, you only ran at twice your speed for >1% of the journey. This is why you need to weight each velocity with the time spent moving at that velocity.

3:30 just don't move

software at 5:05?

What about 2V¹=4Vt/2d

I wasn't sure if there was a smaller cylinder or a ball bearing inside the cylinder.. Funny it's both.

What if you had extended the distance by first running as usual and the second time running zig zag extending the distance?

The no2 riddle is always possible , I can prove that...Where can I prove it?

at the speed of light - wouldn't you have to be (or behave like) a photon? and therefore not experience any time - so from your frame of reference, t_total equals t1. so, to double your average speed, you'd "just" have to turn into light.

Correct me if I am wrong, but I think it is mathematically possible to solve the second riddle. The average speed equals 2d/(t1+t2). In other words, in order to achieve an average speed of 2V1, t1 must be much higher than t2, so instead of running at lightening speed at the second round, simply run with very low speed at the first round to maximize t1. Thankfully the distance (d) isn't specified, so you can reduce it to almost zero to complete the 2 rounds within your life span.

YOU GUYS ARE FOOL it's is a bicycle riding tutorial

At the speed of light there is no time.

but 2d/t = d/0.5t so running faster should work ?

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