Will the billiard ball pop the confetti? (Engineered Bet #5)

[@instinctualhavoc]

Thank you so much for not making this a 10 minute video with 2 ad breaks

[@aaronsteindler3245]

Barring negligible errors, the physics presented is sound. Based on the calc, the pool ball would need a mass of just above 0.4 kg for 3mg to exceed the 12.3 N of force measured by the spring scale. A pool ball's mass is well below half that value, so it makes sense that it would not have enough force to pop the confetti

[@mathieuaurousseau100]

The physics presented is not sound at all. He used the acceleration of a CONSTANT SPEED circular rotation, which is obviously not the case here, since it starts at 0 speed In this case you need to do like we do when deriving the pendulum's equations and observer that the length of the line doesn't, meaning that there is no acceleration in the direction of the line. This gives that the maximum of tension will be mg, instead of 3mg (the maximum still happens at the bottom)

[@robertlitman2661]

@@mathieuaurousseau100no, you are mistaken. You are discounting the centripetal force, because the ball rotating about an arc is the acceleration in question.

[@CliffHanger-fg6uy]

@@mathieuaurousseau100 You went further in the wrong direction. You used the equation for a constant velocity (set acceleration to 0). However, moving along a circular arc requires an acceleration perpendicular to the path. If it is speeding up (which it actually isn’t at its lowest point), there is another component to the acceleration that it tangent to the path. Even if this component were present, it would not impact the tension as it’s orthogonal. Remember: a change in the direction of the velocity requires a nonzero acceleration.

[@OptimusPhillip]

On the intuition poll, I said it would pop because I knew there would be additional tension from the centripetal force. But seeing the figures for how much force there would be compared to the force required, I'm on the side of no pop.

[@altofdish]

in hindsight…

[@AdamHill42]

Usually you make it so the basic theory is attempted exactly in practice, so the only wild card is complex effects, so I was disappointed that it wasn’t the case here 😩

[@jabarzua]

How was it not the case?

[@lerarosalene]

​@@jabarzuait wasn't in poll.

[@theyadersh5263]

I would have liked to see the strength that the pressure gauge would show with the ball attached as well

[@1st2nd2]

I am attempting to assist and educate. I am not attempting to insult anyone. The item used was a force gauge, not a pressure gauge.

[@theyadersh5263]

@@1st2nd2 I appreciate the way you have corrected me. Thank you for the observation

[@RonaldSL-]

@@1st2nd2 Gentlest internet comment I've seen ever.

[@RedScaledKnight1]

To piggy back off 1st2nd2, I want to mention that measuring the force required to pop the popper while the ball was attached would not be helpful. This is because the guage would, at that point, be taking a contaminated reading. If the ball were attached, when measuring with the guage, we would be measuring (force required to pull string) - (force exerted by ball). We want to know (force required to pull string), so involving the ball would complicate that.

[@Dondoesstuffofficial]

@@1st2nd2wonderful human you are

[@danielrhouck]

I’ll go with the favored option here; I trust your calculation. It also happens to agree with me intuition.

[@edurom2111]

I love the dedication put in the making of the video and everything that goes behind a simple result like shown. Cool video and nice work with all the big brain stuff :3

[@Clipper68419]

me too but *BRO SHOULD'NT BE DESTROYING THE WORLD* 💀

[@cca1834]

just found your videos, and i would love to see some longer form content! please don't be afraid to break the 5 minute mark :D

[@bhuvangunessee]

Nice to see the calculations in the video, when I "guessed" from the photo I did not take into account the potential energy converted to downwards acceleration (totally due to laziness) and guessed entirely from the weight the ball puts on the cord at rest which was so ridiculously small I assumed given that the string was taut it would be enough. Good to see you include also the freefall test, I was really curious if it was enough as string length does matter for this one (fall height changes), and really nice to see the Newton meter test on the popper to show it was 12 N. I'm sold on this series, but do please keep the calculations in.

[@drewbryant38]

I just took dynamics last semester. I love seeing this stuff

[@ajrey8457]

Thank you for doing the vertical drop to prove that the billiard ball can produce the force to make the popper go off.

[@randykitchleburger2780]

It's totally gonna pop.

[@randykitchleburger2780]

Awww.

[@kazboven9744]

​@@randykitchleburger2780 lets go gambling Ah dang it Ah dang it Ah dang it

[@bobsmoothie6445]

Yeah I don’t know what he’s talking about, it’s gonna pop

[@leglock916]

2:53 here’s the time if your to lazy to watch it all

[@thereaintnojusti]

That’s how you end up with an ear full of cider.

[@gaijininja]

I assumed it wouldn’t pop because as the ball falls, the “pull” on the string is quite small as it is only minutely changing the direction of the ball. Most of the potential energy of the ball is moving in the x direction (Horizontal.) at the bottom of the arc, so very little force. I’ll make a hypothesis that if a second string was joined to the ball, and was connected to a popper horizontally to the right of the lowest point of the arc, the horizontal velocity would produce enough force to pop the second popper.

[@mrpyro333]

I like your funny words magic man

[@empmachine]

Cool video, but it would have been more fun if it was closer-to-call. Maybe next time you could intentionally weight the ball so that it's super close to the needed weight to cause the pop.

[@ldkutellme]

would have also been interesting to let it drop freely from the height of the popper

[@meirbookatz8304]

But isn't centrifugal force perpendicular to the string? Wouldn't that make the downwards force only the weight of the bowling ball?

[@mr.e822]

Technically, the “centrifugal force” as most people describe it is a misnomer. It is not an actual force, merely an interpretation that feels like a force due to one’s inertia. In other words, if you were to ride on a merry-go-round, the feeling of being pulled outwards is not actually a force acting upon you. You have inertia, and that is trying to carry you in a straight line (which is always opposite of the centripetal force, which is the actual force that is pulling you in). In the example of the merry-go-round, the centripetal force is the friction on your feet. In the video’s example, there is a centripetal force accelerating the ball towards the center of the circle - the tension of the string. The only other force acting on the ball is gravity (air resistance is in there too, but that is negligible). With this in mind, in order to find the tension on the string, you simply calculate what force would provide the proper acceleration towards center to keep that arc. The method in the video is an accurate method to do so, since rotational dynamics describe centripetal acceleration as a function of tangential velocity, and velocity at an instantaneous point of time for a pendulum can be described using conservation of energy.

[@amobilway1032]

No, centrifugal force doesn't exist in inertial reference frames. During rotational motion, there must be a net inwards force and acceleration to maintain circular motion due to the changing velocity (constant speed but changing direction)

[@carolinesherman]

I love this channel

[@dawgohio]

hey dude, im a kid of 12 and i want to learn about science. I'm currently on 7th grade level of science. How do I find out more on these equations and concepts. Should I borrow the next years textbooks because i don't know where to start. I tried using youtube videos but none started at my level. Can someone please give me some suggestions.

[@Hadeshinai1]

I cant figure it out quite well that h=r, was this set up curated so that the string length when the ball is at the bottom is the same as the height when the ball falls if the string is somehow detached?

[@B58_G20]

Should’ve raised it by 5° until it broke, then calculated the theoretical force from that height and compared to the experimental force required to set off the popper

[@peterkn2]

When I was a kid, I would use these as alarms for my doors and windows. I taped or tied them to the door or windows edges and the wall so that they would pop if someone opened the door or window 😊🎉

[@cheetahman515]

You used all of this math, while I used common tuition of momentum

[@Edo_Luna_]

Love this series

[@Kara_Kay_Eschel]

Good use of an infinity ball.

[@harrisonbaylor1432]

Saw the short and came to watch the video because I thought it wouldn’t pop but if it was free falling the snap of the string would pop it.

[@krstev29]

Can I ask you for next time, to put all of the parameters on the side, so as an exercise we can also calculate the result, and see in the end if we were right?

[@gregoryhost5188]

I don't understand the last part why does it pop when free falling, if I understand correctly, the tension force at the bottom would simply be F=mg, but if you use that it's not greater than the required force to pop the confetti

[@melimsah]

What a cool idea!

[@ChristianBurke-pb3wy]

Will you run an open bet for the next ones?

[@bot24032]

I mean here the only thing I thought wasn't considered was air friction and yhat would pnly slow down the ball and reduce the force. Maybe the string's mass too but it really *is* negligible compared to the ball's 180 g

[@icycrusader1947]

I mean, the theoretical newton value was under half the required force to set of the confetti.

[@Cobb]

Was right but not because of math. Swinging the ball doesn't create a large force on the popper as it's just a smooth swing with probably some minor variance in force applied to the charge at any given moment. Whereas dropping the ball gives a large sudden moment of force applied to the charge likely allowing it to be set off.

[@amobilway1032]

Good observation, the impulse is greater since the ball's motion is opposite to the string and experiences a change in momentum opposite rather than perpendicular to the string. Also involves complete stoppage rather than some energy still in the form of kinetic energy

[@Sammypuppy611]

I used to open a bunch of these and light up the powder all together

[@BarbarisII]

For some reason this makes me want to go watch marble racing

[@Hanuansh807]

i like this style of videos

[@gerardogamez291]

2:52 is what youre looking for

[@Deadfish83]

What do you use for animations

[@paulmeynell8866]

Very interesting thank you

[@totally_not_a_bot]

Oddly relevant to rigging and climbing safety. Slack in the line is bad, for the reasons demonstrated.

[@awesomegaming9455]

I need more of this 👍

[@railworksamerica]

dude playin crazy machines 2 irl

[@Drizzydre254]

He is smart

[@user-white007]

It’s popping I like my odds

[@rizalardiansyah4486]

Can anyone explain to me why the free fall case popped? If I draw an FBD of it, the only force applied is gravity which is even smaller than the swinging case. Yes there's initial speed but where does it go in the FBD?

[@Ben-kg9gp]

Today, I learned I really can't trust my assumptions . 😅

[@AdrianHearn1460]

Great video

[@catburner1896]

This made sense to me because most of the force would end up perpendicular to the direction of the popper.

[@drdca8263]

Force ≠ momentum

[@catburner1896]

@@drdca8263 Force is what’s relevant here though… And they are connected causally as momentum is simply the retaining of force.

[@drdca8263]

@@catburner1896 Both of those are true, yes. But at the bottom of the arc, the momentum is mostly horizontal, while the force is mostly vertical. Edit: well, actually it think momentum is probably more fundamental than force?

[@catburner1896]

@@drdca8263 you are right, I messed up my thought process a bit. But the force is minimal since there is only minor angular acceleration.

[@MissMyah37]

Efc = mac or kfc ≠ McDonalds 1:31

[@Homiloko2]

The results are unclear, because you dropped it from a larger height at the end. My question is, at the same height, which drop applies a greater downward force on the object, a free fall or a circular arc? It would be cooler if you solved the math for both cases and compared them.

[@tropixMw2]

Different force for braking the ball aswell

[@jelleidzenga528]

In the free fall case you have to do different physics calculation. You have the velocity when the string is fully extended and then it needs to stop all that velocity in almost an instant. So a lot of force for a very short time, it will pop.

[@amobilway1032]

Free fall will be more force

[@ElizabethWalters-yk4ui]

these are too fun i just *chef's kiss love it. B')

[@IsntPhoenix]

Wow i thought it woupd definitely pop

[@josemarialaguinge]

Great Video.

[@Primetime-sl3gm]

ALL THE MATH HURT MY BRAIN

[@vxcbrtgsvrt]

Banger video

[@davismorehead2494]

You increased the potential energy when dropped to ball since you raised it up. Would have been better to drop from the same height

[@Snowfats]

thanks!

[@Greg_Bal]

Finally, I got one right

[@skeome]

Would it pop if you gave the string slack?

[@Don_of_the_Dad]

Did I just get tricked into learning math?

[@BN99239]

To the 8%. I'm sorry.

[@LiLGhostPlays]

Ok, but now do it with a magic 8 ball

[@meerkatmatrix]

Got it and the bonus right lol

[@heatherhatton2282]

This will definitely pop even if that maths says no

[@heatherhatton2282]

I stand corrected

[@judahmaman]

Hey, a fan of the content!

[@cringevids9971]

Pov: me scrolling at 3am I NEED ANSWERS

[@SillyBilly-j3j]

Brie wall looks like Swiss cheese

[@GruvisMutt]

I guessed it would not pop. 🎉

[@DvsArtemis]

For those who have a short attention span or has ADD or ADHD; No it does not pop if you just drop the billiard ball at a 90° angle.

[@freshcut4754]

The eight ball still has weight

[@eazygamer70]

I was wrong

[@jacobtrang833]

i stepped on one when i was younger and it didn't hurt that much but i went to the doctor and they did a xray and it showed i splinter my pinky toe i had socks on but no shoes

[@jjonestown]

guess: it will pop

[@sober667]

i wold say not if i saw it lets see

[@Lone_Vaquero684]

See I was listening to you explain the math but then i realized you were letting it swing from that height instead of drop, swinging from height and assuming that all of your math is correct then obviously it won’t pop, but if it dropped from that height and thus had to get caught by the string then I think it would’ve popped from the jerk to a stop

[@gdude2775]

I guessed it would pop 😢

[@tymz-r-achangin]

Good grief! Just drop the ball already! From now on I know its time for me to pass on videos that have titles saying 'Engineered" , so I'm not bored to death

[@metern]

I was right all along. I voted NO and commented that the ball won't pull the string because of how hard you really have to pull the string. Anyone who has "fired" a popper knows how hard you have to pull the string. So, to anyone who answered on my comment how wrong i was. I say only this. "I told you so" 😂.

[@metern]

kzbin.info5IBseSqfWwnnjt8b

[@Brittish_people_are_scary]

Short answer: it didn’t pop

[@ojdidit69]

Answer is no

[@matthewbuczkowski2242]

I got it

[@JakkzOfficial]

Write

[@cachamoney3046]

I dont think it will pop

[@TheTDK80]

It won't pop

[@capsfederation3154]

I loss

[@KiddiesDickson]

Hi

[@JaDroppingScience]

hello!

[@KiddiesDickson]

@JaDroppingScience Your videos are always so interesting. You're my favorite science youtuber

[@TomsBackyardWorkshop]

I thing the force you calculated is significantly higher than the actual force felt on the string. By the time the ball reaches its lowest point the velocity is 90 degrees to the string direction. That means the force on the string in the downward direction is pretty much equal to the weight on the ball.

[@thatsomeone3818]

You should look up centripetal acceleration and centrifugal force

[@AK56fire]

How have 6 people up voted to your comment. I understood everything in the video, & nothing of what you wrote. Just because the velocity is at 90 degrees to the string, you concluded that there's no centripetal acceleration ?

[@1islam1]

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[@thatsomeone3818]

@@1islam1 Why are you replying to me bot lmao

[@TinyFord1]

@@1islam1it’s not the religion preached by Moses and Abraham, because if it was then “it could not have been corrupted later”. So either the message was corrupted (and you believe allah’s words can’t be corrupted) or Moses spoke about another God. The Christian God.

[@KadenTokesAlot]

I watched the crap video it doesn't pop

[@DNautSun]

Certified yapper, Get to the point