Loop the Loop (B95) [1M40.20]

[@vidcreations]

I don't know all the formulas, but I did the same experiments with my Hot Wheels when I was 10.

[@Joshuaxiong2]

Jimmy Mecannic Ha, funny, same here.

[@noreenelizabeth6617]

My 8 year old son wants to answer the question "Why does a car need to go fast to go through a loop de loop" ... So I am trying to find a way to teach him in a simple age appropriate way besides "because of gravity". LOL

[@9130cris]

@@noreenelizabeth6617 well its a bit more complicated than that haha, yes, it must not be affected by gravity but the "explanation" as i would call it is that the Normal force must exist in an appropiate way (the theoretical condition is N > 0) and that one directly depends on velocity. Sorry if I confused you a bit, I find amazing that your son has those question by the way, he actually must be very smart. Greetings.

[@luzonnexusx9638]

Omg I love that subtle Zestiria reference in the title XD

[@jpd321]

The moment of inertia of the ball depends on the radius of the ball, r. The velocity needed to stay on the track depends on the radius of the loop, R. They do not cancel when calculating the rotational KE. The rotational KE will be quite small; it depends on (r/R)^2. Thus h is not much larger than 2.5R. I think the experimental problem has to do with friction. If friction were minimal, at h=2R, the ball should reach v=0 at the top of the loop. It doesn't.

[@DonSimone1996]

Didn't know Ryan Gosling studied physics.

[@GoofyHatesNi-]

💀

[@thesocialisticshow3605]

MIT videos are always good

[@joeteichert1726]

If the track were ice, that is, if there were no friction on the ball, starting the ball from a height of 2.5 R would be sufficient, as you say. If the track were sandpaper, that is if the ball did not slide at all, then 2.7 R is all that's required. But even in situation between these two extremes, even if the ball can slide a bit on the track, as it may well do in this excellent video, 2.7 R is enough. But it is not enough in every situation between these extremes. There is a case in which 2.7 R is not enough, even if we ignore air resistance and heat, and only attend to the variables in the equations that were mentioned. . As the ball climbs to the top of the loop, it is moving along the track slower and slower, just like any ball rolling up a hill due to its inertia. But there is no similarly universal reason for it to rotate slower as it climbs. Assume that it wasn't sliding at all when it reached the lowest point of its track. Now that the ball is climbing and moving slower, it will only rotate slower if the track exerts a friction force on it, as it does in this video. And as the track does this, the energy the ball had from rotating changes into translational kinetic energy, helping to keep the ball moving fast as it climbs the loop, and ultimately, ensuring that it will remain in contact with the track at the top of the loop. But Suppose that the ramp were sandpaper, and the loop were ice. That is, suppose only the loop part of the track were frictionless. Then 2.7 R would be insufficient, because not enough rotational energy would get converted into translational energy to keep the ball in contact with the track at the top of the loop. The ball would be spinning like the wheels of a race car when the light turns green, by which I mean that the ball would go faster if there were friction. That is, none of the rotational energy would help the ball to keep up its speed. So as it climbed up the loop, it would slow down to much to stay on the track at the top of the loop. So the ball would look like it does when released from a height of 2.5 R in this video, and the only difference would be that it would be spinning faster as it left the track. It seems clear then that for 2.7 R to be sufficient, the ball must either not be sliding at the top of the loop, or, if it is sliding, it must be sliding like a car squealing as it slows down, and not like a race car accelerating. That is, its rotational energy at the top of the loop must be equal to or less than what it would be if it were not sliding at all at that point. This is because any potential energy that does not become rotational energy becomes translational energy, and whether the ball stays or doesn't stay in contact with the track at the top of the loop depends only on its translational energy at that point, So it's clear that if the ramp were ice and the loop were wood (or something with more friction then ice but less than sandpaper,) the ball would stay in contact at the top of the loop. Because in this case, the rotational energy would be less than what it would be if the ball weren't sliding, so the ball would be sliding like the rolling wheels of a car skidding as it slowed. The angular acceleration of the ball would be positive and its linear acceleration would be negative (it would be slowing down), but it would still be going fast enough to stay on the track. Finally, how much higher would the ball have to be if the ramp were sandpaper and the loop were ice? The ball would have to be 2.9 R, because the energy from 0.2 R that made the ball rotate as it went down the ramp would be wasted since it would never be converted into forward motion (translational kinetic energy). so an extra 0.2 R of potential energy would be required to make up the difference.

[@joeteichert1726]

I found it intriguing that you could change the friction of the track to make even 2.7 R insufficient, and hoped that just a few others would too. You make a good point, though, the comment is wordy and unpolished. We do our readers a favor to write as clearly and likeably as possible.

[@bunana902]

woah that's a whole entire essay bro

[@ihaveacold4446]

Bunana hahaha🤣

[@Metal-Possum]

It all depends how big your balls are to accomplish such a manoeuvre.

[@baqi1]

Where can I make or buy a model similar to this?

[@indoororchidsandtropicals358]

How do we find velocity at the top and bottom of the loop if x is zero and vo is zero? I keep getting an irrational number.

[@sabretan3126]

Can someone explain how to get the KErot at 1/5 mg R? Thanks

[@LG-wn7ou]

For rotation, KE = 1/2 (moment of inertia)(angular velocity squared); for a sphere, moment of inertia = 2/5(mass of ball)(radius of ball squared); angular velocity of ball = velocity of ball/(radius of ball); at minimum velocity to complete loop, (velocity at top of loop squared) = (g)(radius of loop)

[@joeteichert1726]

Interestingly, the minimum potential energy the ball must have to complete the loop if the track conditions change is more than 2.7R. If the loop part of the track were frictionless, none of the rolling potential energy would be converted to translational potential energy. Without that extra bit of speed, the ball would lose contact with the track before reaching the top.

[@gauravop4456]

KZbin recommended me this after 9 years xD

[@latinxcombo3612]

I like how you did a video for a lay audience than expected that the audience would know what "mg" or "h0" refers to. Excuse my sarcasm -- it is a good video, and I enjoyed it. But one of the reasons average people confusing is not because of the math, but because you're using jargon that doesn't mean anything to people not in the business. Perhaps next time an explanation of some of the variables behind the math in the video's description would be helpful.

[@coachking5208]

Good point.

[@correodelalberto]

What if you continue the track until it reaches the same height at the other side, could it be kinda perpetual?

[@highspeedfanatics]

Due to Newtonian mechanics, yes it technically should. But you would need a frictionless track, and have to perform it in a perfect vacuum. Both of those would be very difficult to achieve.

[@r3xtez987]

how was the loop built?

[@thegr8844]

happy birthday 🎉🎉🎉🎂🎂🎂🎂

[@phyarth8082]

2.5 - ball air resistance, track surface friction, ball roundness factor must be accounted into experiment, experiment versus ideal paper and pen calculation :)

[@jay_biswas]

This saves me 👍🔥❤️

[@user-uq1fq6gs3i]

Doesn’t the angle of the slope have an effect?

[@Joshuaxiong2]

James Holloway Yes, I thought this also.

[@semerhi]

No. the angle doesn't matter. just the height. Follow the equations.

[@coachking5208]

@@semerhi So if the slope extends over a mile. What then? Surely it matters.

[@Elvis-tx7iy]

How and what materials were used to make the experiment?

[@cheekibreeki7157]

My professor used small steal balls, it’s slipping the whole time and actually requires 5.1 R to make the loop

[@usamachaudary4627]

Is speed of waves depends upon the number of loops?

[@geometryjosh21]

Is that a 45° angle?

[@diatribeeverything]

No, just think energy

[@parkercazier7118]

What does the slope have to be for this to work?

[@geometryjosh21]

I'd say around 45°

[@Test-mm6wl]

Please, what are the materials? : ´´

[@virgilioborlotti1057]

Grazie.

[@brawlbro7384]

We had a competition to see who could drop it from the lowest and we ended up calculating 2.7r but we dropped it from 3.1r to be safe and the ball still looked like when you dropped it from 2r

[@alexandrabaluti7306]

sorry but where did u get the material from?

[@rkguptasmart442]

why so serious?

[@dianaf4047]

where did you get the materials?

[@rexconsul2730]

What are the materials you have used on your track and also the height of the track

[@SportsHajwala20980]

I tryed this from a bucket

[@brendapaint]

what point on the track is the potential energy?

[@amarsai22]

where is air resistance!!!

[@notarealgamer07]

up your ass, who cares?

[@chambalgator]

Tysm

[@sydnistarshine5558]

good day mate 😇

[@CristobalAlbarran-we3em]

😂❤

[@arquimedesjaviersriverodia2859]

Half LOOP up

[@andrewturner2014]

I think this guy just really likes to play fetch, he needs a dog.

[@Joshuaxiong2]

Obviously.