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2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math


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>> No.16190343 [View]

>>16190080
idk if you're esl or not, because i cannot parse your sentence at all. the equations of motion don't reflect the velocities being disproportionately gained. furthermore i fail to see how energy could be conserved in your scenario.

>> No.16190727 [View]

>>16190343
I am too lazy to check exactly what you wrote. But it is obvious that they lose contact way before y=r.

Imagine epsilon before this moment. At that point the ball is basically vertical at the contact point, so it is moving down dows not increase its width at that height. However, the boxes are moving outwards, so moving further away from the ball. So if the ball is touching the boxes at some time eps before y=r, it can't also be touching them then.

Basically what you can see is that at some point, even though the ball is getting faster, its rate of "widening" at the contact height start getting slower, while the boxes only start moving apart faster and faster. Therefore, at that moment they must not be in contact.

Others' answers have determined the point when they lose contact. I'd be interested to see what your mistake is once you find it.

>> No.16190764 [View]

>>16189845
It could happen. Like a rally car can bounce off from the road, and lagragian mechs goes unpredictable

>> No.16190924 [View]

>>16190727
the other results don't say what the speed of the blocks are, and i suspect they violate conservation of energy

>> No.16190943 [View]

>>16190924
I computed stuff from scratch following the described approaches and got the same thing. Can you find the error there?

>> No.16191096 [View]

>>16190924
>other results don't say what the speed of the blocks are
Because it isn't needed.
Use >>16187558
>h=2r/3
>h' = -sqrt(10gr/27)
and plug them in
>x'^2 + h'^2 = gh
x'^2 = 8gr/27

When all of the potential (2mgr) is converted to kinetic, you get:
(m/2)*(100gr/27 + 8gr/27) = 2mgr

>> No.16191106 [View]
File: 4 KB, 275x183, 1698096807572433.png [View same] [iqdb] [saucenao] [google]
16191106

>>16189845
What if the cubes were much lighter than the ball?

>> No.16191136 [View]
File: 97 KB, 486x580, 1696561980800310.png [View same] [iqdb] [saucenao] [google]
16191136

Retard here. So obviously (r-h)^2 + x^2 = r^2. But why can't we differentiate this equation? If you do, you get something wrong that says when h = r, dx/dt = 0

>> No.16191158 [View]

>>16190924
>dont say the speed of the blocks
see >>16188262

>> No.16191162 [DELETED]  [View]

>>16191136
constant + variable = constant?
so x is constant? of course the derivative is 0
There are a couple eq. already posted that give the right equation u want

>> No.16191171 [View]

>>16191136
the block has already left the ball by h=r, so the eq aint valid there

>> No.16191463 [View]

>>16191106
the two blocks combined have the same mass as the sphere so that's where symmetry comes from

>> No.16191671 [View]

>>16185263
OP already stated it was a high school competition. Tried some common sense, lately?

>> No.16191740 [View]

>>16191171
ok but what about h = r - q where q is very small

>> No.16191801 [View]

>>16191740
read the thread to find the final speed for both objects, and when it detaches, and how to find it

>> No.16191816 [View]

>>16191801
There is nothing in the thread that answers my question.

>> No.16191845 [View]
File: 10 KB, 732x476, Lose contact.png [View same] [iqdb] [saucenao] [google]
16191845

>>16187558
>Let h be the height of the ball center above the top of boxes.
>h=2r/3
i find this sus. i've drawn the image to scale, so that you all can visualize the claim being made for when the sphere detaches from the cubes.

>> No.16192008 [View]

>>16185250
>there is no friction
>sitting on midpoint of 2 planes
Oof

>> No.16192050 [View]

>>16191845
>i find this sus
Then you aren't good at finding.

>> No.16192199 [View]

The ball is inert

>> No.16192259 [View]

>>16191816
You have no question

>> No.16192294 [View]

>>16192259
You just don't understand my question. You're an idiot.

>> No.16192411 [View]

>>16191845
The boxes accelerate much faster than the ball, because they have to move further sideways than the ball has to move down.
In your image, the ball has only travelled by r/3, but the boxes have gone way more to the side than that. So the boxes must be moving faster.

>> No.16192507 [View]

>>16192411
this can't be right. the constraint was supposed to be valid for h > 2r/3, meaning the ball keeps contact with the cubes until precisely that height. therefore the picture i drew must be valid (they're in contact). and that's why i think the proposed solution is wrong, and mine is right (and consistent).
>>16189571

>> No.16192534 [DELETED]  [View]

>>16185206
v = 2 * sqrt(g * r)

lol

>> No.16192909 [View]

>>16192507
>and mine is right
What is the definition of the lagrangian?
What did you write here >>16189571 ?
Cope and seethe

>> No.16192917 [View]

>>16189571
How is y'' positive? Isn't gravity pulling down?

>> No.16193126 [View]

>>16192917
He triple checked it, stop questioning him, you esl or something?

>> No.16193294 [View]

>>16192909
change the potential energy to a negative sign then, i still get the same EOM keeping y'' the same sign as y.
>>16192917
who said y'' is positive? it has the same sign as y (as it should).

>> No.16193302 [View]

>>16185206
gravity is splitting between the squares applying a force on the corners creating a rotational spin, but the square wont spin because it's on the ground so only the perpendicular forces will act
this is just my guess and I don't know how to actually compute an exact analytical solution.

>> No.16193390 [View]

>>16193302
Don't try to solve the DE. You just need to find what everything is when x'' is 0. >>16187558

>> No.16193637 [View]

0 m/s

>> No.16193789 [View]

>>16193302
>applying a force on the corners creating a rotational spin
That would only happen with friction.

>> No.16193792 [View]

>>16192294
If you are so smart, then explain what you meant

>> No.16193800 [View]

>>16193789
Torque is applied since the force between the ball and the boxes is not vertical. In fact it could be enough to tip the boxes outward if the ball is heavy enough (perhaps not with the masses specified in OP).

>> No.16193842 [View]

>>16191136
>>16191740
You can differentiate the equation, it just becomes invalid when the ball loses contact. This happens when the contact force is 0, which implies x’’ = 0. When you differentiate the equation, apply conservation of energy, and solve for x’, you get a function for x’ that increases from 0, reaches a maximum, and then decreases back to 0 at x = r. The maximum of x’ is where x’’=0 and the ball loses contact with the cubes. So the equation is invalid after the maximum, where x’’ < 0 and x’ is decreasing. The force on the cubes is only ever outwards. They accelerate until they lose contact with the ball, after which they continue on with constant velocity.

>> No.16193912 [View]
File: 70 KB, 1280x720, 1696733820969682.jpg [View same] [iqdb] [saucenao] [google]
16193912

>>16193842
Thanks, I geddit now

>> No.16194269 [View]

>>16193792
I’m the guy who responded to his question. I already understand it. Stop responding to him, his comments are dumb and aren’t worth typing something up.

>> No.16195617 [View]

Wow thread is finally done with schizo posters.

>> No.16197354 [View]

>>16185206
by energy conservation,
mg(3r)=0.5 * m * v^2
v= sqrt(6*g*r)

>> No.16197369 [View]

>>16185206
Why do soientists think they can know everything and why do they try to brainwash everyone's children into their delusional cults?

>> No.16197702 [View]
File: 1.40 MB, 818x815, Screenshot 2023-11-04 221308.png [View same] [iqdb] [saucenao] [google]
16197702

>>16185206
Trick question. The ball will never hit the ground because none of the objects are under the influence of any forces.
>source: gravity, electromagnetism, etc, etc, weren't mentioned in the question

>> No.16197718 [View]

>>16193390
I don't understand it's doesn't start that way though
>>16193789
what does friction have to do with inertia?

>> No.16198507 [View]

>>16185206
g(4r^2)

>> No.16198721 [View]
File: 29 KB, 2532x1274, physics and shit.png [View same] [iqdb] [saucenao] [google]
16198721

>>16185696
>>16185696
Theres a space epsilon between the squares so they would slide out from under the ball

>> No.16198730 [View]
File: 61 KB, 2532x1274, did i do it.png [View same] [iqdb] [saucenao] [google]
16198730

>>16198507
>>16185206
did i do it right?

>> No.16198731 [View]

>>16198507
the inititial height is 2r because it stops when the bottom of the ball hits the ground

>> No.16198746 [DELETED]  [View]

>Nobody in the thread has my answer yet.

because its fucking wrong

>> No.16198747 [DELETED]  [View]

>>16189571
>Nobody in the thread has my answer yet.

because its wrong

>> No.16198979 [View]

>>16198721
>>16198730
The right box would rotate counter-clockwise while it moves right and the left box would rotate clockwise while it moves left as you have a non-zero torque.
With epsilon near zero the torque for each box is something like mgr/2. You can't ignore that.