🧵 Untitled Thread
Anonymous at Mon, 20 May 2024 22:36:41 UTC No. 16185206
This was a high school competition problem in my country. Can /sci/ solve it?
Anonymous at Mon, 20 May 2024 23:03:12 UTC No. 16185250
question does not specify the nature of the force which would cause any change in position, so the question is unanswerable without assumptions. should we assume classical universal gravitation, general relativity, constant acceleration from gravity near the Earth's surface or some other source, ...? in the limit as the downward force approaches zero, the velocity approaches zero as well, by simple conservation of energy. or we can simply assume no force is applied (none is mentioned) and thus it's a trick question and the ball never hits. now what's my prize?
Anonymous at Mon, 20 May 2024 23:06:25 UTC No. 16185251
>>16185250
That's a lot of cope for not being able to solve it. Obviously the force is constant acceleration from gravity near the Earth's surface, even if it's not explicitly stated.
Anonymous at Mon, 20 May 2024 23:09:43 UTC No. 16185257
>>16185250
Op here, constant downwards gravitational accelation g is implied in such problems.
Anonymous at Mon, 20 May 2024 23:12:23 UTC No. 16185260
>>16185251
>obvious
source?
Anonymous at Mon, 20 May 2024 23:13:24 UTC No. 16185263
>>16185257
>implied
source?
Anonymous at Mon, 20 May 2024 23:15:55 UTC No. 16185265
>>16185260
>>16185263
Keep coping because you can't solve it lmao
Anonymous at Tue, 21 May 2024 01:18:33 UTC No. 16185411
>>16185206
Apply assumption that if all objects are perfectly rigid, and the implication of a collision with the phrasing of "hits the ground," the ground must be an impermeable entity, and by virtue of this, any object in contact with the ground moving towards it will have 0 velocity in the component orthogonal to the ground's surface, as the ball cannot tunnel into the ground, or deform at the moment of impact
Anonymous at Tue, 21 May 2024 03:46:37 UTC No. 16185539
>>16185411
Why does everyone in this thread feel the need to act smart. This is a standardly phrased mechanics question. Clearly we are talking about the instant right before it hits the ground, i.e. epsilon time before.
Also you are wrong, for a perfectly rigid body, it transitions from velocity V right before hitting to velocity -V right after hitting. There is never an instant when the velocity is 0, as that would violate convervation of energy. If anything it is undefined at the moment when it contacts the ground.
Anonymous at Tue, 21 May 2024 04:16:02 UTC No. 16185569
>>16185206
Sum of energies constant.
Simple pythag relating ball height and block displacement.
Eliminate variable.
Solve differential equation.
Check when contact is lost.
Simple after that.
Anonymous at Tue, 21 May 2024 04:18:07 UTC No. 16185572
>>16185569
Okay, do it?
Anonymous at Tue, 21 May 2024 04:28:39 UTC No. 16185580
>>16185206
the ball will free fall r distance. becouse after r distance the 2 blocks are shoved away for free fall.
so its root(2r/g)
i ignore start velocity becouse its a highschool question.
Anonymous at Tue, 21 May 2024 04:31:45 UTC No. 16185582
>>16185580
wait i used wrong formel. thats the fall time.
Anonymous at Tue, 21 May 2024 04:32:40 UTC No. 16185583
>>16185580
How do you know it won't lose contact before then?
Anonymous at Tue, 21 May 2024 04:34:15 UTC No. 16185586
>>16185583
i dont. maybe moving masses without friction without changing high dont even need energy. then would it free fall from start.
Anonymous at Tue, 21 May 2024 04:35:46 UTC No. 16185587
>>16185586
It will keep adding kinetic energy to the blocks while it is in contact so it won't be free fall from the start.
Anonymous at Tue, 21 May 2024 04:38:14 UTC No. 16185589
>>16185587
maybe that the trick the ball transfers his fall energy 100% in the first r length fall.
So that you have the remaining r free fall without start energy.
Anonymous at Tue, 21 May 2024 04:57:18 UTC No. 16185609
>>16185587
>>16185589
maybe the kinetic energy from the ball equals the kinetic energy from the 2 quebes.
🗑️ Anonymous at Tue, 21 May 2024 05:45:32 UTC No. 16185676
>>16185206
√6gr
faggot
Anonymous at Tue, 21 May 2024 05:47:12 UTC No. 16185681
>>16185206
maybe its a trick question. isnt the max fall speed in vaccum very soon gained ?
so the answer is 9.8 m/s
Anonymous at Tue, 21 May 2024 06:42:59 UTC No. 16185720
>>16185539
People do this a lot when faced with any kind of riddle or brainteaser, not just science/mathematics problems. They'd rather "outsmart" the question by nitpicking some small detail or loophole, thereby absolving themselves of having to make a good-faith attempt at the problem and risk the worst fate imaginable (getting the wrong answer).
Anonymous at Tue, 21 May 2024 07:56:02 UTC No. 16185774
>>16185676
Why?
Anonymous at Tue, 21 May 2024 08:09:21 UTC No. 16185791
>>16185206
[math]\sqrt((1 +\pi/2)mgr)[/math]
Anonymous at Tue, 21 May 2024 08:18:47 UTC No. 16185809
>>16185250
Crazy cope. Goes without saying the ball is affected by gravity and with the starting distance not being 0 and there being no friction it will hit the ground.
>>16185206
So uh, he ball falls 2r - eps. It will horizontally displace both cubes by r-eps/2. You do see how the eps causes alot of unnecessary trouble ?
At first almost all force is used to accelerate the cubes. At the time the ball has fallen r the force to accelerate the cubes has reached 0, the cubes will continue their outward trajectories and the ball will fall as normal.
This would mean one needs to come up with a term for the speed the ball is starting it's free fall with after dropping 1r.
Anonymous at Tue, 21 May 2024 09:07:43 UTC No. 16185876
>>16185206
I'm getting sqrt(4gr) unless there is something tricky happening.
Basically >>16185569 except you do not need to solve the differential equation.
Barkon at Tue, 21 May 2024 09:22:38 UTC No. 16185903
>>16185791
No.
Anonymous at Tue, 21 May 2024 09:30:38 UTC No. 16185915
>>16185696
That is true if epsilon is 0. If epsilon is not 0 the ball exerts a slight horizontal force on the boxes based on the angle of the point of contact, meaning the boxes will move away from eachother and increase this angle at the point of contact, making the ball fall.
Anonymous at Tue, 21 May 2024 09:48:20 UTC No. 16185943
>>16185206
2sqrt(11gr/15) probably doing something wrong?
Anonymous at Tue, 21 May 2024 09:49:23 UTC No. 16185945
>>16185251
But earths atmosphere has air resistance. I think this is more likely on the moon
Anonymous at Tue, 21 May 2024 09:54:24 UTC No. 16185955
>>16185696
Gravity.
Barkon at Tue, 21 May 2024 09:56:41 UTC No. 16185957
>>16185955
IQ low
Anonymous at Tue, 21 May 2024 09:59:36 UTC No. 16185960
>>16185206
>This was a high school competition problem in my country
what, are you 12 or something? Imagine caring about high school comp problems kek
Anonymous at Tue, 21 May 2024 10:00:37 UTC No. 16185962
>>16185539
>OP posts a high school math problem
>surprised that underage smartasses flood the thread
Anonymous at Tue, 21 May 2024 10:45:01 UTC No. 16186008
question is inconsistent with itself. on the one hand, you're lead to believe that the ball will push the squares aside, thus causing them to have a velocity. but if you write down the generalized lagrangian, you'll see the squares have no horizontal component of potential energy, and thus the euler lagrange equations imply their velocities are constant. the initial condition is their velocity is zero, so the squares don't move and the ball never hits the ground. the only way out is to assume quasistatic motion of the squares such that the ball is moving at infinitesimal speeds at all times and that it hits the ground with speed zero, which is patently absurd. you need to include friction for this problem to work..
Anonymous at Tue, 21 May 2024 11:23:09 UTC No. 16186041
>>16186008
Sure they will move because the ball pushes them apart it's just obvious. The problem is to ask what is the limit of the end velocity of the ball when the gap size approaches zero.
Anonymous at Tue, 21 May 2024 11:26:33 UTC No. 16186044
>>16186008
Amateur.
I reject the premise because it implies a sphere has equal mass to a cube. This is absurd because nobody could reproduce a cube to the tolerance of pi. Had they included some error metric, then I would be more inclined. As it stands, I don't waste time solving problems in universes that don't exist.
Anonymous at Tue, 21 May 2024 11:34:35 UTC No. 16186051
>>16185943
I integrated the force along the path the ball takes which I then set be to kinetic energy to obtain this value. KE = mgr + mgr*(integral{0, pi/2} of cos(t)-cos^5(t)dt).
Anonymous at Tue, 21 May 2024 12:15:07 UTC No. 16186086
uuh... what's the inverse of the Jacobi amplitude function?
Anonymous at Tue, 21 May 2024 15:23:31 UTC No. 16186257
(2+sqrt(2))*sqrt(gr)
Anonymous at Tue, 21 May 2024 15:26:16 UTC No. 16186260
>>16186086
>what's the inverse of the Jacobi amplitude function?
The noitcnuf edutilpma ibocaJ
Anonymous at Tue, 21 May 2024 15:37:03 UTC No. 16186269
>>16186257
shouldn't any value larger than 2sqrt(gr) not be possible as it implies the system gains energy?
Anonymous at Tue, 21 May 2024 15:57:23 UTC No. 16186304
>>16186269
Did you account for dark energy?
Anonymous at Tue, 21 May 2024 16:04:08 UTC No. 16186318
>>16185206
0. The ball will never hit the ground because they are all at rest and no force is acting on them. There's no gravity in this problem and everyone who assumed otherwise just got filtered.
Anonymous at Tue, 21 May 2024 16:05:33 UTC No. 16186324
>>16185945
Even space has air resistance.>>16185943
Anonymous at Tue, 21 May 2024 16:40:33 UTC No. 16186369
>>16186269
oops you're right of course, that was retard algebra.
How about sqrt(8/3*gr)?
Anonymous at Tue, 21 May 2024 16:54:58 UTC No. 16186381
>>16186369
your result is the same as this one >>16186051
but with cos^3 instead of cos^5 i think. I may have been applying the forces wrong.
Anonymous at Tue, 21 May 2024 16:56:25 UTC No. 16186383
>>16186381
*almost the same
Anonymous at Tue, 21 May 2024 17:00:10 UTC No. 16186386
>>16186318
There is mass. This implies gravitational potential between all three. Sorry pseud, you failed to see that the system is in a vacuum and we are looking for the final orientation of the composite elements.
The tendencies are the following:
1. The sphere drives the cubes apart, to centralize the mass
2. The cubes rotate against the sphere and stagger / lock into eachother
3. The edges / corners never perfectly balanced against one another so the system remains dynamic, the ball consistently pushes into the cubes.
4. This cyclic interaction builds rotation energy into the system which eventually balances against this.
5. The chirality ultimately determines the final configuration, but the ball is going to lean into a cube going in the same direction of the rotating mass.
6. Notice that in a zero friction state, the rotation is going to be very slow, otherwise displacement will kick the system into orbital mechanics and elastic collision. This change in behavior is controlled by proportion of m to r.
Anonymous at Tue, 21 May 2024 17:20:12 UTC No. 16186410
>>16186386
>This implies gravitational potential between all three.
okay. apply einstein's law of gravitation. there is no earth in the problem.
Anonymous at Tue, 21 May 2024 18:59:24 UTC No. 16186581
>>16185809
Well clearly the question is asking about final velocity an not fall time because that approaches a limit as epsilon goes to 0.
>>16185791
>>16185876
>>16185943
>>16186369
Show your work?
>>16186008
It is completely obvious the ball will push them aside. What are you on?
>>16186386
Schizo
Anonymous at Tue, 21 May 2024 19:34:23 UTC No. 16186609
>>16185206
It's just going to fall to the ground. Nothingburger.
Anonymous at Tue, 21 May 2024 20:29:42 UTC No. 16186667
All of those retards in the thread feverishly doing OP's homework. Kek, they'll never learn.
Anonymous at Tue, 21 May 2024 20:31:25 UTC No. 16186672
>>16185206
Insufficient information regarding to change in conditions from the OP question so here I am to give you three velocity when it hits the ground.
If you want to be more technical then the first v = 0 isn't quite answering the "hits the ground" question because it will always be there. Fuck this poorly worded stupid bait ass question.
Anonymous at Tue, 21 May 2024 20:56:54 UTC No. 16186732
>>16186672
Schizo. Take the limit of the answer as epsilon goes to 0. There are standard conventions for succinctly communicating such problems to non-retards.
Anonymous at Tue, 21 May 2024 21:30:55 UTC No. 16186789
>>16186672
i came to the same conclusion here >>16186008
>>16186581
>It is completely obvious the ball will push them aside. What are you on?
meds. the point of my post was that despite this "obviousness", the information given in the problem indicates the cubes have a constant zero velocity. don't believe me? write the lagrangian and solve the euler lagrange equations.
Anonymous at Tue, 21 May 2024 21:37:43 UTC No. 16186804
>>16186789
Clearly impossible. If everything is stationary then the ball must be getting pushed up by the boxes with force G/2 from each. But since epsilon is positive, there must be a positive component of the force in the x direction, so the boxes accelerate.
Anonymous at Tue, 21 May 2024 21:47:13 UTC No. 16186825
>>16186804
here let's try this instead. in reality, the sphere will transfer some of its stored gravitational potential energy to the kinetic energy of the cubes. this is done via friction. mathematically, friction couples the cubes to the sphere and gives a way for the cubes to acquire kinetic energy. now remove friction from the problem. what physical parameter couples the sphere to the cubes? nothing. so there's no way the cubes can acquire kinetic energy, which is why the euler-lagrange equations indicate the cubes have a constant zero-velocity. ironically enough by removing friction from the problem, you've forced the system into static equilibrium lol
Anonymous at Tue, 21 May 2024 22:04:02 UTC No. 16186856
>>16185206
this gets really ugly when you try to account for the fact the ball makes the boxes rotate
Anonymous at Tue, 21 May 2024 22:14:44 UTC No. 16186871
>>16186825
>this is done via friction.
lol, retard alert. The force acts on the edges that touch the sphere's curvature in one point at a time. It has as little to do with friction as billiard balls transferring momentum to each other. The lack of friction ensures the edges move along the sphere till they stop touching it, and that the boxes move apart.
Anonymous at Tue, 21 May 2024 22:58:52 UTC No. 16186930
all the potential energy of the ball is converted into kinetic energy of the cubes. now you can calculate how much would a force transfering the same energy to the ball slow it down
Anonymous at Tue, 21 May 2024 23:04:31 UTC No. 16186939
>>16185206
If I'm not mistaken, this mostly reduces to a pendulum starting out inverted. If [math]\theta(t)[/math] is the angular distance between the bottom of the ball and the box corners (with [math]\theta(-\infty) = 0[/math]), we get
[math]
\dot{\theta}^2 = \frac{2g}{r}(1 - \cos \theta).
[/math]
This holds as long as the velocity of a box (which is [math]v = \partial_t(r \sin \theta)[/math]) is increasing: for [math]\dot{v} > 0[/math], the ball is pushing the boxes, [math]\dot{v} < 0[/math] would correspond to the ball pulling the boxes closer together. The separation point [math]\theta_1[/math] is thus at [math]\dot{v} = 0[/math], which you can solve for
[math]
\cos \theta_1 = \frac{2}{3}.
[/math]
Then you calculate the energy of the ball at separation time (after which it is in free fall), and eventually get a final velocity of
[math]
\frac{10}{3\sqrt{3}} \sqrt{gr}.
[/math]
Or something like that, I haven't really double-checked anything.
Anonymous at Tue, 21 May 2024 23:06:03 UTC No. 16186943
>>16186386
>There is mass. This implies gravitational potential between all three.
Mass and weight are completely separate things and mass does not imply gravity. Sorry pseud, you failed to see that the system is in a vacuum.
Anonymous at Tue, 21 May 2024 23:16:03 UTC No. 16186961
>>16185539
>Why does everyone in this thread feel the need to act smart.
It's because barely anyone on this board actually is smart. You can see this pretty much any time a problem is posted, 80% of the thread will be faggots like this.
One look at the catalog should've told you about the average intelligence of this board.
Anonymous at Wed, 22 May 2024 00:43:02 UTC No. 16187082
>>16186581
2sqrt(11gr/15) comes from integrating the forces acting on the ball up until it hits the ground, the integral is split into two parts one when it has contact with the two balls and one when it doesnt, the (r-length) integral when it doesnt have contact is just mgr. I then integrate across the r-length segment where it does have contact which I express in terms of the angle from the vertical.
so the integral for that segment is int{0 to pi/2} of rcos(theta)f(theta) where f gives the force acting on the ball. Im not sure if my f is correct so feel free to criticize this reasoning, but heres what I did: the downwards force on the (both)blocks from the ball at angle theta should mgcos^2(theta) which is what the normal force resists so then we find the blocks are pushing mgcos^2(theta)cos^2(theta) = mgcos^4(theta) upwards on the ball in return. and then I just add the force of gravity mg acting on the block so for some given angle theta the force acting on the ball (might) be mg - mgcos^4(theta) so our integral is rcos(theta)(mg - mgcos^4(theta)). I then took the integral and set it to be the kinetic energy of the ball and just rearranged in terms of v.
someone else in the thread got a similar result but with my cos^5 term changed to cos^3 so Im not sure.
Anonymous at Wed, 22 May 2024 00:44:57 UTC No. 16187085
>>16187082
*two blocks
Anonymous at Wed, 22 May 2024 02:13:55 UTC No. 16187198
0 m/s.
I however cannot tell you its speed immediately before it hits the ground.
Anonymous at Wed, 22 May 2024 03:17:58 UTC No. 16187269
>>16186856
It doesn't though? If you work out a solution, you will see that at no point do they stop touching the ground with their whole bases.
Anonymous at Wed, 22 May 2024 03:24:34 UTC No. 16187277
>>16187082
Why mgcos^2(theta)?
Anonymous at Wed, 22 May 2024 03:36:34 UTC No. 16187299
>>16186581
The fall time tends to infinity -- the limit does not exist.
Anonymous at Wed, 22 May 2024 04:59:04 UTC No. 16187354
>>16187299
Fall time yes, but velocity at impact does not.
Anonymous at Wed, 22 May 2024 06:00:48 UTC No. 16187398
>>16185206
[math] \sqrt{4gr - gr \cdot \tfrac{2}{3} + gr \cdot \tfrac{10}{27} } [/math]
The ball is slowed down during the first r/3 units of distance falling, then it falls at normal acceleration.
If there is no blocks, then at most the speed would've been [math] \sqrt{4gr} [/math], which yall prob already know.
The first term in the sqrt is the normal speed, the second term is what was lost since it's not accelerating normally during the first r/3 units, and the last term is the gain of the weird acceleration during the first r/3 units.
Anonymous at Wed, 22 May 2024 06:02:34 UTC No. 16187399
>>16187398
man, this reminds me of the stuff i had to do for college phys. Don't miss it, but sorta fun
Anonymous at Wed, 22 May 2024 06:44:36 UTC No. 16187429
>>16185569
no diff eq
>>16185791
units are bad
>>16185876
this is the same if the block isnt there
>>16186051
integral over 0 -> pi/2? Also, idk where ur getting mgcos^2 from. Im picturing the diagram in my head and I dont see what's the point of using a vector of that length.
>>16186008
Suppose that the left block is removed. The ball is in unstable eq., so suppose it falls to the left. Ur saying the block aint gonna move to the right to counter the fall to the left?
>>16186672
u need to learn about equilibriums and stability
Anonymous at Wed, 22 May 2024 06:45:37 UTC No. 16187431
>>16185250
You're the type to write walls of text on an exam question that asks for an explanation of an answer where two sentences would suffice. Please, worthless bitch, let's not be stupid here.
Anonymous at Wed, 22 May 2024 06:49:13 UTC No. 16187434
>>16185206
I would just to assume that the ball moves slowly until its center reaches 2r, then it will be free falling for r.
Converting its potential energy mgr into kinetic energy gives me ½mv2.
Then solving for v gives me sqrt(2gr).
Anonymous at Wed, 22 May 2024 07:22:07 UTC No. 16187446
>>16185582
It looks like to find the fall time, you would have to solve a nonlinear differential equation of the form y'' = ay^2 + by. I cant see any other way outside of a numerical approx to evaluate this, at least according to my google search
Anonymous at Wed, 22 May 2024 08:05:04 UTC No. 16187486
The boxes absolutely do start to rotate while getting pushed since the force is not applied to the center of mass of the boxes.
Anonymous at Wed, 22 May 2024 08:34:20 UTC No. 16187509
>>16187486
That would be true if there was no ground on which they're sitting.
Anonymous at Wed, 22 May 2024 08:35:56 UTC No. 16187511
>>16187486
>>16187509
To clarify. Obviously it is possible to apply some sort of force to the boxes that would make them spin even though there is a ground, but you can compute what happens and it doesn't happen here. IT is pretty intuitive, so you can just assume it and then check once you have the answer for the trajectories.
Anonymous at Wed, 22 May 2024 09:06:30 UTC No. 16187526
>>16187429
Here's a diagram (the symmetry is implied). i may probably be double counting some cos thetas, not sure.
Anonymous at Wed, 22 May 2024 09:38:19 UTC No. 16187558
Let h be the height of the ball center above the top of boxes.
Let x be the distance of the boxes from the center line.
From energy:
2g(r-h) = x'^2 + h'^2
-gh' = x'x'' + h'h''
From geometric constraint:
r^2 = x^2 + h^2.
xx' + hh' = 0
xx'' + hh'' + x'^2 + h'^2 = 0
Using intuition from >>16186939 we want to find the state where x''=0.
Let's impose this on the equations above.
-gh' = h'h''
hh'' + x'^2 + h'^2 = 0
This gives x'^2 + h'^2 = gh
Plugging this into the first equation gives:
2(r-h)=h
h=2r/3
using this you can get h' = -sqrt(10gr/27)
Do free fall from there to get v^2 = 100gr/27.
The only thing I'm not sure of is if the ball can re-contact the boxes or not.
Anonymous at Wed, 22 May 2024 09:47:34 UTC No. 16187570
>>16185250
retarded pseud
Anonymous at Wed, 22 May 2024 10:03:21 UTC No. 16187590
>>16187558
I guess you can just check whether the didtance is ever < r after the free fall starts. If not, I guess the collisions become discrete events anyway, so you just have to enumerate them and compute the trajectories of stuff bouncing off.
Anonymous at Wed, 22 May 2024 13:43:16 UTC No. 16187851
>>16187429
>u need to learn about equilibriums and stability
And you, a massive retard on the left of the dunning-kruger curve, need to get a bachelor's in physics and a brain for basic reading comprehension to understand the OP is a shitty ass incomplete bait question with no conditions specified for epsilon.
Gosh really /sci/ is full of retarded STEM dropouts
Anonymous at Wed, 22 May 2024 13:50:19 UTC No. 16187862
>>16187851
?
u solve in terms of epsilon tardy
Anonymous at Wed, 22 May 2024 13:55:37 UTC No. 16187872
Anonymous at Wed, 22 May 2024 13:57:00 UTC No. 16187877
>>16187872
Now give me my pip
Anonymous at Wed, 22 May 2024 13:58:29 UTC No. 16187879
>>16187877
I know how to whip up a storm - simple in some eyes - judging by the size of my cock - my attempts at sex - and theirs in comparison. You distinguish it this way. This is falsehood.
Anonymous at Wed, 22 May 2024 14:00:37 UTC No. 16187884
>>16187879
I suffered quite immensely for 15 years. Prior to that I suffered characteristically for 10. You claim the profit I earned from this position must be lawed in response, even a simple reversal. Come on...
Anonymous at Wed, 22 May 2024 14:01:53 UTC No. 16187887
>>16187884
They would have never let go, if the tables were reversed it would be the max of immorality. That's your only real argument, that immorality doesn't match morality.
Anonymous at Wed, 22 May 2024 14:02:54 UTC No. 16187888
>>16187884
Hell doesn't scorn me, neither does heaven. Life does, life does not know all yet. So we should suffer for the unchanging?
Anonymous at Wed, 22 May 2024 14:03:55 UTC No. 16187890
>>16187888
Your forgiven - (joke) - I'll take your hell - etc.
End.
Anonymous at Wed, 22 May 2024 14:26:01 UTC No. 16187924
>>16187526
Ok, so from that diagram, starting at Fn*cos, what length is Fn? So now, you have an Fn and an Fn*cos^2 that are in the same direction. What are each supposed represent? Do u see the problem here?
Anonymous at Wed, 22 May 2024 14:59:59 UTC No. 16187974
>>16187526
you're multiplying by cos twice to get [math]cos^2 [/math] for no reason, I think you can directly decompose [math]F_N = F_x + F_g = F*cos(\theta) + mg [/math]. You're not creating a triangle whose hypothenuse is Fn, you're projecting Fn over cos and sine and solving back using Fg
Anonymous at Wed, 22 May 2024 15:58:43 UTC No. 16188049
>>16185206
Given there have been 94(95 now) posts and not one agreed upon answer, plus OPs faggot 'You should be able to solve this' addition means there is no real way to solve this since there is a variable or something else missing. Gj op at being a faggot, as per usual.
Anonymous at Wed, 22 May 2024 16:09:38 UTC No. 16188065
>gets answer with forces
>gets contradiction with lagrangian mechanics
>lagrangian mechanics equivalent to force analysis
see the problem?? write the lagrangian and you'll see the problem cannot be solved.
Anonymous at Wed, 22 May 2024 17:04:21 UTC No. 16188152
>>16188065
Post your lagrangian, you ridiculous pseud. I bet it's wrong.
Anonymous at Wed, 22 May 2024 17:27:44 UTC No. 16188179
>>16188152
it isn't a hard lagrangian to write. x represents the blocks' kinetic energy while y represents the sphere's kinetic energy.
[math]\mathcal{L} = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right) - mgy [/math]
Anonymous at Wed, 22 May 2024 18:03:17 UTC No. 16188230
>>16188179
yeah, thanks for this lagrangian for a free-falling sphere and blocks randomly moving in some place, now try the one for the actual problem. Maybe look up holonomic constraints.
Anonymous at Wed, 22 May 2024 18:31:26 UTC No. 16188262
>>16187558
>recontact
ooooh did not even think about this. Ok i did the calc. and no, it does not rehit, but it fucking narrowly misses by a little less than r/8.
>>16186369
I noticed that this is is 3 times the final speed of the blocks
>>16187851
>massive retard on the left of the dk curve
>need to get a bach in phys.
this is really ironic, lol. Dude, at least 3 people itt already got the answer, looks like u got a skill issue
Anonymous at Wed, 22 May 2024 18:52:26 UTC No. 16188296
>>16185251
peer reviewed fully funded journal study citation for that, chuddy?
Anonymous at Wed, 22 May 2024 19:07:49 UTC No. 16188328
>>16188179
this guy >>16188230 is right btw. This is what makes it a harder HS problem
Anonymous at Wed, 22 May 2024 19:21:02 UTC No. 16188350
>>16188328
>>16188230
yeah i forgot about that.
[math]f(x,y,\lambda)=y - r \sqrt{r^2-x^2}[/math]
which yields an ODE for the EOM of the cubes as
[math]\ddot{x} = xg\sqrt{1-\left(\frac{x}{r}\right)^
find the velocity of the cubes when the sphere loses contact. apply conservation of energy to find velocity of sphere when it loses contact. then apply free fall.
Anonymous at Wed, 22 May 2024 19:38:10 UTC No. 16188372
>>16188350
ur units are off, but close enough
Anonymous at Wed, 22 May 2024 19:40:24 UTC No. 16188375
>>16188372
forgot a negative sign is all.
[math]f(x,y,\lambda)=y - r -\sqrt{r^2-x^2} [/math]
i mean it changes a lot. i probably fucked up a constant somewhere since i found energy isn't conserved while the ball is in contact with the cubes.
Anonymous at Wed, 22 May 2024 19:51:25 UTC No. 16188387
>>16188375
>energy isn't conserved in this frctionless problem
wrong
Anonymous at Wed, 22 May 2024 20:20:02 UTC No. 16188450
>>16188387
esl? since my answer violated conservation of energy, i made an error. i'll look at it tomorrow with a fresher mind.
Anonymous at Wed, 22 May 2024 20:40:46 UTC No. 16188490
It's something between 2 and 300 m/s
Anonymous at Wed, 22 May 2024 20:55:41 UTC No. 16188510
>>16185206
Solving in terms of an arbitrary initial [math]\epsilon[/math], I get [math]v = \sqrt{2g[r+\sqrt{4r^2-{\epsilon^2}}
Anonymous at Wed, 22 May 2024 23:18:50 UTC No. 16188678
>>16185206
Why is the ball going to hit the ground? Do the cubes vanish? What idiot write this question?
Anonymous at Wed, 22 May 2024 23:37:46 UTC No. 16188692
>>16185206
Why would it hit the ground?
Anonymous at Thu, 23 May 2024 00:47:32 UTC No. 16188773
Ignore the other answers in the thread
Do the same problem but with an upside down equilateral triangle with a side length of r.
>>16188678
>>16188692
Anonymous at Thu, 23 May 2024 00:51:42 UTC No. 16188778
>>16187558
>>16188510
Yeah these are correct.
>>16188049
You are retarded if you think there isn't enough info. The full initial configuration is there. How could you think it's not enough?
Anonymous at Thu, 23 May 2024 01:47:14 UTC No. 16188825
>>16185206
i think it's
[math]
\sqrt{ 4 g r}
[/math]
tho i may have messed up some vectors, or forgotten my vector calculus
Anonymous at Thu, 23 May 2024 02:11:08 UTC No. 16188851
When it hits the ground, it's speed will be reduced to 0, followed either by a bounce or a lack thereof depending on whether or not the collision is elastic or inelastic.
Anonymous at Thu, 23 May 2024 03:44:07 UTC No. 16188934
Well I wrote down the energy conservation equation. That's gotta count for something
Anonymous at Thu, 23 May 2024 06:39:20 UTC No. 16189061
>>16185206
The ball never hits the ground
Anonymous at Thu, 23 May 2024 14:30:49 UTC No. 16189571
I gave this problem a closer look and triple checked my results. Nobody in the thread has my answer yet. We use Lagrangian mechanics. The relevant Lagrangian is
[math]\mathcal{L} = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right) + mgy[/math]
with a holonomic constraint of
[math]f(x,y,\lambda) = y - r - \sqrt{r^2 - x^2} = 0[/math]
here x is the horizontal distance traveled by one block, and y is the height of the sphere with respect to the cubes' centers of masses. The Euler-Lagrange equations read
[math]\begin{align}
m&\ddot{x} - x\lambda = 0 \\
m&\ddot{y} - mg - (y-r)\lambda = 0
\end{align}
[/math]
Some ridiculous simplifications occur, giving
[math]\begin{align}
\ddot{x} &= xg/r \\
\ddot{y} &= yg/r
\end{align}
[/math]
the implication is that the cubes acquire the same acceleration as the sphere, which should make sense given the symmetry of the masses. One could solve these directly, or invoke symmetry. Nevertheless you find the speed of each block and the sphere are all the same:
[math] v_{\text{block}} = v_{\text{cube}} = \sqrt{rg} [/math]
note this is when the sphere loses contact with the cubes, which is at [math]y=r[/math]
The rest free fall, or energy conservation yielding a final speed of the sphere before impact as
[math]v=\sqrt{3gr}[/math]
note that energy is conserved at every stage here.
Anonymous at Thu, 23 May 2024 14:41:59 UTC No. 16189590
>>16189571
by v_block, of course i mean v_sphere.
Anonymous at Thu, 23 May 2024 15:23:26 UTC No. 16189638
>>16189571
>y-r-y=0
Anonymous at Thu, 23 May 2024 15:32:55 UTC No. 16189650
>>16189638
[math]y\neq\sqrt{r^2-x^2)[/math]
to be more precise, the holonomic constraint is simply recasting
[math]y = r + \sqrt{r^2-x^2}[/math]
notice that when x = 0, y = 2r as expected and when x = r, y = r again as expected.
Anonymous at Thu, 23 May 2024 17:40:12 UTC No. 16189798
>>16189571
>note this is when the sphere loses contact with the cubes, which is at y=r
This has been deboonked.
Anonymous at Thu, 23 May 2024 18:25:21 UTC No. 16189845
>>16189798
only true if the cubes acquire speed at a faster rate than the sphere. which i don't think is true, or at least isn't reflected from the equations of motion.
Anonymous at Thu, 23 May 2024 20:18:37 UTC No. 16190037
>>16185206
Trick question. The ball is sitting on two cubes, not travelling toward the ground. The ball doesn't hit the ground. The ball can't have any velocity when it hits the ground because it's not going to hit the ground anyway. Why would it?
Anonymous at Thu, 23 May 2024 20:45:27 UTC No. 16190080
>>16189845
If they were still touching at y=r, the ball would have to be much faster than the box.
The sides of the ball would be pretty much vertical at the connection point.
Anonymous at Thu, 23 May 2024 23:31:22 UTC No. 16190343
>>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.
Anonymous at Fri, 24 May 2024 06:29:09 UTC No. 16190727
>>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.
Anonymous at Fri, 24 May 2024 07:33:01 UTC No. 16190764
>>16189845
It could happen. Like a rally car can bounce off from the road, and lagragian mechs goes unpredictable
Anonymous at Fri, 24 May 2024 10:38:55 UTC No. 16190924
>>16190727
the other results don't say what the speed of the blocks are, and i suspect they violate conservation of energy
Anonymous at Fri, 24 May 2024 11:01:47 UTC No. 16190943
>>16190924
I computed stuff from scratch following the described approaches and got the same thing. Can you find the error there?
Anonymous at Fri, 24 May 2024 14:22:39 UTC No. 16191096
>>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
Anonymous at Fri, 24 May 2024 14:29:45 UTC No. 16191106
>>16189845
What if the cubes were much lighter than the ball?
Anonymous at Fri, 24 May 2024 15:06:52 UTC No. 16191158
>>16190924
>dont say the speed of the blocks
see >>16188262
🗑️ Anonymous at Fri, 24 May 2024 15:08:28 UTC No. 16191162
>>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
Anonymous at Fri, 24 May 2024 15:12:27 UTC No. 16191171
>>16191136
the block has already left the ball by h=r, so the eq aint valid there
Anonymous at Fri, 24 May 2024 19:07:45 UTC No. 16191463
>>16191106
the two blocks combined have the same mass as the sphere so that's where symmetry comes from
Anonymous at Fri, 24 May 2024 21:28:00 UTC No. 16191671
>>16185263
OP already stated it was a high school competition. Tried some common sense, lately?
Anonymous at Fri, 24 May 2024 22:31:10 UTC No. 16191740
>>16191171
ok but what about h = r - q where q is very small
Anonymous at Fri, 24 May 2024 23:15:26 UTC No. 16191801
>>16191740
read the thread to find the final speed for both objects, and when it detaches, and how to find it
Anonymous at Fri, 24 May 2024 23:27:22 UTC No. 16191816
>>16191801
There is nothing in the thread that answers my question.
Anonymous at Fri, 24 May 2024 23:48:38 UTC No. 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.
Anonymous at Sat, 25 May 2024 03:12:32 UTC No. 16192008
>>16185250
>there is no friction
>sitting on midpoint of 2 planes
Oof
Anonymous at Sat, 25 May 2024 04:27:29 UTC No. 16192050
>>16191845
>i find this sus
Then you aren't good at finding.
Anonymous at Sat, 25 May 2024 06:40:30 UTC No. 16192199
The ball is inert
Anonymous at Sat, 25 May 2024 07:28:40 UTC No. 16192259
>>16191816
You have no question
Anonymous at Sat, 25 May 2024 07:52:44 UTC No. 16192294
>>16192259
You just don't understand my question. You're an idiot.
Anonymous at Sat, 25 May 2024 10:03:30 UTC No. 16192411
>>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.
Anonymous at Sat, 25 May 2024 12:27:06 UTC No. 16192507
>>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
🗑️ Anonymous at Sat, 25 May 2024 12:51:38 UTC No. 16192534
>>16185206
v = 2 * sqrt(g * r)
lol
Anonymous at Sat, 25 May 2024 18:34:34 UTC No. 16192909
>>16192507
>and mine is right
What is the definition of the lagrangian?
What did you write here >>16189571 ?
Cope and seethe
Anonymous at Sat, 25 May 2024 18:39:07 UTC No. 16192917
>>16189571
How is y'' positive? Isn't gravity pulling down?
Anonymous at Sat, 25 May 2024 21:26:06 UTC No. 16193126
>>16192917
He triple checked it, stop questioning him, you esl or something?
Anonymous at Sat, 25 May 2024 23:02:57 UTC No. 16193294
>>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).
Anonymous at Sat, 25 May 2024 23:14:19 UTC No. 16193302
>>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.
Anonymous at Sun, 26 May 2024 00:40:59 UTC No. 16193390
>>16193302
Don't try to solve the DE. You just need to find what everything is when x'' is 0. >>16187558
Anonymous at Sun, 26 May 2024 06:00:17 UTC No. 16193637
0 m/s
Anonymous at Sun, 26 May 2024 08:56:14 UTC No. 16193789
>>16193302
>applying a force on the corners creating a rotational spin
That would only happen with friction.
Anonymous at Sun, 26 May 2024 09:03:04 UTC No. 16193792
>>16192294
If you are so smart, then explain what you meant
Anonymous at Sun, 26 May 2024 09:09:53 UTC No. 16193800
>>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).
Anonymous at Sun, 26 May 2024 09:51:42 UTC No. 16193842
>>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.
Anonymous at Sun, 26 May 2024 10:22:48 UTC No. 16193912
>>16193842
Thanks, I geddit now
Anonymous at Sun, 26 May 2024 15:35:06 UTC No. 16194269
>>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.
Anonymous at Mon, 27 May 2024 11:05:58 UTC No. 16195617
Wow thread is finally done with schizo posters.
Anonymous at Tue, 28 May 2024 08:02:41 UTC No. 16197354
>>16185206
by energy conservation,
mg(3r)=0.5 * m * v^2
v= sqrt(6*g*r)
Anonymous at Tue, 28 May 2024 08:22:00 UTC No. 16197369
>>16185206
Why do soientists think they can know everything and why do they try to brainwash everyone's children into their delusional cults?
Anonymous at Tue, 28 May 2024 13:50:42 UTC No. 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
Anonymous at Tue, 28 May 2024 14:00:13 UTC No. 16197718
>>16193390
I don't understand it's doesn't start that way though
>>16193789
what does friction have to do with inertia?
Anonymous at Wed, 29 May 2024 01:41:34 UTC No. 16198507
>>16185206
g(4r^2)
Anonymous at Wed, 29 May 2024 05:48:35 UTC No. 16198721
>>16185696
>>16185696
Theres a space epsilon between the squares so they would slide out from under the ball
Anonymous at Wed, 29 May 2024 05:52:59 UTC No. 16198730
>>16198507
>>16185206
did i do it right?
Anonymous at Wed, 29 May 2024 05:54:34 UTC No. 16198731
>>16198507
the inititial height is 2r because it stops when the bottom of the ball hits the ground
🗑️ Anonymous at Wed, 29 May 2024 06:04:55 UTC No. 16198747
>>16189571
>Nobody in the thread has my answer yet.
because its wrong
Anonymous at Wed, 29 May 2024 10:58:23 UTC No. 16198979
>>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.