This was a high school competition problem in my country. Can IQfy solve it?

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# This was a high school competition problem in my country. Can?

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This was a high school competition problem in my country. Can IQfy solve it?

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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?

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.

>obvious

source?

Keep coping because you can't solve it lmao

But earths atmosphere has air resistance. I think this is more likely on the moon

Even space has air resistance.

peer reviewed fully funded journal study citation for that, chuddy?

Op here, constant downwards gravitational accelation g is implied in such problems.

>implied

source?

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

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.

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.

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 b***h, let's not be stupid here.

moronic pseud

>there is no friction

>sitting on midpoint of 2 planes

Oof

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

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.

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).

>OP posts a high school math problem

>surprised that underage smartasses flood the thread

>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 homosexuals like this.

One look at the catalog should've told you about the average intelligence of this board.

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.

Okay, do it?

I'm getting sqrt(4gr) unless there is something tricky happening.

Basically except you do not need to solve the differential equation.

no diff eq

units are bad

this is the same if the block isnt there

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.

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?

u need to learn about equilibriums and stability

Here's a diagram (the symmetry is implied). i may probably be double counting some cos thetas, not sure.

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?

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

>u need to learn about equilibriums and stability

And you, a massive moron 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 IQfy is full of moronic STEM dropouts

?

u solve in terms of epsilon tardy

Now give me my pip

I know how to whip up a storm - simple in some eyes - judging by the size of my wiener - my attempts at sex - and theirs in comparison. You distinguish it this way. This is falsehood.

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...

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.

Hell doesn't scorn me, neither does heaven. Life does, life does not know all yet. So we should suffer for the unchanging?

Your forgiven - (joke) - I'll take your hell - etc.

End.

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.

wait i used wrong formel. thats the fall time.

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

How do you know it won't lose contact before then?

i dont. maybe moving masses without friction without changing high dont even need energy. then would it free fall from start.

It will keep adding kinetic energy to the blocks while it is in contact so it won't be free fall from the start.

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.

maybe the kinetic energy from the ball equals the kinetic energy from the 2 quebes.

maybe its a trick question. isnt the max fall speed in vaccum very soon gained ?

so the answer is 9.8 m/s

I genuinely don't understand the question. Nothing is moving, so the ball doesn't fall? Did you forget to specify that someone pushes the ball or something?

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.

Gravity.

IQ low

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

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.

[math]sqrt((1 +pi/2)mgr)[/math]

No.

2sqrt(11gr/15) probably doing something wrong?

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).

>This was a high school competition problem in my country

what, are you 12 or something? Imagine caring about high school comp problems kek

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..

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.

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.

uuh... what's the inverse of the Jacobi amplitude function?

>what's the inverse of the Jacobi amplitude function?

The noitcnuf edutilpma ibocaJ

(2+sqrt(2))*sqrt(gr)

shouldn't any value larger than 2sqrt(gr) not be possible as it implies the system gains energy?

Did you account for dark energy?

oops you're right of course, that was moron algebra.

How about sqrt(8/3*gr)?

your result is the same as this one

but with cos^3 instead of cos^5 i think. I may have been applying the forces wrong.

*almost the same

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.

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.

>This implies gravitational potential between all three.

okay. apply einstein's law of gravitation. there is no earth in the problem.

>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.

learn calculus they said

then you end up like this. frick maths

It's just going to fall to the ground. Nothingburger.

All of those morons in the thread feverishly doing OP's homework. Kek, they'll never learn.

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. Frick this poorly worded stupid bait ass question.

Schizo. Take the limit of the answer as epsilon goes to 0. There are standard conventions for succinctly communicating such problems to non-morons.

i came to the same conclusion here

>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.

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.

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

>this is done via friction.

lol, moron 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.

this gets really ugly when you try to account for the fact the ball makes the boxes rotate

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.

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

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}{3sqrt{3}} sqrt{gr}.

[/math]

Or something like that, I haven't really double-checked anything.

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 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.

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.

>recontact

ooooh did not even think about this. Ok i did the calc. and no, it does not rehit, but it fricking narrowly misses by a little less than r/8.

I noticed that this is is 3 times the final speed of the blocks

>massive moron 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

Solving in terms of an arbitrary initial [math]epsilon[/math], I get [math]v = sqrt{2g[r+sqrt{4r^2-{epsilon^2}}(frac{23}{54}+frac{1}{54}frac{epsilon^2}{r^2}})][/math]. As [math]epsilonto0[/math], [math]vtosqrt{frac{100}{27}gr}approxsqrt{3.703gr}<sqrt{4gr}[/math], which agrees with

. Also, as [math]epsilonto2r[/math], [math]vtosqrt{2gr}[/math], which is the velocity expected if the sphere starts free falling with its center at the same height as the top of the cubes. The units also work, so hopefully it's correct.

Yeah these are correct.

You are moronic if you think there isn't enough info. The full initial configuration is there. How could you think it's not enough?

>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.

>i find this sus

Then you aren't good at finding.

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.

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).

>and mine is right

What is the definition of the lagrangian?

What did you write here

?

Cope and seethe

change the potential energy to a negative sign then, i still get the same EOM keeping y'' the same sign as y.

who said y'' is positive? it has the same sign as y (as it should).

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.

*two blocks

Why mgcos^2(theta)?

0 m/s.

I however cannot tell you its speed immediately before it hits the ground.

The fall time tends to infinity -- the limit does not exist.

Fall time yes, but velocity at impact does not.

[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.

man, this reminds me of the stuff i had to do for college phys. Don't miss it, but sorta fun

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).

The boxes absolutely do start to rotate while getting pushed since the force is not applied to the center of mass of the boxes.

That would be true if there was no ground on which they're sitting.

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.

Given there have been 94(95 now) posts and not one agreed upon answer, plus OPs homosexual '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 homosexual, as per usual.

>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.

Post your lagrangian, you ridiculous pseud. I bet it's wrong.

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}mleft(dot{x}^2 + dot{y}^2right) - mgy [/math]

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.

this guy

is right btw. This is what makes it a harder HS problem

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} = xgsqrt{1-left(frac{x}{r}right)^2}[/math]

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.

ur units are off, but close enough

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 fricked up a constant somewhere since i found energy isn't conserved while the ball is in contact with the cubes.

>energy isn't conserved in this frctionless problem

wrong

esl? since my answer violated conservation of energy, i made an error. i'll look at it tomorrow with a fresher mind.

It's something between 2 and 300 m/s

Why is the ball going to hit the ground? Do the cubes vanish? What idiot write this question?

Ignore the other answers in the thread

Do the same problem but with an upside down equilateral triangle with a side length of r.

Why would it hit the ground?

i think it's

[math]

sqrt{ 4 g r}

[/math]

tho i may have messed up some vectors, or forgotten my vector calculus

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.

Well I wrote down the energy conservation equation. That's gotta count for something

The ball never hits the ground

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}mleft(dot{x}^2 + dot{y}^2right) + 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} - xlambda = 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.

by v_block, of course i mean v_sphere.

>y-r-y=0

[math]yneqsqrt{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.

>note this is when the sphere loses contact with the cubes, which is at y=r

This has been deboonked.

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.

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.

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.

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.

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

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

>other results don't say what the speed of the blocks are

Because it isn't needed.

Use

>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

>dont say the speed of the blocks

see

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

What if the cubes were much lighter than the ball?

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

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

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

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?

moron 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

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

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

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

There is nothing in the thread that answers my question.

You have no question

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

If you are so smart, then explain what you meant

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.

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.

Thanks, I geddit now

The ball is inert

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.

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

I don't understand it's doesn't start that way though

what does friction have to do with inertia?

>applying a force on the corners creating a rotational spin

That would only happen with friction.

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).

0 m/s

Wow thread is finally done with schizo posters.

by energy conservation,

mg(3r)=0.5 * m * v^2

v= sqrt(6*g*r)

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

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

g(4r^2)

did i do it right?

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