Eastern European Physics Olympiad

[Anonymous No.10122962]

It seems that it would be an easy problem, but is one of the most difficult mechanics problems I've ever encountered. Am I just a tard or?

[Anonymous No.10122969]

Conservation of energy does the trick right? In the beginning there is only the potential energy of the ball. In the end the ball is stationary and the cubes are moving away with the same speed.

[Anonymous No.10122974]

Just use lagrangian mechanics

[Anonymous No.10122976]

Id just set it up as a 2 part action, just set up a motion integral for both boxes to move 1/2D with the equally distributed downward force as a function of gravity. Physically the angle of the sphere (lets say with respect to the downward y axis) is gonna be directly related to the displacement of the box, since itll be in contact with the corner of the box up until displacement is 1/2D. The force on the box will be in the same direction as that angle. Idk this is all pseudo solving so I'm probably oversimplifying it.
>>10122969
Ball wouldn't be stationary at the end, you'd want to figure out the velocity of the ball at the exact moment that it can fall straight down without interacting with the boxes. Uo = KEbox + KEball i guess

[Anonymous No.10122980]

>>10122976
Oh, youd just want the potential energy between the starting position and the position 1/2 D down in space, so just Uo = mg(1/2D)

[Anonymous No.10122987]

>>10122969
>the two balls are stationary
They aren't.

Imagine the ball has fallen half the way, and its diameter follows into the boxes on both sides. This in fact happens.
Work until this point is given by F.d and minimizes the time it takes for this situation to be reached. From then on the rest of the work goes onto the ball's speed.

[Anonymous No.10122992]

>>10122974
>>10122976
>>10122980
Try it, it is far harder than it appears. You 1. have to take into account that the boxes accelerate as well, and they might even do so fast enough to stop touching the ball. At which point, though, they'll stop accelerating and the ball might reach them again and so on. And there are numerous problematic places in the solution even before that.

[Anonymous No.10123001]

>>10122992
>and they might even do so fast enough to stop touching the ball.
No they wont, theyll only be moving when the ball is applying a force to them, which would be equal to the force applied by gravity at that instant. Youd simply minimize action with the assumption that the ball is the only object with a potential energy. U for both boxes would be 0. The acceleration on the boxes will only increase proportional to 1/2 KEball

[Anonymous No.10123003]

>>10122987
Didn't think it was easy to understand from the text.
One of the great sufferings of life is how GIMP doesn't let me draw lines.

[Anonymous No.10123007]

>>10122992
>>10123001
>theyll only be moving
Sorry that wasnt accurate, theyll only be accelerating when the ball is touching them. Theyll have a kinetic energy at the time t(f) when the ball displaces the boxes 1/2D thats exactly equal to your 1/2Uo if you set up Uo correctly

[Anonymous No.10123011]

JESUS FUCKING CHRIST I AM THE MOST COMPLETE GODDAMN RETARD ON THIS PLANET.
IT'S A FUCKING CIRCLE.
THE BALL AND THE SQUARES HAVE TO HAVE THE SAME FUCKING ACCELERATION TO STAY TOUCHING.
THE FUCKING THING IS LITERALLY THIS SIMPLE FUCKING KILL ME.

[Anonymous No.10123012]

>>10123011
lol, u funny bro

[Anonymous No.10123013]

>>10123011
Hey hey, are you upset?

[Anonymous No.10123018]

>>10123013
Nah, just sorta disappointed in myself. I mean, I'm not as dumb as OP, but I actually thought for a few minutes on this. I can't really afford to be this stupid.

[Anonymous No.10123023]

>>10123011
Thanks for the spoiler

[Anonymous No.10123025]

>>10123011
I mean yeah

[Anonymous No.10123052]

>>10122962
This is an extremely trivial problem that can be solved without Calculus.
First, we'll apply a Newtonian shift of perspective, that is, instead of thinking of the ball moving the box, we'll consider the box moving along the ball. More specifically, the edge of the box as a single point moving along the ball. Since the whole of the force is put into this movement, the point moves in a circular path of ninety degrees with centripet force F=g/2=mv^r/r, where r is the radius of the ball. The dots final speed upwards is the ball's speed downwards under the original perspective, and since it's diameter closes in the two balls it has a D/2 distance left to go as a free fall, and the Torricelli equation can finish the job.

[Anonymous No.10123056]

Not op but ive not seen any of you produce a solution yet.

[Anonymous No.10123066]

>>10123001
>>10123007
>>10123052
I suspect that this is what >>10122992 meant:
The force between the objects will have a gravity component(since there's action from the sphere's weight on the cubes) and also, because the entire thing is moving, there would be an inertial force. If we add up those we might get that the objects possibly lose contact at an earlier point, since the total force between them might reach 0 earlier. There might also be another condition for two objects to be in contact, since this one seems insufficient, because it looks like the problem wants us to deal in energies/velocities instead of forces/accelerations.

[Anonymous No.10123067]

>>10123056
Uhhhh, righty.
For g=10, v=sqrt(10D/m+10D)
Logic here >>10123052
Might've made a mistake tho, huge pain working with fifteen different variables.

[Anonymous No.10123072]

>>10123066
>objects might lose contact
Nah. Not in the mood for proving it but it "isn't efficient for the ball to push the boxes enough to lose contact with them" is enough.

[Anonymous No.10123350]

>>10123067
Friend of OP here. We can look up the answer tomorrow and tell you whether this is correct. I don't remember it right now.

[Anonymous No.10123362]

>>10123072
There are easily imaginable possible scenarios where this happens depending on the force acting on the ball over time. Whether this is one of them is not trivial to prove or disprove.

[Anonymous No.10123494]

I'll give it a go:

Let h and j denote the initial horizontal distance between the two boxes and depth of the chordal section of the circle respectively.

The sum of forces on the ball are:

sum Fx=mgcosq/2-mgcosq/2=0

sum Fy=mg-2sinqm/2g,

Where q is the angle subtended by the point of contact of the ball and box and the center of the circle.

Work is defined both as the difference between the final and initial kinetic energy and the line integral of force. Hence, we have:

Kef-Kei=Kef=lim(j》0) int [D-j,3/2D] dy (mg-msinqg).

Splitting the integrals and equating sinq=y/d, we get mgD/2+mg5D/8=9mgD/8. Finally, we have:

Kef=mvf^2\2=9mgD/8.

So, vf= 1/2sqrt(9gD).

[Anonymous No.10123496]

>>10122962
The basic idea is that the only net force acting on the ball is due to the y components of the normal forces acting on the ball from the blocks and gravity. Since the y components of the normal forces are variable the acceleration will be variable as well.

>>10123066
The centripetal forces are the normal forces from the blocks on the ball. Since we are concerned with the ball’s acceleration and velocity then we are only concerned with the y components of the centripetal forces as the x components cancel. These forces are variable and decrease as the block falls. Then the acceleration of the block is variable as well. You would need calculus to figure out the velocity of the ball. I think this is the basic idea behind the problem but im too lazy to figure the whole thing out. Hopefully op isnt a fag and posts the solution so we can check soon. Im not much of a physics guy though so i dont have much weight in my ideas here.

[Anonymous No.10123677]

>>10123494
The speed without the boxes is [math] v = \sqrt{2 g D} [/math], you made a mistake somewhere.
>>10122962
I think it's [math] v = \sqrt{\frac{3}{2} g D} [/math]

[Anonymous No.10123700]

I keep getting sqrt(gD) no matter how much I chec for errors. I guess we'll see the answer tomorrow

[Anonymous No.10123712]

>>10123700
That gives you an energy that's equivalent to going halfway without contact so I think it has to be higher. I believe the result has to be bounded between [math] \sqrt{gD} \lt v \lt \sqrt{2 g D} [/math]

[Anonymous No.10123719]

It's a pendulum you idiots

[Anonymous No.10123735]

>>10123052
This is wrong. The problem is much more serious than people are making it out to be. You do not know when the ball and the boxes lose contact with each other. If the mass of the boxes is zero, then just barely touching them and imparting any momentum whatsoever will send them flying off to infinity, far before the ball actually gains any momentum.

[Anonymous No.10124056]

>>10123735
This. It really depends on their masses. Btw Im OPs friend will now check with him for the answer.

[Anonymous No.10124060]

>>10122980
>>10122976
>>10123001
>>10123007
>>10123052
>>10123067
>>10123494
>>10123677
>>10123712
This is the answer, none of you got it so I you might now be convinced it's way harder than it appears.

[Anonymous No.10124088]

>>10123011
>>10123012
>>10123013
>>10123023
>>10123025

this isn't even true, wtf /sci/

[Anonymous No.10124089]

>>10123052
kkkkkkkkkkkkkkkkkkkkkkkk

[Anonymous No.10124108]

we can do this tho

let's start trying to solve for when the blocks leave contact with the ball

first, if the blocks move distance x and the ball of radius r drops distance y, then
(r - y)^2 + x^2 = r^2
or x^2 + y(y - 2r) = 0

also 2x dx + (2y - 2r)dy = 0
or x v_block = (r - y) v_ball (when in contact)

and so also v_block^2 + x v_block' = -v_ball^2 + (r - y)v_ball'

now we use other stuff and make sure v_block' = mg

how long is this problem supposed to take?

[Anonymous No.10124110]

this guy was right tho >>10123712

anyway we can do this

let's start trying to solve for when the blocks leave contact with the ball

first, if the blocks move distance x and the ball of radius r drops distance y, then
(r - y)^2 + x^2 = r^2
or x^2 + y(y - 2r) = 0

also 2x dx + (2y - 2r)dy = 0
or x v_block = (r - y) v_ball (when in contact)

and so also v_block^2 + x v_block' = -v_ball^2 + (r - y)v_ball'

now we use other stuff and make sure v_block' = mg

how long is this problem supposed to take?

[Anonymous No.10124111]

this guy was right tho >>10123712

anyway we can do this

let's start trying to solve for when the blocks leave contact with the ball

first, if the blocks move distance x and the ball of radius r drops distance y, then
(r - y)^2 + x^2 = r^2
or x^2 + y(y - 2r) = 0

also 2x dx + (2y - 2r)dy = 0
or x v_block = (r - y) v_ball (when in contact)

and so also v_block^2 + x a_block = -v_ball^2 + (r - y)a_ball

now we use other stuff and make sure v_block' = mg

how long is this problem supposed to take?

[Anonymous No.10124112]

this guy was right tho >>10123712

anyway we can do this

let's start trying to solve for when the blocks leave contact with the ball

first, if the blocks move distance x and the ball of radius r drops distance y, then
(r - y)^2 + x^2 = r^2
or x^2 + y(y - 2r) = 0

also 2x dx + (2y - 2r)dy = 0
or x v_block = (r - y) v_ball (when in contact)

and so also v_block^2 + x a_block = -v_ball^2 + (r - y)a_ball

now we use other stuff and make sure a_block = mg

how long is this problem supposed to take?

[Anonymous No.10124115]

>>10124111
>>10124112
sorry for double post

also this looks bad enough that it might just be worth it to use the lagrangian just to get when a_ball = mg

[Anonymous No.10124130]

If it’s a sphere, only one point will end up touching the box. What if we simplify the notion of the sphere to instead be a rod which is first bent in the middle, tracing out equilateral sides of a triangle, and then eventually becomes the 2x the base of that triangle. The transformation of the triangle would be purely based on g.

Just my two cents, didn’t try to calculate anything.

[Anonymous No.10124132]

>>10124130
i think this is the same concept but with a different shape so it would be a different result

[Anonymous No.10124278]

>>10122962
Is it 10 m/s?

[Anonymous No.10124283]

>>10124060
I got [math]v=\sqrt{\frac{46}{27}gD}[/math]
I probably fucked up somewhere, meh

[Anonymous No.10124298]

>>10122962
sqrt(2gd).

[Anonymous No.10124304]

>>10124298
>2gd
1gd

[Anonymous No.10124323]

<math>/sqrt(2g(pi/2-1))</math>

[Anonymous No.10124342]

>>10124060
>>10124283
Now I got 50/27, it was just a dumb mistake. I'm too lazy to explain how I did it though.

[Anonymous No.10124488]

>>10124342
Yeahh suure. Can you not just post a photo of your scrap notes or whatever...

[Anonymous No.10124554]

>>10122962
>At the start:
Potential energy of ball = m g D
Kinetic energy of ball = 0
Kinetic energy of left box = 0
Kinetic energy of right box = 0
>When the sphere loses contact with the boxes:
Potential energy of ball = m g D/2
Kinetic energy of ball = m g D/4
Kinetic energy of left box = m g D/8
Kinetic energy of right box = m g D/8
>When the sphere hits the ground:
Potential energy of ball = 0
Kinetic energy of ball = 3 m g D/4
Kinetic energy of left box = m g D/8
Kinetic energy of right box = m g D/8

To calculate the speed you use the Newton formula for kinetic energy
m v^2/2 = 3 m g D/4
v^2 = 3 g D/2
v = sqrt(3 g D/2)

[Anonymous No.10124669]

>>10124488
It's basically like >>10123052 said. However, you have to take into account that the ball will lose contact with the boxes. We get
[eqn] \frac{1}{4}mv^2 = -\frac{mgy}{2} = \frac{m}{2}g(1-\cos \varphi)\frac{D}{2} \\
v^2 = gD(1-\cos \varphi) [/eqn]
Normal force
[eqn] F_N = -\frac{2mv^2}{D} + mg\cos \varphi = mg(3\cos \varphi - 2) [/eqn]
The ball and boxes lose contact when F_N is 0 and thus [math]\varphi_0 = \arccos{\frac{2}{3}}[/math]. The velocity of the point in the vertical direction is [math]v_y(\varphi) = v(\varphi)\sin \varphi[/math]

Torricelli's equation
[math] v^2_{y,\mathrm{End}} = v^2_y(\varphi_0) + 2g\Delta y [/math], with [math] \Delta y = D-\frac{D}{2}(1-\cos \varphi_0) [/math]

Plug everything in
[math] v^2_{y,\mathrm{End}} = gD(1-\cos(\arccos\frac{2}{3}))(1-(\frac{2}{3})^2) + 2g(\frac{5}{6}D) = \frac{5}{27}gD + \frac{10}{6}gD = \frac{50}{27}gD [/math]

[Anonymous No.10124785]

>>10122962
brute force works fine here. Fuck you op. I thought you would need a clever trick.

[Anonymous No.10124845]

>>10124785
Beautiful mouse writing, seems correct though

[Anonymous No.10124966]

>>10124785
Hard to read but this looks right.

[Anonymous No.10125808]

>>10124785
Where is the aswer though?

[Anonymous No.10125835]

>>10122962
correct me if I'm being dumb, but why won't this work?
>you have conservation of momentum between the boxes and ball
>you have a constraint on final total kinetic energy
>solve for v

[The Landlord No.10125863]

LAGRANGIAN MECHANICS

Potential E:
U=mgh

Kinetic E is divided between ball and two squares:
T_ball = 0.5 m v^2
When the ball has fallen by D,each box will move D/2 so the v of the box is 0.5 times the v of the ball
T_box = 0.5 (m/2) (v/2)^2
T = T_ball + 2 T_box

Euler-Lagrange says the acceleration is 0.8 g.

Use
x(t) = x_0 + v_0 t + 0.5 a t^2 to get the time when the ball hits the ground

Plug that into
v(t) = a t

[Anonymous No.10125895]

>>10122962

If there is not friccion the force that the boxes aplies to the ball in the x axis is equal to zero, wich means as the ball wants to go down the boxes just move around without any resistance, its a simply free fall problem

a(t)=-9.8 [m/s^2]
V(t)=-9.8*t + Vo[m/s]
X(t)=-4.9*t^2 + Vo*t + Xo

Vo=0
Xo= D

X(tf)=0
-4.9*(tf)^2 + 0 + D = 0

tf = sqrt(D/4.9)


V(tf) = Vf = 9.8*(D/4.9)

[Anonymous No.10125910]

>>10125895
Reading other answers ive realized ive ignored y, fuck me. Well at least you know i like to give it a go before reading the solution

[Anonymous No.10126025]

>>10125910
(;

[Anonymous No.10126066]

>>10125895
the boxes have inertia

[Anonymous No.10126150]

>>10122962
just some thoughts. you can definitely relate the normal force of the box onto the sphere using law of cosines, that would take care of the angles as the motion ensues. I would then probably focus on the momentum of the two boxes being equal to the integral of the weight with respect to time.

[Anonymous No.10126154]

>>10125910
it was a novel approach

[Anonymous No.10126229]

>>10125808
It's an exercise for the reader.

[Anonymous No.10126240]

>>10126229
got em

[Anonymous No.10126265]

>>10125808
>>10124785

shit. The bottom was cut out. but you can solve for the angle where contact is broken in the last eq. (cos(phi)=2/3). then the eq above gives you the velocity at that point. Use energy conservation to get the total velocity after the additional freefall period.
Whole exercise is basically pic related. tedious but straight forward.

[Anonymous No.10126441]

>>10126154
A parker solution