PM

the volumes of displaced water on each side...

was it your riddle?

>was it your riddle?
yes
convincing yourself will require a bit of calculation

>It was obvious form my post what part of gravity was I disregarding.

But you can't just conveniently disregard "part of gravity" if you're trying to account for all forces acting on a structure.

At first I gave you the benefit of the doubt and tried to give an honest answer to your original question. Now I can see that you're either a troll or you're just being argumentative for the sake of it.

I think I got the main idea now. It has something to do with pressure difference of the water acting on the lids that cancels out the difference in buoyancy. Aint got time for calculations.

>It has to do with buoyancy cancelling buoyancy
mhhhh

> the lid isn't.
The lid is physically attached to the arm. The arm will impart a force on the spring and subsequently the lid.

Tough one

My initial idea is that since the one on the right is being compressed, it's density will increase, but since the other end's density is decreasing, the buoyant force increases. I think these two forces would be equal and opposite and so the net torque is zero.

Nigger stop it, the wheel isn't free fall.
You are literally the least intelligent poster in this thread and it's urgent you go back to high school physics.

Since buoyancy is related to volume, whenever an object is compressed it will have a reduced buoyant force. Thus the containers on the top will have the largest force pulling them upwards and those at the bottom having a smaller force pulling them up. Therefore the wheel wouldn't even move to begin with, maybe wiggle with passing currents.

How will a top bottom imbalance cancel a left right imbalance?

So assuming someone came and spun the wheel by hand, there would be no force imparted on the lid by the spring? There has to be an acceleration to start the process, otherwise we wouldn't have movement.

Not to mention the lids and springs are moving in a circle. The inertia of the spring will definitely cause the spring to bend and subsequently imparting a force on the lid from the motion of the wheel.

What am i missing?

>inertia of the spring
why the fuck are you overcomplicating this, just take the spring to be massless.

And nobody is asking you to solve the system dynamically, just integrate the work over a full rotation of a wheel arm.

Nope, still lost. At least it makes sense to me how it wouldnt work dynamically even if i cant prove it in a vacuum

Let me present it this way: take it as an argument ad absurdum.
Either the wheel is rotating, in which case free energy yay!
Or it isn't, in which case we suppose the wheel is stationary, and thus its parts are at equilibrium. And then you average the torque over the full [math]2\pi[/math] arc (which in this case is the same as averaging the work by a factor) and see if it's 0. If it's not then it can't be stationary.

Lmao, this is what happens when you read through a physics book but bother solving any of the problems

>At least it makes sense to me how it wouldnt work dynamically
>I can prove the wheel doesn't spin, but only if it spins
mein gott

Gravity pulls things down. Buoyancy counters gravity and pushes up. That's why ships float. The reason this does not work is because gravity may pull the lid increasing/decreasing the volume of the cylinder. This in turn increases the offsetting buoyancy (based on volume of water displaced). Because of conservation of energy the two forces directly oppose each other each.

Basically a single cylinder will either i) find an equilibrium point at a certain depth ii) float to the surface or iii) sink to the bottom. Basically connecting 8 cylinders in a circle simply complicates the static position analysis. You may get some initial rotating. But water friction will eventually force this to a static equilibrium position.

It's often easiest to start with a method that we know would respect energy conservation, derive the forces claimed in the picture, and note what's missing.

Letting
y = depth of tube below water (increasing downward)
[math]\theta[/math] = orientation of tube (clockwise from lid pointing to the left)
x = displacement of the spring
h = height of tube at spring equilibrium without external forces
A = cross-sectional area of tube
g = gravitational field
[math]\rho[/math] = water density
m = mass of lid
k = spring constant (including effect of air compression)

the tube's potential energy is
[math]V = g \rho A (h + x) y + g m (- y + (h + x) \sin \theta) + \frac12 k x^2[/math].

It's a nice fact that if x is chosen to minimize V (physically, the spring is in the equilibrium state), then [math]\partial V/\partial y[/math] with x constant is equal to [math]\partial V/\partial y[/math] with x chosen to minimize V. Same with [math]\partial V/\partial \theta[/math]. This means we can eliminate the x degree of freedom and still compute forces and torques from V in the usual way.

When the spring is in equilibrium,
[math]x = - \frac1k (g \rho A y + g m \sin \theta)[/math]
and
[math]V = g \rho A h y + g m (- y + h \sin \theta) - \frac1{2k} (g \rho A y + g m \sin \theta)^2[/math].

The force on the tube (measured downward) is
[math]F = - \partial V/\partial y = - g \rho A h + g m + g \rho A \frac1{k} (g \rho A y + g m \sin \theta) = - g \rho A (h + x) + g m[/math]
as expected. Decreasing [math]\sin \theta[/math] increases the upward force, in agreement with the original argument.

continued...

The missing part is the torque (measured clockwise)
[math]T = - \partial V/\partial \theta = - g m h \cos \theta + g m \cos \theta \frac1{k} (g \rho A y + g m \sin \theta) = - g m (h + x) \cos \theta[/math].
At the bottom where [math]\cos \theta = -1[/math], it is positive, increasing the desired torque on the wheel, but at the top where [math]\cos \theta = +1[/math], it is negative, against the desired motion. And at the top it is larger due to the increased length of the spring as water pressure decreases.

To fix the original argument, we can work backwards, starting from
[math]F = - g \rho A (h + x) + g m[/math]
[math]T = - g m (h + x) \cos \theta[/math]
[math]x = - \frac1k (g \rho A y + g m \sin \theta)[/math],
all of which we can work out without considering the energy of the system.
The work done on the wheel during a cycle is
[math]\int dt (F \dot{y} + T \dot{\theta})[/math]
which can be rewritten in terms of the convenient function
[math]V(y, \theta) = g \rho A h y + g m (- y + h \sin \theta) - \frac1{2k} (g \rho A y + g m \sin \theta)^2[/math]
as
[math]\int dt ((- \partial{V}/{\partial y}) \dot{y} + (- \partial{V}/{\partial \theta}) \dot{\theta})[/math]
which is obviously zero over a cycle.

are you autistic?

...

well dun

You should use this knowledge and dedication to solving problems in mathematics.
Explain, using intrgrals, how General relativity can be differentiable with Quantum Mechanics. Feel free to right a hypothesis with variable terms just as you did for this problem. Hint- relate the curvature of space to energy gained in motion. Essentially, explain why GR can't predict QM.

holy crackers!

Nice. An interesting follow up question would be, for small perturbations, finding the frequency of oscillation.

Tubes on the right have a smaller volume because gravity pulls their lids to the arms. Hence, they are less buoyant and the wheel is balanced. It's that simple.

I understand most of this proof but where did the
[eqn]V=gρA(h+x)y[/eqn]
come from?

Potential attached to buoyancy.
A(h+x) is the volume.

Are you stupid or something?

faggot
you're the problem the lesswrong guy mentions here lesswrong.com/lw/ip/fake_explanations/
You just say some bullshit, and because you introduce some scientific vocabulary in there you think it has scientific value.
You could have said "it's magic" and it would have had the same value.

Props to for his very elegant answer, which completely circumvents the computational burden of writing everything in terms of torque and provides a satisfying explanation.