Gravitation and Archimedes principle as a spring?

[uriwolln]

Hey all,

I have seen that these two forces combined together have the potential energy of a spring with a constant 'K' as a combination of some sort of the two forces.
I have no idea how am I to find this K.
All I know
Mgh + ρVgh gives the potential energy of a given system let's say.

help?

 

[PeterO]

Hey all,

I have seen that these two forces combined together have the potential energy of a spring with a constant 'K' as a combination of some sort of the two forces.
I have no idea how am I to find this K.
All I know
Mgh + ρVgh gives the potential energy of a given system let's say.

help?

Can you try to re-phrase what you are having difficulty with please?

 

[uriwolln]

The problem was:
A cylinder is placed inside water, and it is in equilibrium when half of the cylinder height is submerged. So I Know there is the archimedes principle which causes a force upwards and then there is gravity downwards. This problem was part of question somewhere, but in the explanation of the answer concerning the energies of the system when it is in motion, they somehow did not use mgh of either the ρVgh. They did a combination of those to act as a spring, which I did not get a clear answer on how to do that.

 

[tiny-tim]

hi uriwolln!

if the restoring force is proportional to the distance from the equilibrium position,

the the equation is exactly the same as the equation for a spring

 

[uriwolln]

:)
Did not think of it this way.
But then, what will be 'k' equal to?
How do I work the algebra for it?
and will the potential energy of it all will indeed be k(x^2)/2?

 

[tiny-tim]

But then, what will be 'k' equal to?
How do I work the algebra for it?

try it and see

 

[uriwolln]

So let's see:
kx=mg - ρVg ==> k = (mg - ρVg)/2 ?
and then potential energy: k(x^2)/2?

 

[tiny-tim]

kx=mg - ρVg

but V depends on x

 

[uriwolln]

:)
true... forgot that for a sec... but is this the equation I should go for?
K being a function of x? Then the whole potential energy is going wrong, because K is not constant, or does it? ( I think K should be constant to use k(x^2)/2)

 

[tiny-tim]

erm … first, find the equation

then analyse it! ​

 

[uriwolln]

Here goes:

Kx = (pir²H)g - ρ(pir²(H-x)g/2)
K = (pir²Hg/2x) + ρ(pir²g)/2

 

[tiny-tim]

Kx = …

no, you mean mx'' = …

carry on from there

 

[uriwolln]

mx?
why am I meaning mx? m as in mass? and x as displacement?
Then where does K come in?

(I am not being disrespectful I hope, I really liked your comment, which made me laugh, because I really do not know why should I use mx)

 

[tiny-tim]

no, mx''

(ie md²x/dt²)

 

[uriwolln]

:)
ρ_cpir²Hx'' = ρ_cpir²Hg - ρ_l(pir²(H/2 - x)
x'' + x(ρ_lg/ρ_cH) = g(1-ρ_l/2ρ_c)
Which x then I know is oscillating.
But then where does K come in?

 

[tiny-tim]

x'' + x(ρ_lg/ρ_cH) = g(1-ρ_l/2ρ_c)

you have x'' = -Ax + B

if it was just x'' = -Ax, what would k be? ​

 

[uriwolln]

then K should be:
k = (ρ_lg/ρ_cH)
So potential energy is as well 0.5Kx²?

 

[tiny-tim]

then K should be:
k = (ρ_lg/ρ_cH)

yes , but I'm not sure whether you're just guessing

how did you get rid of the B?

So potential energy is as well 0.5Kx²?

maybe … what's x?

 

[uriwolln]

:)
actually I did guess.
Well, I mean, I figured because it was the coefficient of x, then it would be reasonable to assume that, that is K.
X would be the answer for the differential equation I got previous.

 

[tiny-tim]

:)
actually I did guess.

Well, I mean, I figured because it was the coefficient of x, then it would be reasonable to assume that, that is K.

ok, sensible guess for k , but it doesn't tell you how to combine it with the distance (in kx or 1/2 kx²)

when you have x'' = -Ax + B,

rewrite that as (x - B/A)'' - -A(x - B/A) …

now you have it in the same form as the standard spring equation

 

[uriwolln]

AWESOME!
Thanks for that.
Tell me please though, if I did write x to be x - B/A then I think I should be using x - B/A for the potential energy as the x there: 0.5Kx^2. Correct?

 

[tiny-tim]

Correct.

x = B/A is the equilibrium position, and all displacements must be measured from that.

(you get the same effect if you take a horizontal spring and place it vertically, with a weight on the end … the new equilibrium position is different from the original equilibrium position, and all displacements must be measured from the new position instead)

 

[uriwolln]

:)
Thank you so much tiny!