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Cylinder oscillating in water.

 
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Jul14-12, 01:12 PM   #1
 

Cylinder oscillating in water.


ω1. The problem statement, all variables and given/known data

A cylinder of diameter d floats with l of its length submerged. The total height is L. Assume no damping. At time t = 0 the cylinder is pushed down a distance B and released.


What is the frequency of oscillation?
2. Relevant equations

[itex] f = ω/2\pi [/itex]
[itex] Ma = F_{(bouyancy)} [/itex]
Writing this in our differential form, making proper substitutions, and noting that bouyancy is affected by the distance that our cylinder is submerged we come to.

[itex] dx^2 (M_{(cylinder)}) + x (\rho_{(water)} g Area_{(cylinder face)})= 0 [/itex]

we know that [itex]M_{(cylinder)} = V_{(cylinder)}\rho_{(cylinder)} [/itex]

hence we should have
[itex] ω^2 = (\rho_{(water)} g Area_{(cylinder face)}) / V_{cylinder}\rho_{(cylinder)} = g\rho_{(water)} / l \rho_{(cylinder)}[/itex]

however the solution in my problem set has ω^2 = g/l. Can anyone shed some light on why the densities may cancel??
 
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Jul14-12, 01:30 PM   #2
 
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hi ozone!
Quote by ozone View Post
[itex]g\rho_{(water)} / l \rho_{(cylinder)}[/itex]
(that's the same as g/L)

i haven't followed what you've done, but i'd guess you've used the wrong expression for the mass of the cylinder

just use a (vertical) force equation for the cylinder (at depth l + x)
 
Jul15-12, 10:42 PM   #3
 
the mass of the cylinder is the density of the cylinder times the area.. but the force from the water depends only on the density of water.. that is why i don't understand how the densities are cancelled out.
 
Jul15-12, 11:00 PM   #4
 
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Cylinder oscillating in water.


In the denominator of your final expression for ω2, is that a small l or a capital L?

You can find an expression for the ratio of the two densities in terms of the ratio of l and L by considering the condition for equilibrium when length l of the cylinder is submerged.
 
Jul16-12, 09:07 AM   #5
 
that would be the lower case l in the solution set answer. You can view it for yourself at this link on page 4.

http://ocw.mit.edu/courses/physics/8...ents/sol1b.pdf
 
Jul16-12, 09:41 AM   #6
 
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But in the expression that you derived: ω2 = gρw/lρc, you should have a capital L rather than a lower case l in the denominator. Then you should be able to show that this expression reduces to the correct answer.
 
Jul18-12, 07:28 AM   #7
 
We never learned about fluid dynamics in my mechanics class, but I am guessing that the water displaced in equilibrium is equal to the mass of the cylinder?
 
Jul18-12, 08:12 AM   #8
 
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Quote by ozone View Post
We never learned about fluid dynamics in my mechanics class, but I am guessing that the water displaced in equilibrium is equal to the mass of the cylinder?
Yes. Archimedes Principle: Buoyant force = weight of fluid displaced

In equilibrium, Buoyant force equals the weight of the floating object.
 
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