How to calculate the height submerged of cylinder

  • Thread starter lakmalp
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  • #1
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A hollow cylinder (but not open ended) is allowed to float on a liquid. Can anyone please help me how to calculate the height (h) it submerge? I want to get a relationship between height and cylinder weight.

I can calculate the volume of submerged part. But no idea how to derive a formula with weight and height.
 

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  • #2
Simon Bridge
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The formula you have derived for the volume of the submerged part should have an h in it - that's the unknown you want to find right? So the volume submerged is a function of height.

V=f(h)

Presumably you also know the mass of the cylender M and the density p of the water?

By Archimedes: the cylinder floats when M=pV ... solve for h.
 
  • #3
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I think you misunderstood what I said. Sorry for being not descriptive.

Since I know the weight of cylinder, I can calculate volume of the water displaced. But it doesn't contain 'h' in it.

I am struggling to get 'h' into this volume,i.e. I want to know f(h)
 
  • #4
Simon Bridge
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Show me your formula for the volume of the submerged part of the cylinder.
 
  • #5
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Vwaterwater.g = mcylinder.g
Vwater = mcylinderwater
 
  • #6
Simon Bridge
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Oh I see what you mean ...
That is only the volume of water that has the same mass as the cylinder - which is only the volume of cylinder submerged if the cylinder floats. That's the condition you need to find h. You need to also find the volume in terms of h.

The submerged volume of the cylinder is AL (the area of the cylinder end that is under water multiplied by the length of the cylinder). The area will depend on R (radius of the cylinder) and h.

You can work that out by calculus or just look it up.
Then you put the volume of the cylinder found this way equal to the volume of water you already have.
 
  • #7
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Yeah, that's where I am stuck at.

V = L[itex]\int A(h)dh[/itex]

I was not able to find A(h) so far. Anyway I'll have a look again whether I can find any relationship. I'll post what I have done so far a little later.
 
  • #8
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I couldn't go much far...

If [itex]l[/itex] is length of cylinder,

[itex]y·sin\theta=2r-x[/itex]

[itex]y=\frac{2r-x}{sin\theta}[/itex]

[itex]Area=2·y·cos\theta·l[/itex]

[itex]=2·\frac{2r-x}{sin\theta}·cos\theta·l[/itex]

[itex]=2·(2r-x)·cot\theta·l[/itex]

[itex]Volume=\int ^{0}_{\frac{\pi}{2}} 2(2r-x)·cot\theta·l·d\theta[/itex]

How can I get a relationship between [itex]x[/itex] and [itex]\theta[/itex]? Please let me know whether my approach is correct and what should I do from hereon.
 

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  • #9
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I think I found the answer. It is so simple than I thought :)

Circular Segment

Equation (9) is the solution :)
 
  • #10
Simon Bridge
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Yep - that's the one ... you should have done the integral in rectangular coords.
But you should also check the proportion of the cylinder that is submerged ... which is an easier calculation.
 
  • #11
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Thank Simon. I'll look into that and will post what I found.
 

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