Sinking a cylinder with varying hole sizes

In summary, the rate of descent of the cylinder is inversely proportional to the size of the hole. The level inside the cylinder will be less than outside. These parts of the equation system are probably a lot more important than any external drag.
  • #36
lloydthebartender said:
Is this correct?
No. A and h are constants, so what is the change in Ah, d(Ah), in time dt?
You would have seen something is wrong if you had checked dimensional consistency.
lloydthebartender said:
##\frac{1}{2}\rho (0) + \rho gz_{1}+P_{1}=\frac{1}{2}\rho (v+u) + \rho gz_{2}+P_{2}##
Since ##z_{1}=z_{2}##
##P_{1}=\frac{1}{2}\rho (v+u) +P_{2}##
I rewrite pressure in terms of the depth from the surface of water?
##P_{1} = \rho gy## and ##P_{2} = \rho gh## so
##\rho gy=\frac{1}{2}\rho (v+u)+\rho gh##
##gy=\frac{1}{2} (v+u)+ gh##
Yes, except that you have omitted the power of 2 on the velocity term. Again, dimensionally inconsistent.
 
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  • #37
haruspex said:
No. A and h are constants, so what is the change in Ah, d(Ah), in time dt?
You would have seen something is wrong if you had checked dimensional consistency.

Yes, except that you have omitted the power of 2 on the velocity term. Again, dimensionally inconsistent.
Oh...
##\frac{d(Ay)-d(Ah)}{dt}=va##
##\frac{d(Ay)}{dt}-\frac{d(Ah)}{dt}=va##
##\frac{dy}{dt}=\frac{va}{A}##
haruspex said:
Yes, except that you have omitted the power of 2 on the velocity term. Again, dimensionally inconsistent.
##gy=\frac{1}{2}(v+u)^2+gh##
##2g(y-h)=(v+u)^2##
##\sqrt{2g(y-h)}=v+u##
So the downwards velocity of the cylinder is
##u=\sqrt{2g(y-h)}-v##?
 
  • #38
haruspex said:
No. A and h are constants, so what is the change in Ah, d(Ah), in time dt?
You would have seen something is wrong if you had checked dimensional consistency.

Yes, except that you have omitted the power of 2 on the velocity term. Again, dimensionally inconsistent.
At this point do I return to the force equations? I'm still not sure how this all connects back to it.
 
  • #39
lloydthebartender said:
At this point do I return to the force equations? I'm still not sure how this all connects back to it.
Sorry, I forgot to reply...
Your post #37 is correct, but note the dy/dt and u are the same thing, so you can combine those equations to eliminate u.
 
  • #40
haruspex said:
Sorry, I forgot to reply...
Your post #37 is correct, but note the dy/dt and u are the same thing, so you can combine those equations to eliminate u.
##\frac{va}{A}=\sqrt{2g(y-h)}-v##
##a=\frac{A(\sqrt{2g(y-h)}-v)}{v}##

So now I need to make this equation in terms of ##t## and ##a##, right?
 
  • #41
haruspex said:
Sorry, I forgot to reply...
Your post #37 is correct, but note the dy/dt and u are the same thing, so you can combine those equations to eliminate u.
Since
##F_{buoyancy}=\rho gV## and ##V=Ah##
## F_{buoyancy}=\rho gAh ##
Because ##frac_{dy}{dt}=frac_{va}{A}##
##t=\int \frac{va}{A}dy##
Since ##v##, ##a##, ##A## are constants
##t= \frac{vay}{A}##, so ##A=\frac{vay}{t}##
##V=\frac{vay}{t}h##

##F_{weight}=F_{buoyancy} + F_{drag}##
##(m_{water}+m_{cylinder})g=\rho g \frac{vay}{t}h -bv##
##(\rho V+m_{cylinder})g=\rho g \frac{vay}{t}h -bv##I suppose I do some rearranging magic for the ##v## of the drag force using the equation I got in #40?
 
  • #42
I could see something was going wrong. Just traced it to this:
lloydthebartender said:
##P_{2} = \rho gh##
That is not correct. What is the water depth just above the hole?
 
  • #43
haruspex said:
I could see something was going wrong. Just traced it to this:

That is not correct. What is the water depth just above the hole?
Is it the ##P_{2}=\rho gy## because they're at the same height (##z_{1}=z_{2}##)? But that would mean ##P_{1}=P_{2}##?
 
  • #44
lloydthebartender said:
Is it the ##P_{2}=\rho gy## because they're at the same height (##z_{1}=z_{2}##)? But that would mean ##P_{1}=P_{2}##?
No. Look at your diagram in post #24. What is the depth of water inside the cylinder?
 
  • #45
haruspex said:
No. Look at your diagram in post #24. What is the depth of water inside the cylinder?
Right, so ##y-h##; ##P_{2}=\rho g(y-h)##
 
  • #46
lloydthebartender said:
Right, so ##y-h##; ##P_{2}=\rho g(y-h)##
Yes.
 
  • #47
haruspex said:
Yes.
So I get
##\rho gy=\frac{1}{2}ρ(v+u)+ρg(y-h)##, and rearrange for ##u##?
 
  • #48
lloydthebartender said:
So I get
##\rho gy=\frac{1}{2}ρ(v+u)+ρg(y-h)##, and rearrange for ##u##?
You have forgotten to square the velocity term again.
After fixing that, rearrange for v and eliminate v using the u=dy/dt=va/A equation you had in post #37.
 
  • #49
haruspex said:
You have forgotten to square the velocity term again.
After fixing that, rearrange for v and eliminate v using the u=dy/dt=va/A equation you had in post #37.
##gy=\frac{1}{2}(v+u)^2+g(y-h)##
##2gy-2g(y-h)=(v+u)^2##
##2gh=(v+u)^2##
##\sqrt{2gh}=v+u##
##v=\sqrt{2gh}-u##
##u=\frac{va}{A}##, ##v=\frac{uA}{a}##
##\frac{uA}{a}=\sqrt{2gh}-u##
##uA=a\sqrt{2gh}-au##
##u(A+a)=a\sqrt{2gh}##
##u=\frac{a\sqrt{2gh}}{(A+a)}##
 
  • #50
lloydthebartender said:
##gy=\frac{1}{2}(v+u)^2+g(y-h)##
##2gy-2g(y-h)=(v+u)^2##
##2gh=(v+u)^2##
##\sqrt{2gh}=v+u##
##v=\sqrt{2gh}-u##
##u=\frac{va}{A}##, ##v=\frac{uA}{a}##
##\frac{uA}{a}=\sqrt{2gh}-u##
##uA=a\sqrt{2gh}-au##
##u(A+a)=a\sqrt{2gh}##
##u=\frac{a\sqrt{2gh}}{(A+a)}##
Yes!
Last step is to find the time taken. You need to think a bit here about the initial and final values of y. Start by imagining what would happen if there were no hole.
 
  • #51
haruspex said:
Yes!
Last step is to find the time taken. You need to think a bit here about the initial and final values of y. Start by imagining what would happen if there were no hole.
##y## would be 0 if there were no hole, no matter the time because it would be on the surface of the water. But also ##y## varies from 0 to the height of the cylinder if there is a hole.
Since ##\frac{dy}{dt}=\frac{va}{A}## couldn't I integrate this with ##t=\int{\frac{va}{A}dy}##?
 
  • #52
lloydthebartender said:
y would be 0 if there were no hole
No, the cylinder has weight. Apply Archimedes.
 
  • #53
haruspex said:
No, the cylinder has weight. Apply Archimedes.
I'm really not sure...##F_{b}## would be 0 to ##\rho g y##.
 
  • #54
lloydthebartender said:
I'm really not sure...##F_{b}## would be 0 to ##\rho g y##.
If you put a dish on a bucket of water does it sit on the surface, not immersed at all?
 
  • #55
haruspex said:
If you put a dish on a bucket of water does it sit on the surface, not immersed at all?
It would displace the volume of ##Ay## on the surface and the volume equal to that of the cylinder when underwater.
 
  • #56
lloydthebartender said:
It would displace the volume of ##Ay## on the surface and the volume equal to that of the cylinder when underwater.
y is a variable.
Let the mass of the empty cylinder be M. Seal up the hole and place the cylinder on the surface. How far will it sink into the water? Use Archimedes.
 
  • #57
haruspex said:
y is a variable.
Let the mass of the empty cylinder be M. Seal up the hole and place the cylinder on the surface. How far will it sink into the water? Use Archimedes.
Well it displaces it's own weight so the buoyancy is equal to ##\rho g \frac{Mg}{\rho}##
 
  • #58
lloydthebartender said:
Well it displaces it's own weight so the buoyancy is equal to ##\rho g \frac{Mg}{\rho}##
Too many g's. The buoyant force will be Mg. So what volume is displaced?
 
  • #59
haruspex said:
Too many g's. The buoyant force will be Mg. So what volume is displaced?
##\frac{M}{\rho}##?
 
  • #60
lloydthebartender said:
##\frac{M}{\rho}##?
Right.
So now let's think about what will happen when a cylinder with a small hole is placed on the water and released. The flow through the hole will be relatively slow, so the initial behaviour will be much the same as with no hole, namely, that it will immediately sink until it displaces volume ##\frac{M}{\rho}##.
What value of y will result?
 
  • #61
haruspex said:
Right.
So now let's think about what will happen when a cylinder with a small hole is placed on the water and released. The flow through the hole will be relatively slow, so the initial behaviour will be much the same as with no hole, namely, that it will immediately sink until it displaces volume ##\frac{M}{\rho}##.
What value of y will result?
So if it displaced volume ##\frac{M}{\rho}## that means volume ##A(y-h)## equals to that, so ##y=\frac{M}{\rho A}+h##.
 
  • #62
lloydthebartender said:
So if it displaced volume ##\frac{M}{\rho}## that means volume ##A(y-h)## equals to that,
No, ##A(y-h)## is the volume of water inside the cylinder at some later time.
What is the volume of water displaced by the cylinder when hardly any water has come through the hole?
 
  • #63
haruspex said:
No, ##A(y-h)## is the volume of water inside the cylinder at some later time.
What is the volume of water displaced by the cylinder when hardly any water has come through the hole?
Right so is it ##Ay##? Making ##y=\frac{M}{\rho A}##
 
  • #64
lloydthebartender said:
Right so is it ##Ay##? Making ##y=\frac{M}{\rho A}##
Yes. So that is in effect the initial value of y for the time taken to sink. The point is that it will take very little time to reach that depth. Only after that is the rate of sinking constrained by the rate of flow of the water through the hole.
 
  • #65
haruspex said:
Yes. So that is in effect the initial value of y for the time taken to sink. The point is that it will take very little time to reach that depth. Only after that is the rate of sinking constrained by the rate of flow of the water through the hole.
So can I do ##\frac{dy}{dt}=u##, and this with the equation of #49?
 
  • #66
lloydthebartender said:
So can I do ##\frac{dy}{dt}=u##, and this with the equation of #49?
Yes.
 
  • #67
haruspex said:
Yes.
But ##\frac{d(\frac{M}{\rho A})}{dt}=0##, right?
 
  • #68
lloydthebartender said:
But ##\frac{d(\frac{M}{\rho A})}{dt}=0##, right?
Yes, but why is that a problem? ##\frac{M}{\rho A}## is just the initial value of y.
Thereafter, ##\frac {dy}{dt}=u=\frac{a\sqrt{2gh}}{(A+a)}##.
 
  • #69
haruspex said:
Yes, but why is that a problem? ##\frac{M}{\rho A}## is just the initial value of y.
Thereafter, ##\frac {dy}{dt}=u=\frac{a\sqrt{2gh}}{(A+a)}##.
But isn't the whole term of ##\frac{M}{\rho A}## constant?
 
  • #70
lloydthebartender said:
But isn't the whole term of ##\frac{M}{\rho A}## constant?
Initial values generally are constant.
 
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