- #71
lloydthebartender
- 50
- 1
That would mean ##\frac{a\sqrt{2gh}}{(A+a)}## would equate to 0?haruspex said:Initial values generally are constant.
haruspex said:Thereafter, dydt=u=a√2gh(A+a)
That would mean ##\frac{a\sqrt{2gh}}{(A+a)}## would equate to 0?haruspex said:Initial values generally are constant.
haruspex said:Thereafter, dydt=u=a√2gh(A+a)
No.lloydthebartender said:That would mean ##\frac{a\sqrt{2gh}}{(A+a)}## would equate to 0?
So ##t=\int u dy## with the range ##t=\int_{\frac{M}{\rho A}}^{L} u dy##.haruspex said:No.
You have an equation for dy/dt. This describes how y changes with time from some initial value to its final value L, the length of the cylinder.
The initial value of y (i.e. the value at t=0) is M/(ρA).
Solve the differential equation for y and plug in the initial and final values of y as the bounds on the integral of dy/dt.
Since u is constant, yes. (It would not be that simple otherwise.)lloydthebartender said:So ##t=\int u dy## with the range ##t=\int_{\frac{M}{\rho A}}^{L} u dy##.
This gives ##u(L-\frac{M}{\rho A})##
So I got an equation for d^2 against 1/t...and experimentally this was supported. All is well except for one thing. In post #25 you say ##h## is constant. Why is this?haruspex said:No.
You have an equation for dy/dt. This describes how y changes with time from some initial value to its final value L, the length of the cylinder.
The initial value of y (i.e. the value at t=0) is M/(ρA).
Solve the differential equation for y and plug in the initial and final values of y as the bounds on the integral of dy/dt.
In post #22 you found that ##h=\frac m{\rho\pi r^2}##, which is constant.lloydthebartender said:So I got an equation for d^2 against 1/t...and experimentally this was supported. All is well except for one thing. In post #25 you say ##h## is constant. Why is this?
Alright I'll come back after getting everything sortedharuspex said:In post #22 you found that ##h=\frac m{\rho\pi r^2}##, which is constant.
Suppose it is not constant and it increases a little. Now the buoyancy exceeds the weight of the cylinder so it slows its descent. But the pressure difference at the hole is increased, increasing the rate of inflow of water and reducing h again.
But #22 doesn't take into account the drag force...and surely experimentally H diminishes to 0 when submerging.haruspex said:In post #22 you found that ##h=\frac m{\rho\pi r^2}##, which is constant.
Suppose it is not constant and it increases a little. Now the buoyancy exceeds the weight of the cylinder so it slows its descent. But the pressure difference at the hole is increased, increasing the rate of inflow of water and reducing h again.
Feel free to repeat the analysis taking drag into account. You may find it cannot be solved analytically, though. As I recall, it can be modeled as a reduced hole size, but that could be wrong.lloydthebartender said:But #22 doesn't take into account the drag force...and surely experimentally H diminishes to 0 when submerging.
Thanks I get it now. One more thing...in the Bernoulli equation am I taking two points right above and below the hole, or at the surface of water and below the hole. Because it seems to me that if ##z_{1}=z_{2}## the pressure terms would be the same?haruspex said:Feel free to repeat the analysis taking drag into account. You may find it cannot be solved analytically, though. As I recall, it can be modeled as a reduced hole size, but that could be wrong.
h will only diminish when the water outside reaches the top of the cylinder and flows in. Total submersion will follow swiftly, so the time taken for it to reduce to zero can be ignored.
There are three phases:
1. On placing the cylinder on the surface of the water and releasing, it quickly sinks to the point where its weight is balanced by the buoyancy.
2. It sinks slowly, limited by the rate at which water can flow through the hole. In this phase, h is nearly constant.
3. The water outside reaches the top and flows in, quickly reducing the airspace inside to zero.
For the purposes of the time to sink, only the second phase is interesting. The others are too brief.
What drives the water through the hole (at a much higher speed than that at which the cylinder sinks) is a pressure difference over a short distance, from just below the hole to just above it.lloydthebartender said:Thanks I get it now. One more thing...in the Bernoulli equation am I taking two points right above and below the hole, or at the surface of water and below the hole. Because it seems to me that if ##z_{1}=z_{2}## the pressure terms would be the same?
But how can there be a difference in the depth (h or y-h) and the height from the bottom (##z_{1}## or ##z_{2}##)?haruspex said:What drives the water through the hole (at a much higher speed than that at which the cylinder sinks) is a pressure difference over a short distance, from just below the hole to just above it.
The z2-z1 term in Bernoulli is the change in altitude of the flow of water. It is not quite zero here, but it is small. Most of the work the pressure difference does is in accelerating the water; only a small amount goes into lifting the water the short distance through the hole.
I forget which is z1... say z is the height of the hole from some common baseline, z1 is at z-d and z2 is at z+d, where d is small.lloydthebartender said:But how can there be a difference in the depth (h or y-h) and the height from the bottom (##z_{1}## or ##z_{2}##)?
I don't understand why the pressure above the hole is ##\rho g(y-h)##. Why isn't it ##\rho g(y)##. if ##z_{1}=z_{2}##?haruspex said:I forget which is z1... say z is the height of the hole from some common baseline, z1 is at z-d and z2 is at z+d, where d is small.
The pressure under the hole is ##\rho g(y+d)## and above the hole it is ##\rho g(y-h-d)##.
The pressure difference is ##\rho g(h+2d)##. The height difference is 2d.
Plug all that into Bernoulli and obtain an expression for v. You will see that
terms involving d are second order, so can be ignored.
As I just posted, z1 and z2 are not quite equal. In the gap between them there is a large pressure gradient pushing the water up through the hole.lloydthebartender said:I don't understand why the pressure above the hole is ##\rho g(y-h)##. Why isn't it ##\rho g(y)##. if ##z_{1}=z_{2}##?
So since the pressure of water within the cylinder is equal we use (y-h)? But wouldn't that be atmospheric pressure?haruspex said:As I just posted, z1 and z2 are not quite equal. In the gap between them there is a large pressure gradient pushing the water up through the hole.
Above z2 not much is happening. There is a stationary column of water of height y-h-z2, so the pressure there is that height multiplied by ##\rho g##.
Similarly, each side of z1 there is little movement, so the pressure there is the same as elsewhere at that depth below the level of water outside the cylinder.
What you seem not to be grasping is the large pressure gradient through the hole, making for a substantial pressure difference over a short distance.
I don't understand your question.lloydthebartender said:So since the pressure of water within the cylinder is equal we use (y-h)? But wouldn't that be atmospheric pressure?
Apparently:under-the-sheet said:Did I miss something here?
No help was requested regarding the experiment itself.lloydthebartender said:I'm trying to write up some theory for this experiment
Did you mean post #73?cwinqi said:In #75, how would you integrate the equation you substitute in for u when you are integrating with respect to y and there aren't any y terms?
I apologise if I've totally misunderstood.
haruspex said:Did you mean post #73?
You don't need y terms in order to integrate wrt y; you just need that the only variable present is y. In this case, the integrand is constant.
Aside these three phases, is there a chance that in the process of sinking, the cylinder will attain equilibrium and therefore would not sink further to a complete submersion?haruspex said:Feel free to repeat the analysis taking drag into account. You may find it cannot be solved analytically, though. As I recall, it can be modeled as a reduced hole size, but that could be wrong.
h will only diminish when the water outside reaches the top of the cylinder and flows in. Total submersion will follow swiftly, so the time taken for it to reduce to zero can be ignored.
There are three phases:
1. On placing the cylinder on the surface of the water and releasing, it quickly sinks to the point where its weight is balanced by the buoyancy.
2. It sinks slowly, limited by the rate at which water can flow through the hole. In this phase, h is nearly constant.
3. The water outside reaches the top and flows in, quickly reducing the airspace inside to zero.
For the purposes of the time to sink, only the second phase is interesting. The others are too brief.
Water will continue to flow into the cylinder through the hole until either the top of the cylinder falls below the water line (entering stage 3) or the water inside reaches the same level as the water outside. Either way, if the material of the cylinder is denser than water the cylinder will sink.Finder said:Aside these three phases, is there a chance that in the process of sinking, the cylinder will attain equilibrium and therefore would not sink further to a complete submersion?
Alright, great, thank you.haruspex said:Water will continue to flow into the cylinder through the hole until either the top of the cylinder falls below the water line (entering stage 3) or the water inside reaches the same level as the water outside. Either way, if the material of the cylinder is denser than water the cylinder will sink.
Finder said:Alright, great, thank you.
haruspex said:Water will continue to flow into the cylinder through the hole until either the top of the cylinder falls below the water line (entering stage 3) or the water inside reaches the same level as the water outside. Either way, if the material of the cylinder is denser than water the cylinder will sink.
Well,yes, i will start a separate thread. I was trying to know if there can be another phase aside the three phases @haruspex stated for a sinking cylinder. either way, all the derivations for this thread applies to mine. the weight of the cylinder as used for this thread can be translated as the total weight for my question.kuruman said:The original thread is more than 3 years old and the problem as @Finder defined it is not the original in which the cylinder is fully submerged. The current question should be in a separate thread. I have reported it.