Sinking a cylinder with varying hole sizes

[lloydthebartender]

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 $t=\int{\frac{va}{A}dy}$?

 

[haruspex]

y would be 0 if there were no hole

No, the cylinder has weight. Apply Archimedes.

 

[lloydthebartender]

No, the cylinder has weight. Apply Archimedes.

I'm really not sure...$F_{b}$ would be 0 to $\rho g y$.

 

[haruspex]

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?

 

[lloydthebartender]

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.

 

[haruspex]

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.

 

[lloydthebartender]

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 $\rho g \frac{Mg}{\rho}$

 

[haruspex]

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?

 

[lloydthebartender]

Too many g's. The buoyant force will be Mg. So what volume is displaced?

$\frac{M}{\rho}$?

 

[haruspex]

$\frac{M}{\rho}$?

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?

 

[lloydthebartender]

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 $y=\frac{M}{\rho A}+h$.

 

[haruspex]

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?

 

[lloydthebartender]

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 $y=\frac{M}{\rho A}$

 

[haruspex]

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.

 

[lloydthebartender]

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?

 

[haruspex]

So can I do $\frac{dy}{dt}=u$, and this with the equation of #49?

Yes.

 

[lloydthebartender]

Yes.

But $\frac{d(\frac{M}{\rho A})}{dt}=0$, right?

 

[haruspex]

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, $\frac {dy}{dt}=u=\frac{a\sqrt{2gh}}{(A+a)}$.

 

[lloydthebartender]

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?

 

[haruspex]

But isn't the whole term of $\frac{M}{\rho A}$ constant?

Initial values generally are constant.

 

[lloydthebartender]

Initial values generally are constant.

That would mean $\frac{a\sqrt{2gh}}{(A+a)}$ would equate to 0?

Thereafter, dydt=u=a√2gh(A+a)

 

[haruspex]

That would mean $\frac{a\sqrt{2gh}}{(A+a)}$ would equate to 0?

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.

 

[lloydthebartender]

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.

So $t=\int u dy$ with the range $t=\int_{\frac{M}{\rho A}}^{L} u dy$.
This gives $u(L-\frac{M}{\rho A})$

 

[haruspex]

So $t=\int u dy$ with the range $t=\int_{\frac{M}{\rho A}}^{L} u dy$.
This gives $u(L-\frac{M}{\rho A})$

Since u is constant, yes. (It would not be that simple otherwise.)
Now plug in the expression you found for u in post #49.

 

[lloydthebartender]

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.

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]

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?

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.

 

[lloydthebartender]

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.

Alright I'll come back after getting everything sorted

 

[lloydthebartender]

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]

But #22 doesn't take into account the drag force...and surely experimentally H diminishes to 0 when submerging.

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.

 

[lloydthebartender]

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.

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]

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?

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.

 

[lloydthebartender]

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.

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]

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 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.

 

[lloydthebartender]

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.

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]

I don't understand why the pressure above the hole is $\rho g(y-h)$. Why isn't it $\rho g(y)$. if $z_{1}=z_{2}$?

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.

 

[lloydthebartender]

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.

So since the pressure of water within the cylinder is equal we use (y-h)? But wouldn't that be atmospheric pressure?

 

[haruspex]

So since the pressure of water within the cylinder is equal we use (y-h)? But wouldn't that be atmospheric pressure?

I don't understand your question.

We can ignore atmospheric pressure and just deal with gauge pressure. That's because we only care about the pressure difference through the hole, and atmospheric pressure contributes equally outside and in.

The gauge pressure inside the cylinder, at its bottom, is due to the depth of water in the cylinder, y-h.
The gauge pressure just under the cylinder is the ambient pressure at that depth in the water outside the cylinder, depth y.

 

[under-the-sheet]

Did I miss something here? Did anyone measure the time it took for the cylinder to become completely submerged (with varying small hole sizes) as required in the Homework Statement? The Homework did not ask for some theory based on the Homework Equations or to even develop a hypothesis based on the observations of test results.

 

[haruspex]

Did I miss something here?

Apparently:

I'm trying to write up some theory for this experiment

No help was requested regarding the experiment itself.

 

[cwinqi]

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]

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.

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.

 

[cwinqi]

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.

I missed that part, thank you. I need to get my eyes checked. All makes sense now.

 

[Finder]

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.

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]

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?

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]

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.

Alright, great, thank you.

 

[Finder]

Alright, great, thank you.

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.

In this particular case, the material of the cylinder is not denser than water but a certain mass (load) is to be added to the cylinder to cause it to sink just a little below the outside water level. i.e to the point where the total weight (added mass+cylinder itself) balances the Buoyant force and from there it starts sinking slowly. will it continue to sink till the water inside the cylinder reaches the same level as the water outside?

NB: the cylinder's height is more than the entire column of water it is being submerged into so only the second situation you stated can occur. (the cylinder cannot fall below the outside water line)

 

[kuruman]

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.

 

[Finder]

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.

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.