Sinking a cylinder with varying hole sizes

[haruspex]

$d(A(y-h))=vadt$
$dt=\frac{d(A(y-h))}{va}$?

Right.
But A and h are constants, so can you rewrite that in the form dy/dt=...?

 

[haruspex]

The next step is the Bernoulli equation. Need to be careful, though. The v I defined is relative to the cylinder, and the cylinder is moving. Bernoulli will tell you the gained velocity in the lab frame.

 

[lloydthebartender]

Right.
But A and h are constants, so can you rewrite that in the form dy/dt=...?

$\frac{dy}{dt}=\frac{va}{A-h}$?

The next step is the Bernoulli equation. Need to be careful, though. The v I defined is relative to the cylinder, and the cylinder is moving. Bernoulli will tell you the gained velocity in the lab frame.

So if $u$ is the negative, downwards velocity of the cylinder [at the hole = above the hole]
$\frac{1}{2}\rho (v+u)^2+\rho gz_{1}+P_{1}=\frac{1}{2}\rho (\frac{a(v+u)}{A})^2+\rho gz_{2}+P_{2}$?

 

[haruspex]

$\frac{dy}{dt}=\frac{va}{A-h}$?

No, you've made some mistake in the algebra. Try again. A-h makes no sense, you cannot subtract a distance from an area.

So if $u$ is the negative, downwards velocity of the cylinder [at the hole = above the hole]
$\frac{1}{2}\rho (v+u)^2+\rho gz_{1}+P_{1}=\frac{1}{2}\rho (\frac{a(v+u)}{A})^2+\rho gz_{2}+P_{2}$?

Not sure which is 1 and which is 2, but the water is stationary below the hole.
As already noted, z₁=z₂, so you can cancel those.
What expressions do you have for P₁ and P₂?

 

[lloydthebartender]

No, you've made some mistake in the algebra. Try again. A-h makes no sense, you cannot subtract a distance from an area.

$\frac{d(Ay)−d(Ah)}{dt}=va$
$\frac{Ad(y)−Ah}{dt}=va$
$\frac{d(y)−h}{dt}=\frac{va}{A}$
$\frac{d(y)}{dt}=\frac{va}{A}+h$
Is this correct?

No, you've made some mistake in the algebra. Try again. A-h makes no sense, you cannot subtract a distance from an area.

Not sure which is 1 and which is 2, but the water is stationary below the hole.
As already noted, z₁=z₂, so you can cancel those.
What expressions do you have for P₁ and P₂?

I see.
Below the hole = Above the hole
$\frac{1}{2}\rho (0) + \rho gz_{1}+P_{1}=\frac{1}{2}\rho (v+u) + \rho gz_{2}+P_{2}$
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
$\rho gy=\frac{1}{2}\rho (v+u)+\rho gh$
$gy=\frac{1}{2} (v+u)+ gh$

 

[haruspex]

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.

$\frac{1}{2}\rho (0) + \rho gz_{1}+P_{1}=\frac{1}{2}\rho (v+u) + \rho gz_{2}+P_{2}$
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.

 

[lloydthebartender]

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...
$\frac{d(Ay)-d(Ah)}{dt}=va$
$\frac{d(Ay)}{dt}-\frac{d(Ah)}{dt}=va$
$\frac{dy}{dt}=\frac{va}{A}$

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
$u=\sqrt{2g(y-h)}-v$?

 

[lloydthebartender]

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.

 

[haruspex]

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.

 

[lloydthebartender]

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?

 

[lloydthebartender]

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?

 

[haruspex]

I could see something was going wrong. Just traced it to this:

$P_{2} = \rho gh$

That is not correct. What is the water depth just above the hole?

 

[lloydthebartender]

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 $P_{1}=P_{2}$?

 

[haruspex]

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?

 

[lloydthebartender]

No. Look at your diagram in post #24. What is the depth of water inside the cylinder?

Right, so $y-h$; $P_{2}=\rho g(y-h)$

 

[haruspex]

Right, so $y-h$; $P_{2}=\rho g(y-h)$

Yes.

 

[lloydthebartender]

Yes.

So I get
$\rho gy=\frac{1}{2}ρ(v+u)+ρg(y-h)$, and rearrange for $u$?

 

[haruspex]

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.

 

[lloydthebartender]

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.

$gy=\frac{1}{2}(v+u)^2+g(y-h)$
$2gy-2g(y-h)=(v+u)^2$
$2gh=(v+u)^2$
$\sqrt{2gh}=v+u$
$v=\sqrt{2gh}-u$
$u=\frac{va}{A}$, $v=\frac{uA}{a}$
$\frac{uA}{a}=\sqrt{2gh}-u$
$uA=a\sqrt{2gh}-au$
$u(A+a)=a\sqrt{2gh}$
$u=\frac{a\sqrt{2gh}}{(A+a)}$

 

[haruspex]

$gy=\frac{1}{2}(v+u)^2+g(y-h)$
$2gy-2g(y-h)=(v+u)^2$
$2gh=(v+u)^2$
$\sqrt{2gh}=v+u$
$v=\sqrt{2gh}-u$
$u=\frac{va}{A}$, $v=\frac{uA}{a}$
$\frac{uA}{a}=\sqrt{2gh}-u$
$uA=a\sqrt{2gh}-au$
$u(A+a)=a\sqrt{2gh}$
$u=\frac{a\sqrt{2gh}}{(A+a)}$

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.

 

[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?