# Sinking a cylinder with varying hole sizes

#### lloydthebartender

$dV=vA\frac{1}{dt}$ where A is the area cross-sectional area of the hole?

#### haruspex

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$dV=vA\frac{1}{dt}$ where A is the area cross-sectional area of the hole?
Almost, but 1/dt?? That doesn't mean anything. Also, I specified a as the cross section of the hole and used A for that of the cylinder. Please stick to that.

#### lloydthebartender

Almost, but 1/dt?? That doesn't mean anything. Also, I specified a as the cross section of the hole and used A for that of the cylinder. Please stick to that.
Since $Q=\frac{dV}{dt}$, and $Q=va$, so $dV=vadt$?

#### haruspex

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Since $Q=\frac{dV}{dt}$, and $Q=va$, so $dV=vadt$?
Right.
So what is the increase in the depth of water (y-h) in the cylinder in dt?

#### lloydthebartender

Right.
So what is the increase in the depth of water (y-h) in the cylinder in dt?
$d(A(y-h))=vadt$
$dt=\frac{d(A(y-h))}{va}$?

#### haruspex

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$d(A(y-h))=vadt$
$dt=\frac{d(A(y-h))}{va}$?
Right.
But A and h are constants, so can you rewrite that in the form dy/dt=...?

#### haruspex

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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

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

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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

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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

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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

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

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#### lloydthebartender

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

#### haruspex

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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

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

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