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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.##dV=vA\frac{1}{dt}## where A is the area cross-sectional area of the hole?
Since ##Q=\frac{dV}{dt}##, and ##Q=va##, so ##dV=vadt##?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.
Right.Since ##Q=\frac{dV}{dt}##, and ##Q=va##, so ##dV=vadt##?
##d(A(y-h))=vadt##Right.
So what is the increase in the depth of water (y-h) in the cylinder in dt?
Right.##d(A(y-h))=vadt##
##dt=\frac{d(A(y-h))}{va}##?
##\frac{dy}{dt}=\frac{va}{A-h}##?Right.
But A and h are constants, so can you rewrite that in the form dy/dt=...?
So if ##u## is the negative, downwards velocity of the cylinder [at the hole = above the hole]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.
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{dy}{dt}=\frac{va}{A-h}##?
Not sure which is 1 and which is 2, but the water is stationary below the hole.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}##?
##\frac{d(Ay)−d(Ah)}{dt}=va ##No, you've made some mistake in the algebra. Try again. A-h makes no sense, you cannot subtract a distance from an area.
I see.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, z_{1}=z_{2}, so you can cancel those.
What expressions do you have for P_{1} and P_{2}?
No. A and h are constants, so what is the change in Ah, d(Ah), in time dt?Is this correct?
Yes, except that you have omitted the power of 2 on the velocity term. Again, dimensionally inconsistent.##\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##
Oh...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.
##gy=\frac{1}{2}(v+u)^2+gh##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.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.
Sorry, I forgot to reply...At this point do I return to the force equations? I'm still not sure how this all connects back to it.
##\frac{va}{A}=\sqrt{2g(y-h)}-v##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.
SinceSorry, 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.
That is not correct. What is the water depth just above the hole?##P_{2} = \rho gh##
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}##?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?
No. Look at your diagram in post #24. What is the depth of water inside the cylinder?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}##?
Right, so ##y-h##; ##P_{2}=\rho g(y-h)##No. Look at your diagram in post #24. What is the depth of water inside the cylinder?
Yes.Right, so ##y-h##; ##P_{2}=\rho g(y-h)##
So I getYes.
You have forgotten to square the velocity term again.So I get
##\rho gy=\frac{1}{2}ρ(v+u)+ρg(y-h)##, and rearrange for ##u##?
##gy=\frac{1}{2}(v+u)^2+g(y-h)##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.
Yes!##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)}##