A little phsyics/Calculus help please

[Deathslayer7]

There are two seperate questions to these two problems. With both of them, I got to the final part, but am unable to solve the freaking integral. :smallannoyed:

Question 2:

A cylindrical open tank needs cleaning. The tank is filled with water to a height meter, so you decide to empty it by letting the water flow steadily from an opening at the side of the tank, located near the bottom. The cross-sectional area of the tank is square meters, while that of the opening is square meters.

How much time t(1/2) does it take to empty half the tank? (Note: A useful antiderivative is integral of (x^(-1/2)=2x^(1/2)).)
Express your answer numerically in seconds. Take the free-fall acceleration due to gravity to be 9.81 meters per second per second.

t(1/2)=????

For this i have dh/dt= -sqrt(2*g*h/((A1/A2)^2-1)). (which is correct)

Note that if A(hole)/A(tank) is very small, then dh/dt is very small. For this reason, the water level is considered constant in many problems that involve fluid flowing from a tank through an opening in the bottom or at the side of the tank, or fluid flowing from a siphon tube.



Any help is appreciated. Thanks.

[MethosH]

Thank you for reminding me that I have a test about that in 20 days and I need to study :smalltongue:

Hm... I'll do that as soon as I finish studying the rest... If you wait for 20 days I may be able to solve this :smallbiggrin:

[Assassin89]

For question 1, divide both sides of the equation by p and take the antiderivative of both sides. Assuming that the equation is dp/p = w^2(r dr), this results in ln p = 1/2(w^2)*r^2. or is it taking the integral of w^(2r)? The form in which the equation is written confuses me.

[Deathslayer7]

see that is what i did (and you forgot +C)

ln(p)=1/2*w^2*r^2+C

where we exponentiate both sides to get.

p=C*e^(.5w^2*r^2)

but what im having trouble establishing is finding C.

[Ponce]

Insert a particular solution, ie, do

p(0) = C*e^(.5w^2*(0)^2)

C = p(0)

It says your answer can be expressed in terms of p(0) so you can stop there.

[Mando Knight]

Alright... let me see...

1.) I'm assuming you meant to write dp= ρω^2*rdr. This integrates to p(r)=(ρ(ωr)^2)/2-p(0), as ρω^2 is constant and independent from r, which in turn integrates to r^2/2.

[Deathslayer7]

Insert a particular solution, ie, do

p(0) = C*e^(.5w^2*(0)^2)

C = p(0)

It says your answer can be expressed in terms of p(0) so you can stop there.

then you get p=p(0)*e^(.5w^2*r^2)

what about r(0)? :smallconfused:

[Mando Knight]

then you get p=p(0)*e^(.5w^2*r^2)

what about r(0)? :smallconfused:

Don't worry about that: the integral Ponce used makes no physical sense.

Pressure is measured in Pascals, which is Newtons per square meter. Rho is density, or kilograms per cubic meter. Omega is angular velocity, measured in radians per second. R is the radius, which is in meters. All of these numbers have units, which means that using them as an exponent is automatically wrong unless they have a coefficient that negates the units.

[Deathslayer7]

Don't worry about that: the integral Ponce used makes no physical sense.

Pressure is measured in Pascals, which is Newtons per square meter. Rho is density, or kilograms per cubic meter. Omega is angular velocity, measured in radians per second. R is the radius, which is in meters. All of these numbers have units, which means that using them as an exponent is automatically wrong unless they have a coefficient that negates the units.

ok. but im asking you the same question what about r(0)?

and don't say it doesnt matter cause the question explicitly states Express your answer in terms of p(rho), w(omega), r, r(0), and p(0).

Bolded for emphasis.

[SilentDragoon]

The full question isn't requiring you to use every single term in the answer, it is instead asking you to use only those terms. Mando's answer is correct. If you wish to include r(0), then p(0)=f(r(0)) as is stated in the original question.

[Deathslayer7]

The full question isn't requiring you to use every single term in the answer, it is instead asking you to use only those terms. Mando's answer is correct. If you wish to include r(0), then p(0)=f(r(0)) as is stated in the original question.

this where i have to disagree. All terms need to be used to get a right answer. I know this to be true. So at one point I need to use all the variables given, not just some of them.

so at the moment i have p(r)= .5*p*w^2*r^2 -p(0)

yet somehow r(0) should be in there.
edit: i even checked it by trying to submit it. It said r(0) is needed.

[SilentDragoon]

so at the moment i have p(r)= .5*p*w^2*r^2 -p(0)

yet somehow r(0) should be in there.
edit: i even checked it by trying to submit it. It said r(0) is needed.

For clarity's sake and to make sure I understand...
p(r(t))= .5*rho*omega^2*r(t)^2 -p(r(0)) is the equation you have currently have, correct?

[Mando Knight]

...Out of curiosity, what's your online homework system? This kind of technicality sounds familiar...

For clarity's sake and to make sure I understand...
p(r(t))= .5*rho*omega^2*r(t)^2 -p(r(0)) is the equation you have currently have, correct?
Time never entered the equation, and the radius is independent from it, as the problem was presented. I don't even know what r(0) is, either. :smallconfused:

[SilentDragoon]

My word... I think I should stop trying to... help? and just go to bed. Hence the asking for clarity, as I'd usually write it r0. Presumably, if I'm correct this time around, r(0) or r0 is the point radially closest to the center of rotation of the centrifuge.

Edit: For actual clarity, I'd assume it's the distance from the center of rotation to the nearest point of the test tube, such p(rnaught) is pressure at the top of the test tube and p(rmax) is pressure furthest out.

[Deathslayer7]

r(0) = the intial distance the water is from the point of roation.

and i have

p(r)= (1/2)*rho*omega squared*r^2 - p(0)

edit: and its masteringphysics.com

[Deathslayer7]

heres a pic.


http://i198.photobucket.com/albums/aa125/death_slayer7/centrifuge.jpg


where the rotational axis is omega (forgot to put it there).

[Mando Knight]

edit: and its masteringphysics.com

The arch-enemy of us all. :smallannoyed:

I would chalk it up to the idiotic program, but not everyone feels like giving up on a bad system as quickly as I do.

[The Extinguisher]

Talk to your prof about them. See if you get to the same conclusion. Could be that the site is just broken.

[SilentDragoon]

Going back a bit, what you have is:
dp = rho * omega^2 * r * dr
Let asdf = rho * omega^2
dp = asdf * r * dr
INT[dp] = asdf * INT[r * dr]
p - pnaught = asdf* [ .5*r^2 - .5*rnaught^2 ]
p = .5 * asdf * (r^2 - rnaught^2) + pnaught

Something like that? (Now... must sleep....)

[Deathslayer7]

after lots of "educated" guessing. I got the right answer on the last attempt.

we had integral of dp=integral of rho*omega^2*r*dr

Integrate both sides to get

p-p(0)= .5*p*w^2*r^2 where r is evaluated from r to r(0).

thus you get

p-p(0)= .5*p*w^2*(r^2-(r(0))^2)

then p=.5*p*w^2*(r^2-(r(0))^2) + p(0)

now for the tank problem. >.>

edit: ninjaed but yes, like that :smallsmile:

[SilentDragoon]

Well... along the same lines

dh / dt = Z * h^.5
where Z = -1 * SQRT(2*g*((A1 / A2)^2-1)^-1)
dh / (h^.5) = Z * dt
2 * h^.5 - 2 * hnaught^.5 = Z * t
h= ((Z * t + 2 * hnaught^.5)/2)^2
Z is negative, so as t increases (Zt+2hnaught^.5) decreases, thus h decreases
find the time for half to drain by -or- h = .5hnaught, t=?
or
2 * (.5 * hnaught)^.5 - 2 * (hnaught)^.5 = Z * t


blargh... sleep time.