Time it takes for submerged object to rise to surface

[bgizzle]

Is there a formula for this? if object is 300m deep, 30m in diameter, 50,000kg, in a ball shape how long will it take to reach surface?

 

[xodin]

The upward force acting on an object submerged in a fluid is equal to the weight of the fluid displaced--that is archimedes' principle.

So, you would have to use the volume of a sphere V=4*(pi)*r^3/3 to find the volume of water that is displaced by your object, and since density p = m/V->m = p*V, where the density of water is about 1000kg/m^3, you can find out the mass of water displaced, which you can take times gravity to get the weight. That will be the buoyant force.

Now, there is also the weight of the object itself dragging it down, in that case, you would just take its mass times gravity to find the downward force due to gravity: F=m*g (or F=m*a more appropriately, since g is gravitational acceleration). So long as you take the buoyant force to be positive, and the downward force/weight of the object to be negative, the sum of these two will give you the net force on the object, assuming no other forces are present, such as tension from a cable.

Using the net force, you can use F=m*a to solve for the acceleration of the object, which, if the buoyant force is greater will be upward (positive), and if it's less, it will be downward (negative).

Once you have the acceleration, you can use the linear motion formulas to solve for the time it takes to reach the surface. Assuming zero initial velocity that would be distance d = a*t^2/2 -> t = sqrt(2*d/a).

Hope this helps.

 

[bgizzle]

yes, that helps a lot. 300m depth, 10m diameter, 50,000kg, 1000 density of water

volume of water for sphere (10m diameter, 5m radius) ~523

mass of water displaced: ~523,000

buoyant force of weight: 9.81*523,000= ~5.1M (this is correct right?)

downward force due to gravity: 50,000*9.81= 490,500

net force: 5.1m-490k=4.6m

acceleration: 4.6m/50,000kg= 92.92 (is this correct?)

time (in minutes?): sqrt(2*300m/92.92)=2.54 minutes

There is tension from a cable, I have no clue how much :( physics is hard

 

[voko]

The derivation above does not take the resistance of the medium into account, which in this case will be significant.

 

[xodin]

Your calculations are correct except for a few things:

1. In your original problem statement you stated 30m for diameter not 10m, which is it?

2. The time is in seconds, not minutes, so 2.54 seconds.

3. If there is tension in the cable, then that totally changes the game, but based on the problem statement I'm thinking you aren't suppose to take any tension into account. Let me guess: Does it say that there is a cable tying the object to the bottom of the water, and at the start of the problem is the cable cut? If so, then the problem really begins after the cable is cut and therefore there would be no tension.

4. Voko is correct, the viscosity of the water would come into play in reality, but you'll have to decide whether that was meant to be ignored. Does it say to ignore it? Does it say to think of the water as an "ideal fluid"? Have you not talked about viscosity in class before? If the answer to any of those is yes, you should probably assume it shouldn't be taken into account.

PS. Most equations (unless explicitly stated otherwise) use MKS values for measurements: meters, kilograms, seconds; these equations are no exception, so that's why t is in seconds.

 

[bgizzle]

this isn't for a problem in a book, i am trying to mimic the real world as closely as I can. the object would be attached to a wire which would be attached to an object that would be spinning as the ball rises. I'm not sure if that counts as a significant amount of tension because the object would be allowed to spin.

the diameter is 10m

2.54 seconds seems fast for a 10 meter buoyant object to travel 300m upwards. that's faster than the object would fall (8 seconds to fall).

 

[xodin]

It is fast, but, relatively speaking, you are displacing a HUGE amount of water with a very light object. Have you ever tried to hold something like a basketball under water? The force can be huge. I assure you it's accurate for this ideal scenario.

If you need to take viscosity into account, I can't help you there as I'm not that far in my own studies yet--it's fairly advanced as I believe it depends on the velocity vectors of the fluid flowing around the object, and the area of the object causing the flow, or something. For a sphere I believe that would require some complex integration, and I'm guessing would require significant approximation just to calculate. In reality you'd also have to take into consideration the flow or current already present in the water in addition to the flow caused by the object itself.

In the scenario you describe with the tension, you wouldn't have to take that into account, because it is circling constantly and therefore the tension is effectively equal in all directions and so the net effect would be zero. I was thinking of tension more as a cable fixed to a boat or anchored to the ground, etc.

FYI: You can also use linear motion formulas to determine the velocity of the object at any point on it's journey, including after it flies out of the water (though the acceleration would change from that found above to simply gravity after it leaves the surface). For the portion before it leaves the water, it would be v = a*t, so at the moment it reaches the surface it would be traveling at 236 m/s! That's pretty darn fast. After it leaves the surface it would then be v = Vi - at -> v = 236 - (9.81)*t, with t resetting at the surface. It would keep moving upward into the air for 24.057 seconds reaching a max height of 2,841.6 meters in the air before coming back down, or nearly 3 kilometers! Once again this is still ignoring the viscosity, which would, in reality, slow that all down significantly.

 

[bgizzle]

ok, good to know the calculations i used are correct. Maybe I'm off on this but a 30 ft in diameter sphere weighing 50,000 kg doesn't seem that light or buoyant to me. maybe i am missing something.

 

[xodin]

It is compared to the 523,000 kg of water it's displacing, and that's what matters in terms of physics :)

The scenario is quite likely impossible. The pressure at that depth would be ~3,000,000 Pa or N/m^2 which, given the relatively huge size and small weight of the object would likely result in it being crushed if made from any known materials, thereby reducing its volume. In addition, the force required to bring the object down that far would be extraordinary, and if it was done with a cable of any sorts, the tension in the cable used to pull it down would quite possibly snap it due to the buoyant force, again, using any known materials.

 

[xodin]

Just for fun a did a couple more calculations.

Let's say the solid parts of the object were made of iron. It would only take 6.35 m^3 of iron to reach 50,000 kg, whereas the volume of your sphere is 523 m^3. If the object was lined on the outside with iron, and the inside were completely hollow, you could only have a shell 2.03 cm thick for it to weigh only 50,000 kg, which I'm guessing would easily be crushed at that depth.

 

[Alesak]

Well, to submerge the sphere you could fill it with mercury and then when it hit the bottom you could just pull the plug. That would be fun. You would also need hose for air from surface to top of the sphere.

 

[bgizzle]

ok, makes sense.

i'm struggling to reach a realistic size for this ball. if its made out of iron and i want it to withstand 300m depth, what would be a realistic diameter and weight? i want it buoyant but it does not need to be ultra buoyant.

how about 250,000 kg and 10m diameter. that gives me a time of 200 seconds... how did you figure out the density of the iron shell?

 

[VantagePoint72]

As I told you in your other thread, whether an object is buoyant or not depends only on its density. If it's made of a uniform material, its buoyant acceleration will not change by changing its size. What would be a realistic diameter and weight depends entirely on the sort of materials you want to use.

You don't really seem to have any fixed problem parameters since you keep saying that such-and-such can be changed if necessary. Why you don't just tell us what the real-world scenario you're trying to figure out is? It's difficult to help you when you state a very vague problem like this since we can't really tell what it is you're trying to do.

 

[voko]

ok, good to know the calculations i used are correct. Maybe I'm off on this but a 30 ft in diameter sphere weighing 50,000 kg doesn't seem that light or buoyant to me. maybe i am missing something.

Your own calculation implies that the net acceleration is about 10g. A typical space rocket has about 3g. So your ball will be raising very fast - so fast, that the viscosity of water will need to be taken into account if you want to model the reality closely.

 

[xodin]

As for how thick the shell would need to be before it crushes, I'm not sure. Of course, it varies material to material, and it also depends on the shape and size (sphere is the best I'm pretty sure, in this situation). An example I found, which might not help but is at least something, is a submarine that made it to almost 11,000 meters: it was 2.16 meters in diameter and had 12.7 cm thick walls of steel. The reduced diameter plays a major role in maintaining the structural integrity.

As for how I calculated the thickness of the iron (the density of the shell is the density of iron):

Iron density (roughly): 7874 kg/m^3

Target mass was 50,000 kg, so V = m/p = 50,000/7874 = 6.35 m^3

The volume of your sphere of radius 5m is 523 m^3. Subtract 6.35 from that, gives: 516.65 m^3.

Now, reverse the volume of a sphere equation to solve for the remaining radius:

r = cube root(3*V/[(pi)*4]) = cube root(3*516.65/[(pi)*4]) = 4.977 m

The original radius of the sphere 5m - 4.977m of hollow space = 0.0222 = 2.22 cm. I must have used different rounding the first time, but you get the gist.

 

[xodin]

Your own calculation implies that the net acceleration is about 10g. A typical space rocket has about 3g. So your ball will be raising very fast - so fast, that the viscosity of water will need to be taken into account if you want to model the reality closely.

Yep, exactly. Though it's fun to play around our ideal scenario, all of this is horribly inaccurate without taking viscosity into account. Does someone know how to do that? From the little I know about it, I believe it's very complex.

 

[bgizzle]

awesome, thanks xodin, that was exactly what i needed. 250,000 of iron would give a density of ~10cm which should hold at 1k ft depth when compared with your example. Also the ball would be rising at an average leisurely speed of 3.6mph (I am aware that none of this includes viscosity)

If someone would like to share how to include viscosity that would be fine, if not or if I am being too vague, please don't bother. I can accept this estimate is the best I can do for now and assume that my estimates are inaccurate but hopefully within the ballpark.

 

[xodin]

awesome, thanks xodin, that was exactly what i needed. 250,000 of iron would give a density of ~10cm which should hold at 1k ft depth when compared with your example. Also the ball would be rising at an average leisurely speed of 3.6mph (I am aware that none of this includes viscosity)

If someone would like to share how to include viscosity that would be fine, if not or if I am being too vague, please don't bother. I can accept this estimate is the best I can do for now and assume that my estimates are inaccurate but hopefully within the ballpark.

Sure thing, glad to help. The slower speed should definitely help reduce the effect of viscosity. I should point out again though, that the 10cm is not the density, but the thickness of the shell. Also, did you convert to mph? Otherwise the speed would be m/s. In addition, you can't fairly compare the 10 cm as holding up simply because the sub can, because the sub is so much smaller and therefore more structurally stable.

Sorry, I always edit a lot as I keep thinking of more stuff. I got to thinking that the resistance caused by the viscosity of the fluid might be very similarly related to the effect of air resistance on an object in free fall, and I've heard that calculating that involves differential equations, which I don't know yet.

 

[bgizzle]

i meant thickness of shell not density, your right.

i calculated speed based on the fact that i know the object travels ~984 feet and i know it takes ~200 seconds... is that a correct options for estimating speed?

 

[xodin]

i meant thickness of shell not density, your right.

i calculated speed based on the fact that i know the object travels ~984 feet and i know it takes ~200 seconds... is that a correct options for estimating speed?

That would be 4.92 ft/s (feet per second), which would translate to 3.35 mph (miles per hour) or 1.5 m/s (meters per second). You can use whichever units you like, so long as the formulas are always used with MKS (meters, kilograms, seconds) units and then converted after the calculations are made.

 

[bgizzle]

ya i was sloppy and just guestimated my conversion of seconds to minutes... i changed it to be exact and got 3.35. again thanks for the help and double checking me.

 

[xodin]

ya i was sloppy and just guestimated my conversion of seconds to minutes... i changed it to be exact and got 3.35. again thanks for the help and double checking me.

Cool, I saw your number was close so I figured you probably did convert and just rounded somewhere. No problem at all. Good luck with your idea! :)

 

[xodin]

No need to reply to this, but one thing I thought I'd add again was the fact that the velocity you calculated above is the average velocity. In reality, the velocity of the object will continually increase as it is accelerating from zero initially to its max velocity at the surface. The actual velocity at any point would be its instantaneous velocity and can be calculated using the formulas I put on the first page that involve time.

 

[voko]

Yep, exactly. Though it's fun to play around our ideal scenario, all of this is horribly inaccurate without taking viscosity into account. Does someone know how to do that? From the little I know about it, I believe it's very complex.

It is not very complex, but it is somewhat more complex. The equation is:$$m\frac {dv} {dt} = (\rho V - m)g - \frac 1 2 \rho v^2 c_D A$$ $c_D$ is the drag coefficient, and $A$ is the reference area, which for the sphere is simply the area of its greatest planar section, i.e., $A = \pi r^2$, and $V = \frac 4 3 \pi r^3$ so we get $$m\frac {dv} {dt} = (\rho \frac 4 3 \pi r^3 - m)g - \frac {\pi \rho c_D r^2} {2} v^2$$ $$\frac {dv} {dt} = (\rho \frac {4 \pi r^3} {3m} - 1)g - \frac {\pi \rho c_D r^2} {2m} v^2= a - b v^2 = a(1 - \alpha^2v^2)$$ This can be integrated, yielding $$v = \frac {\tanh {\alpha a t}}{\alpha}$$ This can be integrated further, yielding $$x = x_0 + \frac {\ln {\cosh {\alpha at}}} {\alpha^2a}$$ Solving for time, $$t = \frac {\cosh^{-1} { e^{\alpha^2a(x - x_0)}}} {\alpha a}$$

With $x_0 = -300, x = 0, r = 5, m = 50,000, \rho = 1000, g = 9.8$ we get $a = 92.8, b = 0.785 c_D, \alpha = 0.092 \sqrt{c_D}$

$$t = \frac {\cosh^{-1} { e^{235.5 \sqrt {c_D}}}} {8.5}$$
The drag coefficient depends on the so-called Reynolds number, which in this case is greater than $10^6$, i.e., fully turbulent flow, and the coefficient is about $0.5$. So $$t = \frac {\cosh^{-1} { e^{235.5 \cdot 0.707}}} {8.5} = \frac {\cosh^{-1} { 2.039E72}} {8.5} = \frac {167.19} {8.5} = 19.7$$ Which is about an order of magnitude greater than estimated without the drag.

 

[xodin]

That is awesome to see, thanks for sharing that voko. I was able to follow your math for everything except the right side of the equation for the two integration steps. I'm pretty familiar with integration in general and just finished calc II. We did cover hyperbolic functions and their inverses in differentiation back in calc I, but we didn't really have any problems involving them in integration. If I remember correctly, the derivative of tanh(x) is sech^2(x) and the derivative of cosh(x) is sinh(x), so I'm not seeing what identities you used to reverse the process. Would you mind briefly explaining the identities you used to integrate?

PS. Is this a differential equation? I've always heard such problems involved differential equations.

PPS. Is there a name for the original formula you used?

 

[xodin]

Actually, I found the formula for drag that you are using, so I see your first formula is really just Fnet = Fb - W - Fd, where Fb is the buoyant force, W is the weight of the object, and Fd is the force due to drag, so I'm guessing there is no explicit name for the equation.

Reading more here now:

http://en.wikipedia.org/wiki/Drag_(physics )

Still though, if you could explain the integration identities and whether this is a differential equation, that would be greatly appreciated.

In addition to not knowing the identities, one of the things I'm confused by is where v^2 goes when you integrate on the right side of the equation. As shown on the left side, the integral of v is x, so I don't understand why x isn't present on the right side of the integrated equation.

 

[VantagePoint72]

xodin, yes, that is a differential equation: it involves both $v$ and at least one of its derivatives. Strictly speaking, only the latter condition is necessary (i.e. technically $\frac{dv}{dt} = f(t)$ where $f$ has no explicit $v$ dependence is also a differential equation; it's just a trivial one solved by usual integration). In particular, this is an ordinary ($v$ is a function of only one variable) first-order (only first derivatives appear) non-linear (because $v^2$ appears) differential equation.

The reason it looks a bit mysterious is that you can't just integrate both sides of the equation with respect to time. The right has a term with $v$ in it, so how do you integrate that when that's what we're trying to find? Its anti-derivative if it were alone would, as you said, be $x(t)$, but it's part of a larger equation so we don't know how to do that integral. So, when voko said, "This can be integrated...," that was really sweeping a few details under the rug. The solution uses a technique called separation of variables. Informally (or the mathematicians will shoot us), we can rewrite:

$\frac {dv} {dt} = a(1 - \alpha^2v^2)$

as

$\frac {dv}{a(1 - \alpha^2v^2)}= dt$

where we've perversely treated differentials as simple numbers that can be algebraically manipulated willy-nilly. It's a bit dodgy, and there's a more formal way of doing this, but for well-behaved functions this is OK. Then we simply integrate each side with respect to their differentials:

$\int \frac {dv}{a(1 - \alpha^2v^2)}= \int dt = t + t_0$.

Now you can do the integral on the left in the usual way (with trigonometric substitution, in this case) and invert the resulting equation to get voko's answer (where, presumably, he or she set $t_0 = 0$).

Or if that's too much bother, you can just plug the result for $v(t)$ into the original equation and see that it all works out—I did, and it does.

 

[xodin]

That was VERY enlightening, thank you so much! You also answered several other curiosities I've had about differential equations simultaneously :)

If you don't mind helping out one more time, would you mind elaborating how the trigonometric substitution results in a hyperbolic function? I'm still trying to work this out and know I still would need to replace theta with a function of v in the equation below and reformulate the equation, but I decided to post my work mid-way because I'm not even sure I'm taking the right path here:

(I'm taking a and $\alpha$ as constant, is that correct?)

$\frac {1}{a}\int \frac{dv}{1-\alpha^{2} v^{2}}=t_{0} + t$

Let $v=\frac {sin{\theta}}{\alpha} \rightarrow dv = \frac {cos{\theta}}{\alpha} d\theta$

$\frac {1}{a \alpha}\int \frac {cos{\theta}}{1-sin^{2}{\theta}} d\theta=t_{0} + t$

$\frac {1}{a \alpha}\int \frac {cos{\theta}}{cos^{2}{\theta}} d\theta=t_{0} + t$

$\frac {1}{a \alpha}\int sec{\theta} d\theta=t_{0} + t$

$\frac {ln{\left|sec{\theta} + tan{\theta}\right|}}{a \alpha}=t_{0} + t$

I'm guessing I'm using the wrong substitution here based on my intermediate result not involving hyperbolics? Thanks again for your help.

 

[VantagePoint72]

Side note: an interesting consequence of this result is the terminal velocity. Since the limit of $tanh(x)$ as $x \rightarrow \infty$ is 1, this means that with increasing time the velocity approaches a finite value of $\frac{1}{\alpha} = \sqrt{\frac{a}{b}} = \sqrt{\frac{(\rho \frac {4 \pi r^3} {3m} - 1)g}{\frac {\pi \rho c_D r^2} {2m}}}$.

In fact, we didn't need all this calculus to tell us that. Since the drag force increases with velocity with no upper limit, at some point the combined drag and gravitational force must balance out the buoyant force. Thus, the net force will be zero and the acceleration will be zero. To find the terminal velocity this way, we just set the net force to zero:

$F_{net} = 0 = (\rho \frac 4 3 \pi r^3 - m)g - \frac {\pi \rho c_D r^2} {2} v^2_{terminal}$

Solving for $v_{terminal}$ in this equation gives us the same result as we got from the differential equation! This a nice way to check the answer voko worked out for us. One insight we gain from the differential equation that we couldn't get immediately from the above equation (though we could probably reason out anyways) is that terminal velocity is in fact a limit: the object will never quite reach its terminal velocity. It will get very, very close though; close enough that the distinction won't really matter.

 

[VantagePoint72]

If you don't mind helping out one more time, would you mind elaborating how the trigonometric substitution results in a hyperbolic function?

Actually, I lied: you just have to refer to a table of standard derivatives to see that: $\frac{d}{dx}arctanh(x) = \frac{1}{1 - x^2}$. You can check this if you like. Your calc textbook should also have a proof of this in it. Now just rescale the variables appropriately and you get the result.

 

[xodin]

Actually, I lied: you just have to refer to a table of standard derivatives to see that: $\frac{d}{dx}arctanh(x) = \frac{1}{1 - x^2}$. You can check this if you like. Your calc textbook should also have a proof of this in it. Now just rescale the variables appropriately and you get the result.

Ah, ok, I think I'm getting close now! Great stuff on the terminal velocity too, that part makes total sense, thanks again!

$\frac {1}{a} \int \frac {dv}{1-\alpha^{2} v^{2}} = t_{0} + t$

$\frac {tanh^{-1}\left(\alpha v\right)}{a} = t_{0} + t$

Ignoring $t_{0}$:

$tanh^{-1}\left(\alpha v\right) =a t$

$\alpha v = tanh\left(a t\right)$

$v = \frac {tanh\left(a t\right)}{\alpha}$

Somewhere I lost an alpha! :P

 

[VantagePoint72]

Let's just focus on:
$\int \frac {dv}{1-\alpha^{2} v^{2}}$

Set $u=\alpha v, du = \alpha dv$. Thus we have:
$\frac{1}{\alpha}\int \frac {du}{1-u^{2}} = \frac{1}{\alpha}arctanh(u)$

There's your missing alpha. Not just sub u back in and multiply the $\frac{1}{a}$ I ignored.

Incidentally, if you want to show that $arctanh(x) = \int \frac{dx}{1 - x^2}$, you can use trig substitution like I said—but with hyperbolic trig functions, not regular ones like you tried. Substitute $x = tanh(u)$ and away you go. Of course, if you knew to do this off the top of your head, you could have just solved the original differential equation by inspection without all this fiddling. Solving diff eq's is as much an art as a science. Such is life.

 

[xodin]

Let's just focus on:
$\int \frac {dv}{1-\alpha^{2} v^{2}}$

Set $u=\alpha v, du = \alpha dv$. Thus we have:
$\frac{1}{\alpha}\int \frac {du}{1-u^{2}} = \frac{1}{\alpha}arctanh(u)$

There's your missing alpha. Not just sub u back in and multiply the $\frac{1}{a}$ I ignored.

Incidentally, if you want to show that $arctanh(x) = \int \frac{dx}{1 - x^2}$, you can use trig substitution like I said—but with hyperbolic trig functions, not regular ones like you tried. Substitute $x = tanh(u)$ and away you go. Of course, if you knew to do this off the top of your head, you could have just solved the original differential equation by inspection without all this fiddling. Solving diff eq's is as much an art as a science. Such is life.

You are awesome, thank you once again. I was just coming back to look and see if I should have used u substitution there to get my missing alpha, and apparently should have. I should have known better on that one, but rushed it! I can't wait to take the course, but will be taking multivariable first for intro to mechanics followed by differential equations next semester.

I really appreciate both your's and voko's posts, but especially your added explanations! Enjoy the rest of your evening.

 

[VantagePoint72]

My pleasure, good luck in the course.

 

[voko]

$$\frac {1} {a} \int_0^v \frac {dv} {1 - \alpha^2 v^2} = \frac {1} {2a} \int_0^v \left[ \frac {1} {1 - \alpha v} + \frac {1} {1 + \alpha v}\right]dv = \frac {1} {2\alpha a} \left[ -ln (1 - \alpha v) + ln(1 + \alpha v)\right]_0^v = \frac {1} {2\alpha a} \ln \frac {1 + \alpha v} {1 - \alpha v}$$ Using Euler's formula for hyperbolic functions, we can see immediately that the latter is $\frac {1} {\alpha a} \tanh^{-1} \alpha v$, so $$\frac {1} {\alpha a} \tanh^{-1} \alpha v = \int_0^t dt = t$$