Water sloshing in a bathtub

[mbigras]

This is a multipart question, I'm working on part (b) but the full question can be found on page 85 at http://www.scribd.com/doc/50357722/Vibrations

Homework Statement

(b) Assuming that the water flow is predominantly horizontal, its speed $v$ must vary with $x$, being greatest at $x = 0$ and zero at $x = \pm L/2$. Because water is incompressible (more or less) we can relate the difference of flow velocitites at $x$ and $x + dx$ to the rate of change $dy/dt$ of the height of the water surface at $x$. This is a continuity condition. Water flows in at $x$ at the rate $vhb$ ($b$ is the width of the tub) and flows out at $x+dx$ at the rate $(v+dv)hb$. (We are assuming $y_{0} << h$.) The difference must be equal to $(b dx)(dy/dt)$, which represents the rate of increase of the volume of water contained between $x$ and $x+dx$. Using this condition, show that
$$v(x) = v(0) - \frac{1}{hL}x^{2}\frac{dy_{0}}{dt}$$
where
$$v(0) = \frac{L}{4h}\frac{dy_{0}}{dt}$$

The Attempt at a Solution

The question gives an explanation about what the continuity continuity condition is. Conceptually I think I understand what going on: Water is flowing into the imaginarily thin container at some rate and is flowing out of it and some slower rate so the water has to go somewhere and ends up flowing up at some related rate. I feel confused about how to actually use the continuity condition to get the equation for the velocity as a function of x. I'm also not sure if all my integration and wiggling around of the continuity condition is allowed.
$$vbh - (v+ dv)hb = b dx \left(\frac{dy}{dt}\right) \\ -dv h = dx \left(\frac{dy}{dt}\right) \\ dv = - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ \int dv = \int _{0} ^{\frac{L}{2}} - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ v = - \frac{1}{h} \left(\frac{dy}{dt}\right) \frac{L}{2}$$

 

[Redbelly98]

This is a multipart question, I'm working on part (b) but the full question can be found on page 84 at http://www.scribd.com/doc/50357722/Vibrations

Homework Statement

(b) Assuming that the water flow is predominantly horizontal, its speed $v$ must vary with $x$, being greatest at $x = 0$ and zero at $x = \pm L/2$.

Not quite, there is non-zero flow at x=±L/2. Look at the hint in the book:

"... the product velocity × cross section must have the same value everywhere along the tube"

So, velocity would only go to zero if the cross section became infinitely large, which does not happen. The radius varies linearly from r to 2r along the horizontal section, so that can be used to get the cross section as a function of x.

Also, you know that the velocity in the left-hand vertical part is dy/dt, where the tube radius is r.

Because water is incompressible (more or less) we can relate the difference of flow velocitites at $x$ and $x + dx$ to the rate of change $dy/dt$ of the height of the water surface at $x$. This is a continuity condition. Water flows in at $x$ at the rate $vhb$ ($b$ is the width of the tub) and flows out at $x+dx$ at the rate $(v+dv)hb$.

Nope. Water in the vertical sections is flowing vertically, and "wraps around" the right-angle bend when it enters or exits the horizontal part on either side. Effectively, it is a circular cross section everywhere.

(We are assuming $y_{0} << h$.) The difference must be equal to $(b dx)(dy/dt)$, which represents the rate of increase of the volume of water contained between $x$ and $x+dx$. Using this condition, show that
$$v(x) = v(0) - \frac{1}{hL}x^{2}\frac{dy_{0}}{dt}$$
where
$$v(0) = \frac{L}{4h}\frac{dy_{0}}{dt}$$

The Attempt at a Solution

The question gives an explanation about what the continuity continuity condition is. Conceptually I think I understand what going on: Water is flowing into the imaginarily thin container at some rate and is flowing out of it and some slower rate so the water has to go somewhere and ends up flowing up at some related rate. I feel confused about how to actually use the continuity condition to get the equation for the velocity as a function of x. I'm also not sure if all my integration and wiggling around of the continuity condition is allowed.
$$vbh - (v+ dv)hb = b dx \left(\frac{dy}{dt}\right) \\ -dv h = dx \left(\frac{dy}{dt}\right) \\ dv = - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ \int dv = \int _{0} ^{\frac{L}{2}} - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ v = - \frac{1}{h} \left(\frac{dy}{dt}\right) \frac{L}{2}$$

 

[mbigras]

I think that I may have mislead you because I wrote a different page number than I meant down, I should have said page 85 instead of 84. I'm working on question 3-18. I've edited it above.

 

[Redbelly98]

I think that I may have mislead you because I wrote a different page number than I meant down, I should have said page 85 instead of 84. I'm working on question 3-18. I've edited it above.

Aha!

The Attempt at a Solution

The question gives an explanation about what the continuity continuity condition is. Conceptually I think I understand what going on: Water is flowing into the imaginarily thin container at some rate and is flowing out of it and some slower rate so the water has to go somewhere and ends up flowing up at some related rate. I feel confused about how to actually use the continuity condition to get the equation for the velocity as a function of x. I'm also not sure if all my integration and wiggling around of the continuity condition is allowed.
$$vbh - (v+ dv)hb = b dx \left(\frac{dy}{dt}\right) \\ -dv h = dx \left(\frac{dy}{dt}\right) \\ dv = - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ \int dv = \int _{0} ^{\frac{L}{2}} - \frac{1}{h} \left(\frac{dy}{dt}\right) dx \\ v = - \frac{1}{h} \left(\frac{dy}{dt}\right) \frac{L}{2}$$

You're fine up until setting up the integrals. I see some issues with how you are treating some of these quantities from then on.

For starters, you are doing the integral from 0 all the way to L/2, which would be fine if you were asked for v at x=L/2. But they are asking for v as a function of any value of x. So the integral's upper limit should be x.

Second, the integral over dv is not simply v. It is the difference,
$$v(x \text{ at upper limit of integration}) - v(x \text{ at lower limit of integration})$$
So, I recommend making the lower limit of integration correspond to an x value where you know what the velocity is. (There are two such values.)

Finally, it looks like you are treating dy/dt as a constant that can be factored out of the integral. That is wrong -- dy/dt is a function of x and should remain inside the integral. But, notice that the final answer is in terms of dy₀/dt. So rewrite dy/dt in terms of x and dy₀/dt (and any other quantities, as necessary), then take another stab at evaluating the integral.

Hope that helps, and that you can follow it all.

 

[mbigras]

I like how you used the term "treating quantities" because that's where I feel most of my confusion lies.

$$-dv h = \left( \frac{dy}{dt} \right) dx \\ dv = -\frac{1}{h} \left( \frac{dy}{dt} \right) dx \\$$
I don't see how I can integrate the left-hand side with respect to $x$ if there isn't a $dx$ tacked onto the end. One thing I was thinking about is maybe:
$$dv = d \left(\frac{dx}{dt}\right) = \frac{d}{dt} dx$$
But I'm not sure what that would mean or if it's allowed. The $dv$ is one of those quantites that I don't see how to treat.

Working with the right-hand side:
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}\\ y = \frac{2y_{0}}{L} x \\ \frac{dy}{dx} = \frac{2 y_{0}}{L}\\ \frac{dx}{dt} = v\\ \frac{dy}{dt} = \frac{2 y_{0}}{L}v$$
I'm also not sure what that would mean or if it's allowed. The $y_{0}$ is another quantity that I don't see how to treat. When I used it to get get $y = \frac{2y_{0}}{L}x$ I treated it like a constant but in the context of this question it doesn't seem to be because you see things like $\frac{dy_{0}}{dt}$ scattered about.

 

[Redbelly98]

Sorry about the delay in replying. During the work week, I have less time for the forum than on weekends, and I am having to think out a good explanation that will be helpful to you.

I like how you used the term "treating quantities" because that's where I feel most of my confusion lies.

$$-dv h = \left( \frac{dy}{dt} \right) dx \\ dv = -\frac{1}{h} \left( \frac{dy}{dt} \right) dx \\$$
I don't see how I can integrate the left-hand side with respect to $x$ if there isn't a $dx$ tacked onto the end.

Well, you don't actually integrate with respect to x, you do still integrate with respect to v. I was just referring to what the limits of integration should be in the integral.

What I meant was, if the limits of integration on the right-side -- the integral w.r.t. x -- are x1 and x2, then the limits of integration on the left side (integral w.r.t. v) should correspond to those limits. Meaning, you would use the velocity to x=x1 for the lower limit, and likewise the velocity at x=x2 for the upper limit.

Your lower limit for x was zero in your original attempt, meaning the lower limit in the velocity integral should be the velocity at x=0. But we don't know that velocity, so x=0 perhaps isn't a good choice for a lower limit (in the x-integral).

Instead, I suggest using -L/2 for the lower limit in the x-integral. Then for the v-integral, the lower limit is the velocity at x=-L/2, which we know is zero. So, you can use -L/2 and 0 as the lower limits in the integrals w.r.t. x and v, respectively.

Then, for upper limits, just use x and v. This v would be the velocity at x, which is what we are trying to find an expression for.

One thing I was thinking about is maybe:
$$dv = d \left(\frac{dx}{dt}\right) = \frac{d}{dt} dx$$
But I'm not sure what that would mean or if it's allowed. The $dv$ is one of those quantites that I don't see how to treat.

Working with the right-hand side:
$$\frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dt}\\ y = \frac{2y_{0}}{L} x \\ \frac{dy}{dx} = \frac{2 y_{0}}{L}\\ \frac{dx}{dt} = v\\ \frac{dy}{dt} = \frac{2 y_{0}}{L}v$$
I'm also not sure what that would mean or if it's allowed. The $y_{0}$ is another quantity that I don't see how to treat. When I used it to get get $y = \frac{2y_{0}}{L}x$ I treated it like a constant but in the context of this question it doesn't seem to be because you see things like $\frac{dy_{0}}{dt}$ scattered about.

The above seems to go off on a wrong track, so let me try another explanation.

We have this $\frac{dy}{dt}$ term in the integral w.r.t. x.

Looking at the figure in the problem, we can see that y is a linear function of x, going from h-y₀ at x=-L/2 to h+y₀ at x=+L/2. So that means we have
$$y=mx+b,$$
and you can figure out what this slope m is in terms of h, L, and/or y₀, or some combination of those.

Since y₀ is changing with time, that will mean that the slope m is changing in time as well. So, at some given, fixed-in-time value of x, we can use the product rule on
$$y=mx+b$$
to get
$$\frac{dy}{dt}=\frac{dm}{dt}x+m\frac{dx}{dt}+\frac{db}{dt}$$
Since we are looking at a fixed-in-time x-coordinate, the $\frac{dx}{dt}$ term is zero. See if you can work out the next part. You'll need to substitude whatever you come up with for m into there. Keep in mind that, while h and L are constants, y₀ is not constant. y₀ can change with time, and its time-derivitive is not zero.

Also, you'll need to think a little (just a little) about what b is in terms of h, L, and/or y₀, and so what db/dt would be.

Hope that helps. Again, sorry for the delay in replying.

 

[mbigras]

Man so many parts of your responses have helped me. A couple key ones that stand out are:

$y_{0}$ is a function of time thus use the product rule for $\frac{dy}{dt}$

using limits of integration that correspond. This is when $v(x)$ and $v(0)$ jumped out, and shed light on an earlier post that you made.

$$vbh - (v+dv)bh = b dx \left( \frac{dy}{dt} \right) \\ dv = -\frac{1}{h} \frac{dy}{dt} dx \\$$
$$y = \frac{2}{L} y_{0} x \\ \frac{dy}{dt} = \frac{2}{L} \frac{dy_{0}}{dt} x + \frac{2}{L} y_{0} \frac{dx}{dt} \\ \frac{dy}{dt} = \frac{2}{L} \frac{dy_{0}}{dt} x \\$$
$$dv = -\frac{1}{h} \frac{2}{L} \frac{dy_{0}}{dt} x dx \\ \int _{v \ when \ x = 0} ^{v \ when \ x = x} dv = -\frac{1}{h} \frac{2}{L} \frac{dy_{0}}{dt} \int _{0} ^{x} x dx \\ v(x) - v(0) = -\frac{1}{hL} \frac{dy_{0}}{dt} x^{2} \\ v\left(\frac{L}{2}\right) = v(0) -\frac{1}{hL} \frac{dy_{0}}{dt} \frac{L^{2}}{4} = 0 \\$$
$$v(x) = v(0) - \frac{1}{hL} x^{2} \frac{dy_{0}}{dt}\\ v(0) = \frac{L}{4h}\frac{dy_{0}}{dt}\\$$
Also here's a video that I think provides a visual of what's going on here http://www.youtube.com/watch?v=-2JwUSD3x2s

 

[Redbelly98]

You got it

Also here's a video that I think provides a visual of what's going on here http://www.youtube.com/watch?v=-2JwUSD3x2s

Cool video. I watched it, wondering "what are they doing to make the water slosh around?" -- then I read the "earthquake" in the video title. Wow.

http://www.scribd.com/doc/50357722/Vibrations

It made me smile to see that books by A.P. French are still being used. My freshman physics class at Case Western used both his Newtonian Mechanics and his Special Relativity book. That was about 30+ years ago.