The dropped beer mug.

Cecil's Columns/Staff Reports
1

Cecils’ answer seems weak. Not wrong, just weak.

The bubbles in a beer glass rise because they are bouyant, i.e. less dense than the surrounding fluid. Gravity effect.

The bubbles would remain stationary within the glass as it dropped, because the entire system is in free fall (the accelaration due to gravity is not materially counteracted).

As the glass approached terminal velocity, the “free fall” condition would gradually cease to exist. This glass ain’t falling in a vacuum.

I guarantee you that the bubbles in the glass (providing the thing stayed upright) would be rising just as they would on a counter-top just before the thing hit the pavement.

I aint doin’ the math.

2

I too think something is wrong with Cecil’s answer. His saying ‘gravity would act uniformly on the fluid surrounding the bubbles’ is odd. Doesn’t gravity ‘act uniformly on the fluid’ when it’s perched atop my kitchen counter? Why would the bubbles not still float to the top of the glass?

3

Wow that was a bad answer.

As exgineer has pointed out - atomosphere definetly plays a role here, but more than one.

however for now let’s ignore them for now

an object in freefall is, simplified, without gravity’s effects. Hence the same behavior would be observed in, say, a space shuttle circling the earth. The beer-mug (terminal velocity aside - which we won’t ever get into just yet) would be as well “weightless” and its existance can be safely ignored.

Along that line of thought - we can conclude that any bubbles produced in the middle of this “glob” of liquid will remain in the middle, not flowing up / down / whatever. (think space shuttle)

we are of course wondering about the bubbles on top. without any reason to go into the liquid, it would stay on top, but notice that bubbles are usually not perfectly spherical due to buyancy (a side-effect of gravity) when at rest. however in the absence of gravity, the surface tention of the beer will make all the bubbles round (if non collpses). since round bubbles requires more liquid in between the bubbles, the “line” where beer end and foam start will become less distinct, and move toward the bottom - in the other words the foam will become taller and thicker (but the total volumn - beer + air - does not change - so nothing overflows.

NOW let’s talk about the effect of atmosphere.

the changing air pressure as the beer goes down (it’s a fairly rapid change) will cause less bubbles to be produced, and the current bubbles to become smaller. assuming nothing happens except this force, smaller bubbles means bubbles are not as likely to burst. This is mentioned because if you take beer and rapidly decrease the environmental pressure in which it reside (bring up an express elevator, say), the bubbles will burst (become bigger until the film of beer that forms the bubble can no longer hold it), and many bubbles will hastily come into existence (think opening a bottle of soda).

the mug encloses the beer on three sides so they can be assumed the same terminal velocity (if the mug does not tip over). without accounting for turbulence and the low-pressure zone right above the mug (see above paragraph for the low-pressure zone), they will indeed become somewhat affected by gravity (or, at least, exhibit signs thereof) before reaching term.v and exhibit gravity’s full effect after reaching term.v.

these are really all the important factors that affect the beer and mug and bubbles (i mean, neutrinos are only SO important in this scenario). I have analized them separately, but together I am sure y’all are smart enough to draw a more complete conlusion.

4

Well, in all fairness to Cecil, the statement of the original question implies that L.T. was not asking for each and every effect present in the beer glass, but rather only about the motion of the bubbles. “Would the bubbles shoot to the top of the glass, the bottom, or stay motionless?” Still, good comments all. And welcome to the Straight Dope, lingqi.

5

If you factor in the air resistance, it would at no time be in a free fall condition. OTOH, a human body gets to terminal velocity much faster than a mug of beer (don’t ask me how I know), and I’m not certain that even a human body would have time to reach terminal velocity in just the short drop from the Empire State building.

Don’t come in here shouting weak about a twenty-year-old answer, and say that.

6

Terminal velocity is reached when the drag force on the body reaches the force applied by gravity.
Drag force on a body = 1/2[symbol]r[/symbol]V[sup]2[/sup]AC[sub]D[/sub]
where:
[symbol]r[/symbol] = air density (~0.0023 slug/ft[sup]3[/sup])
V = velocity
A = frontal area (~0.049 ft[sup]2[/sup] for a 3" diameter beer mug)
C[sub]D[/sub] = drag coefficient (~0.8 for a blunt-nosed cylinder).

For a ~3 lb. beer mug,

3lb = 1/2[symbol]r[/symbol]V[sup]2[/sup]AC[sub]D[/sub] = 1/2(0.0023)V[sup]2/sup(0.8) = 0.000056V[sup]2[/sup]

V = 230 ft/sec

The empire state building is 1250 ft high. Ignoring air resistance, the mug would reach a velocity of ~280 ft/sec in that distance. In truth, the velocity of the mug would asymptote to its terminal velocity, so I’m not quite sure what the velocity-time graph would look like, or what the mug speed would be when it hit the pavement, but seems like it would be some pretty substantial fraction of terminal velocity.

7

In the interest of clarifying my OP (which, admittedly, wasn’t terribly clear or coherent) I post the following:

When I said, “I ain’t doin’ the math,” I meant, “I really don’t feel like looking up the drag coefficient for a circular aspect.”*

Fortunately for me, zut did it gratis.

The beer mug would reach terminal velocity long before it hit the pavement.** And that’s all I was trying to say.

My initial complaint stands. Cecil’s answer was incomplete.

As for this:

Hey, it was on the front page, and I call 'em like I see 'em.

[sub]* We’re talking aerodynamic drag, and therefore frontal cross-sectional area here, and ignoring the handle of the mug. I trust this is a sufficient approximation.

** This is ignoring the actual architecture of the ESB. It was built with multiple setbacks to comply with the building code at the time. Our hypothetical beer mug would have smashed on a ledge before getting to the pavement.[/sub]

8

That’s all I was trying to say :slight_smile:

zut’s post seems to say differently.

9

Slight mistake in my earlier post: I mis-multiplied, and the terminal velocity ought to be 257 ft/sec.

To determine mug velocity when dropped, figure out the total force on the mug, which is force of gravity minus air resistance:

F = weight - 1/2[symbol]r[/symbol]V[sup]2[/sup]AC[sub]D[/sub] = 3lb - 4.5E-05V[sup]2[/sup]

The acceleration is then the force equation divided by mass (a=F/m):

acc = 32.2 - 4.2E-06V[sup]2[/sup]

Realizing that acceleration is the time derivative of velocity gives the differential equation:

dV/dt = 32.2 - 4.2E-06V[sup]2[/sup]

which, as far as I can tell, is kind of nasty to solve in closed form. I went ahead and did a numerical time-step solution, though, and came up with the following:



**Time Accel Vel  Dist**
0   32.2    0     0
1   31.7   32.0  16.8
2   30.4   63.1  65.4
3   28.2   92.4 144
4   25.5  119   251
5   22.4  143   383
6   19.3  164   537
7   16.4  182   711
8   13.6  196   900
9   11.2  209  1103
9.7  9.7  216  1250

Impact off a 1250 ft high building is at 9.7 sec. The velocity at that time = 216 ft/sec, which is 84% of terminal velocity. (Just for reference, the beer mug takes about 16 sec and 2900 ft to reach 250 ft/sec velocity.)

Anyway, the speed of 216 ft/sec looks like the “substantial fraction” I thought it would be. Although the beer mug doesn’t really “reach terminal velocity long before it hits the pavement”, part of the reason is that, mathematically, the speed is going to asymptote to the terminal velocity, so one needs to define how close one needs to get before “reaching” terminal velocity.

All conclusions subject to revision upon review of mathematical accuracy, of course, but the numbers seem ballpark correct to me.

10

Fast enough, but guarantees aside, Cecil’s answer still seems to be appropriate for when the mug is dropped–and in the spirit of the question.

11

It’s appropriate but not well worded. “Gravity would act uniformly”? Gravity is always working uniformly.

That said, I’m not sure what would be a better explanation. You could say that normally the gravity pushes the beer against the mug, creating pressure which pushes the bubbles up. If you drop it, there is no longer any pressure since there is nothing for the beer to push against.

Or you could just say that the mug is in free fall, and being in free fall is indistinguishable from weightlessness. So there is no “up” towards which the bubbles could rise.

12

Okay, color me embarrassed.*

I screwed up my shirt-cuff calculation of terminal velocity (dropped a decimal). I actually spent valuable time trying to work out how the handle on a standard beer mug would affect aerodynamic drag.**

Our beer mug would not, in fact approach terminal velocity when dropped from that height. Drag would tend to counteract acceleration due to gravity, but equilibrium would not be reached over this distance.

I screwed up.

Correct. I’ll slink away in shame now, if it’s all the same to you.

*[sub]I know that doesn’t make any sense. It’s evidently par for the course.[/sub]

**[sub]Danged thing would just tip over and spill its contents.[/sub]

13

Exgineer, clearly math is the problem, we should have been doing field work. Let’s meet for beers, we can drop them off the Royal Gorge. No, wait, let’s just drink 'em.

scr4, I’m trying hard to make sense of it myself, and I think I’ve got it. Clearly, Cecil was talking about gravity in spacetime, so uniform gravity implies flat spacetime, or an inertial reference frame–that is, freefall. WTF is he talking about sloshing, though?