Cylindrical Physics
September 22, 2004 5:23 AM   Subscribe

Physicsfilter: How does the centre of gravity of a metal cylinder filled with liquid vary as the liquid is drained from the cylinder? [MI]

Assume that the cylinder is a light metal with walls of constant thickness and density, and with a disc of metal at either end to keep the liquid in. For example as one might find with an aluminium can of delicious foamy wife-beater. (I was sat on the train last night wondering about the stability of a tin on the table)
posted by biffa to Science & Nature (14 answers total)
It depends on the density of the cylinder. If the cylinder was weightless, the center of gravity of the system would lower as the amount of liquid was drained. It would stay on the axis of the fluid cylinder directly midway from top to bottom.

The more the cylinder weighs, the less affect the removal of fluid has on the center of mass.

For a soda or beer can, the dominant force for stability might actually be friction (which is gravity related, but not really center of mass related). As the beer empties, it weighs less and has less frictional sideways force, so is more likely to topple? Just a theory.

Actually, as I think about it, say the liquid weighs more than the cylinder to start. Twice as much or more. The center of mass starts in the center of the can. As the liquid drains, the center of mass moves down. As the mass of the remaining liquid gets to be less than that of the can, the center of mass will eventually start to creep back up again until it's back in the center. I think. It's been a while.
posted by RustyBrooks at 5:40 AM on September 22, 2004

The more you drank, the less likely it was to fall over, until you reached close to the last drops of Stella, when it's no longer a great factor in the gravity of the can, and as a result, the centre of gravity moves back up a little, towards the centre.

I think.
posted by armoured-ant at 6:54 AM on September 22, 2004

ok, if...
  • the can has height H (assuming the end caps are of negligable thickness)
  • the mass of the can is Mc
  • the initial mass of the liquid is Ml
  • the fraction height of the liquid in the can is f(t) (so f(0) = 1 and lim t->inf f(t) = 0)
then the height of the centre of mass is
H (Mc + Ml f (t))
- x (-------------)
2 (Mc + Ml f(t) )

posted by andrew cooke at 7:01 AM on September 22, 2004

if the two initial masses are equal then i reckon it's most stable when f(t)=sqrt(2)-1 (or about two fifths full).
posted by andrew cooke at 7:08 AM on September 22, 2004

(and that formula describes what other people have said - the centre of mass moves down then up again)
posted by andrew cooke at 7:09 AM on September 22, 2004

Yup, Rusty's basically got it right. Let h be the height of the can, x be the height of the total liquid left in it, m be the mass of the can, and M be the mass of the liquid in the can when it's full. The center of mass of the can is at a height h/2; the center of mass of the liquid is at a height x/2, and the liquid remaining in the can has a mass of x M/h. If we calculate the CM of these two objects, we get the CM of the entire system. It turns out to be

CM = (h/2) (m/M + (x/h)2) / (m/M + x/h)

As x decreases from h to 0, CM initially decreases as well; but at a certain point CM reaches a minimum and increases back up to h/2 when the can is empty. The lowest point of the CM is when

x0 = h ( sqrt( m/M (m/M+1) ) - m/M )

at which point the CM is at

CM0 = (h/2) [ (2 (m/M)2 + m/M) / sqrt( (m/M) (m/M + 1) ) - 2].

I think this is all correct, but the coffee hasn't yet kicked in so I might be making some sort of egregious error.

On preview, andrew cooke got the same result I did for the position of the CM, so I feel a little better.
posted by Johnny Assay at 7:09 AM on September 22, 2004

Rusty is right.

Let H be the cylinder height, T be the wall thickness, R be the cylinder radius, RHOc be the cylinder density, RHOl be the liquid density, and Hl be the height of the liquid.

Assume that the CG lies along the long axis of the cylinder.

The mass of the cylinder (Mc) is then RHOc*H*pi*(R^2-T^2).
The mass of the liquid (Ml) is RHOl*Hl*pi*(R^2-T^2).

The CG of the cylinder (CGc) is always H/2.
The CG of the liquid (CGl)is always Hl/2.

To find the CG of the system, we find the weighted average of the cylinder CG and the liquid CG:

CGsys = (Mc*CGc + Ml*CGl)/(Mc + Ml)

Algebraic reduction of this equation is left as an exercise for the reader.

So for the CG of the system as a function of the liquid height we get:

CGsys(Hl) = 1/2*(RHOc*H^2 + RHOl*Hl^2)/(RHOc*H + RHOl*Hl)

Assume a steel (RHOc = 7.8 g/cm^3) cylinder, 15cm tall, filled with water (RHOl = 1.0 g/cm^3).

Plugging in we get

CGsys(Hl) = (1/2)*(1755 + Hl^2)/(117 + Hl)

When the cylinder is full, Hl = 15 and CGsys = 7.5
When the cylinder is empty, Hl = 0 and CG sys = 7.5

Finding the minimum is also left as an exercise for the reader, but it can be shown that it occurs at a liquid height of 7.27 cm in our example.

(God, I hope I did that right. Nice exercise to start the morning with.)
posted by mbd1mbd1 at 7:10 AM on September 22, 2004

Crap, I didn't think I was that slow....
posted by mbd1mbd1 at 7:11 AM on September 22, 2004

Now see, why'd you guys have to go and bring math into it ;) I prefer physics by the seat of my pants, and if I must, the back of the envelope.
posted by RustyBrooks at 7:12 AM on September 22, 2004

For a soda or beer can, the dominant force for stability might actually be friction.

Actually stability relates to the Center of Gravity being directly over the base of the can (that is, the projection of the center of gravity on the plane of the base lies within the base). The angle (measured from the vertical) the can has to be at in order to move the CG off of the base is inversely proportional to the height of the CG from the base, so that's why the can is most stable when it CG is lowest.
posted by signal at 7:38 AM on September 22, 2004

Do these calulations take into consideration the fact that the top and bottom of an aluminum can are of differing thickness from the sides?

(My math-fu is failing me)
posted by kamylyon at 7:52 AM on September 22, 2004

yes - it's only necessary for the ends to be the same as each other (although they don't take into account the shape of the ends - they're assuming flat ends of negligable thickness).
posted by andrew cooke at 8:14 AM on September 22, 2004

in practice, friction may be important because otherwise the can may slide across the train table and tip over the edge. the force that stops it sliding is proportional to the total weight (assuming, probably incorrectly, that the table and can bottom and dry and clean).
posted by andrew cooke at 8:17 AM on September 22, 2004

Well, this has been even more interesting than I thought the question was. Thanks to all who've done the maths, especially the bits I understood. I'm tempted to go and get another tin and do some weighing.
posted by biffa at 11:26 AM on September 22, 2004

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