A balloon filled with air is completely submerged under water, a weight attached to it.

In article <399hql$9eo@mozo.cc.purdue.edu>, Eric P Burke writes:

[...], volume must decrease as pressure goes up.  When the volume of air
decreases, the volume it displaces decreases.  Thus, when you pull it
down, it displaces LESS water, and will then be out of equilibrium.  It
should rise until the air expands again to its original volume so that
the system in equilibrium once again.

David Karr replies:

The fallacy here is that not all equilibria are stable equilibria.
Some are unstable equilibria.  Unfortunately, most physics courses
teach you plenty about stable equilibria and nothing (or next to
nothing) about unstable equilibria, so it's easy to forget that the
latter exist.

Consider a pendulum that hangs from a pivot so that it can rotate a full
360 degrees in a vertical plane.  This system has two points of
equilibrium.  One is the familar equilibrium where the pendulum bob is
directly underneath the pivot.  This equilibrium is stable: if you push
the pendulum a little bit to one side and let go, it will immediately
swing back toward the equilibrium.  (Unfortunately it then starts to
oscillate around that equilibrium, but that's another matter.  Still,
it will never travel farther away than the distance you pushed it.)

The other equilibrium is where the pendulum bob is directly above the
pivot.  If you do indeed place it directly above, the forces of gravity
and the mechanical forces all act along the vertical line, with nothing
to push the pendulum right or left.  So it will stay there if not
disturbed.  Of course you have to get it dead spot on and be careful
not to jiggle the apparatus, because if you displace that bob ever so
slightly from the equilibrium point, it is *not* going to come right
back.  This is an unstable equilibrium.

I believe the balloon thought experiment is intended to represent an
unstable equilibrium.  As I recall, this instability is a real problem
for fish that use air bladders for buoyancy compensation.  Scuba divers'
air is handled by mechanics that are quite different from fish's
(divers have hard-shelled tanks and regulators), and so I don't recall
this problem being raised in scuba training.

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: further material on this puzzle, edited separately from other collected postings (batch 18); it may repeat some of the above. ]

[ Editor: problem as stated there: ]
A balloon filled with air is completely submerged under water, a weight
attached to it.  The balloon is in equilibrium.  I pull it a foot lower
and release the string.  Does the balloon

a)  stay a foot lower,
b)  sink, or
c)  come back up to the earlier equilibrium?


b) it sinks

At equilibrium, the density of the balloon-weight-system is equal to the
density of water.  (That's equilibrium)
The air is compressed when the balloon is pushed down.
The mass is constant, the volume decreases, therefore the density increases.
Something more dense than its medium sinks.

-- Rick Banghart <banghar4@studentb.msu.edu>

Ask anyone who has ever gone scuba diving. The balloon will sink.

If the total system is in equilibrium, the weight equals the volume of
water displaced. If the balloon drops a foot, the increased water pressure
compresses the air, making the whole system displace less water. Since its
weight hasn't changed, it will sink.

Conversely, if you raise the balloon a foot, the air will expand, it will
displace more water than it weighs and it will rise.

An expandable air-filled container under water is an unstable system. Any
motion away from equilibrium will drive it further away from equilibrium.

Ask any scuba diver who's ever had trouble deflating his BC when coming to
the surface. This phenomenon is a matter of life and death to them.

-- Timothy Reed <treed@ball.com>