Buoyancy and Weightlessness
If I put a polystyrene ball into a test tube that is full
of water, and then put a cork on the test tube, the ball will float
to the top of the test tube. What would happen if I were to launch
that test tube, say, 100 metres into the air, and then have it turn
over, and free fall (straight) back to Earth? Would the fact that
the entire test tube is falling mean that everything would fall at
the same rate, meaning that the balls would remain at what is now
the lower end of the test tube (now that the test tube has reversed
itself for the return journey back to Earth) or would the
differences in the relative densities between the polystyrene balls
and the water push the polystyrene balls up, away from Earth?
Consider the forces at work during the fall. Consider what causes
pressure within the liquid. Buoyancy occurs because the water pressure
below the object is greater than the pressure above the object. This
net force is just enough to counter the weight of the displaced water.
If the ball is less dense than the water, making the weight less than
that of the displaced water, the buoyancy force upward on the ball is
greater than the gravitational force downward on the ball.
If the water is in free fall, the net force down on the displaced water
due to gravity would have to be larger than the upward force due to
buoyancy by just enough to allow for gravitational acceleration
downward. The buoyancy force would equal zero. Pressure throughout the
liquid would be constant. It would equal the gas pressure within the
tube. The ball would feel gravitational force downward and no buoyancy
upward, giving it a downward acceleration of "g". The ball remains
where it is. This is what would happen if the tube never rotates.
If the tube rotates while falling, you have another factor: centripetal
acceleration. During the rotation around its center of mass, some parts
of the tube will move faster than others. Any portion of material not
at the center of mass of the tube will have to feel a net force toward
the center equal to m(v^2)/r in order to maintain its position within
the tube. This will be larger near the ends. Liquid near the ends will
therefore have greater pressure. The increase in pressure will not be
uniform as with gravity, because the centripetal acceleration will be
different at different radii, greatest at the ends of the tube. A
likely outcome is some liquid at either end of the tube. The ball will
be in contact with either portion of water, down as far as necessary to
make that net buoyant force equal to m(v^2)/r. "r" will be the distance
from the tube's axis of rotation.
If the tube feels outside forces causing it to start rotating at the top
of the path and then causing it to stop rotating, exactly what happens
will depend on these specific forces and the location of the rotation
axis. If the tube is not rotating as it falls, the ball will remain
where close to where it is after the rotation has stopped.
Dr. Ken Mellendorf
Illinois Central College
Buoyancy is due to a difference in water pressure between the top
and bottom of the body floating in the water. No net weight (free
fall), no pressure, no floating.
R. W. "Bob" Avakian
B.S. Earth Sciences; M.S. Geophysics
Oklahoma State Univ. Inst. of Technology
You are correct in saying all things fall at the same rate. The
foam ball would remain wherever it was in the tube at the moment of
free fall. As soon as the tube was caught and free fall was ended,
the polystyrene ball would continue its journey to the top of the tube.
You can do this in the class room with a plastic tube corked on both
ends filled with water and a ping pong ball. (Get whatever sized
ball will fit into your tube.) Stand on your lab table. Let the
ball start drifting toward the top. In mid rise, drop it. Have a
student catch it. Have the other students watch where the ball is
when it is falling. It will begin to move immediately after it is
caught. So, drop high and get the longest clear bit of plastic
cylinder you can find.
It is a cool demo.
Free-fall means there is no local experience of gravity inside the tube.
During this time the water will not care whether it is over or under a lower-density object.
So the arrangement would mostly stay the same,
and the ball would tend to remain nearest to the mouth of the tube.
The water will care to remain stuck to the object by wetting forces,
which are usually pretty weak compared to gravity,
but when gravity is gone, they are about the strongest force operating inside the tube.
If no other forces were operating, the lowest energy state would be with
the water hugging itself into the bottom end of the test-tube,
with the plastic bead embedded in it's air-interface surface,
probably stuck lightly against a side wall.
A corked test-tube free-falling for minutes in vacuum would definitely do this.
The smaller the tube diameter, the faster it would settle into this state
and the harder it will "stick" there.
You may have noticed it can be difficult to shake water out
of the bottom end of a narrow tube.
For large-diameter test-tubes, such as 1cm or larger,
momentum of vibrations and swirls existing at the moment of release
will be pretty disruptive, unless one throws very carefully.
Some water probably will go around the ball during the free-fall because of them.
Question: what force will cause your tube to turn over at the top of it's arc?
I.E., does it suddenly turn, or was it always turning from the beginning?
If the tube has rocket-fins that provide a new push
when outside air-velocity changes direction,
then due to inertia, the water will try not to follow this new motion,
and the ball may trade places with the water in the tube's frame of reference.
In other words the ball would go nearer to the bottom of the tube.
Sounds like your question was getting at this.
If on the other hand the tube was gently started tumbling before it was released,
then the water and ball would be set into rotation around each other,
and the rotating motion would continue at the same speed throughout the flight.
The imparted tumble rate could be just right to coincidentally land mouth down,
and the ball could easily remain nearest to the mouth of the tube.
Viscous drag (linear with flow-speed) and dynamic drag (square-law with flow speed),
will try to limit the speed of any changes.
Of course water's viscosity is pretty low, so this will not prevent much.
And I have neglected air-drag on the outside of the tube,
which will make the tube slow down as it travels upwards,
so it would make a small negative perceived gravity in the tube just after release,
pushing water and ball towards the mouth of the tube.
This might make the real outcome difficult to predict.
You might also check how strong wetting-stiction is.
That is the degree of impediment to the advance of wettedness at the very tip of
the water meniscus.
If the wettedness will not advance as fast as the water is tumbling,
a little not-too-wettable plastic ball would try to stay stuck at a dry spot while
water goes around it.
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Update: June 2012