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Re: [Phys-L] Dirigible Flight Question

Sounds like a plan. Of course for the Goodyear blimp you may have to wait
minutes for the oscillations to die down. And of course as already pointed
out this is a problem where you are sweeping things under the rug, but all
physics problems do that. There is another factor which might change things
and that is the extra weight on the bottom of the blimp may cause some
deformation which could change the result, but that is probably a second
order problem. As stated it is a reasonable problem to give to beginning
physics students after they have done a McDermott tutorial on buoyancy. But
I would make it a rigid balloon just to keep some problems away. But for
advanced students a blimp could be used as a second problem and aske them to
make a qualitative graph of the velocity & acceleration of the hanging mass
when first released. They should realize it would fall freely initially,
oscillate, and eventually come to a terminal velocity. The regular students
should be able to draw a velocity graph starting at zero, increasing, and
eventually leveling off.

The NTN2 analysis is really a good way of brining home the idea that the
total mass is the mass of the balloon, helium, and the hanging mass.
Students have to learn that the buoyant force is not caused by the Helium
pushing the balloon up from the inside. This is misconception is a
consequence of (a) not understanding interactions and (b) not realizing that
the pressure on the bottom is higher than on the top of a balloon, hence the
buoyant force.

I might use this problem with some suitable cooked up numbers.

John M. Clement
Houston, TX



Ok, John, just for you I'll give it a little push up when I
attach the weight so that the oscillation dies down and all
of the new tensions around the balloon have time to
equilibrate before the vertical velocity goes to zero. What
is the acceleration at zero velocity?

Paul

On Nov 5, 2012, at 12:18 PM, "John Clement"
<clement@hal-pc.org> wrote:

This analysis is probably good for a perfectly stiff dirigible, but
remember a blimp is elastic and will stretch initially so
immediately
after depends on how immediately. So immediately
afterwards the mass
will accelerate at g and the blimp not at all. Then a little later
(how long?) the analysis would be correct, except for
corrections due
to the oscillations of the mass.

I presume you could try this with a Helium filled balloon and using
some video analysis you could confirm the idea. But a
balloon would
have a short oscillation period so that might not be noticeable.

I have often seen the Goodyear blimp, but sufficiently far
away as to
be unable to see the flexing of the balloon. But I seem to recall
there are some videos showing it. I missed riding in it
when it was
stationed outside Houston, but perhaps someone who has been
in it can come up with an account.
I suspect the dirigibles were a bit smoother in their "flight".

John M. Clement
Houston, TX

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Paul
Nord
Sent: Monday, November 05, 2012 11:49 AM
To: Phys-L@Phys-L.org
Subject: [Phys-L] Dirigible Flight Question

If I've got a blimp inflated to neutral buoyancy and I
hang a small
mass from it, what will the acceleration of the blimp be
immediately
after I attach the mass?

Since we're still at zero velocity we can ignore the
viscous effects
of the air for just a moment. I believe that I need to know the
un-inflated mass of the balloon and the payload, the volume of the
helium, the mass of the helium, and the mass of the displaced air.
Let's assume a very small pressure is held by the balloon
so that we
can think of it as simply a volume of helium at the
ambient pressure.
The mathematics of a simple Attwood's Machine would seem to apply.
The total mass going down:
balloon and payload
helium mass
extra ballast weight (call this 'm')
Total mass going up:
mass of displaced air

Let's call the sum of all of the mass except for the
ballast weight
'M'.
M = balloon + payload + helium + displaced air

The acceleration of the balloon is then:
a = g * m / (m + M)

Question 1: Of course Pascal's Principle says that air
pressure will
distribute itself equally on all sides. In the static case I can
ignore the effects of pressure and air mass. The net
force is zero
(ignoring the vertical pressure gradient of the air, yes).
However,
for a balloon to move down, an identical volume of air
needs to move
up the same distance. The mass of that air cannot be ignored. Is
this a valid assumption?

Question 2: Flow through a vicious fluid is typically
modeled with a
term which rises as a function of the square of the
velocity. There
is a force resisting the passage of a moving object. If
it use the
mass of the displaced fluid in the calculation above, am I already
accounting for some of the "drag" force which is normally
accounted
for in this velocity squared term?

Paul
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Forum for Physics Educators
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http://www.phys-l.org/mailman/listinfo/phys-l