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# Motion in a circle watch

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1. The following question gave rise to an interesting discussion and I thought you might be interested
in commenting.

If a mass is suspended from a string in say a railway carriage then when the carriage moves in a
circle the string becomes inclined to the vertical by an angle that is simple to calculate given the
necessary radius and speed. The mass moves away from the centre of the circle until the angle is
such that the component of tension in the string gives the necessary force towards the centre of the
circle. So far OK.

Now consider a balloon filled with helium floating in the carriage above the point of attachment.
Which way will the balloon move when the carriage goes round a circle?

2. I would say that the balloon moves outwards, away from the centre of the circle.

It's really just the same problem as the suspended mass, turned on its head. Granted, there are now
three vertical forces on the balloon (weight, 'uplift' and a component of tension) but these will
simplify to give one resultant vertical force - just as in the suspend mass case. The only
horizontal force on the balloon is a component of tension, just as in the suspend mass case.

I'm ignoring all effects of air resistance on the balloon though. They're probably quite
significant!

Dan.
3. In article <[email protected]>, jwbradley <[email protected]>
writes snip
[q1]>Now consider a balloon filled with helium floating in the carriage above the point of attachment.[/q1]
[q1]>Which way will the balloon move when the carriage goes round a circle?[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]>[/q1]
I think we need to consider the air in the carriage; It has a shape of a cuboid, (for simplicity),
with a balloon shaped hole. When the train moves round the curved track, the centre of mass of the
air tends to keep moving in a straight line (Newton's first Law) and so to an observer in the train
would seem to move to the outside of the curve. For this to happen, the balloon shaped hole must
appear, to the observer in the train, to move to the inside of the curve.

Another approach would be to consider that the apparent direction of 'down' has shifted away from
vertical, and now points downwards and outwards. Apparent value of 'g' is now sqrt(g^2 +
(v^2/r)6^2). Apparent 'g' now acts at an angle of tan^-1((v^2/r)/g) to the vertical (downward and
outwards).

This modified magnitude and direction of 'g' should be used to calculate the upthrust on the balloon
using Archimedes' principle. (Upthrust = weight of air displaced; weight of air + mass of air x
modified 'g' )
--
Chris Holford
4. In article <[email protected] on.co.uk>, Chris Holford <[email protected]>
writes OOPS!
[q1]>outwards. Apparent value of 'g' is now sqrt(g^2 + (v^2/r)^2)).[/q1]

--
Chris Holford
5. "jwbradley" <[email protected]> wrote in message
news:[email protected]...
[q1]> The following question gave rise to an interesting discussion and I[/q1]
thought
[q1]> you might be interested in commenting.[/q1]
[q1]>[/q1]
[q1]> If a mass is suspended from a string in say a railway carriage then when[/q1]
the
[q1]> carriage moves in a circle the string becomes inclined to the vertical by[/q1]
an
[q1]> angle that is simple to calculate given the necessary radius and speed.[/q1]
The
[q1]> mass moves away from the centre of the circle until the angle is such that the component of[/q1]
[q1]> tension in the string gives the necessary force towards[/q1]
the
[q1]> centre of the circle. So far OK.[/q1]
[q1]>[/q1]
[q1]> Now consider a balloon filled with helium floating in the carriage above[/q1]
the
[q1]> point of attachment. Which way will the balloon move when the carriage[/q1]
goes
[q1]> round a circle?[/q1]

Centrifugal forces would surely move it away from the centre of the circle.

--
MESSAGE ENDS. John Porcella
6. John Porcella <[email protected]> wrote:
[q1]>"jwbradley" <[email protected]> wrote in message[/q1]
[q2]>> Now consider a balloon filled with helium floating in the carriage above[/q2]
[q1]>the[/q1]
[q2]>> point of attachment. Which way will the balloon move when the carriage[/q2]
[q1]>goes[/q1]
[q2]>> round a circle?[/q2]
[q1]>[/q1]
[q1]>Centrifugal forces would surely move it away from the centre of the circle.[/q1]

Centrifugal forces push everything in the carriage away from the centre of the circle. Since the air
is denser than the helium-filled balloon, the air is pushed out harder, and the balloon is displaced
towards the centre of the circle.

Or, to put it another way: 'down' acquires a component pointing out from the centre of the circle,
and since the balloon is trying to go 'up', it heads in towards the centre.

--
Rob. http://www.mis.coventry.ac.uk/~mtx014/
7. In article <[email protected] om>,
"John Porcella" <[email protected]> wrote:

[q2]> > Now consider a balloon filled with helium floating in the carriage above[/q2]
[q1]> the[/q1]
[q2]> > point of attachment. Which way will the balloon move when the carriage[/q2]
[q1]> goes[/q1]
[q2]> > round a circle?[/q2]
[q1]>[/q1]
[q1]> Centrifugal forces would surely move it away from the centre of the circle.[/q1]
[q1]>[/q1]
[q1]>[/q1]
[q1]> --[/q1]
[q1]> MESSAGE ENDS. John Porcella[/q1]

If the balloon were floating in a vacuum, you would be right, but since it is floating in a denser
mass of air (or it would not be floating at all), the denser mass of air moves away from the
centre of rotation with a greater impetus than the balloon, so the balloon is floats towards the
center of rotation.
8. Robert Low wrote:
[q1]>[/q1]
[q1]> John Porcella <[email protected]> wrote:[/q1]
[q2]> >"jwbradley" <[email protected]> wrote in message[/q2]
[q2]> >> Now consider a balloon filled with helium floating in the carriage above[/q2]
[q2]> >the[/q2]
[q2]> >> point of attachment. Which way will the balloon move when the carriage[/q2]
[q2]> >goes[/q2]
[q2]> >> round a circle?[/q2]
[q2]> >[/q2]
[q2]> >Centrifugal forces would surely move it away from the centre of the circle.[/q2]
[q1]>[/q1]
[q1]> Centrifugal forces push everything in the carriage away from the centre of the circle. Since the[/q1]
[q1]> air is denser than the helium-filled balloon, the air is pushed out harder, and the balloon is[/q1]
[q1]> displaced towards the centre of the circle.[/q1]
[q1]>[/q1]
[q1]> Or, to put it another way: 'down' acquires a component pointing out from the centre of the circle,[/q1]
[q1]> and since the balloon is trying to go 'up', it heads in towards the centre.[/q1]
[q1]>[/q1]
That's a neat explanation. You can do something similar with a helium filled ballon in a car just
starting from rest. Which way will the ballon move as you accelerate? I recall trying to explain to
my five-year-old daughter why the balloon she held on a string (as she was securely strapped in
behind me) moved the way it did. I'm really not sure whether I succeeded or not. (Though she's now
studying physics at Padova (Galileo's alma mater) - whether as a result of frustration at my lack of
clarity in explaining or the opposite I suppose I'll never know. Probably neither.)

It's a good thought experiment ... which way will the ballon move? Then check your answer (but keep
an eye on the road!).

Bob

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