From: ram@zedat.fu-berlin.de
Phillip Helbig writes:
>In article ,
>ram@zedat.fu-berlin.de (Stefan Ram) writes:
>> Imagine a free particle is moving leftwards without forces
>> tangentially to a circle around a center:
>>
>> C_____B_____A
>> .-' '-.
>> .' '.
>> / \
>> ; ;
>> | o |
>> ; ;
>> \ /
>> '. .'
>> '-._____.-'
>>
Thank you to the moderation for making me aware of the
problems with posts containing Unicode characters encoded
using UTF-8!
>The DISTANCE has changed, but not the VELOCITY, so I see no acceleration
>here.
I was thinking about a situation where one /only considers/
the radius coordinate r.
With the cartesian coordinate of the center of the circle
being (0,0), the cartesian x,y-coordinates of the moving
object are (x,y)=(vt,d) with a constant distance d of the
line of movement to the center of the circle and a
constant speed v. x=vt, y=d.
_________C_____B_____A___________________ <--- object moving
.-' ^ '-. y direction ^ straight in this
.' | '. | line, from the
/ d \ d right to the left
; | ; |
| <----o | ---
; x direction ;
\ /
'. .'
'-._____.-'
Square brackets, as in "[x]", will denote the square root
from now on, e.g., [4]=2. Mnemonic: "square" as in "square
bracket" means "square" as in "square root".
The distance r(t) from the moving object to the center of
the circle depends on the time t and is r(t) = [xx+yy] =
[vvtt+dd]. ("xx" is the product of "x" with "x", and so on.)
"^r" is the derivative or "r" with respect to its sole
argument t, and "^^r" is the derivative of "^r" with
respect to its sole argument t.
r = [vvtt+dd]
^r = dr/dt = d[vvtt+dd]/dt = vvt/[vvtt+dd].
assuming v = 1 m/s and d = 1 m (m = meter, s = second):
r = [tt+1], curve shape: V
^r = t/[tt+1], curve shape: ___/¯¯¯¯
^^r = 1/[1+tt]^3, curve shape: ___/\___
I call "r" the "radius", "^r" the "radial velocity",
and "^^r" the "radial acceleration".
In this sense, there is an "acceleration" ("radial
acceleration", ^^r = 1/[1+tt]^3). So, this is my reply to:
>The DISTANCE has changed, but not the VELOCITY, so I see no acceleration
>here.
When t = 0s, we have,
r = d = 1 m,
^r = 0 m/s, and
^^r = 1 m/(ss).
When t = 1s, we have,
r = [2] m ~ 1.41 m,
^r = 1/[2] m/s ~ 0.70 m/s, and
^^r = [2]/4 m/(ss) ~ 0.35 m/(ss).
Now, the observer who only observes the radius can define
a (repulsive) "radial force" F=m(^^r) = m/[1+tt]^3.
Concluding Remarks
What is the relation of all this to the OP?
The OP wanted to investigates how bodies seem to move
according to "apparent forces" when their distance to
a center ("moves outwards") is investigated. I gave a
simplified model of a simple, linear movement.
Inhabitants of a one-dimensional world
Now, imagine inhabitants of a one-dimensional world who
only can observe the distance of objects from their
center, but not the angle coordinate. (In fact, my model
/is/ such a world as there is no physical means of how
to measure the angel between two bodies, when there are
only those two bodies.)
Those observers would come to the conclusion that their
center must have a "repulsive force" that gets stronger
as the moving object comes nearer, because, eventually,
it does not get any nearer but starts to move away again.
However, they eventually might have an "Einstein" who
will find out that one can add another "hypothetical"
dimension to the description of the movement, where the
movement is just straight and uniformly and no force
exists at all.
--- SoupGate-Win32 v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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