# Hmm, Force on two rotating systems.

1. Aug 16, 2007

### K.J.Healey

Assume you have some universal force that acts as a function F = k/r^2

Now in our little simulation picture point particle PAIRS. Almost like a H-H molecule, where its 0----0

And picture one more of these.

Consider them to each be an independent system for the moment. Each one rotates about its center of "mass" or whatever you'd like to call it.

O--~--O O--~--O

Now I want to compare the total force in all directions of each system on the other.

The two scenarios are:
1. Both rotate Counter-Clock_wise(CCW) with angle Theta.
2. The Left one rotates Clockwise, the right one CCW w/ theta.

They rotate at the same rate (about ~)

Below I will write out what I have so far. This is NOT HOMEWORK. This is somethign I thought about while falling asleep and just wanted to solve it. Unfortunately my computer doesn't even have excel let alone mathematica. This is where you all come in to check my work and help me :)

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We'll call them
1--~--2,,,,,,,,,,,,,,3--~--4
where the distance between the ~'s are D and the radius of each is "r".

Well since they both rotate at the same rate this scenario should be trivial.
For the the 1-3 I get the distance maintains at D, as is 2,4, and the Y seperation is 0.
1,3 = X = L = D
2,4 = X = L = D
And for the other I get:
1,4 = L =>
___X = D+2rCos(T)
___Y = 2rSin(T)

2,3 = L =>
___X = D+2rCos(T+pi)
___Y = 2rSin(T+pi)

Now on to Case #2 Where they spin opposite:
1--~--2,,,,,,,,,,,3--~--4

1,3 :
X = D
Y = 2rSin(T)
2,4 :
X = D
Y = 2rSin(T+pi)
1,4 :
X = D+2rCos(T)
Y = 0
2,3 :
X = D+2rCos(T+pi)
Y = 0

Now with these what should I do next?
What is a worthy attribute to compare? Net force? Assuming the ~'s are mounted and immovable, what is there to compare that would be different?

I guess what I'm seeing is if you don't mount them, you can basically just list the different forces from each starting configuration, and perhaps get an average?
What do you think is my next step in comparing these two situations?