# Uniform circular motion; two balls, two cables, and a post; no gravity

1. Jun 6, 2010

### emr13

1. The problem statement, all variables and given/known data

In space in the absence of gravity, there is a post of "infinite mass," meaning that the post is fixed or "nailed down" and cannot move. A ball of mass m1 is connected to the post by a cable of negligible mass of length L1. A second ball of mass m2 is connected to the first ball by a second cable of negligible mass of length L2. The entire contraption (except for the post) is moving around the post in a perfect circle with a constant angular speed ω2 (in radians per unit time). Find the tension in each cable.

2. Relevant equations

F = ma= mv2/r = mω2r

3. The attempt at a solution

So there is a cable tension (T1) acting on the ball with mass m1 and then a cable tension (T2) acting on the ball with mass m2.

This is what I have, and I'm wondering if it is correct. There is no movement or force in the y-direction at all, so I think only the x-components matter...but if I include cosθ, then there's a denominator of zero, so I'm guess that's not supposed to be there. Are these correct then?

For cable of length L1, T=m1ω2L1

For cable of length L2, T=m2ω2(L1+L2)

2. Jun 6, 2010

### aim1732

For inner mass Fnet = mw2L1 but there are two forces on it.

3. Jun 6, 2010

### emr13

Do you mean the force from the post?

4. Jun 7, 2010

### cartonn30gel

Draw a free body diagram for each ball. Then it will be easier to calculate the centripetal force for each object.

Last edited: Jun 7, 2010
5. Jun 7, 2010

### aim1732

I don't see how a post can exert a force on a faraway mass!

6. Jun 7, 2010

### emr13

So then you must mean the force from the other cable...I'm not sure how to include that, to be honest.

Maybe for cable of length L1, T1=m1ω2L1-m2ω2(L1+L2)? And then for cable of length L2, it would still be T2=m2ω2(L1+L2)?

I honestly did that...

7. Jun 7, 2010

### aim1732

Yes I was.
The force on the outer mass is determined by the second law as there is just one force there. So tension in outer cable is now known. Then you can apply the second law to the inner mass- one force is known and the net force is known.

8. Jun 7, 2010

### emr13

Okay. I think I've got it now; thanks.