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

[emr13]

Homework Statement

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.

Homework Equations

F = ma= mv²/r = mω²r

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)

 

[aim1732]

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

 

[emr13]

Do you mean the force from the post?

 

[cartonn30gel]

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

 

[aim1732]

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

 

[emr13]

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

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)?

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

I honestly did that...

 

[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.

 

[emr13]

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