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

AI Thread Summary
In a scenario involving two balls connected by cables in uniform circular motion around a fixed post in the absence of gravity, the tension in each cable must be calculated based on centripetal force principles. The tension T1 in the first cable is determined by the formula T1 = m1ω²L1, while the tension T2 in the second cable is given by T2 = m2ω²(L1 + L2). The discussion emphasizes the importance of drawing free body diagrams for both masses to clarify the forces acting on them. It is noted that the force exerted by the post does not affect the outer mass directly, and the inner mass's tension must account for the force from the outer cable. Ultimately, the calculations hinge on understanding the net forces acting on each mass in the system.
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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.

post.jpg


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)
 
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For inner mass Fnet = mw2L1 but there are two forces on it.
 
Do you mean the force from the post?
 
Draw a free body diagram for each ball. Then it will be easier to calculate the centripetal force for each object.
 
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I don't see how a post can exert a force on a faraway mass!
 
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...
 
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.
 
Okay. I think I've got it now; thanks.
 

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