# Tension, centrifugal force, two objects rotating

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1. Jan 5, 2016

### Cozma Alex

1. The problem statement, all variables and given/known data there's an image of the situation

2. Relevant equations F= mω^2 R

3. The attempt at a solution so, in the image there are two spheres that exerts a tension on the string using their centrifugal, since the centrifugal bodies are two I expect that the tension is the sum in intensity of the two forces, do you agree? Another question is: why the centrifugal forces acts to the string? Aren't these forces supposed to act just on the spheres?

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Last edited by a moderator: Jan 5, 2016
2. Jan 5, 2016

3. Jan 5, 2016

### Cozma Alex

Ahh thanks, but, doesn't both of the spheres acts on the string? However, is the centrifugal force the reaction to the centripetal force (third law of dynamics)? In general is the third law implied in the fictitious forces? (I noticed that all the fictitious force are opposite and with the same magnitude of the "real one" )example a guy in a bus that is stopping in the non inertial frame of system of the bus the guy is accelerating because of a force equal but opposite of the force that stops the bus

4. Jan 5, 2016

### haruspex

Yes, but as I wrote in that article, tension is not exactly a force - it is more like a pair of equal and opposite forces.
Consider the string as made up of a number of short sections. A sphere at one end pulls on an end section with force T. That section pulls on the next section with the same force, and the reaction from that section pulls on the first section with force T. And so on all along the way - it' always T, not 2T.
No. The centripetal force is not an applied force. It is the component of the resultant force orthogonal to the direction of motion. The action and reaction law is only for applied forces.
Yes. Suppose the net force on a mass m is $\vec F$. That produces an acceleration $\frac 1m\vec F$. In the frame of reference of the mass m, there is no acceleration. To explain this in that reference frame, we must introduce a fictitious force to balance $\vec F$, so that will be $-\vec F$.