Tension, centrifugal force, two objects rotating

AI Thread Summary
In a discussion about two spheres exerting tension on a string through centrifugal force, it was clarified that the tension is not the sum of the forces from both spheres but rather a single force, T, transmitted through the string. The centrifugal force does not act on the string directly; instead, it is a fictitious force that helps explain the motion from a non-inertial frame. The centripetal force is not an applied force but a component of the resultant force acting orthogonally to the motion. The action-reaction principle applies only to applied forces, not to fictitious forces like centrifugal force. Overall, the tension in the string remains constant along its length, regardless of the number of spheres involved.
physics user1

Homework Statement

[/B]there's an image of the situation

Homework Equations

[/B]F= mω^2 R

The Attempt at a Solution

[/B]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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haruspex said:
haruspex said:

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
 
Cozma Alex said:
doesn't both of the spheres acts on the string?
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.
Cozma Alex said:
is the centrifugal force the reaction to the centripetal force
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.
See https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-forces/.
Cozma Alex said:
I noticed that all the fictitious force are opposite and with the same magnitude of the "real one"
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##.
 
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haruspex said:
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.
See https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-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##.

Thanks
 
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