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Posted (edited)

A sphere of iron is stuck to the end of a ruber band at a point A. On the other end of the ruber band it is fixed to a point O that will act as the center of rotation.

Questions:

 

1- If the sphere is pushed so that it achieves a speed V, with no friction and no outside forces, is it possible to achieve equilibrium, so that the length of the (stretched) ruber band and the magnitude of V both remain constant?

 

3- Under sufficient magnitudes and proportions, would the sphere experience shearing? Specifically, would the diameter between the point attached to the ruber band A and the opposite point B be greater than the diameter between the leading point C and the trailing point D (along the direction of revolution), and would B lag behind A in the direction of motion?

Simplifying:

Does AO and BO form an angle at O?

 

Reasons why I think the sphere could experience shearing:

The ruber exerts a constant centripetal force (acceleration). The circumference described by B is necessarily greater than the one described by A, hence, if AO and BO coincide (remain aligned) the speed of B would have to be greater than the speed of A. Since the ruber band = AO, it pulls the point A, which in turn pulls the point B transferring the centripetal force through the iron. This energy transfer takes time and also loses strength until it reaches the point B. I conclude the sphere would deform all around, stretching along AB, compressing along CD and also along the "poles", while the point A will experience a centripetal force directly from the ruber band. Because iron has elasticity and the transfer of momentum from A to B takes time, the centripetal force exerted on A will not be transmitted to B instantaneously, and some energy loss should be expected, so I doubt that B can be accelerated enough (due to torque from A) to keep AO and BO perfectly aligned, with the angle at O between AO and BO set to zero.

Edited by altergnostic

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