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A non-rigid object (such as a beam, rod, blade, etc.) experiences elongation due to the (fictitious) centrifugal force when rotated (as opposed to rigid bodies and simple mass points). This elongation can be easily viewed in the inertial frame of reference and in the rotating frame of reference. There are sources which claim that the centripetal force and the centrifugal force, which act on a mass element of length dx at any position along the longitudinal axis of the object, are the same in magnitude and opposite in direction. In force equilibrium, how is it possible that the length of a non-rigid object becomes greater as a result of rotation? To my understanding, a resultant force is required so that any deformation of the non-rigid object can occur.

Please correct me if any of my above statements are wrong. I would appreciate and be thankful if someone explained the matter to me by using formulas, please.

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The core of the question seems to be a misconception here:

There are sources which claim that the centripetal force and the centrifugal force, which act on a mass element of length dx at any position along the longitudinal axis of the object, are the same in magnitude and opposite in direction. In force equilibrium, how is it possible that the length of a non-rigid object becomes greater as a result of rotation? To my understanding, a resultant force is required so that any deformati

I.e. if the centrifugal and centripetal forces are equal and opposite, so they add to zero.

But those two forces cannot be summed, because they exist in different frames. One is in the inertial frame and the other is in the rotating frame. In fact, they’re the same force in those two separate frames.

In any particular frame, only one force acts, and that determines the motion.

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  • $\begingroup$ Thanks! Unfortunately, this information is often not properly communicated in the literature. Can you recommend a book on this, please? $\endgroup$
    – ExOrbitant
    Commented Feb 8, 2020 at 17:26
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To understand what is happening, forget about mathematical tricks like "centrifugal force" and go back to Newton's laws of motion.

Start by considering a mass on the end of a (flexible) string of length $r$ rotating with angular velocity $\omega$. The mass is accelerating towards the center of rotation. Therefore there must be a real force acting on it, equal to mass $\times$ acceleration, i.e. $m\omega^2 r$. That force is applied by the tension in the string.

If a flexible string is in tension, it stretches.

Now think about a rotating flexible rod (with no extra mass at the tip)

If you imagine cutting the rod at any point, the outer part is accelerating towards the center of rotation just like the mass in the first example, and the force to make it accelerate comes from the tension in the rod at the point where you cut it. Again, the tension in a flexible rod causes it to stretch.

The only difference between the two examples is that the tension in the rod varies along its length, from zero at the tip to a maximum at the center of rotation, and therefore calculating the amount of extension is a bit more complicated (and requires calculus)

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  • $\begingroup$ By tension do you mean tensile stress (N/m^2), right? The force should be at its maximum at the tip as the acceleration is there the largest according to your formula (assuming mass is equally distributed). When the tensile stress is the largest at the center of rotation, then the dx element closest to the center of rotation should experience the largest elongation, right? $\endgroup$
    – ExOrbitant
    Commented Feb 8, 2020 at 17:56
  • $\begingroup$ The is no mass outside the tip, so even though the tip is accelerating, no mass is being accelerated, so the tension at the tip is zero. But, you are correct about the tensile stress and strains near the center of rotation. $\endgroup$
    – Chet Miller
    Commented Feb 8, 2020 at 18:36
  • $\begingroup$ How can something accelerate that has no mass? You said that the tip is accelerating...I would have thought that tension (internal force in the material) is at the maximum where the centrifugal force in the rotating reference frame is at its maximum. So are tension (internal force), strain and tensile stress at their maxima at the center of rotation whilst the centrifugal force has its at the tip? I thought tension is proportional to the external force (centrifugal force)... $\endgroup$
    – ExOrbitant
    Commented Feb 8, 2020 at 19:00
  • $\begingroup$ Re. the string example, by Newton's third law the mass exerts and equal and opposite force on the string. As the string is flexible, it stretches. Now the centripetal force varies as the inverse of the radius, so as the mass moves outward because of the stretching, the force it exerts on the string diminishes until a point is reached where it is no longer able to stretch it. $\endgroup$
    – Not_Einstein
    Commented Feb 10, 2020 at 0:10
  • $\begingroup$ "diminishes until a point"...I think we have to be careful here. The centripetal force is $m\omega{}^2r$ so it increases with distance to the centre of rotation. $\endgroup$
    – ExOrbitant
    Commented Feb 11, 2020 at 13:13

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