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Imagine a body with moment of inertia, $I$ and with angular velocity $\omega$. If no torque is applied, and moment of inertia is reduced to $I/2$, the angular velocity goes to $2\omega$. Thereby, angular momentum stays conserved. But it is evident that rotational kinetic energy is doubled.

So there should be a work done to explain the scenario. I assume, $KE+Work=2KE$. Here no rotational work is being done as there is no torque. So the work must be some translational work done to change moment of inertia.

Am I correct? So the rotational kinetic energy is not always conserved in conserved momentum scenario? And is rotational kinetic energy a type mechanical energy?

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Yes, in the scenario that you describe work is done.

To sustain circumnavigating motion a centripetal force is required. In the case of a single solid object the structural integrity of the object is providing that force.

What you describe is a device with a mechanism such that the whole can contract to a smaller diameter.

In order to achieve that contraction a surplus of centripetal force must be provided. The larger the surplus the faster the contraction, but a minute surplus is already enough.

The following diagram represents a point mass that is in circumnavigating motion, being pulled closer to the axis of rotation.

Contraction of a rotating system

The dark grey arrow represents the centripetal force.

The two lighter grey arrows represent a decomposition of the centripetal force into two perpendicular components, one component perpendicular to the instantaneous velocity vector, the other component parallel to the instantaneous velocity vector.


The force component parallel to the instantaneous velocity vector is increasing the rotational kinetic energy of the point mass.

General property of rotational motion:
When a rotating system contracts the centripetal force is doing work, increasing the rotational kinetic energy.


When the contraction phase has finished the motion is again perfect circumnavigating motion. That is, after the contraction phase the instantaneous velocity vector is once again perpendicular to the direction of the centripetal force.

The point is: during the contraction phase the instantaneous velocity vector is not perpendicular to the centripetal force. That angle gives the opportunity for the centripetal force to increase the rotational kinetic energy of the circumnavigating object.



A vivid example of acceleration due to contraction is the orbit of Halley's comet.

Check out the animation in the wikipedia article.

perihelion: where the celestial body is closest to the Sun
aphelion: where the celestial body is farthest away from the Sun

From aphelion to perihelion the Sun's gravity is accelerating the comet all the time. Then the comet sweeps around the Sun, and from perihelion to aphelion the Sun's gravity is doing negative work, decelerating the comet. When all the radial velocity has been depleted the comet passes aphelion. That is how the orbit is cycling.


General statement:
Anytime you see that a system with circumnavigating motion is contracting you know that work is being done, increasing the kinetic energy of the system. Conversely, when the diameter of the system-with-circumnavigating-motion is increasing the centripetal force is doing negative work.

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There are two ways to think about the work done, but neither of them is translational. The infinitesimal work $dW$ is not equal to $N\,d\theta$ except in special cases, such as single-axis rotation with constant $I$. More generally (but still for uniaxial rotation), $dW=L\,d\omega$.* In this case, $L$ is conserved but the angular frequency $\omega$ is increasing. So the change to the kinetic energy after the moment of interia changes from $I$ to $I/C$ is proportional to $\int_{\omega}^{C\omega}d\omega'=(C-1)\omega$.

The other way to do the analysis is to work in the rotating frame of the object. In this frame there is a (fictitious) centrifugal force. This is directed away from the axis, with a magnitude per unit mass of $f_{c}=\omega^{2}$. To decrease $I$, it is necessary to pull the mass elements of the rotating body closer to the rotation axis, which requires net positive work to be done in opposition to the centrifugal force.

*The key distinction is that when $L$ and $\omega$ are directly proportional, $L=I\omega$, the differential an be applied to either $L$ or $\omega$, since $dW=L\,d\omega=dL\,\omega=(dL/dt)(\omega\,dt)=N\,d\theta$.

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  • $\begingroup$ How is dW = Ldω? If dW= Torque * dθ, then dW should equal to dL*dω right? $\endgroup$ Commented Dec 25, 2024 at 19:40
  • $\begingroup$ Can there be work with no torque applied? $\endgroup$ Commented Dec 25, 2024 at 19:41
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    $\begingroup$ @Himalayan Where does a spinning figure skater's energy go when she slows down? $\endgroup$ Commented Dec 25, 2024 at 19:58

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