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In this video, at around 12:00, it is said that spinning about the axis with the smallest moment of inertia gives the most kinetic energy. But, isn’t rotational kinetic energy equal to $(1/2)(I)(ω)^2$ . Thus, shouldn’t kinetic energy increase with increase in moment of inertia?

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  • $\begingroup$ Your formula would indicate minimal kinetic energy when spinning about the axis with smallest moment of inertia if $\omega$ were constant as the axis adjusts over time. However, as Bhavay's answer indicates, the equations of motion preserve $I \omega$ in this process rather than just $\omega$. $\endgroup$
    – jawheele
    Apr 28 at 21:36
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It is because the angular momentum is conserved while the kinetic energy is not.

So: $$I_1\omega_1=I_2\omega_2$$

When $I_2$ decreases, $\omega_2$ increases. $\omega$ is squared in the expression of kinetic energy and hence the net kinetic energy increases.

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  • $\begingroup$ If the kinetic energy is not conserved, what is it converted to/from (in this case)? $\endgroup$ Apr 28 at 21:07
  • $\begingroup$ @PeterMortensen it depends on the setup. In the video it's converted to kinetic energy within the contained fluid and, ultimately, heat via viscous dissipation. $\endgroup$
    – Kyle
    Apr 28 at 21:11

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