The problem says:

A cylindrical chandelier is hung on a wire and when it rotates around its siege an angle $\theta$ the twisting moment acting on it is $\tau$ = $−\theta$. The moment of inertia is unitary and its angular velocity in the equilibrium position is unitary. Calculate the maximum twist angle $\theta_0$, ...

And the suggested solution is:

From the conservation of mechanical energy: $\frac{1}{2} = \frac{1}{2}\theta_0^2$ so $\theta_0 = 1$

But where does this formula came from?

  • $\begingroup$ It would be more accurate to say "From the work energy theorem" instead of "conservation of mechanical energy". You can find the KE at $\theta =0$ and $\theta = \theta_0$, and the work done by the twisting moment. $\endgroup$
    – alephzero
    Jun 30 '20 at 10:29

The chandelier is in an SHM since $\tau=-\theta$, where $\kappa=1$. Since the kinetic energy in the equilibrium position equals the potential energy at the amplitude position, $\frac12I\omega^2=\frac12\kappa\theta_0^2$.

  • $\begingroup$ Then why $I \omega^2 = 1$? $\endgroup$
    – Andre
    Jun 30 '20 at 10:49
  • $\begingroup$ It is given that $I=1$kgm$^2$ and $\omega=1$rad/s at the equilibrium position. $\endgroup$ Jun 30 '20 at 10:52

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