0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
2
$\begingroup$

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, $\frac12I\omega^2=\frac12\kappa\theta_0^2$.

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.