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The period T is equal to $2\pi/w_n$ where the natural frequency can be found from $w_n^2{\theta}+\ddot{\theta} = 0$. Since $\tau = I\alpha $, as there is no net torque about point P because gravity acts at the center of mass and equals the normal force, $\ddot{\theta} = 0$ so $w_n = 0$. Does that mean the period is infinite?

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    $\begingroup$ Think about where the centre of mass of the bent rod is. Hint: it is not at P and if the rod is not at its central position then it is not directly below P either. $\endgroup$
    – gandalf61
    Jan 1 at 15:30
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No. Gravity does not act at point $P$ because your CoM is not at point $P$. Your CoM is lower than point $P$. Therefore, gravity produces a torque.

Your normal force will not produce a torque, since it's applied at point $P$, so the torque about point $P$ due to the normal force is zero.

Lastly, I'd like to emphasize that force equilibrium does not imply torque equilibrium. (and this pendulum is not even in force equilibrium, because the CoM will accelerate).

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  • $\begingroup$ Thanks for responding. This means that if the thickness of the pendulum were provided the torque could be calculated. But if the point in question were the center of mass would the period be infinite as the natural frequency would equal 0? $\endgroup$
    – user436788
    Jan 1 at 15:45
  • $\begingroup$ The thickness is not the only thing that affects the CoM. The main reason is the shape. For instance, if I have a straight uniformly dense thin string, then its CoM is right in the middle of the string. However, when I use that string to shape a perfect circle, then the CoM will be at the geometric center of the circle, and no longer on the string. Shape matters. $\endgroup$
    – user256872
    Jan 1 at 15:59
  • $\begingroup$ If the CoM is at the same position as the point of rotation, then gravity will provide no torque. If no other forces provide any torques, then you won't have any oscillations. The object will just continue doing what it was doing before -- if it was rotating with angular speed $\omega$ then it will keep rotating with that same angular speed, if it was not rotating then it will not rotate. $\endgroup$
    – user256872
    Jan 1 at 16:01
  • $\begingroup$ But it will not have an infinite period. See here for more: physics.stackexchange.com/questions/98684/…. $\endgroup$
    – user256872
    Jan 1 at 16:01
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    $\begingroup$ @R.W. Bird I think you'd find that (against expectations) you don't need to know the position of the C of G in order to find the period of small oscillations about P. A little fun with the parallel axes theorem led me to this conclusion. $\endgroup$ Jan 1 at 17:21

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