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In the Feynman Lectures Vol 1. Chapter 41 - The Brownian movement, the following statement is present under Section 1 -

We know the formula for the kinetic energy of rotation—it is given by Eq. (19.8): $T = \tfrac{1}{2}I\omega^2$. That is the kinetic energy, and the potential energy that goes with it will be proportional to the square of the angle—it is $V = \tfrac{1}{2}\alpha\theta^2$. But, if we know the period $t_0$ and calculate from that the natural frequency $\omega_0 = 2\pi/t_0$, then the potential energy is $V = \tfrac{1}{2}I\omega_0^2\theta^2$.

I have a hard time understanding how the potential energy is derived. Can someone please help me understand it ?

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With a potential of the form indicated, we have (Lagrange equation): $I_0\ddot{\theta}=-\alpha\theta$ and therefore ${\omega_0}^2=\alpha/I_0$

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  • $\begingroup$ If this isn't obvious, think of this in analogy to a mass on a spring, where the potential energy is $\frac{1}{2} k x^2$ and $\omega_0^2 = k/m$. $\endgroup$
    – Michael Seifert
    Commented Jul 7, 2021 at 17:31
  • $\begingroup$ Thank you @Michael Seifert. I saw the moment of inertia I, and confused the angular frequency with the angular velocity for omega. It finally makes sense. $\endgroup$
    – Yogesh
    Commented Jul 8, 2021 at 13:53
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All that is being said here is that the potential energy $V$ is proportional to the square of the angular displacement $\theta$. This is true for any situation where you have harmonic oscillation. Thus, there is some proportionality factor $\alpha$ such that $V = \frac{1}{2} \alpha \theta^2$.

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  • $\begingroup$ I understand that, infact in the image just above this statement it is indicated that 𝛼 might be the spring constant of the fibre. Where I am stuck is how he deduced that the potential energy must be V = $\tfrac{1}{2}I\omega_0^2\theta^2$. I am sure it must be something simple but I just cant make the connection. $\endgroup$
    – Yogesh
    Commented Jul 7, 2021 at 16:00

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