# How did Feynman derive the potential energy of the ballistic galvanometer?

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 ?

With a potential of the form indicated, we have (Lagrange equation): $$I_0\ddot{\theta}=-\alpha\theta$$ and therefore $${\omega_0}^2=\alpha/I_0$$
• 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 $\omega_0^2 = k/m$. Commented Jul 7, 2021 at 17:31
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 $$V = \frac{1}{2} \alpha \theta^2$$.
• 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. Commented Jul 7, 2021 at 16:00