For a 2D disk of diameter $l$ spinning with angular velocity $\omega$, what is the velocity auto-correlation in the y-direction averaged over the disk?
So far, I have that the y-component of velocity at a point ($r, \theta_0$) is:
\begin{equation} v_y (r,\theta_0) = \omega r \sin(\theta_0) \end{equation}
and some time $\tau$ later the point on the disk has rotated by $\Delta \theta = \omega \tau $:
\begin{equation} \begin{split} v_y&= \omega r \sin(\theta_0 + \Delta \theta) \newline &= \omega r \sin(\theta_0 + \omega \tau) \end{split} \end{equation}
I think I'm after the integral $A$ to compute the average (over the disk area of $\pi l^2/4$) auto-correlation:
\begin{equation} \begin{split} A &= \frac{4}{\pi l^2} \int_0^{l/2} \int_{-\infty}^{\infty} v_y(\theta_0) \cdot v_y(\theta_0 + \omega \tau) d\theta_0 dr \newline &= \frac{4}{\pi l^2} \int_0^{l/2} \int_{-\infty}^{\infty} \omega r \sin(\theta_0) \cdot \omega r \sin(\theta_0 + \omega \tau) d\theta_0 dr \end{split} \end{equation}
Now this answer shows the autocorrelation of a sine wave is $\frac{1}{2} \cos(\Delta \theta)$. This works for any point on the disk. So now we get:
\begin{equation} \begin{split} A &=\frac{2 \omega^2}{\pi l^2}\cos(\omega \tau) \int_0^{l/2} r^2dr \newline &=\frac{l \omega^2}{ 12 \pi }\cos ( \omega \tau) \end{split} \end{equation}
Are these the right integrals to take, resulting in the auto-correlation being this cosine wave with $\lim_{\tau \rightarrow 0} A = \frac{l \omega^2}{ 12 \pi}$? I'm trying to work out if this working is right and if $A$ represents the auto-correlation of the collective points on the disk.